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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.05%3A_Color_and_Constitution |
Visible light is electromagnetic radiation having a rather narrow range of wavelengths (\(400\)-\(800 \: \text{nm}\)). A black substance absorbs wavelengths of visible light. Selective absorption of visible light by a substance imparts color, but the color is not that of the light absorbed but instead of the residual light that the substance transmits or reflects. For example, a compound that absorbs in the region \(435\)-\(480 \: \text{nm}\) removes blue light from the visible spectrum, and the residual light is recognized by the eye as being yellow. The relationship of the observed color to wavelength of light absorbed is shown in Table 28-1. It is customary to call the color observed the or the to that absorbed. Clearly, the color perceived, its brightness and its intensity, depends on the shape of the electronic spectral curve of the absorbing substance, which in turn depends on the chemical structure of the substance. A change in absorption from the blue to the red end of the spectrum corresponds to a in the energy of the associated electronic transitions. We know also that this trend is associated with increasing conjugation of multiple bonds. For instance, 1,2-diphenylethene is colorless, whereas 1,10-diphenyl-1,3,5,7,9-decapentaene is yellow-orange: The effect of substituents on colors associated with conjugated systems is of particular interest in the study of dyes, because most dyes have relatively short conjugated systems and would not be intensely colored in the absence of substituent groups. (The plant pigments \(\beta\)-carotene, , and lycopene, often used as food coloring, are exceptions.) The conjugated \(\pi\) system common to all four compounds is that of the benzenoid ring, which is described as the absorbing ( ). The hydroxyl and nitro substituents can be seen individually to shift the \(\lambda_\text{max}\) of the chromophore to longer wavelengths. However, the combined effect of the two substituents is much more dramatic, especially if the \(\ce{OH}\) group is converted to the corresponding anion, 4-nitrobenzenolate. Now \(\lambda_\text{max}\) is shifted into the visible region, giving a yellow color, and because \(\epsilon\) is large, the color is intense. Thus, properly chosen substituents can shift the main benzenoid absorption band from the ultraviolet into the visible region of the spectrum. Such substituents are often called . They act by extending the conjugation of the chromophore and are particularly effective in causing large shifts towards the visible when one substituent is a \(\pi\)-electron donor and the other a \(\pi\)-electron acceptor. Thus, with the 4-nitrobenzenolate ion, interaction between the strongly electron-donating \(\ce{-O}^\ominus\) group and the strongly electron-accepting \(\ce{-NO_2}\) group provides significant stabilization: Hence, substitution of an electron-attracting group (such as \(\ce{NO_2}\)) at one end of such a system and an electron-donating group (such as \(\ce{O}^\ominus\)) at the other end should be particularly favorable for stabilization of the excited state (relative to the ground state, where \(17a\), \(17b\), etc., are of lesser importance). At the same time, we should expect that two electron-attracting (or two electron-donating) groups at opposite ends would be nearly as effective. We hope you will understand from the foregoing discussion why it is that many intensely colored substances of natural or synthetic origin have conjugated structures with substituents, often cationic or anionic substituents, that can donate or accept electrons from the conjugated system. Such compounds provide us with many useful dyes, pigments, indicators, and food-coloring agents, as well as conferring color on plants and animals. A few examples follow: Historically, the dye industry has been closely linked with the development of synthetic organic chemistry. Although dyes have been extracted from natural sources for centuries, it was not until 1856 that a synthetic dye was produced commercially. The previous year, William Henry Perkin - at age 17 - oxidized benzenamine (aniline) with potassium dichromate and isolated from the product (which was mostly aniline black; ) a purple compound that was excellent for dyeing silk. Perkin started commercial production of the dye under enormous difficulties. Because there was no organic chemical industry at the time, he had to design and build his own equipment as well as devise efficient syntheses for starting materials. His route to benzenamine stared with crude benzene from coal, which he nitrated and then reduced with iron and acid. He had to make the nitric acid (from nitrate salts and sulfuric acid) because concentrated nitric acid was not available. It was not until 1890 that the structure of Perkin's dye, called mauveine, was established by Otto Fischer. The dye was actually a mixture because the benzene used contained methylbenzene), but the product from the oxidation of benzenamine itself is structurally related to aniline black: Although the mauveine dyes have been replaced with better dyes, they are representative of a group of useful dyes having the general structure in which \(\ce{X}\) and \(\ce{Y}\) can be oxygen, nitrogen, sulfur, or carbon. The rings invariably carry substituents (hydroxyl or amino) that provide enhanced stabilization of the excited states. Examples of these ring systems follow: A large number of useful dyes are substituted triphenylmethane derivatives. Crystal Violet ( ) and phenolphthalein are excellent examples of this kind of dye. Other important dyes are derivatives of the following types of substances: Examples are There is more to a successful dye than just an attractive color.\(^5\) If it is to be useful, say for coloring fabrics, some simple means must be available for introducing the color into the fiber and then, usually of greater difficulty and importance, the color must be reasonably permanent - that is, resistant to normal laundry or cleaning procedures (wash-fast) and stable to light (light-fast). Here again, fundamentally important problems are involved. The scientific approach to improving wash-fastness of fabric dyes has to be based on a knowledge of the structural factors bearing on the intermolecular forces that determine solubilities. Light-fastness is connected with the photochemistry of organic compounds. The distinction between a dye and a pigment is that a dye actually is absorbed by the material to be colored, whereas a pigment is applied with a binding material to the surface. Pigments usually are highly insoluble substances. Many insoluble inorganic substances that would be wholly unsatisfactory as dyes are useful pigments. Copper phthalocyanine is an example of a very important class of organic pigments. These are tetraazatetrabenzo derivatives of the porphyrin compounds discussed in and . Copper phthalocyanine arises from condensation of four molecules of 1,2-benzenedicarbonitrile in the presence of copper metal at \(200^\text{o}\): \(^5\)For a good account of dyes, see L. F. Fieser and M. Fieser, , D. C. Health & Co., Lexington, Mass., 1956, Chapter 36. and (1977) | 7,192 | 3,160 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.02%3A_Chemistry_of_Dioxygen |
The reduction potential for the four-electron reduction of dioxygen (Reaction 5.1) is a measure of the great oxidizing power of the dioxygen molecule. However, the reaction involves the transfer of four electrons, a process that rarely, if ever, occurs in one concerted step, as shown in Reaction (5.2). \[O_{2} \xrightarrow{e^{-}} O_{2}^{-} \xrightarrow{e^{-}, 2H^{+}} H_{2}O_{2} \xrightarrow{e^{-}, H^{+}} H_{2}O + OH \xrightarrow{e^{-}, H^{+}} 2 H_{2}O \tag{5.2}\] \[dioxygen \xrightarrow{e^{-}} superoxide \xrightarrow{e^{-}, 2H^{+}} hydrogen\; peroxide \xrightarrow{e^{-}, H^{+}} water + hyrdoxyl\; radical \xrightarrow{e^{-}, H^{+}} water\] Since most reducing agents can transfer at most one or two electrons at a time to an oxidizing agent, the thermodynamics of the one- and two-electron reductions of dioxygen must be considered in order to understand the overall mechanism. In aqueous solution, the most common pathway for dioxygen reduction in the absence of any catalyst is one-electron reduction to give superoxide. But this is the least favorable of the reaction steps that make up the full four-electron reduction (see Table 5.1) and requires a moderately strong reducing agent. Thus if only one-electron pathways are available for dioxygen reduction, the low reduction potential for one-electron reduction of O to O presents a barrier that protects vulnerable species from the full oxidizing power of dioxygen that comes from the subsequent steps. If superoxide is formed (Reaction 5.3), however, it disproportionates quite rapidly in aqueous solution (except at very high pH) to give hydrogen peroxide and dioxygen (Reaction 5.4). The stoichiometry of the overall reaction is therefore that of a net two-electron reduction (Reaction 5.5). It is thus impossible under normal conditions to distinguish one-electron and two-electron reaction pathways for the reduction of dioxygen in aqueous solution on the basis of stoichiometry alone. \[2O_{2} + 2e^{-} \rightarrow 2 O_{2}^{-} \tag{5.3}\] \[2O_{2}^{-} + 2 H^{+} \rightarrow H_{2}O_{2} + O_{2} \tag{5.4}\] \[O_{2} + 2 e^{-} + 2 H^{+} \rightarrow H_{2}O_{2} \tag{5.5}\] The thermodynamics of dioxygen reactions with organic substrates is also of importance in understanding dioxygen reactivity. The types of reactions that are of particular interest to us here are hydroxylation of aliphatic and aromatic C—H bonds and epoxidation of olefins, since these typical reactions of oxygenase enzymes are ones that investigators are trying to mimic using synthetic reagents. Some of the simpler examples of such reactions (plus the reaction of H for comparison) are given in the reactions in Table 5.2. It is apparent that all these reactions of dioxygen with various organic substrates in Table 5.2 are thermodynamically favorable. However, reactions of dioxygen with organic substrates in the absence of a catalyst are generally very slow, unless the substrate is a particularly good reducing agent. To understand the sluggishness of dioxygen reactions with organic substrates, we must consider the kinetic barriers to these reactions. The principal kinetic barrier to direct reaction of dioxygen with an organic substrate arises from the fact that the ground state of the dioxygen molecule is triplet, i.e., contains two unpaired electrons. Typical organic molecules that are representative of biological substrates have singlet ground states, i.e., contain no unpaired electrons, and the products resulting from their oxygenation also have singlet ground states. Reactions between molecules occur in shorter times than the time required for conversions from triplet to singlet spin. Therefore the number of unpaired electrons must remain the same before and after each elementary step of a chemical reaction. For these reasons, we know that it is impossible for Reaction (5.6) to go in one fast, concerted step. \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.6}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] The arrows represent electron spins: \(\downarrow \uparrow\) represents a singlet molecule with all electron spins paired; \(\uparrow \uparrow\) represents a triplet molecule with two unpaired electrons; and \(\uparrow\) (which we will see in Reaction 5.13) represents a doublet molecule, also referred to as a free radical, with one unpaired electron. The pathways that do not violate the spin restriction are all costly in energy, resulting in high activation barriers. For example, the reaction of ground-state triplet dioxygen, i.e., O , with a singlet substrate to give the excited triplet state of the oxygenated product (Reaction 5.7) is spin-allowed, and one could imagine a mechanism in which this process is followed by a slow spin conversion to a singlet product (Reaction 5.8). \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{3}XO \tag{5.7}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \qquad \quad \uparrow \uparrow\] \[ \;^{3}XO \xrightarrow{slow} \;^{1}XO \tag{5.38}\] \[ \; \uparrow \uparrow \qquad \qquad \downarrow \uparrow\] But such a reaction pathway would give a high activation barrier, because the excited triplet states of even unsaturated molecules are typically 40-70 kcal/mol less stable than the ground state, and those of saturated hydrocarbons are much higher. Likewise, a pathway in which O is excited to a singlet state that then reacts with the substrate would be spin-allowed (Reactions 5.9 and 5.10). The high reactivity of singlet dioxygen, generated by photochemical or chemical means, is well-documented. However, such a pathway for a reaction of dioxygen, which is initially in its ground triplet state, would also require a high activation energy, since the lowest-energy singlet excited state of dioxygen is 22.5 kcal/mol higher in energy than ground-state triplet dioxygen. \[\;^{3}O_{2} + 22.5\; kcal/mol \rightarrow \;^{1}O_{2} \tag{5.9}\] \[ \uparrow \uparrow \qquad \qquad \qquad \qquad \qquad \downarrow \uparrow\] \[\frac{1}{2} \;^{1}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.10}\] \[\quad \downarrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] Moreover, the products of typical reactions of singlet-state dioxygen with organic substrates (Reactions 5.11 and 5.12, for example) are quite different in character from the reactions of dioxygen with organic substrates catalyzed by oxygenase enzymes (see Section V): \(\tag{5.11}\) \(\tag{5.12}\) One pathway for a direct reaction of triplet ground-state dioxygen with a singlet ground-state organic substrate that can occur readily without a catalyst begins with the one-electron oxidation of the substrate by dioxygen. The products of such a reaction would be two doublets, i.e., superoxide and the oneelectron oxidized substrate, each having one unpaired electron (Reaction 5.13). These free radicals can diffuse apart and then recombine with their spins paired (Reaction 5. 14). \[\;^{3}O_{2} + \;^{1}X \rightarrow \;^{2}O_{2}^{-} + \;^{2}X^{+} \tag{5.13}\] \[\uparrow \uparrow \qquad \downarrow \uparrow \quad \quad \uparrow \qquad \uparrow\] \[\;^{2}O_{2}^{-} + \;^{2}X^{+} \rightarrow \;^{2}O_{2}^{-} + 2X^{+} \rightarrow \;^{1}XO_{2} \tag{5.14}\] \[\uparrow \qquad \uparrow \qquad \qquad \uparrow \qquad \downarrow \qquad \qquad \downarrow \uparrow\] Such a mechanism has been shown to occur for the reaction of dioxygen with reduced flavins shown in Reaction (5.15). \(\tag{5.15}\) However, this pathway requires that the substrate be able to reduce dioxygen to superoxide, a reaction that requires an unusually strong reducing agent (such as a reduced flavin), since dioxygen is not a particularly strong one-electron oxidizing agent (see Table 5.1 and discussion above). Typical organic substrates in enzymatic and nonenzymatic oxygenation reactions usually are not sufficiently strong reducing agents to reduce dioxygen to superoxide; so this pathway is not commonly observed. The result of these kinetic barriers to dioxygen reactions with most organic molecules is that uncatalyzed reactions of this type are usually quite slow. An exception to this rule is an oxidation pathway known as free-radical autoxidation. The term free-radical autoxidation describes a reaction pathway in which dioxygen reacts with an organic substrate to give an oxygenated product in a free-radical chain process that requires an initiator in order to get the chain reaction started. (A free-radical initiator is a compound that yields free radicals readily upon thermal or photochemical decomposition.) The mechanism of free radical autoxidation is as shown in Reactions (5.16) to (5.21). Initiation: $$X_{2} \rightarrow 2X \cdotp \tag{5.16}\] \[X \cdotp + RH \rightarrow XH + R \cdotp \tag{5.17}\] Propagation: $$R \cdotp + O_{2} \rightarrow ROO \cdotp \tag{5.18}\] \[\downarrow \qquad \uparrow \uparrow \qquad \quad \uparrow\] \[ROO \cdotp + RH \rightarrow ROOH + R \cdotp \tag{5.19}\] Termination: $$R \cdotp + ROO \cdotp \rightarrow ROOR \tag{5.20}\] \[2 ROO \cdotp \rightarrow ROOOOR \rightarrow O_{2} + ROOR \tag{5.21}\] (plus other oxidized products, such as ROOH, ROH, RC(O)R, RC(O)H). This reaction pathway results in oxygenation of a variety of organic substrates, and is not impeded by the spin restriction, because triplet ground-state dioxygen can react with the free radical R• to give a free-radical product ROO•, in a spin-allowed process (Reaction 5.18). It is a chain reaction, since R• is regenerated in Reaction (5.19), and it frequently occurs with long chain lengths prior to the termination steps, resulting in a very efficient pathway for oxygenation of some organic substrates, such as, for example, the oxidation of cumene to give phenol and acetone (Reaction 5.22). \(\tag{5.22}\) When free-radical autoxidation is used for synthetic purposes, initiators are intentionally added. Common initiators are peroxides and other compounds capable of fragmenting readily into free radicals. Free-radical autoxidation reactions are also frequently observed when no initiator has been intentionally added, because organic substrates frequently contain peroxidic impurities that may act as initiators. Investigators have sometimes been deceived into assuming that a metal-complex catalyzed reaction of dioxygen with an organic substrate occurred by a nonradical mechanism. In such instances, the reactions later proved, upon further study, to be free-radical autoxidations, the role of the metal complex having been to generate the initiating free radicals. Although often useful for synthesis of oxygenated derivatives of relatively simple hydrocarbons, free-radical autoxidation lacks selectivity and therefore, with more complex substrates, tends to give multiple products. In considering possible mechanisms for biological oxidation reactions used for biosynthesis or energy production, free-radical autoxidation is not an attractive possibility, because such a mechanism requires diffusion of highly reactive free radicals. Such radicals, produced in the cell, will react indiscriminately with vulnerable sites on enzymes, substrates, and other cell components, causing serious damage. In fact, free-radical autoxidation is believed to cause certain deleterious reactions of dioxygen in biological systems, for example the oxidation of lipids in membranes. It is also the process that causes fats and oils to become rancid (Reaction 5.23). \(\tag{5.23}\) We see then the reasons that uncatalyzed reactions of dioxygen are usually either slow or unselective. The functions of the metalloenzymes for which dioxygen is a substrate are, therefore, to overcome the kinetic barriers imposed by spin restrictions or unfavorable one-electron reduction pathways, and, for the oxygenase enzymes, to direct the reactions and make them highly specific. It is instructive to consider (1) how these metalloenzymes function to lower the kinetic barriers to dioxygen reactivity, and (2) how the oxygenase enzymes redirect the reactions along different pathways so that very different products are obtained. The first example given below is cytochrome c oxidase. This enzyme catalyzes the four-electron reduction of dioxygen. It overcomes the kinetic barriers to dioxygen reduction by binding dioxygen to two paramagnetic metal ions at the dioxygen binding site, thus overcoming the spin restriction, and by reducing dioxygen in a two-electron step to peroxide, thus bypassing the unfavorable one-electron reduction to form free superoxide. The reaction occurs in a very controlled fashion, so that the energy released by dioxygen reduction can be used to produce ATP. A second example is provided by the catechol dioxygenases, which appear to represent substrate rather than dioxygen activation, and in which dioxygen seems to react with the substrate while it is complexed to the paramagnetic iron center. Another example given below is the monooxygenase enzyme cytochrome PASO, which catalyzes the reaction of dioxygen with organic substrates. It binds dioxygen at the paramagnetic metal ion at its active site, thus overcoming the spin restriction, and then carries out what can be formally described as a multielectron reduction of dioxygen to give a highly reactive high-valent metal-oxo species that has reactivity like that of the hydroxyl radical. Unlike a free hydroxyl radical, however, which would be highly reactive but nonselective, the reaction that occurs at the active site of cytochrome P-450 can be highly selective and stereospecific, because the highly reactive metal-oxo moiety is generated close to a substrate that is bound to the enzyme in such a way that it directs the reactive oxygen atom to the correct position. Thus, metalloenzymes have evolved to bind dioxygen and to increase while controlling its reactivity. | 13,842 | 3,162 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/18%3A_Solubility_and_Complex-Ion_Equilibria/18.7%3A_Solubility_and_pH |
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH. We begin our discussion by examining the effect of pH on the solubility of a representative salt, \(\ce{M^{+}A^{−}}\), where \(\ce{A^{−}}\) is the conjugate base of the weak acid \(\ce{HA}\). When the salt dissolves in water, the following reaction occurs: \[\ce{MA (s) \rightleftharpoons M^{+} (aq) + A^{-} (aq)} \label{17.13a} \nonumber\] with \[K_{sp} = [\ce{M^{+}},\ce{A^{−}}] \label{17.13b} \nonumber\] The anion can also react with water in a hydrolysis reaction: \[\ce{A^{-} (aq) + H2O (l) \rightleftharpoons OH^{-} (aq) + HA (aq)} \label{17.14}\] Because of the reaction described in , the predicted solubility of a sparingly soluble salt that has a basic anion such as S , PO , or CO is increased. If instead a strong acid is added to the solution, the added H will react essentially completely with A to form HA. This reaction decreases [A ], which decreases the magnitude of the ion product \[Q = [\ce{M^{+}},\ce{A^{-}}]\] According to , more MA will dissolve until \(Q = K_{sp}\). Hence an acidic pH dramatically increases the solubility of virtually all sparingly soluble salts whose anion is the conjugate base of a weak acid. In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a strong acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH) is relatively insoluble in water: \[Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+} (aq) + 2OH^− (aq) \label{17.15a}\] with \[K_{sp} = 5.61 \times 10^{−12} \label{17.15b}\] When acid is added to a saturated solution that contains excess solid Mg(OH) , the following reaction occurs, removing OH from solution: \[H^+ (aq) + OH^− (aq) \rightarrow H_2O (l) \label{17.16}\] The overall equation for the reaction of Mg(OH) with acid is thus \[Mg(OH)_{2(s)} + 2H^+ (aq) \rightleftharpoons Mg^{2+} (aq) + 2H_2O (l) \label{17.17}\] As more acid is added to a suspension of Mg(OH) , the equilibrium shown in is driven to the right, so more Mg(OH) dissolves. Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF is a sparingly soluble salt: \[CaF_{2(s)} \rightleftharpoons Ca^{2+} (aq) + 2F^− (aq) \label{17.18a}\] with \[K_{sp} = 3.45 \times 10^{−11} \label{17.18b}\] When strong acid is added to a saturated solution of CaF , the following reaction occurs: \[H^+ (aq) + F^− (aq) \rightleftharpoons HF (aq) \label{17.19}\] Because the forward reaction decreases the fluoride ion concentration, more CaF dissolves to relieve the stress on the system. The net reaction of CaF with strong acid is thus \[CaF_{2(s)} + 2H^+ (aq) \rightarrow Ca^{2+} (aq) + 2HF (aq) \label{17.20}\] Example \(\Page {1}\) shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt. Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution. Lead oxalate (PbC O ), lead iodide (PbI ), and lead sulfate (PbSO ) are all rather insoluble, with values of 4.8 × 10 , 9.8 × 10 , and 2.53 × 10 , respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities? values for three compounds relative solubilities in acid solution Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions. The solubility Equilibria for the three salts are as follows: \[PbC_2O_{4(s)} \rightleftharpoons Pb^{2+} (aq) + C_2O^{2−}_{4(aq)} \nonumber\] \[PbI_{2(s)} \rightleftharpoons Pb^{2+} (aq) + 2I^− (aq) \nonumber\] \[PbSO_{4(s)} \rightleftharpoons Pb^{2+} (aq) + SO^{2−}_{4(aq)} \nonumber\] The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI will not greatly affect its solubility; the acid will simply dissociate to form H (aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO CCO H), which is a weak diprotic acid (p = 1.23 and p = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions: \[C_2O^{2−}_{4(aq)} + H^+ (aq) \rightarrow HO_2CCO^−_{2(aq)} \nonumber\] \[HO_2CCO^−_{2(aq)} + H^+ (aq) \rightarrow HO_2CCO_2H (aq) \nonumber\] These reactions will decrease [C O ], causing more lead oxalate to dissolve to relieve the stress on the system. The p of HSO (1.99) is similar in magnitude to the p of oxalic acid, so adding a strong acid to a saturated solution of PbSO will result in the following reaction: \[ SO^{2-}_{4(aq)} + H^+ (aq) \rightleftharpoons HSO^-_{4(aq)} \nonumber\] Because HSO has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO will be significantly less than for PbC O . Which of the following insoluble salts—AgCl, Ag CO , Ag PO , and/or AgBr—will be substantially more soluble in 1.0 M HNO than in pure water? Ag CO and Ag PO Solubility Products and pH: Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility Equilbria(part (a) in ). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows: \[\ce{CO2(aq) + H2O (l) \rightleftharpoons H^{+} (aq) + HCO^{−}3(aq)} \label{17.21}\] \[\ce{HCO^{−}3(aq) \rightleftharpoons H^{+} (aq) + CO^{2-}3(aq)} \label{17.22}\] \[\ce{Ca^{2+} (aq) + CO^{2−}3(aq) \rightleftharpoons CaCO3(s)} \label{17.23}\] Limestone deposits that form caves consist primarily of CaCO from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO in CO -rich water rises toward Earth’s surface or is otherwise heated, CO gas is released as the water warms. CaCO then precipitates from the solution according to the following equation (part (b) in ): \[Ca^{2+} (aq) + 2HCO^−_{3(aq)} \rightleftharpoons CaCO_{3(s)} + CO_{2(g)} + H_2O (l) \label{17.24}\] The forward direction is the same reaction that produces the solid called scale in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated. When groundwater-containing atmospheric CO ( finds its way into microscopic cracks in the limestone deposits, CaCO dissolves in the acidic solution in the reverse direction of . The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in to shift to the right. A circular layer of solid CaCO is deposited, which eventually produces a long, hollow spire of limestone called a stalactite that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a stalagmite, to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China ( ). One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. There is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements are acidic oxides. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water: \[Cs_2O (s) + H_2O (l) \rightarrow 2Cs^+ (aq) + 2OH^− (aq) \label{17.25}\] \[SO_{3(g)} + H_2O (l) \rightarrow H_2SO_{4(aq)} \label{17.26}\] Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions. The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete M cations and O anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: E –O . The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO , which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO ): \[MoO_{3(s)} + 2OH^− (aq) \rightarrow MoO^{2−}_{4(aq)} + H_2O (l) \label{17.27}\] As shown in Figure \(\Page {3}\), there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic and basic solutions; consequently, they are called (from the Greek ampho, meaning “both,” as in amphiprotic). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in for example, mixing the amphoteric oxide Cr(OH) (also written as Cr O •3H O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH) to dissolve to give a bright violet solution of Cr (aq), which contains the [Cr(H O) ] ion, whereas adding strong base gives a green solution of the [Cr(OH) ] ion. The chemical equations for the reactions are as follows: \[\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \label{17.28}\] Aluminum hydroxide, written as either Al(OH) or Al O •3H O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base. amphoteric compound dissolution reactions in acid and base Using and \(\ref{17.29}\) as a guide, write the dissolution reactions in acid and base solutions. \[Al(OH)_{3(s)} + 3H^+ (aq) \rightarrow Al^{3+} (aq) + 3H_2O (l) \nonumber\] In aqueous solution, Al forms the complex ion [Al(H O) ] . \[Al(OH)_{3(s)} + OH^− (aq) \rightarrow [Al(OH)_4]^− (aq) \nonumber\] Copper(II) hydroxide, written as either Cu(OH) or CuO•H O, is . Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base. \[Cu(OH)_{2(s)} + 2H^+ (aq) \rightarrow Cu^{2+} (aq) + 2H_2O (l) \nonumber\] \[Cu(OH)_{2(s)} + 2OH^− (aq) \rightarrow [Cu(OH)_4]^2_{−(aq)} \nonumber\] Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate. The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations. Suppose, for example, we have a solution that contains 1.0 mM Zn and 1.0 mM Cd and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilbria can be written as follows: \[ZnS (s) \rightleftharpoons Zn^{2+} (aq) + S^{2−} (aq) \label{17.30a}\] with \[K_{sp}= 1.6 \times 10^{−24} \label{17.30b}\] and \[CdS (s) \rightleftharpoons Cd^{2+} (aq) + S^{2−} (aq) \label{17.31a}\] with \[K_{sp} = 8.0 \times 10^{−27} \label{17.31b}\] Because the S ion is quite basic and reacts extensively with water to give HS and OH , the solubility equilbria are more accurately written as \(MS (s) \rightleftharpoons M^{2+} (aq) + HS^− (aq) + OH^−\) rather than \(MS (s) \rightleftharpoons M^{2+} (aq) + S^{2−} (aq) \). Here we use the simpler form involving S , which is justified because we take the reaction of S with water into account later in the solution, arriving at the same answer using either equilibrium equation. The sulfide concentrations needed to cause \(ZnS\) and \(CdS\) to precipitate are as follows: \[K_{sp} = [Zn^{2+},S^{2−}] \label{17.32a}\] \[1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\label{17.32b}\] \[1.6 \times 10^{−21}\; M = [S^{2−}]\label{17.32c}\] and \[K_{sp} = [Cd^{2+},S^{2−}] \label{17.33a}\] \[8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\label{17.33b}\] \[8.0 \times 10^{−24}\; M = [S^{2−}] \label{17.33c}\] Thus sulfide concentrations between 1.6 × 10 M and 8.0 × 10 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H S contains 0.10 M H S at 20°C. The p for H S is 6.97, and p corresponding to the formation of [S ] is 12.90. The equations for these reactions are as follows: \[H_2S (aq) \rightleftharpoons H^+ (aq) + HS^− (aq) \label{17.34a}\] with \[pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \label{17.34b}\] \[HS^− (aq) \rightleftharpoons H^+ (aq) + S^{2−} (aq) \label{17.34c}\] with \[pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \label{17.34d}\] We can show that the concentration of S is 1.3 × 10 by comparing and and recognizing that the contribution to [H ] from the dissociation of HS is negligible compared with [H ] from the dissociation of H S. Thus substituting 0.10 M in the equation for for the concentration of H S, which is essentially constant regardless of the pH, gives the following: Substituting this value for [H ] and [HS ] into the equation for , \[K_{\textrm{a2}}=1.3\times10^{-13}=\dfrac{[\mathrm{H^+},\mathrm{S^{2-}}]}{[\mathrm{HS^-}]}=\dfrac{(1.1\times10^{-4}\textrm{ M})x}{1.1\times10^{-4}\textrm{ M}}=x=[\mathrm{S^{2-}}]\] Although [S ] in an H S solution is very low (1.3 × 10 M), bubbling H S through the solution until it is saturated would precipitate both metal ions because the concentration of S would then be much greater than 1.6 × 10 M. Thus we must adjust [S ] to stay within the desired range. The most direct way to do this is to adjust [H ] by adding acid to the H S solution (recall ), thereby driving the equilibrium in to the left. The overall equation for the dissociation of H S is as follows: \[H_2S (aq) \rightleftharpoons 2H^+ (aq) + S^{2−} (aq) \label{17.36}\] Now we can use the equilibrium constant for the overall reaction, which is the product of and , and the concentration of H S in a saturated solution to calculate the H concentration needed to produce [S ] of 1.6 × 10 M: \[K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \label{17.37}\] Thus adding a strong acid such as HCl to make the solution 0.94 M in H will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H S. A solution contains 0.010 M Ca and 0.010 M La . What concentration of HCl is needed to precipitate La (C O ) •9H O but not Ca(C O )•H O if the concentration of oxalic acid is 1.0 M? values are 2.32 × 10 for Ca(C O ) and 2.5 × 10 for La (C O ) ; p = 1.25 and p = 3.81 for oxalic acid. concentrations of cations, values, and concentration and p values for oxalic acid concentration of HCl needed for selective precipitation of La (C O ) Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox for oxalate, we write the solubility product expression for calcium oxalate as follows: \[K_{sp} = [Ca^{2+},ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \nonumber\] \[[ox^{2−}] = 2.32 \times 10^{−7}\; M \nonumber\] The expression for lanthanum oxalate is as follows: \[K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \nonumber\] \[[ox^{2−}] = 2.9 \times 10^{−8}\; M \nonumber\] Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between \(2.9 \times 10^{−8} M\) and \(2.32 \times 10^{−7} M\). To prevent Ca from precipitating as calcium oxalate, we must add enough H to give a maximum oxalate concentration of 2.32 × 10 M. We can calculate the required [H ] by using the overall equation for the dissociation of oxalic acid to oxalate: \[HO_2CCO_2H (aq) \rightleftharpoons 2H^+ (aq) + C_2O^{2−}_{4(aq)}\] = = (10 )(10 ) = 10 = 8.7×10 Substituting the desired oxalate concentration into the equilibrium constant expression, \[\begin{align}8.7\times10^{-6}=\dfrac{[\mathrm{H^+}]^2[\mathrm{ox^{2-}}]}{[\mathrm{HO_2CCO_2H}]} &= \dfrac{[\mathrm{H^+}]^2(2.32\times10^{-7})}{1.0} \\ [\mathrm{H^+}] &=\textrm{6.1 M}\end{align} \nonumber\] Thus adding enough HCl to give [H ] = 6.1 M will cause only La (C O ) •9H O to precipitate from the solution. A solution contains 0.015 M Fe and 0.015 M Pb . What concentration of acid is needed to ensure that Pb precipitates as PbS in a saturated solution of H S, but Fe does not precipitate as FeS? values are 6.3 × 10 for FeS and 8.0 × 10 for PbS. 0.018 M H The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. Acidic oxides either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. Basic oxides either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called amphoteric oxides. Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation. | 20,141 | 3,163 |
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The of a solution are those which depend on the number of particles (and hence the amount) of solute dissolved in a given quantity of solvent, irrespective of the chemical nature of those particles. We have already seen from that the vapor pressure of a solution depends on the mole fraction of solute (amount of solute), and now we are in a position to see how this affects several other properties of solutions. In the image below, there are two solutions. Both have a dark blue solvent and the same number of solute particles dissolved in said solvent, light blue particles on the left and green particles on the right. A colligative property is a property that would be the same for both of the solutions below, even though they contain different solutes. Since they contain the same of solute particles, any colligative properties of the solutions would be identical. | 890 | 3,164 |
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Equation (1) from can tell us a lot more than this about gases, however. If both sides are multiplied by , we have \[PV=\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}} \label{1} \] is ½ ( ) , and so the average kinetic energy ( ) of a collection of molecules, all of the same mass is \[( E_{k})_{\text{ave}} = ( \frac{1}{2} m u^{2})_{\text{ave}} = \frac{1}{2} m (u^{2})_{\text{ave}} \nonumber \] The total kinetic energy is just the number of molecules times this average: \[E_{k} = N \times (E_{k})_{\text{ave}} = N \times \frac{1}{2} m (u^{2})_{\text{ave}} \nonumber \] or, multiplying both sides by 3/3 (i.e., by 1) \[E_{k}=\tfrac{\text{3}}{\text{3}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{2}}Nm(u^{\text{2}})_{\text{ave}}=\tfrac{\text{3}}{\text{2}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}} \nonumber \] Substituting from Eq. \[E_{k}=\tfrac{\text{3}}{\text{2}}PV \nonumber \] or \[PV=\tfrac{\text{2}}{\text{3}}E_{k} \nonumber \] . Now we can understand why comes out in joules—it is indeed energy. According to postulate 4 of the , gas molecules have constant total kinetic energy. This is reflected on the macroscopic scale by the constancy of , or, in other words, by . The kinetic theory also gives an important insight into what the of gas means on a microscopic level. We know from the that = . Substituting this into Eq. \(\ref{6}\), \[nRT = \tfrac{2}{3} E_{k}\label{6} \] If we divide both sides of Eq. \(\ref{7}\) by and multiply by \(\tfrac{3}{2}\), \[\frac{E_{k}}{n}=\tfrac{\text{3}}{2}RT\label{7} \] The term / is the total kinetic energy divided by the amount of substance, that is, the . Representing molar kinetic energy by we have \[E_{\text{m}}=\tfrac{\text{3}}{2}RT \nonumber \] The molar kinetic energy of a gas is proportional to its temperature, and the proportionality constant is \(\tfrac{3}{2}\) times the gas constant . The video below demonstrates the relationship between molar kinetic energy and temperature. The demonstration highlights the fact that a higher temperature means a higher molar kinetic energy. When food coloring is placed into water of different temperatures, it behaves differently. Food coloring in hot water is rapidly dispersed because of its high molar kinetic energy/temperature. The cold water on the other hand, has a low molar kinetic energy and therefore the food color spreads slowly through it. | 2,453 | 3,165 |
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Because enthalpy is a , the enthalpy change for a reaction depends on only two things: It does depend on the path by which reactants are converted to products. If you climbed a mountain, for example, the altitude change would not depend on whether you climbed the entire way without stopping or you stopped many times to take a break. If you stopped often, the overall change in altitude would be the sum of the changes in altitude for each short stretch climbed. Similarly, when we add two or more balanced chemical equations to obtain a net chemical equation, \(ΔH\) for the net reaction is the sum of the \(ΔH\) values for the individual reactions. This principle is called , after the Swiss-born Russian chemist Germain Hess (1802–1850), a pioneer in the study of thermochemistry. Hess’s law allows us to calculate ΔH values for reactions that are difficult to carry out directly by adding together the known \(ΔH\) values for individual steps that give the overall reaction, even though the overall reaction may not actually occur via those steps. We can illustrate Hess’s law using the thermite reaction. The overall reaction shown in Equation \(\ref{12.7.1}\) can be viewed as occurring in three distinct steps with known \(ΔH\) values. As shown in Figure \(\Page {1}\): As you can see in Figure \(\Page {1}\), the overall reaction is given by the longest arrow (shown on the left), which is the sum of the three shorter arrows (shown on the right). Adding Equations \ref{12.7.1}a-\ref{12.7.1}c} gives the overall reaction, shown Equation \ref{12.7.1}d: \[ \begin{matrix}
2Al\left ( s, \; 25 ^{o}C \right ) + Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right )& \rightarrow & 2Fe\left ( l, \; 1758 ^{o}C \right ) + Al_2O_3 \left ( s, \; 1758 ^{o}C \right ) & \Delta H=-732.5 \; kJ& \left ( reaction \,a \right ) \\
2Fe\left ( l, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( s, \; 1758 ^{o}C \right ) & \Delta H=-\;\; 27.6 \; kJ & \left (reaction\, b \right )\\
2Fe\left ( s, \; 1758 ^{o}C \right ) + 2Al\left ( s, \; 1758 ^{o}C \right ) & \rightarrow & 2Fe\left ( s, \; 25 ^{o}C \right ) + 2Al\left ( s, \; 25 ^{o}C \right ) & \Delta H=-\;\; 91.0 \; kJ & \left ( reaction \,c \right )\\ \hline
2Al\left ( s, \; 25 ^{o}C \right ) + Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \rightarrow & 2Al\left ( s, \; 25 ^{o}C \right ) + 2Fe_{2}O_{3}\left ( s, \; 25 ^{o}C \right ) & \Delta H=-852.2 \; kJ & \left ( total \,reaction \, d \right )
\end{matrix} \label{12.7.1} \] By Hess’s law, the enthalpy change for part (d) is the sum of the enthalpy changes for parts (a), (b), and (c). In essence, Hess’s law enables us to calculate the enthalpy change for the sum of a series of reactions without having to draw a diagram like that in Figure \(\Page {1}\). Comparing parts (a) and (d) in Equation \(\ref{12.7.1}\) also illustrates an important point: The magnitude of ΔH for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution). When the product is liquid iron at its melting point (part (a) in Equation \(\ref{12.7.1}\)), only 732.5 kJ of heat are released to the surroundings compared with 852 kJ when the product is solid iron at 25°C (part (d) in Equation \(\ref{12.7.1}\)). The difference, 120 kJ, is the amount of energy that is released when 2 mol of liquid iron solidifies and cools to 25°C. It is important to specify the physical state of all reactants and products when writing a thermochemical equation. When using Hess’s law to calculate the value of ΔH for a reaction, follow this procedure: When carbon is burned with limited amounts of oxygen gas (\(\ce{O2}\)), carbon monoxide (\(\ce{CO}\)) is the main product: \[ \ce{ 2C (s) + O2 (g) -> 2CO (g)} \quad\quad \Delta H=-221.0 \; kJ \label{reaction 1} \tag{reaction 1}\] When carbon is burned in excess O , carbon dioxide (CO ) is produced: \[\ce{ C (s) + O2(g) \rightarrow CO2 (g)} \quad\quad \Delta H=-393.5 \; kJ \label{reaction 2} \tag{reaction 2} \] Use this information to calculate the enthalpy change per mole of \(\ce{CO}\) for the reaction of \(\ce{CO}\) with \(\ce{O2}\) to give \(\ce{CO2}\). two balanced chemical equations and their \(ΔH\) values enthalpy change for a third reaction We begin by writing the balanced chemical equation for the reaction of interest: \[ \ce{ CO (g) + 1/2O2 (g) -> CO2 (g)} \quad\quad \Delta H_{rxn}=? \label{reaction 3} \tag{reaction 3}\] There are at least two ways to solve this problem using Hess’s law and the data provided. The simplest is to write two equations that can be added together to give the desired equation and for which the enthalpy changes are known. Observing that \(\ce{CO}\), a reactant in \ref{reaction 2} and a product in Equation \ref{reaction 1}, we can reverse \ref{reaction 1} to give \[ 2CO\left ( g \right ) \rightarrow 2C\left ( s \right ) + O_{2}\left ( g \right ) \; \;\ \; \Delta H=+221.0 \; kJ \nonumber\] Because we have reversed the direction of the reaction, the sign of ΔH is changed. We can use \ref{reaction 2}, as written because its product, CO , is the product we want in \ref{reaction 3},: \[\ce{C (s) + O2(g) -> CO2 (s)} \quad \quad \Delta H=-393.5 \; kJ \] Adding these two equations together does not give the desired reaction, however, because the numbers of C(s) on the left and right sides do not cancel. According to our strategy, we can multiply the second equation by 2 to obtain 2 mol of C(s) as the reactant: \[\ce{ 2C(s) + 2O2 (g) -> 2CO2 (s)} \quad\quad \Delta H=-787.0 \; kJ \] Writing the resulting equations as a sum, along with the enthalpy change for each, gives \[ \begin{matrix}
2CO\left ( g \right ) & \rightarrow & \cancel{2C\left ( s \right )}+\cancel{O_{2}\left ( g \right )} & \Delta H & = & -\Delta H_{1} & = & +221.0 \; kJ \\
\cancel{2C\left ( s \right )}+\cancel{2}O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & -\Delta 2H_{2} & = & -787.0 \; kJ \\
2CO\left ( g \right ) + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = & & -566.0 \; kJ
\end{matrix} \] Note that the overall chemical equation and the enthalpy change for the reaction are both for the reaction of 2 mol of \(\ce{CO}\) with \(\ce{O2}\), and the problem asks for the amount per mole of \(\ce{CO}\). Consequently, we must divide both sides of the final equation and the magnitude of \(ΔH\) by 2: \[ \begin{matrix}
CO\left ( g \right ) + \frac{1}{2}O_{2}\left ( g \right ) & \rightarrow & CO_{2} \left ( g \right ) & \Delta H & = & & -283.0 \; kJ
\end{matrix} \nonumber \] An alternative and equally valid way to solve this problem is to write the two given equations as occurring in steps. Note that we have multiplied the equations by the appropriate factors to allow us to cancel terms: \[ \small \begin{matrix}
\left ( A \right ) & 2C\left ( s \right ) + O_{2}\left ( g \right ) & \rightarrow & \cancel{2CO\left ( g \right )} & \Delta H_{A} & = & \Delta H_{1} & = & +221.0 \; kJ \\
\left ( B \right ) &\cancel{2CO\left ( g \right )} + O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H_{B} & & & = & ? \\
\left ( C \right ) & 2C\left ( s \right ) + 2O_{2}\left ( g \right ) & \rightarrow & 2CO_{2} \left ( g \right ) & \Delta H & = 2\Delta H_{2} & =2\times \left ( -393.5 \; kJ \right ) & =-787.0 \; kJ
\end{matrix} \] The sum of reactions A and B is reaction C, which corresponds to the combustion of 2 mol of carbon to give CO . From Hess’s law, ΔH + ΔH = ΔH , and we are given ΔH for reactions A and C. Substituting the appropriate values gives \[ \begin{align*} -221.0 \; kJ + \Delta H_{B} &= -787.0 \; kJ \\[4pt] \Delta H_{B} &= -566.0 \end{align*} \] This is again the enthalpy change for the conversion of 2 mol of CO to CO . The enthalpy change for the conversion of 1 mol of CO to CO is therefore −566.0 ÷ 2 = −283.0 kJ/mol of CO, which is the same result we obtained earlier. As you can see, there may be more than one correct way to solve a problem. The reaction of acetylene (C H ) with hydrogen (H ) can produce either ethylene (C H ) or ethane (C H ): \[ \begin{matrix}
C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right )& \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \\
C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2}
\end{matrix} \nonumber\] What is ΔH for the reaction of C H with H to form C H ? −136.3 kJ/mol of C H Hess's law is that t For a chemical reaction, the (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of ΔH are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\). ( ) | 8,758 | 3,166 |
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The Arrhenius definition of an acid as a compound that dissolves in water to yield hydronium ions (H O ) and a base as a compound that dissolves in water to yield hydroxide ions (\(\ce{OH-}\)). This definition is not wrong; it is simply limited. We extended the definition of an acid or a base using the more general definition proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry. Their definition centers on the proton, \(\ce{H^+}\). A proton is what remains when a normal hydrogen atom, \(\ce{^1_1H}\), loses an electron. A compound that donates a proton to another compound is called a Brønsted-Lowry acid, and a compound that accepts a proton is called a Brønsted-Lowry base. An acid-base reaction is the transfer of a proton from a proton donor (acid) to a proton acceptor (base). In a subsequent chapter of this text we will introduce the most general model of acid-base behavior introduced by the American chemist G. N. Lewis. Definitions of Acids and Bases: Acids may be compounds such as HCl or H SO , organic acids like acetic acid (\(\ce{CH_3COOH}\)) or ascorbic acid (vitamin C), or H O. Anions (such as \(\ce{HSO_4^-}\), \(\ce{H_2PO_4^-}\), \(\ce{HS^-}\), and \(\ce{HCO_3^-}\)) and cations (such as \(\ce{H_3O^+}\), \(\ce{NH_4^+}\), and \(\ce{[Al(H_2O)_6]^{3+}}\)) may also act as acids. Bases fall into the same three categories. Bases may be neutral molecules (such as \(\ce{H_2O}\), \(\ce{NH_3}\), and \(\ce{CH_3NH_2}\)), anions (such as \(\ce{OH^-}\), \(\ce{HS^-}\), \(\ce{HCO_3^-}\), \(\ce{CO_3^{2−}}\), \(\ce{F^-}\), and \(\ce{PO_4^{3−}}\)), or cations (such as \(\ce{[Al(H_2O)_5OH]^{2+}}\)). The most familiar bases are ionic compounds such as \(\ce{NaOH}\) and \(\ce{Ca(OH)_2}\), which contain the hydroxide ion, \(\ce{OH^-}\). The hydroxide ion in these compounds accepts a proton from acids to form water: \[\ce{H^+ + OH^- \rightarrow H_2O} \label{14.2.1}\] We call the product that remains after an acid donates a proton the of the acid. This species is a base because it can accept a proton (to re-form the acid): \[\text{acid} \rightleftharpoons \text{proton} + \text{conjugate base}\label{14.2.2a}\] \[\ce{HF \rightleftharpoons H^+ + F^-} \label{14.2.2b} \nonumber\] \[\ce{H_2SO_4 \rightleftharpoons H^+ + HSO_4^{−}}\label{14.2.2c} \nonumber\] \[\ce{H_2O \rightleftharpoons H^+ + OH^-}\label{14.2.2d} \nonumber\] \[\ce{HSO_4^- \rightleftharpoons H^+ + SO_4^{2−}}\label{14.2.2e} \nonumber\] \[\ce{NH_4^+ \rightleftharpoons H^+ + NH_3} \label{14.2.2f} \nonumber\] We call the product that results when a base accepts a proton the base’s . This species is an acid because it can give up a proton (and thus re-form the base): \[\text{base} + \text{proton} \rightleftharpoons \text{conjugate acid} \label{14.2.3a}\] \[\ce{OH^- +H^+ \rightleftharpoons H2O}\label{14.2.3b} \nonumber\] \[\ce{H_2O + H^+ \rightleftharpoons H3O+}\label{14.2.3c} \nonumber\] \[\ce{NH_3 +H^+ \rightleftharpoons NH4+}\label{14.2.3d} \nonumber\] \[\ce{S^{2-} +H^+ \rightleftharpoons HS-}\label{14.2.3e} \nonumber\] \[\ce{CO_3^{2-} +H^+ \rightleftharpoons HCO3-}\label{14.2.3f} \nonumber\] \[\ce{F^- +H^+ \rightleftharpoons HF} \label{14.2.3g} \nonumber\] In these two sets of equations, the behaviors of acids as proton donors and bases as proton acceptors are represented in isolation. In reality, all acid-base reactions involve the transfer of protons between acids and bases. For example, consider the acid-base reaction that takes place when ammonia is dissolved in water. A water molecule (functioning as an acid) transfers a proton to an ammonia molecule (functioning as a base), yielding the conjugate base of water, \(\ce{OH^-}\), and the conjugate acid of ammonia, \(\ce{NH4+}\): Similarly, in the reaction of acetic acid with water, acetic acid donates a proton to water, which acts as the base. In the reverse reaction, \(H_3O^+\) is the acid that donates a proton to the acetate ion, which acts as the base. Once again, we have two conjugate acid–base pairs: the parent acid and its conjugate base (\(CH_3CO_2H/CH_3CO_2^−\)) and the parent base and its conjugate acid (\(H_3O^+/H_2O\)). In the reaction of ammonia with water to give ammonium ions and hydroxide ions, ammonia acts as a base by accepting a proton from a water molecule, which in this case means that water is acting as an acid. In the reverse reaction, an ammonium ion acts as an acid by donating a proton to a hydroxide ion, and the hydroxide ion acts as a base. The conjugate acid–base pairs for this reaction are \(NH_4^+/NH_3\) and \(H_2O/OH^−\). Conjugate Acid-Base Pairs: Like water, many molecules and ions may either gain or lose a proton under the appropriate conditions. Such species are said to be . Another term used to describe such species is , which is a more general term for a species that may act either as an acid or a base by any definition (not just the Brønsted-Lowry one). Consider for example the bicarbonate ion, which may either donate or accept a proton as shown here: \[\ce{HCO^{-}3(aq) + H_2O(l) <=> CO^{2-}3(aq) + H_3O^{+}(aq)} \nonumber\] \[ \ce{HCO^{-}3(aq) + H_2O(l) <=> H2CO3(aq) + OH^{-}(aq)} \nonumber\] Water is the most important amphiprotic species. It can form both the hydronium ion, \(\ce{H3O^{+}}\), and the hydroxide ion, \(\ce{OH^-}\) when it undergoes : \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber\] Write separate equations representing the reaction of \(\ce{HSO3-}\) Write separate equations representing the reaction of \(\ce{H2PO4-}\) \[\ce{H2PO4- (aq) + HBr(aq) <=> H3PO4(aq) + Br-(aq)} \nonumber\] \[\ce{H2PO4-}(aq)+\ce{OH^-} (aq)\rightleftharpoons \ce{HPO4^2-}(aq)+ \ce{H_2O}_{(l)} \nonumber \] A compound that can donate a proton (a hydrogen ion) to another compound is called a Brønsted-Lowry acid. The compound that accepts the proton is called a Brønsted-Lowry base. The species remaining after a Brønsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Brønsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H O , and the hydroxide ion, \(\ce{OH^-}\) when it undergoes autoionization: \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}(aq)+\ce{OH^-} (aq) \nonumber\] ). | 6,633 | 3,167 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/11%3A_Fluids/11.04%3A_Vapor_Pressure |
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid ( ), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy ( ) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than . The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization) , where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in . When a liquid is introduced into an evacuated chamber (part (a) in ), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with > will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in ). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in . Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no change in a system, constitute a dynamic equilibrium . In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an container, however, most of the molecules that escape into the vapor phase will collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water ( ) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point ( ). It does depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in . Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. The exponential rise in vapor pressure with increasing temperature in allows us to use natural logarithms to express the nonlinear relationship as a linear one.
\[ ln\left ( P \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C \tag{11.4.1} \] where ln is the natural logarithm of the vapor pressure, is the , is the universal gas constant [8.314 J/(mol·K)], is the temperature in Kelvin, and is the -intercept, which is a constant for any given liquid. A plot of ln versus the inverse of the absolute temperature (1/ ) is a straight line with a slope of −Δ / . , called the Clausius–Clapeyron equation , can be used to calculate the Δ of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine Δ is to measure the vapor pressure of a liquid at two temperatures and insert the values of and for these points into , which is derived from the Clausius–Clapeyron equation:
\[ ln\left ( \dfrac{P_{2}}{P_{1}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right) \tag{11.4.2} \] Conversely, if we know Δ and the vapor pressure at any temperature , we can use to calculate the vapor pressure at any other temperature , as shown in Example 6. The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: From these data, calculate the enthalpy of vaporization (Δ ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) vapor pressures at four temperatures Δ of mercury and vapor pressure at 160°C The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using and find the value of Δ from the slope of the line, an alternative approach is to use to obtain Δ directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvins because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( ) and 120.0°C ( ) into gives \[ ln\left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \] \[ ln\left ( 8.398 \right)=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \] \[ 2.13=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \] \[ \Delta H_{vap} =61,400 \; J/mol = 61.4 \; kJ/mol \] We can now use this value of Δ to calculate the vapor pressure of the liquid ( ) at 160.0°C ( ): Using the relationship = , we have \[ ln\left ( \dfrac{P_{2} }{0.0888 \; Torr} \right)=3.86 \] \[ \dfrac{P_{2} }{0.0888 \; Torr} =e^{3.86} = 47.5 \] \[ P_{2} = 4.21 Torr \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. Exercise The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? 1896°C As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure increases, the boiling point of a liquid increases and vice versa. Use to estimate the following. data in , pressure, and boiling point corresponding boiling point and pressure Exercise Use the data in to estimate the following. Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called or , generates a above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via . Eventually, a is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of . Under these conditions, a liquid exhibits a characteristic that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. are liquids with high vapor pressures, which tend to evaporate readily from an open container; have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its . \[ ln\left ( P \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C \tag{11.4.1} \] \[ ln\left ( \dfrac{P_{2}}{P_{1}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{2}}-\dfrac{1}{T_{1}} \right) \tag{11.4.2}\] What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure? What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility? An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol? Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away? If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer. What is the relationship between the vapor pressure of a liquid and At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure. Acetylene (C H ), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of Δ for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K? The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of Δ for water. The Δ of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm? The normal boiling point of sodium is 883°C. If Δ is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C? An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in to identify the liquid. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in to identify the liquid. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid? If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid? The vapor pressure of liquid SO is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K. The vapor pressure of CO at various temperatures is given in the following table: vapor pressure at 273 K is 3050 mmHg; Δ = 18.7 kJ/mol, 1.44 kJ 12.5°C Δ = 28.9 kJ/mol, -hexane Δ = 7.81 kJ/mol, 36°C | 17,421 | 3,168 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.15%3A_Diffusion_and_Effusion_and_Graham's_Law |
We usually cannot see gases, so we need methods to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride \(\left( \ce{NH_4Cl} \right)\), a white solid that appears in the glass tube as a ring. When a person opens a bottle of perfume in one corner of a large room, it doesn't take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse. A related process to diffusion is effusion. is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass. Scottish chemist Thomas Graham (1805-1869) studied the rates of effusion and diffusion of gases. states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham's law can be understood by comparing two gases (\(\ce{A}\) and \(\ce{B}\)) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation \(KE = \frac{1}{2} mv^2\) where \(m\) is mass and \(v\) is velocity. Setting the kinetic energies of the two gases equal to one another gives: \[\frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2\nonumber \] The equation can be rearranged to solve for the ratio of the velocity of gas \(\ce{A}\) to the velocity of gas \(\ce{B}\) \(\left( \frac{v_A}{v_B} \right)\). \[\frac{v_A^2}{v_B^2} = \frac{m_B}{m_A} \: \: \: \text{which becomes} \: \: \: \frac{v_A}{v_B} = \sqrt{\frac{m_B}{m_A}}\nonumber \] For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for \(m\). Calculate the ratio of diffusion rates of ammonia gas \(\left( \ce{NH_3} \right)\) to hydrogen chloride \(\left( \ce{HCl} \right)\) at the same temperature and pressure. Substitute the molar masses of the gases into Graham's law and solve for the ratio. \[\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{36.46 \: \text{g/mol}}{17.04 \: \text{g/mol}}} = 1.46\nonumber \] The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride. Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1. | 3,360 | 3,169 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/Electrophilic_Hydration_to_Make_Alcohols |
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane. Electrophilic hydration is the reverse and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following: The phrase "electrophilic" literally means "electron loving" (whereas "nucleophilic" means "nucleus loving"). Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see for more details). Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. Regiochemistry deals with where the substituent bonds on the product. 's and 's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation In the mechanism for a 3º alcohol shown above, the is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by electrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater . In other words, connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction. If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products: The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, . Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that . Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products: Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity. Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example: The Oxymercuration - Demercuration reaction is a more efficient reaction to make alcohols and does not allow for rearrangements. However, it requires the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include: Predict the product of each reaction. 1) 2) How does the cyclopropane group affect the reaction? 3) (Hint: What is different about this problem?) 4) (Hint: Consider stereochemistry.) 5) Indicate any shifts as well as the major product: 1) This is a basic electrophilic hydration. 2) The answer is additional side products, but (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such). 3) A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed. 4) . For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp hybridization, and when the water is added on, the carbons change their hybridization to sp . This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic. 5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because . There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes. | 5,851 | 3,170 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.05%3A_Cis-Trans_Isomerism_in_Alkenes |
After completing this section, you should be able to Make certain that you can define, and use in context, the key term below. Your previous studies in chemistry may have prepared you to discuss the nature of a carbon-carbon double bond. If not, you should review Section 1.8 of this course before beginning the present section. It is particularly important that you make molecular models of some simple alkenes to gain insight into the geometry of these compounds. Geometric isomerism (also known as cis-trans isomerism or E-Z isomerism) is a form of stereoisomerism. Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. is a form of stereoisomerism, and is dealt with in a separate Module. These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - and that's what this page will concentrate on. Think about what happens in molecules where there is restricted rotation about carbon bonds - in other words where the carbon-carbon bonds are all single. The next diagram shows two possible configurations of 1,2-dichloroethane. These two models represent exactly the same molecule. You can get from one to the other just by twisting around the carbon-carbon single bond. These molecules are isomers. If you draw a structural formula instead of using models, you have to bear in mind the possibility of this free rotation about single bonds. You must accept that these two structures represent the same molecule: But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene? These two molecules are not the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't! Drawing structural formulae for the last pair of models gives two possible isomers: The most likely example of geometric isomerism you will meet at an introductory level is but-2-ene. In one case, the CH groups are on opposite sides of the double bond, and in the other case they are on the same side. It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as CH CH=CHCH If you write it like this, you will almost certainly miss the fact that there are geometric isomers. If there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120°) around the carbon atoms at the ends of the bond. In other words, use the format shown in the last diagrams above. You obviously need to have restricted rotation somewhere in the molecule. Compounds containing a carbon-carbon double bond have this restricted rotation. (Other sorts of compounds may have restricted rotation as well, but we are concentrating on the case you are most likely to meet when you first come across geometric isomers.) If you have a carbon-carbon double bond, you need to think carefully about the possibility of geometric isomers. Think about this case: Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over. You won't have geometric isomers if there are two groups the same on one end of the bond - in this case, the two pink groups on the left-hand end. So there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this: But you could make things even more different and still have geometric isomers: Here, the blue and green groups are either on the same side of the bond or the opposite side. Or you could go the whole hog and make everything different. You still get geometric isomers, but by now the words and are meaningless. This is where the more sophisticated - notation comes in. As discussed in Section 1.8 the double bond in the molecule ethene (H C=CH ) is created by the overlap of two different sets of orbitals. The C-C σ bond is formed when an orbital from each carbon atom overlaps end to end. Also, the C-C pi bond is created by the side-to-side overlap of a p orbital from each carbon atom. Because they are the result of side-by-side overlap (rather then end-to-end overlap like a sigma bond), If rotation about this bond were to occur, it would involve disrupting the side-by-side overlap between the two 2 orbitals that make up the pi bond. If free rotation were to occur the p-orbitals would have to go through a phase where they are 90° from each other, which would break the pi bond because there would be no overlap. Since the pi bond is essential to the structure of ethene it must not break, so there can be no free rotation about the carbon-carbon sigma bond. The presence of the pi bond thus ‘locks’ the six atoms of ethene into the same plane. Restricted rotation about the double bond means that the relative positions of substituent groups above or below the double bond become significant. This leads to a special kind of isomerism in double bonds. Consider the alkene with the condensed structural formula CH CH=CHCH . We could name it 2-butene, but there are actually two such compounds due to this isomerism. The isomer in which the two methyl (CH ) groups lie on the same side of the molecule is called the cis isomer (Latin , meaning “on this side”) and is named cis-2-butene. The isomer with the two (CH ) groups on opposite sides of the molecule is the trans isomer (Latin , meaning “across”) and is named -2-butene. These two compounds are isomers (or geometric isomers), compounds that have different configurations (groups permanently in different places in space) because of the presence of a rigid structure in their molecule. In general these isomers have different physical, chemical, and physiological properties. It is important to note that the presence of a double bond does necessarily lead to isomerism. Being able to tell if a double bond has the possibility of isomerism is a very important skill. can occur whenever both double-bond carbons are directly attached to a carbon and a hydrogen. In this case, interchanging the substituents on one of the double-bond carbons creates a different isomer. If one of the double-bond carbons of an alkene is attached to two identical groups, isomerism is not possible. Here, interchanging the substituents on one of the double-bond carbons forms an identical molecule. Are the following molecules isomers? Although the two molecules different propenes, these two structures are not really different from each other. Because the one of the double-bond carbons is attached to two identical groups (Hydrogens) it is incapable of forming isomers. The interchange of two substituens seen here does not create a new isomer. If either molecule were flipped over top to bottom, the two would you would look identical. Classify each compound as a isomer, a isomer, or neither. a) trans isomer b) neither c) cis isomer d) cis isomer Which of the following compounds could exist as cis/trans isomers? Draw (& label) both of the isomers for the ones that can. a) CH CH=CHCH b) (CH ) C=CHCH c) H C=CHCH d) CH CH CH=CHBr Draw (& label) the cis and trans isomer for each of the following compound names. If no cis/trans isomerism is possible, write none. Name the following compounds using cis/trans nomenclature a) -4-methylhex-2-ene ( -4-methyl-2-hexene) b) -2,5-dibromohex-3-ene ( -2,5-dibromo-3-hexene) | 8,168 | 3,172 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Tollens_Test |
Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a ketone. It exploits the fact that aldehydes are readily oxidized (see oxidation), whereas ketones are not. Tollens’ test uses a reagent known as Tollens’ reagent, which is a colorless, basic, aqueous solution containing silver ions coordinated to ammonia \([Ag(NH_3)^{2+}]\). It is prepared using a two-step procedure. Aqueous silver nitrate is mixed with aqueous sodium hydroxide. \[ \begin{align*} \ce{AgNO_3 + NaOH} &\rightarrow \ce{AgOH + NaHO_3} \\[4pt] \ce{2AgOH} &\rightarrow \ce{Ag_2O + H_2O} \end{align*} \] Aqueous ammonia is added drop-wise until the precipitated silver oxide completely dissolves. \[\ce{Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^{-}} \nonumber\] Tollens’ reagent oxidizes an aldehyde into the corresponding carboxylic acid. The reaction is accompanied by the reduction of silver ions in Tollens’ reagent into metallic silver, which, if the test is carried out in a clean glass test tube, forms a mirror on the test tube. eg: Ketones are not oxidized by Tollens’ reagent, so the treatment of a ketone with Tollens’ reagent in a glass test tube does not result in a silver mirror (Figure \(\Page {1}\); right). | 1,296 | 3,173 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Hard_Water |
Hard water contains high amounts of minerals in the form of ions, especially the metals calcium and magnesium, which can precipitate out and cause problems in water cconducting or storing vessels like pipes. Hard water can be distinguished from other types of water by its metallic, dry taste and the dry feeling it leaves on skin. It is responsible for the scum rings seen in bathtubs, as well as the inability of soap to lather. Hard water is water containing high amounts of mineral ions. The most common ions found in hard water are the metal cations calcium (Ca ) and magnesium (Mg ), though iron, aluminum, and manganese may also be found in certain areas. These metals are water soluble, meaning they will dissolve in water. The relatively high concentrations of these ions can saturate the solution and consequently cause the equilibrium of these solutes to shift to the left, towards reactants. In other words, the ions can precipitate out of the solution. This displacement of minerals from the solution is responsible for the calcination often seen on water faucets, which is a precipitation of calcium or magnesium carbonate. Hard water may also react with other substances in the solution, such as soap, and form a precipitate called "scum." There are two defined types of hard water, temporary and permanent, which are described below. Temporary hard water is hard water that consists primarily of calcium (Ca ) and bicarbonate (HCO ) ions. Heating causes the bicarbonate ion in temporary hard water to decompose into carbonate ion (CO ), carbon dioxide (CO ), and water (H O). The resultant carbonate ion (CO ) can then react with other ions in the solution to form insoluble compounds, such as CaCO and MgCO . The interactions of carbonate ion in the solution also cause the well-known mineral build-up seen on the sides of pots used to boil water, a rust known as "boiler scale." Increasing the temperature of temporary hard water, with its resultant decomposition of the bicarbonate ion, signifies a shift in the equilibrium equation (shown below). The high temperature causes the equilibrium to shift to the left, causing precipitation of the initial reactants. \[CaCO_{3 \; (s)} + CO_{2 \; (aq)} + H_2O_{(l)} \rightleftharpoons Ca^{2+}_{(aq)} + 2HCO^-_{3 \; (aq)} \tag{1}\] This shift is responsible for the white scale observed in the boiling containers described above, as well as the mineral deposits that build up inside water pipes, resulting in inefficiency and even explosion due to overheating. The CaCO or other scale does not completely dissolve back into the water when it is cooled because it is relatively insoluble, as shown by its small solubility constant. For this reason, this type of hard water is "temporary" because boiling can remove the hardness by displacing the offending ions from solution. \[CaCO_{3 \; (s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2-}_{3 \; (aq)} \tag{2a}\] \[K_{(sp)} = 2.8 \times 10^{-9} \tag{2b}\] Permanent hard water consists of high concentrations of anions, like the sulfate anion (SO ). This type of hard water is referred to as "permanent" because, unlike temporary hard water, the hardness cannot be removed simply by boiling the water and thereby precipitating out the mineral ions. However, the name is deceiving as "permanent" hard water can be softened by other means. The scale caused by permanent hard water has detrimental effects similar to those seen with temporary hard water, such as obstruction of water flow in pipes. Permanent hard water is also responsible for the bathtub "ring," or soap scum, seen after showering or bathing. As previously mentioned, permanent hard water contains calcium and magnesium cations.These cations react with soap to form insoluble compounds that are then deposited on the sides of the tub. Additionally, the reaction of these cations with soap is the reason it is difficult for soap to foam or lather well in hard water. The equation below gives an example of the reaction of magnesium ion with components of soap, in this case stearate (C H O ), to form the insoluble compound magnesium stearate, which is responsible for the infamous soap scum. \[2(C_{18}H_{35}O_2)^{2-}_{(aq)} + Mg^{2+}_{(aq)} \longrightarrow Mg(C_{18}H_{35}O_2)_{2 \; (s)} \tag{3}\] Though the taste of hard water may be unpleasant to some, it has many health benefits when compared to soft water. Two of the most prevalent minerals in hard water are calcium and magnesium. Both calcium and magnesium are considered essential nutrients, meaning that they must be provided in the diet in order to maintain healthy body function. Calcium is a critical component of bones, and has many positive effects on the body, such as prevention of serious life-threatening and painful ailments like osteoporosis, kidney stones, hypertension, stroke, obesity, and coronary artery disease. Magnesium also has positive health effects. Inadequate amounts of magnesium in the body increase the risks for some forms of health problems, such as hypertension, cardiac arrhythmia, coronary heart disease, and diabetes mellitus. Studies done on the health effects of hard and soft water have shown that people who drink greater amounts of soft water have much higher incidences of heart disease, as well as higher blood pressure and cholesterol levels, and faster heart rates than those who drink mostly hard water. Furthermore, soft water is corrosive to pipes, which may allow for toxic substances like lead to contaminate drinking water. Some wish to soften hard water to control its irritating, and in many cases damaging, effects. The diminished ability of soap to lather is not only annoying, but can also be potentially harmful economically. Businesses that depend on the foaming of soap, such as car washes and pet groomers, may wish to soften hard water to avoid excessive use of soap due to a decreased ability to lather. Likewise, it is often necessary to soften water that comes into contact with pipes to avoid the destructive and compromising build-up of deposits. Also, many people may find the calcifying effects that hard water has on faucets and other items unfavorable and choose to soften the water to prevent such mineral deposits from forming. Still others may dislike the sticky, dry feeling left by the precipitation of soap scum onto the skin. Whatever the reasons, there are many processes available to soften hard water. One way to soften water is through a process called ion exchange. During ion exchange, the unwanted ions are "exchanged" for more acceptable ions. In many cases, it is desirable to replace the hard water ions, such as Ca and Mg , with more agreeable ions, like that of Na . To do this, the hard water is conducted through a zeolite or resin-containing column, which binds the unwanted ions to its surface and releases the more tolerable ions. In this process, the hard water ions become "fixed" ions because of their attachment to the resin material. These fixed ions displace the desirable ions (Na ), now referred to as counterions, from the column, thus exchanging the ions in the water. This process is illustrated in Figure 1. Unfortunately, this process has the disadvantage of increasing the sodium content of drinking water, which could be potentially hazardous to the health of people with sodium-restricted diets. Another process is called lime softening. In this process, the compound calcium hydroxide, Ca(OH) , is added to the hard water. The calcium hydroxide, or "slaked lime," raises the pH of the water and causes the calcium and magnesium to precipitate into CaCO and Mg(OH) . These precipitates can then be easily filtered out due to their insolubility in water, shown below by the small solubility constant of magnesium hydroxide (the solubility product constant for calcium carbonate is shown above). After precipitation and removal of the offending ions, acid is added to bring the pH of the water back to normal. \[Mg(OH)_{2 \; (s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \tag{4a}\] \[K_{(sp)} = 1.8 \times 10^{-11} \tag{4b}\] can also be used to soften hard water. Polydentate ligands, such as the popular hexadentate ligand EDTA, bind the undesirable ions in hard water. These ligands are especially helpful in binding the magnesium and calcium cations, which as already mentioned are highly prevalent in hard water solutions. The chelating agent forms a very stable ring complex with the metal cations, which prevents them from interacting with any other substances that may be introduced to the solution, such as soap. In this way, chelators are able to diminish the negative effects associated with hard water. A simplified equation representing the chelation of the metal calcium cation (Ca ) with the hexadentate ligand EDTA is shown below. The large value of the formation constant (K ) reflects the tendency of the reaction to proceed to completion in the forward direction. \[Ca^{2+} + EDTA^{4-} \longrightarrow [Ca(EDTA)]^{2-} \tag{5a}\] \[K_f = 4.9 \times 10^{10} \tag{5b}\] The final process, reverse osmosis, uses high pressures to force the water through a semipermeable membrane. This membrane is generally intended to be impermeable to anything other than water. The membrane serves to filter out the larger ions and molecules responsible for the water's hardness, resulting in softened water. During this process, the water is forced from an area with a high concentration of solute in the form of dissolved metal ions and similar compounds, to an area that is very low in the concentration of these substances. In other words, the water moves from a state of hardness to a softer composition as the ions causing the water's hardness are prevented passage through the membrane. Reverse Osmosis does have a disadvantage of wasting wastewater compared to other water treatment methods. This process is shown in Figure 2 below. Note that this figure describes the desalination of salt water. However, the process for softening hard water is the same. 1. Name the two main types of hard water. (Highlight blue area for the answers) 2. What makes "hard" water hard? 3. What are the two most prevalent ions in hard water? How are these important to the proper function of the body? 4. Name the four described processes for softening hard water. It is also important to understand the essential steps of each process. 5. What is a major disadvantage of ion exchange when Na is used as a counterion? | 10,481 | 3,174 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.04%3A_Rates_of_Spontaneous_Processes |
The phrase points up a major difficulty in dealing with spontaneous processes. Some of them occur quite rapidly, but others are so slow as to be imperceptible. A rapid spontaneous process occurs when 2 mol H O is mixed with 2 mol “heavy water,” D O, made from the isotope deuterium, \({}_{\text{1}}^{\text{2}}\text{H}\), or . The two species start to transfer protons and deuterons (D ions) as soon as they are stirred together, and we rapidly obtain a mixture consisting of 2 mol H—O—D and 1 mol each of H—O—H and D—O—D. Assuming that deuterium atoms behave the same chemically as ordinary hydrogen atoms, this is what the laws of probability would predict. There are four equally likely possibilities for a randomly selected water molecule: The shift from the improbable situation of 2 mol H O + 2 mol D O to the more probable 2 mol HDO + 1 mol H O + 1 mol D O occurs rapidly because of the ease with which protons and deuterons can transfer from one water molecule to another. When such a shuffling process is slow, however, the situation is quite different. For example, we would expect that mixing 2 mol H with 2 mol D would produce 2 mol HD and a mole each of H and D . At room temperature, though, nothing happens, even over a period of days, because there is no easy way for H or D atoms to swap partners. Reshuffling requires breaking an H—H or a D—D bond, and this takes some 400 kJ mol . The molecules are stuck in a situation of low probability because there is no pathway by which they can attain higher probability. If such a pathway is provided, by raising the temperature or adding a catalyst, the molecules start exchanging H and D and move toward the most probable situation. The moral of this story is that saying a reaction is spontaneous is not the same as saying it occur if the reactants are mixed. Rather, it means the reaction occur but may be so slow that nothing seems to happen. In the case of a slow spontaneous reaction it is worthwhile to look for a catalyst, but if we know the reaction is nonspontaneous, there is no point in even mixing the reactants, let alone searching for a catalyst. A nonspontaneous reaction cannot occur of itself without outside intervention. | 2,225 | 3,175 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Secondary_Structure%3A_-Pleated_Sheet |
This structure occurs when two (or more, e.g. ψ-loop) segments of a polypeptide chain overlap one another and form a row of hydrogen bonds with each other. This can happen in a parallel arrangement: Or in anti-parallel arrangement: Parallel and anti-parallel arrangement is the direct consequence of the directionality of the polypeptide chain. In anti-parallel arrangement, the C-terminus end of one segment is on the same side as the N-terminus end of the other segment. In parallel arrangement, the C-terminus end and the N-terminus end are on the same sides for both segments. The "pleat" occurs because of the alternating planes of the peptide bonds between amino acids; the aligned amino and carbonyl group of each opposite segment alternate their orientation from facing towards each other to facing opposite directions. The parallel arrangement is less stable because the geometry of the individual amino acid molecules forces the hydrogen bonds to occur at an angle, making them longer and thus weaker. Contrarily, in the anti-parallel arrangement the hydrogen bonds are aligned directly opposite each other, making for stronger and more stable bonds. Commonly, an anti-parallel beta-pleated sheet forms when a polypeptide chain sharply reverses direction. This can occur in the presence of two consecutive proline residues, which create an angled kink in the polypeptide chain and bend it back upon itself. This is not necessary for distant segments of a polypeptide chain to form beta-pleated sheets, but for proximal segments it is a definite requirement. For short distances, the two segments of a beta-pleated sheet are separated by 4+2n amino acid residues, with 4 being the minimum number of residues. | 1,735 | 3,177 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations |
After you have completed Chapter 11, you should be able to In this course, you have already seen several examples of nucleophilic substitution reactions; now you will see that these reactions can occur by two different mechanisms. You will study the factors that determine which mechanism will be in operation in a given situation, and examine possible ways for increasing or decreasing the rates at which such reactions occur. The stereochemical consequences of both mechanisms will also be discussed. Elimination reactions often accompany nucleophilic substitution; so these reactions are also examined in this chapter. Again you will see that two different mechanisms are possible, and, as in the case of nucleophilic substitution reactions, chemists have learned a great deal about the factors that determine which mechanism will be observed when a given alkyl halide undergoes such a reaction.
| 919 | 3,178 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.10%3A_Collision_Theory |
Collision Theory was first introduced in the 1910s by Max Trautz (Trautz, 1916) and William Lewis (Lewis, 1918) to try to account for the magnitudes of rate constants in terms of the frequency of molecular collisions, the collisional energy, and the relative orientations of the molecules involved in the collision. The rate of a reaction, according to collision theory, can be expressed as \[ \text{rate} = Z_{ab} F \label{majorrate} \] where \(Z_{AB}\) is the frequency of collisions between the molecules \(A\) and \(B\) involved in the reaction, and \(F\) is the fraction of those collisions that will lead to a reaction. The factor \(F\) has important contributors, the energy of the collision and the orientation of the molecules when they collide. The first term, \(Z_{AB}\), can be taken from the kinetic molecular theory discussed in Chapter 2. \[ Z_{AB} = \left( \dfrac{8k_BT}{\pi \mu} \right)^{1/2} \sigma_{AB} [A,B] \label{collision} \] Where the first term is the average relative velocity in which \(\mu\) is the reduced mass of the A-B collisional system, \(\sigma_{AB}\) is the collisional cross section, and \([A]\) and \([B]\) are the concentrations of \(A\) and \(B\). The factor \(F\) depends on the activation energy. Assuming a Boltzmann (or Boltzmann-like) distribution of energies, the fraction of molecular collisions that will have enough energy to overcome the activation barrier is given by \[ F= e^{-E_a/RT} \label{arr} \] Combining Equations \ref{collision} and \ref{arr}, the rate of the reaction (Equation \ref{majorrate}) is predicted by \[ \text{rate} = \left( \dfrac{8k_BT}{\pi \mu} \right) ^{1/2} \sigma_{AB} e^{-E_a/RT} [A,B] \nonumber \] So if the rate law can be expressed as a second order rate law \[\text{rate} = k [A,B] \nonumber \] it is clear that the rate constant \(k\) is given by \[k=\left( \dfrac{8k_BT}{\pi \mu} \right)^{1/2} \sigma_{AB} e^{-E_a/RT} \nonumber \] By comparison, the theory predicts the form of the to be \[A= \left( \dfrac{8k_BT}{\pi \mu} \right)^{1/2} \sigma_{AB} \nonumber \] It should be noted that collision theory appears to apply only to , since it takes two molecules to collide. But there are many reactions that have first order rate laws, but are initiated by bimolecular steps in the mechanisms. (Reaction mechanisms will form a large part of the discussion in Chapter 12.) Consider as an example, the decomposition of \(N_2O_5\), which follows the reaction \[2 N_2O_5 \rightarrow 4 NO_2 + O_2 \nonumber \] Under a certain set of conditions, the following concentrations are observed as a function of time. Graphically, these data look as follows: The data for N O can be analyzed empirically to show that the reaction is first order in N O , with a rate constant of 1.697 x 10 s . (The graph is shown below.) So the rate law for the reaction is \[\text{rate} = 1.697 \times 10^{-3}s^{-1} [N_2O_5] \nonumber \] So how can collision theory be used to understand the rate constant? As it turns out, the mechanism for the reaction involves a bimolecular initiation step. The mechanism for the reaction has a bimolecular initiation step \[ N_2O_5 + M \rightleftharpoons N_2O_5^* + M \nonumber \] where \(N_2O_5^*\) is an energetically activated form of \(N_2O_5\) which can either relax to reform \(N_2O_5\) or decompose to form the products of the reaction. Because the initiation step is bimolecular, collision theory be used to understand the rate law, but because the product of the unimolecular step undergoes slow conversion to products unimolecularly, the overall rate is observed to be first order in \(N_2O_5\). The analysis of reaction mechanisms, and reconciliation with observed rate laws, form the subjects of Chapter 12. | 3,735 | 3,179 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Cyanohydrins |
Cyanohydrins have the structural formula of R C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. In order to form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins). Cyanohydrin reactions occurs when an gets treated by a cyanide anion (such as HCN) or a nitrile forming a cyanohydrin product. This special reaction is a nucleophilic addition, where the nucleophilic CN attacks the electrophilic carbonyl carbon on the ketone, following a protonation by HCN, thereby the cyanide anion being regenerated. This reaction is also reversible. Cyanohydrins are also intermediates for the . The preparation of displacements of sulfite by cyanide salts are also followed under cyanohydrins. Acid-catalysed hydrolysis of silylated cyanohydrins has recently been shown to give cyanohydrins instead of ketones; thus an efficient synthesis of cyanohydrins has been found which works with even highly hindered ketones. Acetone cyanohydrins (ACH) have the structural formula of (CH ) C(OH)CN and are extremely hazardous substances, since they rapidly decomposes in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid. Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide. Other cyanohydrins, excluding acetone cyanohydrins, are: mandelonitrile and glycolonitrile. Mandelonitrile have a structural formula of C H CH(OH)CN and occur in pits of some fruits. Glycolonitrile is an organic compound with the structural formula of HOCH CN, which is the simplest cyanohydrin that is derived by formaldehydes. Complete the following reactions for cyanohydrins: 1.) 2.) 3.) True or False: For a cyanohydrin to form, a of strong acid to cyanide salt is carried out to convert the salt into HCN. 4.) True or False: Cyanohydrin reactions are . 5.) What is the product for the overall reaction? 1.) 2.) 3.) False, slow addition 4.) False, reversible 5.) | 2,194 | 3,180 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/06%3A_Risk_Assessment_and_Regulation/6.07%3A_Risk_perception |
: Fred Woudenberg : Ortwin Renn Risk perception, fear, worry, risk, context If risk perception had a first law like toxicology has with Paracelsus' "Sola dosis facit venenum" (see section on ) it would be: "People fear things that do not make them sick and get sick from things they do not fear." People can, for instance, worry extremely over a newly discovered soil pollution site in their neighborhood, which they hear about at a public meeting they have come to with their diesel car, and, when returning home, light an extra cigarette without thinking to relieve the stress. The explanation for this first law is quite easy. The annual risk of getting sick, being injured or to die has only limited influence on the perception of a risk. Other factors are more important. Figure 1 is a model of risk perception in its most basic form. In the middle of this figure, there is a list with several factors which determine risk perception to a large extent. In any given situation, they can each end up at the left, safe side or on the right, dangerous side. The model is a simplification. Research since the late sixties of the previous century has led to a collection of many more factors that often are connected to each other (for lectures of some well-known researchers see examples , , , and ). An example can show this interconnection and the discrepancy between the annual health risks (at the top of Figure 1) and other factors. The risk of harmful health effects for people living on polluted soil is often very small. The factor 'risk' thus ends at the left, safe side. Most of the other factors end up at the right. People do not voluntary choose to have a soil pollution in their garden. They have absolutely no control over the situation and an eventual sanitation. For this, they depend on authorities and companies. Nowadays, trust in authorities and companies is very small. Many people will suspect that these authorities care more about their money than about the health and well-being of their citizens and neighbours. A newly discovered soil pollution will get local media attention and this will certainly be the case if there is controversy. If the untrusted authorities share their conclusion that the risks are low, many people will suspect that they withhold information and are not completely open. Especially saying that there is 'no cause for alarm' will only make people worry more (see a ). People will not believe the conclusion of authorities that the risk is low, so effectively all factors end up at the dangerous side. For smoking a cigarette, the evaluation is the other way around. Almost everybody knows that smoking is dangerous, but people light their cigarette themselves. Most people at least think they have control over their smoking habit as they can decide to stop at any moment (but being addicted, they probably highly overestimate their level of control). For their information or taking measures, people do not depend on others and no information is withheld about smoking. Some smokers suffer from what is called optimistic bias, the idea that misery only happens to other. They always have the example of their grandfather who started smoking at 12 and still ran the marathon at 85. People can be upset if they learn that cigarette companies purposely make cigarettes more addictive. It makes them feel the company takes over control which people greatly resent. This, and not the health effects, can make people decide to quit smoking. This also explains why passive smoking is more effective than active smoking in influencing people's perceptions. Although the risk of passive smoking is 100 times smaller than the risk of active smoking, most factors end up at the right, dangerous side, making passive smoking maybe 100 times more objectionable and worrisome than active smoking. Many people are surprised to find out that the calculated or estimated health risks influences risk perception so little. But we experience it in our own daily lives, especially when we add another factor in the model: advantages. All of us perform risky activities because they are necessary, come with advantages or sometimes out of sheer fun. Most of us take part in daily traffic with an annual risk of dying far higher than 1 in a million. Once, twice or even more times a year we go on a holiday with a multitude of risks: transport, microbes, robbery, divorce. The thrill seekers of us go diving, mountain climbing, parachute jumping without even knowing the annual fatality rates. If the stakes are high, people can knowingly risk their life in order to improve it, as the thousands of immigrants trying to cross the Mediterranean illustrate, or even to give their life for a higher cause, like soldiers at war (Winston Churchill in 1940: "I have nothing to offer but blood, toil, tears and sweat"). An example at the other side can show it maybe even clearer. No matter how small the risk, it can be totally unacceptable and nonsensical. Suppose the government starts a new lottery with an extremely small chance of winning, say one in a billion. Every citizen must play and tickets are free. So far nothing strange, but there is a twitch. The main and only price of the lottery is a public execution broadcasted live on national TV. The government will probably not make itself very popular with this absurd lottery. When government, as still is done, tells people they have to accept a small risk because they accept larger risks of activities they choose themselves, it makes people feel they have been given a ticket in the above mentioned lottery. This is how people can feel if the government tells them the risk of the polluted soil they live on is extremely small and that it would be wiser for them to quit smoking. A main lesson which can be learned from the study of risk perception is that risks always occur in a context. A risk is always part of a situation or activity which has many more characteristics than only the chance of getting sick, being injured or to die. We do not judge risks, we judge situations and activities of which the risk is often only a small part. Risk perception occurs in a rich environment. After 50 years of research a lot has been discovered, but predicting how angry of afraid people will be in a new, unknown situation is still a daunting task. Name at least 5 important determinants of risk perception. Do people judge risks in isolation or the situation or activities they are part of? How large in general is the influence of the risk on risk perception? Do experts react differently from lay people if they encounter risks in their own daily lives? | 6,666 | 3,181 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Colors_of_Coordination_Complexes |
The color for a can be predicted using the . Knowing the color can have a number of useful applications, such as the creation of pigments for dyes in the textile industry. The tendency for coordination complexes to display such a wide array of colors is merely coincidental; their absorption energies happen to fall within range of the visible light spectrum. Chemists and physicists often study the color of a substance not to understand its sheer appearance, but because color is an indicator of a chemical's physical proprieties on the atomic level. The (EM) spectrum is made up of photons of different wavelengths. Photons, unique in displaying the properties of both waves and particles, create visible light and colors in a small portion of the EM spectrum. This visible light portion has wavelengths in approximately the 400-700 nanometer range (a nanometer, “nm,” is 10 meters). Each specific wavelength corresponds to a different color (Figure \(\Page {1}\)), and when all the wavelengths are present, it appears as white light. The wavelength and frequency of a wave are inversely proportional: as one increases, the other decreases; this is a consequence of all light traveling at the same speed. \[\lambda \propto \nu^{-1}\] Because of this relationship, blue light has a much higher frequency and more energy than red light. Color is perceived in two ways, through additive mixing, where different colors are made by combining different colors of light, and through subtractive mixing, where different wavelengths of light are taken out so that the light is no longer pure white. For colors of coordination complexes, subtractive mixing is considered. As shown in Figure \(\Page {2}\), the idea behind subtractive mixing is that white light (which is made from all the colors mixed together) interacts with an object. The object absorbs some of the light, and then reflects or transmits (or both, depending on the object) the rest of the light, which contacts the eye. The object is perceived as whichever color is not absorbed. In Figure \(\Page {2}\), white light (simplified as green, red, and blue bands) is shone through a solution. The solution absorbs the red and green wavelengths; however, the blue light is reflected and passes through, so the solution appears blue. This procedure takes place whenever an object displays visible color. If none of the light is absorbed, and all is reflected back off, the object appears white; if all of the light is absorbed, and there is none left to reflect or transmit through, the object appears black. When ligands attach to a transition metal to form a coordination complex, electrons in the d orbital split into high energy and low energy orbitals. The difference in energy of the two levels is denoted as ∆, and it is a characteristic a property both of the metal and the ligands. This is illustrated in Figure \(\Page {3}\); the "o" subscript on the \(\Page {4A}\) ligands — ligand ligand A photon equal to the energy difference As certain wavelengths are absorbed in this process, subtractive color mixing occurs and the coordination complex solution becomes colored. If the ions have a noble gas configuration, and have no unpaired electrons, the solutions appear colorless; in reality, they still have a measured energy and absorb certain wavelengths of light, but these wavelengths are not in the visible portion of the EM spectrum and no color is perceived by the eye. In general, a larger \(∆_o\) indicates that higher energy photons are absorbed, and the solution appears further to the left on the EM spectrum shown in Figure \(\Page {1}\). This relationship is described in the equation \[∆_o=hc/λ\] where \(h\) and \(c\) are constants, and \(λ\) is the wavelength of light absorbed. Using a color wheel can be useful for determining what color a solution will appear based on what wavelengths it absorbs (Figure \(\Page {6}\)). If a complex absorbs a particular color, it will have the appearance of whatever color is directly opposite it on the wheel. For example, if a complex is known to absorb photons in the orange range, it can be concluded that the solution will look blue. This concept can be used in reverse to determine ∆ for a complex from the color of its solution. According to the Crystal Field Theory, ligands that have smaller \(\Delta\)) values are considered "weak field" and will absorb lower-energy light with longer \(\lambda\) values (ie a "red shift"). Ligands that have larger \(\Delta\)) values are considered "strong field" and will absorb higher-energy light with shorter \(\lambda\) values (ie a "blue shift"). This relates to the colors seen in a coordination complex. Weaker-field ligands induce the absorption of longer wavelength (lower frequency=lower energy) light than stronger-field ligands since their respective \(\Delta_o\) values are than the electron pairing energy. The energy difference, \(\Delta_o\) \(\Delta_o\)) \(\Delta_o\)) To predict which possible colors and their corresponding wavelengths are absorbed, the can be used: /high spin) If a solution with a dissolved octahedral complex appears yellow to the eye, what wavelength of light does it absorb? Is this complex expected to be low spin or high spin? A solution that looks yellow absorbs light that is violet, which is roughly 410 nm from the color wheel. Since it absorbs high energy, the electrons must be raised to a higher level, and \(\Delta_o\) is high, so the complex is likely to be low spin. An octahedral metal complex absorbs light with wavelength 535 nm. What is the crystal field splitting \(\Delta_o\) for the complex? What color is it to the eye? To solve this question, we need to use the equation \[\Delta_o =\dfrac{hc}{\lambda} \nonumber\] with It is also important to remember that 1 nm is equal to \(1 \times 10^{-9}\) meters. With all this information, the final equation looks like this: \[\Delta_o =\dfrac{(6.625 \times 10^{-34}\, J \cdot s)(2.998 \times 10^8\, m/s)}{(535nm) \left(\dfrac{1\,m}{1 \times 10^9\, nm}\right)}= 3.712 \,J/molecule \nonumber\] It is not necessary to use any equations to solve the second part of the problem. Light that is 535 nm is green, and because green light is absorbed, the complex appears red (refer to Figures \(\Page {1}\) and \(\Page {6}\) for this information). Note: the fact that the complex is octahedral makes no impact when solving this problem. Although the splitting is different for complexes of different structures, the mechanics of solving the problem are identical. In order to solve this problem, it is necessary to know the relative strengths of the ligands involved. A sample ligand strength list is given here, but see Crystal Field Splitting for a more complete list: CN > en > NH > H O > F >SCN > Cl From this information, it is clear that NH is a stronger ligand than Cl , which means that the complex involving NH has a greater ∆, and the complex will be low spin. Because of the larger ∆, the electrons absorb higher energy photons, and the solution will have the appearance of a lower energy color. Since orange light is less energetic than blue light, the NH containing solution is predicted to be orange | 7,212 | 3,182 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.05%3A_Concentration_Cells |
The generation of an electrostatic potential difference is dependent on the creation of a difference in chemical potential between two half-cells. One important manner in which this can be created is by creating a concentration difference. Using the Nernst equation, the potential difference for a concentration cell (one in which both half-cells involve the same half-reaction) can be expressed \[E = -\dfrac{RT}{nF} \ln \dfrac{[\text{oxdizing}]}{[\text{reducing}]} \nonumber \] Calculate the cell potential (at 25 °C) for the concentration cell defined by \[Cu(s) | Cu^{2+} (aq,\, 0.00100 \,M) || Cu^{2+} (aq,\, 0.100 \,M) | Cu(s) \nonumber \] Since the oxidation and reduction half-reactions are the same, \[ E_{cell}^o =0\,V \nonumber \] The cell potential at 25 °C is calculated using the Nernst equation: \[E = -\dfrac{RT}{nF} \ln Q \nonumber \] Substituting the values from the problem: \[ \begin{align*} E_{cell} &= - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96484\,C)} \ln \left( \dfrac{0.00100\,M}{0.100\,M} \right) \\[4pt] &= 0.059\,V \end{align*} \] | 1,073 | 3,183 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.02%3A_The_connection_to_G |
Recall that in addition to being used as a criterion for spontaneity, \(\Delta G\) also indicated the maximum amount of non p-V work a system could produce at constant temperature and pressure. And since w is non p-V work, it seems like a natural fit that \[\Delta G = -nFE \nonumber \] If all of the reactants and products in the electrochemical cell are in their standard states, it follows that \[\Delta G^o = -nFE^o \nonumber \] where \(E^o\) is the . Noting that the molar Gibbs function change can be expressed in terms of the reaction quotient \(Q\) by \[\Delta G = \Delta G^o + RT \ln Q \nonumber \] it follows that \[-nFE = -nFE^o + RT \ln Q \nonumber \] Dividing by \(–nF\) yields \[E = E^o - \dfrac{RT}{nF} \ln Q \nonumber \] which is the . This relationship allows one to calculate the cell potential of a electrochemical cell as a function of the specific activities of the reactants and products. In the Nernst equation, n is the number of electrons transferred per reaction equivalent. For the specific reaction harnessed by Volta in his original battery, E = 0.763 V (at 25 C) and \(n = 2\). So if the Zn and H ions are at a concentration that gives them unit activity, and the H gas is at a partial pressure that gives it unit fugacity: \[ E = 0.763\,V - \dfrac{RT}{nF} \ln (1) = 0/763 \nonumber \] | 1,335 | 3,184 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.09%3A_Chain_Reactions |
A large number of reactions proceed through a series of steps that can collectively be classified as a . The reactions contain steps that can be classified as These types of reactions are very common when the intermediates involved are radicals. An example, is the reaction \[H_2 + Br_2 \rightarrow 2HBr \nonumber \] The observed rate law for this reaction is \[ \text{rate} = \dfrac{k [H_2,Br_2]^{3/2}}{[Br_2] + k'[HBr]} \label{exp} \] A proposed mechanism is \[Br_2 \ce{<=>[k_1,k_{-1}]} 2Br^\cdot \label{step1} \] \[ 2Br^\cdot + H_2 \ce{<=>[k_2,k_{-2}]} HBr + H^\cdot \label{step2} \] \[ H^\cdot + Br_2 \xrightarrow{k_3} HBr + Br^\cdot \label{step3} \] Based on this mechanism, the rate of change of concentrations for the intermediates (\(H^\cdot\) and \(Br^\cdot\)) can be written, and the steady state approximation applied. \[\dfrac{d[H^\cdot]}{dt} = k_2[Br^\cdot,H_2] - k_{-2}[HBr,H^\cdot] - k_3[H^\cdot,Br_2] =0 \nonumber \] \[\dfrac{d[Br^\cdot]}{dt} = 2k_1[Br_2] - 2k_{-1}[Br^\cdot]^2 - k_2[Br^\cdot,H_2] + k_{-2}[HBr,H^\cdot] + k_3[H^\cdot,Br_2] =0 \nonumber \] Adding these two expressions cancels the terms involving \(k_2\), \(k_{-2}\), and \(k_3\). The result is \[ 2 k_1 [Br_2] - 2k_{-1} [Br^\cdot]^2 = 0 \nonumber \] Solving for \(Br^\cdot\) \[ Br^\cdot = \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} \nonumber \] This can be substituted into an expression for the \(H^\cdot\) that is generated by solving the steady state expression for \(d[H^\cdot]/dt\). \[[H^\cdot] = \dfrac{k_2 [Br^\cdot] [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \] so \[[H^\cdot] = \dfrac{k_2 \sqrt{\dfrac{k_1[Br_2]}{k_{-1}}} [H_2]}{k_{-2}[HBr] + k_3[Br_2]} \nonumber \] Now, armed with expressions for \(H^\cdot\) and \(Br^\cdot\), we can substitute them into an expression for the rate of production of the product \(HBr\): \[ \dfrac{[HBr]}{dt} = k_2[Br^\cdot] [H_2] + k_3 [H^\cdot] [Br_2] - k_{-2}[H^\cdot] [HBr] \nonumber \] After substitution and simplification, the result is \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2,Br_2]^{1/2}}{1+ \dfrac{k_{-1}}{k_3} \dfrac{[HBr]}{[Br_2]} } \nonumber \] Multiplying the top and bottom expressions on the right by \([Br_2]\) produces \[ \dfrac{[HBr]}{dt} = \dfrac{2 k_2 \left( \dfrac{k_1}{k_{-1}}\right)^{1/2} [H_2,Br_2]^{3/2}}{[Br_2] + \dfrac{k_{-1}}{k_3} [HBr] } \nonumber \] which matches the form of the rate law found experimentally (Equation \ref{exp})! In this case, \[ k = 2k_2 \sqrt{ \dfrac{k_1}{k_{-1}}} \nonumber \] and \[ k'= \dfrac{k_{-2}}{k_3} \nonumber \] | 2,551 | 3,185 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.06%3A_Collisions_with_Other_Molecules |
A major concern in the design of many experiments is collisions of gas molecules with other molecules in the gas phase. For example, molecular beam experiments are often dependent on a lack of molecular collisions in the beam that could degrade the nature of the molecules in the beam through chemical reactions or simply being knocked out of the beam. In order to predict the frequency of molecular collisions, it is useful to first define the conditions under which collisions will occur. For convenience, consider all of the molecules to be spherical and in fixed in position except for one which is allowed to move through a “sea” of other molecules. A molecular collision will occur every time the center of the moving molecule comes within one molecular diameter of the center of another molecule. One can easily determine the number of molecules the moving molecule will “hit” by determining the number of molecules that lie within the “collision cylinder”. Because we fixed the positions of all but one of the molecules, we must use the relative speed of the moving molecule, which will be given by \[ v_{rel} = \sqrt{2} \times \,v \nonumber \] The volume of the collision cylinder is given by \[ V_{col} = \sqrt{2}\, v\,\Delta t \, A \nonumber \] The , which determined by the size of the molecule is given by \[ \sigma = \pi d^2 \nonumber \] Some values of \(\sigma\) are given in the table below: Since the number of molecules within the collision cylinder is given by \[N_{col} = \dfrac{N}{V} V_{col} \nonumber \] and since the number density (\(N/V\)) is given by \[ \dfrac{N}{V} = \dfrac{p}{k_BT} \nonumber \] the number of collisions is given by \[ N_{col} = \dfrac{p}{k_BT} ( \sqrt{2} \, v \Delta t \sigma) \nonumber \] The frequency of collisions (number of collisions per unit time) is then given by \[ Z = \dfrac{\sqrt{2} p \sigma}{k_B T} \langle v \rangle \nonumber \] Perhaps a more useful value is the (\(\lambda\)), which is the distance a molecule can travel on average before it collides with another molecule. This is easily derived from the collision frequency. How far something can travel between collisions is given by the ratio of how fast it is traveling and how often it hits other molecules: \[ \lambda = \dfrac{\langle v \rangle}{Z} \nonumber \] Thus, the mean free path is given by \[ \lambda = \dfrac{k_B T}{\sqrt{2} p \sigma} \nonumber \] The mere fact that molecules undergo collisions represents a deviation from the kinetic molecular theory. For example, if molecules were infinitesimally small (\(\sigma ≈ 0\)) then the mean free path would be infinitely long! The finite size of molecules represents one significant deviation from ideality. Another important deviation stems from the fact that molecules do exhibit attractive and repulsive forces between one another. These forces depend on a number of parameters, such as the distance between molecules and the temperature (or average kinetic energy of the molecules.) | 2,976 | 3,187 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9C%3A_8.10.9C%3A__Weak_and_Strong_Electrolytes |
The serious study of electrolytic solutions began in the latter part of the 19th century, mostly in Germany — and before the details of dissociation and ionization were well understood. These studies revealed that the equivalent conductivities of electrolytes all diminish with concentration (or more accurately, with the square root of the concentration), but they do so in several distinct ways that are distinguished by their behaviors at very small concentrations. This led to the classification of electrolytes as weak, intermediate, and strong. You will notice that plots of conductivities vs. √ start at =0. It is of course impossible to measure the conductance of an electrolyte at vanishingly small concentrations (not to mention zero!), but for strong and intermediate electrolytes, one can extrapolate a series of observations to zero. The resulting values are known as or sometimes as "equivalent conductances at infinite dilution", designated by Λ°. Since ions are the charge carriers, we might expect the conductivity of a solution to be directly proportional to their concentrations in the solution. So if the electrolyte is totally dissociated, the conductivity should be directly proportional to the electrolyte concentration. But this is never observed; instead, the conductivity of electrolytes of all kinds diminishes as the concentration rises. The non-ideality of electrolytic solutions is also reflected in their , especially and . The primary cause of this is the presence of the ionic atmosphere that was introduced above. To the extent that ions having opposite charge signs are more likely to be closer together, we would expect their charges to partially cancel, reducing their tendency to migrate in response to an applied potential gradient. A secondary effect arises from the fact that as an ion migrates through the solution, its counter-ion cloud does not keep up with it. Instead, new counter-ions are continually acquired on the leading edge of the motion, while existing ones are left behind on the opposite side. It takes some time for the lost counter-ions to dissipate, so there are always more counter-ions on the trailing edge. The resulting asymmetry of the counter-ion field exerts a retarding effect on the central ion, reducing its rate of migration, and thus its contribution to the conductivity of the solution. The curvature of the plots for intermediate electrolytes is a simple consequence of the , which predicts that the equilibrium \[MX_{(aq)} = M^+_{(aq)} + X^–_{(aq)}\] will shift to the left as the concentration of the "free" ions increases. In more dilute solutions, the actual of these ions is smaller, but their fractional abundance in relation to the undissociated form is greater. As the solution approaches zero concentration, virtually all of the \(MX_{ becomes dissociated, and the conductivity reaches its limiting value. Dissociation, of course, is a matter of degree. The equilibrium constants for the dissociation of an intermediate electrolyte salt MX are typically in the range of 1-200. This stands in contrast to the large number of weak acids (as well as weak bases) whose dissociation constants typically range from 10 to smaller than 10 . These weak electrolytes, like the intermediate ones, will be totally dissociated at the limit of zero concentration; if the scale of the weak-electrolyte plot (blue) shown above were magnified by many orders of magnitude, the curve would resemble that for the intermediate electrolyte above it, and a value for Λ° could be found by extrapolation. But at such a high dilution, the conductivity would be so minute that it would be masked by that of water itself (that is, by the H and OH ions in equilibrium with the massive 55.6 M L concentration of water) — making values of Λ in this region virtually unmeasurable. | 3,860 | 3,189 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/09%3A_Gases/9.2%3A_Relating_Pressure_Volume_Amount_and_Temperature%3A_The_Ideal_Gas_Law |
During the seventeenth and especially eighteenth centuries, driven both by a desire to understand nature and a quest to make balloons in which they could fly (Figure \(\Page {1}\)), a number of scientists established the relationships between the macroscopic physical properties of gases, that is, pressure, volume, temperature, and amount of gas. Although their measurements were not precise by today’s standards, they were able to determine the mathematical relationships between pairs of these variables (e.g., pressure and temperature, pressure and volume) that hold for an gas—a hypothetical construct that real gases approximate under certain conditions. Eventually, these individual laws were combined into a single equation—the —that relates gas quantities for gases and is quite accurate for low pressures and moderate temperatures. We will consider the key developments in individual relationships (for pedagogical reasons not quite in historical order), then put them together in the ideal gas law. Imagine filling a rigid container attached to a pressure gauge with gas and then sealing the container so that no gas may escape. If the container is cooled, the gas inside likewise gets colder and its pressure is observed to decrease. Since the container is rigid and tightly sealed, both the volume and number of moles of gas remain constant. If we heat the sphere, the gas inside gets hotter (Figure \(\Page {2}\)) and the pressure increases. This relationship between temperature and pressure is observed for any sample of gas confined to a constant volume. An example of experimental pressure-temperature data is shown for a sample of air under these conditions in Figure \(\Page {3}\). We find that temperature and pressure are linearly related, and if the temperature is on the kelvin scale, then and are directly proportional (again, when ); if the temperature on the kelvin scale increases by a certain factor, the gas pressure increases by the same factor. Guillaume Amontons was the first to empirically establish the relationship between the pressure and the temperature of a gas (~1700), and Joseph Louis Gay-Lussac determined the relationship more precisely (~1800). Because of this, the - relationship for gases is known as either or . Under either name, it states that . Mathematically, this can be written: \[P∝T\ce{\:or\:}P=\ce{constant}×T\ce{\:or\:}P=k×T \nonumber \] where ∝ means “is proportional to,” and is a proportionality constant that depends on the identity, amount, and volume of the gas. For a confined, constant volume of gas, the ratio \(\dfrac{P}{T}\) is therefore constant (i.e., \(\dfrac{P}{T}=k\)). If the gas is initially in “Condition 1” (with = and = ), and then changes to “Condition 2” (with = and = ), we have that \(\dfrac{P_1}{T_1}=k\) and \(\dfrac{P_2}{T_2}=k\), which reduces to \(\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\). This equation is useful for pressure-temperature calculations for a confined gas at constant volume. Note that temperatures must be on the kelvin scale for any gas law calculations (0 on the kelvin scale and the lowest possible temperature is called ). (Also note that there are at least three ways we can describe how the pressure of a gas changes as its temperature changes: We can use a table of values, a graph, or a mathematical equation.) A can of hair spray is used until it is empty except for the propellant, isobutane gas. Rearranging and solving gives: \(P_2=\mathrm{\dfrac{360\:kPa×323\cancel{K}}{297\cancel{K}}=390\:kPa}\) A sample of nitrogen, N , occupies 45.0 mL at 27 °C and 600 torr. What pressure will it have if cooled to –73 °C while the volume remains constant? 400 torr If we fill a balloon with air and seal it, the balloon contains a specific amount of air at atmospheric pressure, let’s say 1 atm. If we put the balloon in a refrigerator, the gas inside gets cold and the balloon shrinks (although both the amount of gas and its pressure remain constant). If we make the balloon very cold, it will shrink a great deal, and it expands again when it warms up. These examples of the effect of temperature on the volume of a given amount of a confined gas at constant pressure are true in general: The volume increases as the temperature increases, and decreases as the temperature decreases. Volume-temperature data for a 1-mole sample of methane gas at 1 atm are listed and graphed in Figure \(\Page {2}\). The relationship between the volume and temperature of a given amount of gas at constant pressure is known as Charles’s law in recognition of the French scientist and balloon flight pioneer Jacques Alexandre César Charles. states that . Mathematically, this can be written as: \[VαT\ce{\:or\:}V=\ce{constant}·T\ce{\:or\:}V=k·T \nonumber \] with being a proportionality constant that depends on the amount and pressure of the gas. For a confined, constant pressure gas sample, \(\dfrac{V}{T}\) is constant (i.e., the ratio = ), and as seen with the - relationship, this leads to another form of Charles’s law: \(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\). A sample of carbon dioxide, CO , occupies 0.300 L at 10 °C and 750 torr. What volume will the gas have at 30 °C and 750 torr? Because we are looking for the volume change caused by a temperature change at constant pressure, this is a job for Charles’s law. Taking and as the initial values, as the temperature at which the volume is unknown and as the unknown volume, and converting °C into K we have: Rearranging and solving gives: \(V_2=\mathrm{\dfrac{0.300\:L×303\cancel{K}}{283\cancel{K}}=0.321\:L}\) This answer supports our expectation from Charles’s law, namely, that raising the gas temperature (from 283 K to 303 K) at a constant pressure will yield an increase in its volume (from 0.300 L to 0.321 L). A sample of oxygen, O , occupies 32.2 mL at 30 °C and 452 torr. What volume will it occupy at –70 °C and the same pressure? 21.6 mL Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm . Find the temperature of boiling ammonia on the kelvin and Celsius scales. A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking and as the initial values, as the temperature at which the volume is unknown and as the unknown volume, and converting °C into K we have: \[\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\textrm{ which means that }\mathrm{\dfrac{150.0\:cm^3}{273.15\:K}}=\dfrac{131.7\:\ce{cm}^3}{T_2} \nonumber \] Rearrangement gives \(T_2=\mathrm{\dfrac{131.7\cancel{cm}^3×273.15\:K}{150.0\:cm^3}=239.8\:K}\) Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C. What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm? 635 mL If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in Figure \(\Page {5}\). Unlike the and relationships, pressure and volume are not directly proportional to each other. Instead, \(P\) and \(V\) exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written: \[P \propto \dfrac{1}{V} \nonumber \] or \[P=k⋅ \dfrac{1}{V} \nonumber \] or \[PV=k \nonumber \] or \[P_1V_1=P_2V_2 \nonumber \] with \(k\) being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure \(\left(\dfrac{1}{P}\right)\) versus the volume ( ), or the inverse of volume \(\left(\dfrac{1}{V}\right)\) versus the pressure (\(P\)). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons, scientists often try to find a way to “linearize” their data. If we plot versus , we obtain a hyperbola (Figure \(\Page {6}\)). The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as : The sample of gas has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using: Comment on the likely accuracy of each method. Using and as the known values 13.0 psi and 15.0 mL, as the pressure at which the volume is unknown, and as the unknown volume, we have: \[P_1V_1=P_2V_2\mathrm{\:or\:13.0\:psi×15.0\:mL}=P_2×7.5\:\ce{mL} \nonumber \] Solving: \[P_2=\mathrm{\dfrac{13.0\:psi×15.0\cancel{mL}}{7.5\cancel{mL}}=26\:psi} \nonumber \] It was more difficult to estimate well from the - graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow. The sample of gas has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using: Comment on the likely accuracy of each method. about 17–18 mL ~18 mL 17.7 mL; it was more difficult to estimate well from the - graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow What do you do about 20 times per minute for your whole life, without break, and often without even being aware of it? The answer, of course, is respiration, or breathing. How does it work? It turns out that the gas laws apply here. Your lungs take in gas that your body needs (oxygen) and get rid of waste gas (carbon dioxide). Lungs are made of spongy, stretchy tissue that expands and contracts while you breathe. When you inhale, your diaphragm and intercostal muscles (the muscles between your ribs) contract, expanding your chest cavity and making your lung volume larger. The increase in volume leads to a decrease in pressure (Boyle’s law). This causes air to flow into the lungs (from high pressure to low pressure). When you exhale, the process reverses: Your diaphragm and rib muscles relax, your chest cavity contracts, and your lung volume decreases, causing the pressure to increase (Boyle’s law again), and air flows out of the lungs (from high pressure to low pressure). You then breathe in and out again, and again, repeating this Boyle’s law cycle for the rest of your life (Figure \(\Page {7}\)). The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by : . In equation form, this is written as: \[V∝n\textrm{ or }V=k×n\textrm{ or }\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2} \nonumber \] Mathematical relationships can also be determined for the other variable pairs, such as versus , and versus . Visit this to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws). To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas: Combining these four laws yields the , a relation between the pressure, volume, temperature, and number of moles of a gas: \[PV=nRT \nonumber \] where is the pressure of a gas, is its volume, is the number of moles of the gas, is its temperature on the kelvin scale, and is a constant called the or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol K and 8.3145 kPa L mol K . , and units for the gas constant, R. Gases whose properties of , , and are accurately described by the ideal gas law (or the other gas laws) are said to exhibit or to approximate the traits of an . An ideal gas is a hypothetical construct that may be used along with to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the behavior observed for many gases at relatively high pressures and low temperatures. The ideal gas equation contains five terms, the gas constant and the variable properties , , , and . Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises. Methane, CH , is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH . What is the volume of this much methane at 25 °C and 745 torr? Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car. 350 bar If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\) using units of atm, L, and K. Both sets of conditions are equal to the product of × (where = the number of moles of the gas and is the ideal gas law constant). When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (Figure \(\Page {8}\)). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm? Letting represent the air in the scuba tank and represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have: \[\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}⟶\mathrm{\dfrac{(153\:atm)(13.2\:L)}{(300\:K)}=\dfrac{(3.13\:atm)(\mathit{V}_2)}{(310\:K)}} \nonumber \] Solving for : \[V_2=\mathrm{\dfrac{(153\cancel{atm})(13.2\:L)(310\cancel{K})}{(300\cancel{K})(3.13\cancel{atm})}=667\:L} \nonumber \] (Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.) A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm. 0.193 L Whether scuba diving at the Great Barrier Reef in Australia (shown in Figure \(\Page {9}\)) or in the Caribbean, divers must understand how pressure affects a number of issues related to their comfort and safety. Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization. Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in his expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and she or he begins to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface. We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a for reporting properties of gases: 273.15 K (0.00 °C) and 1 atm (101.325 kPa). At , an ideal gas has a volume of about 22.4 L—this is referred to as the (Figure \(\Page {10}\)). The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law). The equations describing these laws are special cases of the ideal gas law, = , where is the pressure of the gas, is its volume, is the number of moles of the gas, is its kelvin temperature, and is the ideal (universal) gas constant. | 19,576 | 3,190 |
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In order to explain both the ground state and the excited state involved in an absorption band in , it is necessary to look at the electronic structure of molecules in somewhat different terms from the description given in the sections on and . In those chapters we treated electrons either as bonding pairs located between two nuclei, or as lone pairs associated with a single . Such a model of electronic structure is known as the model. It is of very little use in explaining molecular spectra because photons are absorbed by the whole molecule, not an individual atom or bond. Thus we need to look upon electrons in a molecule as occupying orbitals which belong to the molecule as a whole. Such orbitals are called , and this way of looking at molecules is referred to as (abbreviated MO) . The term molecular orbital is mentioned in when we described formation of a covalent bond in an H molecule as a result of overlap of two 1 atomic —one from each H atom. In that section, however, we do not point out that there are two ways in which the 1 of one H atom can combine with the 1 electron wave of another. One of these involves between the two waves and is referred to as . This results in a bigger electron wave (and hence more electron density) between the two atomic nuclei. This attracts the positively charged nuclei together, forming a bond as . A molecular orbital formed as a result of positive overlap is called a . It is also possible to combine two electron waves so that of the waves occurs between the atomic nuclei. This situation is referred to as , and it decreases the probability of finding an electron between the nuclei. In the case of two H atoms this results in a planar node of zero electron density halfway between the nuclei. Without a buildup of negative charge between them, the nuclei repel each other and no chemical bond is possible. A molecular orbital formed as a result of negative overlap is called an . If one or more electrons occupy an antibonding MO, the repulsion of the nuclei increases the energy of the molecule, and so such an orbital is higher in energy than a bonding MO. This is shown in Figure \(\Page {1}\). Electron dot-density diagrams for the 1 electron in each of two separate H atoms are shown on the left and right sides of the figure. The horizontal lines show the energy each of these electrons would have. A dot-density diagram for a single electron occupying the bonding MO formed by positive overlap of the two orbitals is shown in the center of the diagram. This is labeled σ . Above it is a dot-density diagram for a single electron occupying the antibonding MO, σ . (In general, antibonding MO’s are distinguished from bonding MO’s by adding an * to the label.) The energies of the molecular orbitals are indicated by the horizontal lines in the center of the diagram. The Greek letter σ in the labels for these orbitals refers to the fact that their positive or negative overlap occurs directly between the two atomic nuclei. Like an atomic orbital, each molecular orbital can accommodate two electrons. Thus the lowest energy arrangement for H would place both electrons in the a MO with paired spins. This molecular electron configuration is written (σ ) , and it corresponds to a covalent electron-pair bond holding the two H atoms together. If a sample of H is irradiated with ultraviolet light, however, an absorption band is observed between 110 and 170 nm. The energy of such an absorbed photon is enough to raise one electron to the antibonding MO, producing an excited state whose electron configuration is (σ ) (σ ) . In this excited state the effects of the bonding and the antibonding orbitals exactly cancel each other; there is no overall bond between the two H atoms, and the H molecule dissociates. When the absorption of a photon results in the dissociation of a molecule like this, the phenomenon is called . It occurs quite frequently when UV radiation strikes simple molecules. The molecular-orbital model we have just described can also be used to explain why a molecule of He cannot form. If a molecule of He were able to exist, the four electrons would doubly occupy both the bonding and the antibonding orbitals, giving (σ ) (σ ) . However, the antibonding electrons would cancel the effect of the bonding electrons, and there would be no resultant buildup of electronic charge between the nuclei and hence no bond. Interestingly enough, an extension of this argument predicts that if He loses an electron to become the He ion, a bond is possible. He would have the structure (σ ) (σ ) and the single electron in the antibonding orbital would only cancel half the effect of the two electrons in the bonding orbital. This would leave the ion with a “half-bond” joining the two nuclei. The spectrum of He shows bands corresponding to He , and from them it can be determined that He has a bond enthalpy of 322 kJ mol . The molecular-orbital model can easily be extended to diatomic molecules). Three general rules are followed. First, only the core orbitals and the valence orbitals of the atoms need be considered. Second, only atomic orbitals whose energies are similar can combine to form molecular orbitals. Third, the number of molecular orbitals obtained is always the same as the number of atomic orbitals from which they were derived. Applying these rules to diatomic molecules which consist of atoms from the second row of the periodic table, such as N , O , and F , we need to consider the 1 , 2 , 2 , 2 , and 2 atomic orbitals. Since the 1 orbital of each atom differs in energy from the 2s, we can overlap the two 1 orbitals separately from the 2 . This gives a σ and a σ MO, as in the case of H . Similarly, the 2 orbitals can be combined to give σ and σ before we concern ourselves with the higher energy 2 orbitals. There are three 2 orbitals on each atom, and so we expect a total of six molecular orbitals to be derived from them. The shapes of these six molecular orbitals are shown by the boundary-surface diagrams in Figure \(\Page {2}\). Two of them are formed by positive and negative overlap of 2 orbitals directly between the atomic nuclei. Consequently they are labeled σ and σ *. Two more molecular orbitals are formed by sideways overlap of 2 atomic orbitals. These are labeled π and π *, because the molecular orbitals have two parts—one above and one below a nodal plane containing the nuclei. The atomic 2 orbitals also overlap sideways to form π and π * molecular orbitals. These are identical to π and π *, except for a 90° rotation around the line connecting the nuclei. Consequently π and π have the same energy, as do π * and π *. The electron configuration for any homonuclear diatomic molecule containing fewer than 20 electrons can be built up by filling electrons into the molecular orbitals we have just derived, starting with the orbital of lowest energy. The relative energies of the molecular orbitals at the time they are being filled are shown in Figure \(\Page {3}\). Like the energies of atomic orbitals given in , these relative molecular-orbital energies vary somewhat from one diatomic molecule to another. In particular the σ orbital is often lower than π . Nevertheless Figure \(\Page {3}\) gives the correct order of filling the orbitals, and we can use it to determine molecular electron configurations. Find the electronic configuration of the oxygen molecule, O . Starting with the lowest lying orbitals (σ and σ ) we add an appropriate number of electrons to successively higher orbitals in accord with the Pauli principle and Hund’s rule. O has 16 electrons, the first 14 of which are easily accommodated in the following way: The remaining two electrons must now be added to the π * orbitals. Since both these orbitals are of equal energy, one electron must be placed in each orbital and the spins must be parallel. The total electronic structure is thus \[ (\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 ( \pi_{2p})^4 (\sigma_{2p})^2 (\pi_{2px}^*)^1 (\pi_{2py}^*)^1 \nonumber \] As the previous problem shows, the molecular-orbital model predicts that O has two unpaired electrons. Substances whose atoms, molecules, or ions contain unpaired electrons are weakly attracted into a magnetic field, a property known as . (In a few special cases, such as iron, a much stronger magnetic attraction called is also obsewed.) Most substances have all their electrons paired. Such materials are weakly repelled by a magnetic field, a property known as . Hence measurement of magnetic properties can tell us whether all electrons are paired or not. O , for example, is found to be paramagnetic, an observation which agrees with the electron configuration predicted in Example 21.6. Before the advent of MO theory, however, the paramagnetism was a mystery, since the Lewis diagram predicted that all electrons should be paired. The molecular-orbital model also allows us to estimate the strengths of bonds in diatomic molecules. We simply count each electron in a bonding orbital as contributing half a bond, while each electron in an antibonding orbital takes away half a bond. Thus if there are B electrons in bonding orbitals and A electrons in antibonding orbitals, the net bond order is given by \[ \text{ Bond order} = \frac{A-B}{2} \nonumber \] The larger the bond order, the more strongly the atoms are held together. Calculate the bond order for the molecule N . There are 14 electrons, and so the electron configuration is \[ ( \sigma_{1s})^2 (\sigma_{1s}^* )^2 ( \sigma_{2s} )^2 (\sigma_{2s}^* )^2 ( \pi_{2p})^4 (\sigma_{2p})^2 \nonumber \] There are a total of 2 + 2 + 4 + 2 = 10 electrons in bonding MO’s and only 4 in the antibonding orbitals σ and σ . Thus \[ \text{ Bond order} = \frac{10-4}{2} = 3 \nonumber \] The bond orders derived from the molecular-orbital model for stable molecules agree exactly with those predicted by Lewis’ theory. Not only do we find a triple bond for N , but we also find a double bond for O and a single bond for F . The results of such bond-order calculations are summarized in Table 1. * Some molecule-ions such as H are included. Except fur the number of electrons involved, the MO theory is applied to them in exactly the same way as to molecules. Some of the molecules in the table, such as C and B , are only stable at high temperatures or only exist transitorily in discharge tubes, and so you are probably not familiar with them. Nevertheless, their spectra can be studied. Also included in the table are values for the bond enthalpies and bond lengths of the various species obtained from their spectra. Note the excellent qualitative agreement with the MO theory. The higher the bond order predicted by the theory, the larger the bond enthalpy and the shorter the bond length. | 10,890 | 3,192 |
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Chemical theory has not reached the point where it can predict exactly how much of one substance will dissolve in another. The best we can do is to indicate in general terms the relationships between solubility and the microscopic structures of solute and solvent. To begin with, moving particles of any kind tend to become more randomly distributed as time passes. If you put a layer of red marbles in the bottom of a can and cover it with a second layer of white marbles, shaking the can for a short time will produce a nearly random distribution. The same principle applies on the microscopic level. Moving molecules tend to become randomly distributed among one another, unless something holds them back. Thus gases, whose molecules are far apart and exert negligible forces on one another, are all completely miscible with other gases. In liquid solutions, the molecules are much closer together and the characteristics of different types of molecules are much more important. In particular, if the solute molecules exert large intermolecular forces on each other but do not attract solvent molecules strongly, the solute molecules will tend to group together. This forms a separate phase and leaves the solvent as a second phase. Conversely, if the solvent molecules attract each other strongly but have little affinity for solute molecules, solvent molecules will segregate, and two phases will form. A classic example of the second situation described in the previous paragraph is the well-known fact that oil and water do not mix—or if they do, they do not stay mixed for long. The reason is that oil consists of alkanes and other nonpolar molecules, while water molecules are polar and can form strong hydrogen bonds with each other. Suppose that alkane or other nonpolar molecules are randomly dispersed among water molecules, as shown in part of the figure.The constant jostling of both kinds of molecules will soon bring two water molecules together. Dipole forces and hydrogen bonding will tend to hold the water molecules together, but there are only weak London forces between water and nonpolar molecules. Before long, clusters of water molecules like those in part will have formed. These clusters will be stable at room temperature because the energy of interaction between the water molecules will be larger than the average energy of molecular motion. Only an occasional molecular collision will be energetic enough to bump two water molecules apart, especially if they are hydrogen bonded. Given enough time, this process of aggregation will continue until the polar molecules are all collected together. If the nonpolar substance is a liquid, this process corresponds on the macroscopic level to the liquids separating from each other and forming two layers. If instead of mixing substances like oil and water, in which there are quite different kinds of intermolecular attractions, we mix two polar substances or two nonpolar substances, there will be a much smaller tendency for one type of molecule to segregate from the other. Thus two alkanes like -heptane, C H , and -hexane, C H , are completely miscible in all proportions. The C H and C H molecules are so similar (recall the projection formulas of ) that there are only negligible differences in intermolecular forces. Thus the molecules remain randomly mixed as they jostle among one another. For a similar reason, methanol, CH OH, is completely miscible with water. In this case both molecules are polar and can form hydrogen bonds among themselves, and so there are strong intermolecular attractions within each liquid. However, CH OH dipoles can align with H O dipoles, and CH OH molecules can hydrogen bond to H O molecules, and so the attractions among unlike molecules in the solution are similar to those among like molecules in each pure liquid. Again there is little tendency for one type of molecule to become segregated from the other. All the cases just discussed are examples of the general rule that . Two substances whose molecules have very similar structures and consequently similar intermolecular forces will usually be soluble in each other. Two substances whose molecules are quite different will not mix randomly on the microscopic level. In general, polar substances will dissolve other polar substances, while nonpolar materials will dissolve other nonpolar materials. The greater the difference in molecular structure (and hence in intermolecular attractions), the lower the mutual solubility. The following video succiently showcases this principle. In the video a number of mixing events occur. A nonpolar, colored solid is added to CCl . Since CCl is also nonpolar, like dissolves like, and the solid is dissolved. Next, water is added. Since water is polar, it does not mix with the CCl solution, even after vigorous shaking. Two layers remain, with the less dense water on top. Finally, hexane is added. Since it is nonpolar and less dense than water, it forms a third layer, on top of the water. When the test tube is shaken, however, two layers remain. Since hexane is nonpolar, it is miscible with CCl , and so both form a single layer below the water. Predict which of the following compounds will be most soluble in water: Since ethanol contains an group, it can hydrogen bond to water. Although the same is true of hexanol, the OH group is found only at one end of a fairly large molecule. The rest of the molecule can be expected to behave much as though it were a nonpolar alkane. This substance should thus be much less soluble than the first. Experimentally we find that ethanol is completely miscible with water, while only 0.6 g hexanol dissolves in 100 g water. Our discussion of solubility in terms of microscopic structure concludes with one more point. The solubility of other substances in solids are usually small. The constituent particles in a solid crystal lattice are packed tightly together in a very specific geometric arrangement. For one particle to replace another in such a structure is very difficult, unless the particles are almost identical. The most common solid solutions are alloys, in which one essentially spherical metal atom replaces another. Thus alloys are easily made by melting two metals and cooling the liquid solution. In many other cases, however, completely miscible liquids separate when a solid phase forms. A good example of this is benzene and naphthalene: A naphthalene molecule is almost twice as big as a benzene molecule and cannot fit in the benzene lattice. Therefore, even though the liquids are miscible, the solids are not due to the molecular structures of benzene and naphthalene. | 6,683 | 3,193 |
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Electrolysis reactions are the basic foundations of today's modern industry. There are various elements, chemical compounds, and organic compounds that are only produced by electrolysis including aluminum, chlorine, and NaOH. Electrolysis is the process by which an electric current spurs an otherwise non-spontaneous reaction. The electrorefining process refines metals or compounds at a high purity for a low cost. The pure metal can coat an otherwise worthless object. Let's consider the electrorefining process of copper: At the anode, there is an impure piece of copper that has other metals such as Ag, Au, Pt, Sn, Bi, Sb, As, Fe, Ni, Co, and Zn. The copper in this impure ore is oxidized to form Cu at the anode, and moves through an aqueous sulfuric acid-Copper (II) sulfate solution into the cathode. When it reaches the cathode, the Cu is reduced to Cu. This whole process takes place at a fairly low voltage (about .15 to .30 V), so Ag, Au, and Pt are not oxidized at the anode, as their standard oxidation electrode potentials are -.800, -1.36 and -1.20 respectively; these unoxidized impurities turn into a mixture called anode mud, a sludge at the bottom of the tank. This sludge can be recovered and used in different processes. Unlike Ag, Au and Pt, the impurities of Sb, Bi and Sn in the ore are indeed oxidized at the anode, but they are precipitated as they form hydroxides and oxides. Finally, Fe, Ni, Co and Zn are oxidized as well, but they are dissolved in water. Therefore, the only solid we are left with is the pure solid copper plate at the cathode, which has a purity level of about 99.999%. The image below gives an outline about the fate of the main components of an impure iron ore. Electrosynthesis is the method of producing substances through electrolysis reactions. This is useful when reaction conditions must be carefully controlled. One example of electrosynthesis is that of MnO , Manganese dioxide. MnO occurs naturally in the form of the mineral pyrolusite, but this mineral is not easily used due to the nature of its size and lattice structure. However, MnO can be obtained a different way, through the electrolysis of MnSO in a sulfuric acid solution. \(\begin{align} &\textrm{Oxidation: } &&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-} &&\mathrm{\hspace{12px}E^0_{MnO_2/H_2}=-1.23}\\ &\textrm{Reduction: } &&\mathrm{2e^- + 2H^+ \rightarrow H_2} &&\mathrm{{-E}^0_{H^+/H_2}= -0}\\ &\textrm{Overall: } &&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 2H^+ +H_2} &&\mathrm{\hspace{12px}E^0_{MnO_2/H_2} -E^0_{H^+/H_2}= -1.23 - 0= -1.23} \end{align}\) The commercial process for organic chemicals that is currently practiced on a scale comparable to that of inorganic chemicals and metals is the electrohydrodimerization of acrylonitrile to adiponitrile. \(\begin{align} &\textrm{Anode: } &&\mathrm{H_2O \rightarrow 2H^+ + \dfrac{1}{2} O_2 + 2e^-}\\ &\textrm{Cathode: } &&\ce{2CH2=CHCN + 2H2O + 2e- \rightarrow NC(CH2)CN + 2OH-}\\ &\textrm{Overall: } &&\textrm{(acrylonitrile) }\ce{2CH2=CHCN + H2O \rightarrow \dfrac{1}{2} O2 + NC(CH2)4CN}\:\textrm{(adiponitrile)} \end{align}\) The importance of adiponitrile is that it can be readily converted to other useful compounds. This process is the electrolysis of sodium chloride (NaCl) at an industrial level. We will begin by discussing the equation for the chlor-alkali process, followed by discussing three different types of the process: the diaphragm cell, the mercury cell and the membrane cell. We will begin the explanation of the chlor-alkali process by determining the reactions that occur during the electrolysis of NaCl. Because NaCl is in an aqueous solution, we also have to consider the electrolysis of water at both the anode and the cathode. Therefore, there are two possible reduction equations and two possible oxidation reactions. \begin{align} & \mathrm{Na^+_{\large{(aq)}} + 2e^- \rightarrow Na_{\large{(s)}}} && \mathrm{E^0_{Na^+/Na}= -2.71\: V \tag{1}}\\ & \mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}} && \mathrm{E^0_{H_2O/H_2}= -.83\: V \tag{2}} \end{align} \(\begin{align} &\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{-E^0_{Cl_2/Cl^-}= -(1.36\: V) \tag{3}}\\ &\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-} &&\mathrm{-E^0_{O_2/H_2O}= -(1.23\: V) \tag{4}} \end{align}\) As we can see, due to the very much more negative electrode potential, the reduction of sodium ions is much less likely to occur than the reduction of water, so we can assume that in the electrolysis of NaCl, the reduction that occurs is reaction (2). Therefore, we should try to determine what the oxidation reaction that occurs is. . We would get: \(\begin{align} &\textrm{Reduction: } &&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}} &&\mathrm{\hspace{12px}E^0_{H_20/H_2}= -.83\: V \tag{2}}\\ &\textrm{Oxidation: } &&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{-E^0_{Cl_2/Cl^-}= -(1.36\: V) \tag{3}} \\ &\textrm{Overall: } &&\mathrm{2 H_2O_{\large{(l)}} + 2Cl^-_{\large{(aq)}}} &&\mathrm{\hspace{12px}E^0_{H_20/H_2} - E^0_{Cl_2/Cl^-}\tag{5}}\\ & &&\mathrm{\hspace{10px}\rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}} +Cl_2} &&\mathrm{\hspace{22px} = -.83 + (-1.36)= -2.19} \end{align}\) \(\begin{align} &\textrm{Reduction: } &&\mathrm{2\,[2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}]} &&\mathrm{\hspace{12px}E_{H_2O/H_2O}=-.83\: V \tag{2}}\\ &\textrm{Oxidation: } &&\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-} &&\mathrm{-E_{O_2/H_20}^0= -(1.23\: V) \tag{4}}\\ &\textrm{Overall: } &&\mathrm{2H_2O_{\large{(l)}} \rightarrow 2H_{2\large{(g)}} + O_{2\large{(g)}}} &&\mathrm{\hspace{12px}E^0_{H_2O/H_2} - E^0_{O_2/H_20}\tag{6}}\\ & && && \mathrm{\hspace{22px}= -.83\: V -(1.23\: V)= -2.06} \end{align}\) At first glance it would appear as though reaction (6) would occur due to the smaller (less negative) electrode potential. However, O actually has a fairly large overpotential, so instead Cl is more likely to form, making reaction (5) the most probable outcome for the electrolysis of NaCl. Depending on the method used, there can be several different products produced through the chlor-alkali process. The value of these products is what makes the chlor-alkali process so important. The name comes from the two main products of the process, chlorine and the alkali, sodium hydroxide (NaOH). Therefore, one of the main uses of the chlor-alkali process is the production of NaOH. As described earlier, the equation for the chlor-alkali process, that is, the electrolysis of NaCl, is as follows: \(\begin{align} &\textrm{Reduction: } &&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}} &&\mathrm{E_{H_2O/H_2O}= -.83}\\ &\textrm{Oxidation: } &&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{E^0_{Cl_2/Cl^-}= -1.36}\\ &\textrm{Overall: } &&\mathrm{2Cl^- + 2H_2O_{\large{(l)}} \rightarrow 2 OH^- + H_{2\large{(g)}} +Cl_{2\large{(g)}}} &&\mathrm{E^0_{H_2O/H_2}- E^0_{Cl/l2^-}}\\ & && &&\hspace{12px}\mathrm{= -83\:V- (-1.36\: V)= -2.19\: V} \end{align}\) To even further improve the purity of NaOH, a mercury cell can be used for the location of electrolysis, opposed to a diaphragm cell. Figure 2: One final way to make even more pure NaOH is to use a membrane cell. It is preferred over the diaphragm cell or mercury cell method because it uses the least amount of electric energy and produces the highest quality NaOH. For instance, it can produce NaOH with a degree of chlorine ion contamination of only 50 parts per million. An ion-permeable membrane is used to separate the anode and cathode. : General Chemistry, Principles & modern Applications, Petrucci, 2007, 2002, 1997 | 7,915 | 3,194 |
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The most important chemical characteristic of ionic compounds is that . . In other words, an Na ion is quite different from an Na atom, and a Cl ion is unlike an isolated Cl atom or either of the Cl atoms in a Cl molecule. You eat a considerable quantity of Na and Cl ions in table salt every day, but Na atoms or Cl molecules would be quite detrimental to your health. The unique chemical properties of each type of ion are quite evident in aqueous solutions. Most of the reactions of BaCl ( ), for example, can be classified as reactions of the Ba ( ) ion or the Cl ( ) ion. If sulfuric acid, H SO , is added to a solution of BaCl , the solution turns milky and very fine crystals of BaSO ( ) eventually settle out. The reaction can be written as \[\text{Ba}^{2+}(aq) + \text{ H}_{\text{2}}\text{SO}_4(aq)\rightarrow \text{ BaSO}_{\text{4}}(s) + \text{2H}^{+}(aq) \nonumber \] Below is a video of this reaction. The solution of BaCl is clear and colorless, but when H SO is added through the the thin glass tube, the contents become white and opaque, as insoluble BaSO ( ) come out of solution. This reaction is characteristic of the . It will also occur if H SO is added to solutions such as BaI ( ) or BaBr ( ) which contain barium ions but no chloride ions. By contrast, if a solution of silver nitrate, AgNO , [which contains silver ions, Ag ( )] is added to a BaCl solution, a reaction characteristic of the occurs. A white curdy precipitate of AgCl( ) forms according to the equation \[\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)\rightarrow\text{AgCl}(s) \nonumber \] Other ionic solutions containing chloride ions, such as LiCl( ), NaCl( ), or MgCl ( ), give an identical reaction. Below is a video of the reaction of a sodium chloride solution with a silver nitrate solution. Both the NaCl(aq) solution and the AgNO (aq) solution begin clear and colorless. When the NaCl(aq) solution is added to the AgNO (aq) solution, a cloudy white precipitate of AgCl(s) is formed. The same result would have occurred had BaCl been used, as the reaction is only between the Ag and Cl ions, as seen: \[\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)\rightarrow\text{AgCl}(s) \nonumber \] Many binary ionic solids not only dissolve in water, they also react with it. When the compound contains an anion such as N , O , or S , which has more than one negative charge, the reaction with water produces hydroxide ions, OH : \[\text{O}^{2-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{OH}^{-}(aq) + \text{OH}^{-}(aq) \nonumber \] \[\text{S}^{2-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{HS}^{-}(aq) + \text{OH}^{-}(aq) \nonumber \] \[\text{N}^{3-} + \text{ 3H}_{\text{2}}\text{O}\rightarrow\text{ NH}_{\text{3}}(aq) + \text{3OH}^{-}(aq) \nonumber \] Thus, when sodium oxide, Na O, is added to water, the resulting solution contains sodium ions and hydroxide ions but no oxide ions: \[\text{ Na}_{\text{2}}\text{O} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{2Na}^{+}(aq) + \text{2OH}^{-}(aq) \nonumber \] The hydride ion also reacts with water to form hydroxide ions. When lithium hydride, LiH, is dissolved in water, for example, the following reaction occurs: \[\text{LiH}(s) + \text{ H}_{\text{2}}\text{O}\rightarrow\text{Li}^{+}(aq) + \text{OH}^{-}(aq) + \text{ H}_{\text{2}}(g) \nonumber \] Note that hydrogen gas is evolved in this reaction. Lithium hydride crystals provide a very compact, if somewhat expensive, method for storing hydrogen. Among the (F , Cl , Br , I ) only the fluoride ion shows any tendency to react with water, and that only to a limited extent. When sodium fluoride is dissolved in water, for example, faint traces of hydroxide ion can be detected in the solution owing to the reaction \[\text{F}^{-} + \text{ H}_{\text{2}}\text{O}\rightarrow\text{HF} + \text{OH}^{-} \nonumber \] With sodium chloride, by contrast, no such reaction occurs. | 3,915 | 3,198 |
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What is an "aromatic" compound? It is common to start by saying that aromatic compounds are compounds related to benzene. However, as you go on in organic chemistry you will find a variety of compounds called aromatic, even though they are not so obviously benzene derivatives. Defining aromatic in terms of benzene is a useful start in an introductory course. As we will see here, it is not easy to give a more complete definition that is satisfying in an introductory course. First, why is aromaticity an issue? The special feature of benzene is that it is more stable than we might expect. If we write the structure of "1,3,5-cyclohexatriene", it looks like one of the Kekule structures for benzene. But benzene, a real chemical, does not have the properties we would expect for "1,3,5-cyclohexatriene". Why? Benzene has two , and that the actual structure is the resonance hybrid. In general, resonance structures delocalize electrons, and thus stabilize a structure. The resonance stabilization in benzene is considerably more than we might expect for simply having some double bonds near each other. Clearly, there is something special about benzene, which results in an unusual degree of resonance stabilization -- and results in the property called aromaticity. We can discuss this more, but doing so goes beyond course material. The purpose is to help those who are going on in organic chemistry get a sense of what this topic holds. The resonance system in benzene involves six electrons. They occupy a series of p orbitals, one on each C atom. The p orbitals on neighboring C atoms overlap sideways, a type of bonding known as π (pi) bonding. Effectively, the six overlapping p orbitals form one large orbital -- a loop around the entire molecule. This "loop orbital" provides an unusual degree of stabilization, and is part of the secret to aromaticity. That is, one key feature of aromatic compounds is that there is a set of electrons in a loop orbital resulting from overlapping p orbitals around a ring. (sometimes referred to as a "closed loop of six electrons", which he also calls the "aromatic sextet"). Let's look at some examples of compounds that are aromatic, but are less obviously "like benzene". As we discuss these examples, some patterns will emerge. Unfortunately, reasons for not discussing this too much in an introductory class will also emerge; understanding aromaticity even at an elementary level beyond "like benzene" requires understanding orbitals. One simple example is pyridine, C H N, shown below: This example is simple enough, because it is very similar to benzene in structure. Some aromatic compounds with a heteroatom (an atom other than C or H) in the ring have a five-membered ring. These include pyrrole, C H N, and furan, C H O: What do these compounds share with benzene and pyridine that makes them aromatic? They appear to have only two double bonds, thus only four p orbital electrons perpendicular to the ring. But they also have one lone pair that is in a p orbital perpendicular to the ring. Thus they actually have electrons that are in p orbitals perpendicular to the ring. At this point, then, a pattern may be emerging: it seems that a loop of six electrons in overlapping p orbitals perpendicular to the molecular ring is a condition for aromaticity. Let's test this prediction by looking at some other structures. Five membered rings of C. Look at cyclopentadiene, shown below. It has 2 double bonds, hence has 4 electrons in p orbitals perpendicular to the ring. But the other C is "saturated"; it is a -CH - and is sp -hybridized. No loop, not even 6 p electrons. Therefore not aromatic; this is correct. But now look at the anion derived from this chemical -- the ion that would result if this chemical behaved as an acid, and gave off an H ; this ion is also shown below. Note that it now has six electrons in p orbitals perpendicular to the ring -- very similar to pyrrole, above. If pyrrole is aromatic, then maybe this ion should be? Yes, it is. A manifestation of this is that the parent compound, cyclopentadiene, is a rather "strong" acid -- by the standards of H attached to hydrocarbons. The acidity of cyclopentadiene is due to the stabilization of the resulting anion. K for cyclopentadiene is about 10 . That certainly isn't strong compared to compounds commonly discussed as acids, but it is 10 times stronger than for the non-cyclic form of this molecule. Now we have seen a variety of chemical species, both neutral molecules and ions, that are aromatic. They share the common feature that they all have 6 electrons in a continuous loop of overlapping p orbitals. Turns out that this is still not a sufficiently broad description of what makes a chemical aromatic. Now, the non-aromatic character of cyclooctatetraene alone is not too hard to explain. After all, if it were aromatic and planar, it would have bond angles of 135 deg. That is considerably more than the simple sp bond angle of 120 deg. So we might suggest that the bond angle strain of the planar form more than compensates for any gain due to aromaticity. However, there is more to the story. Turns out that it is fairly easy for cyclooctatetraene to gain 2 electrons, forming the dianion, C H . This ion planar, and aromatic. Two drawings of this ion are below, followed by a 3D model. So what do we learn from the story of cyclooctatetraene and its dianion? Prior to considering this, all our examples of aromaticity had electrons in the π electron loop. Cyclooctatetraene has , and is not aromatic; its dianion has , and is aromatic. So clearly, six is not the only allowed number. Study of many molecules and ions, as well as theoretical work that is well beyond our course, have indicated that a species will be aromatic if there are 4n+2 electrons in a planar π electron loop -- where n is any integer, starting with 0. (This is known as Hückel's rule.) 0? That would mean that a π electron loop with two electrons is aromatic. In fact, the cation derived from cyclopropene, shown below, is unusually stable, and is considered aromatic. As another example, with a larger n, consider a particular isomer of 18-annulene -- the isomer with every third double bond cis. The following set of figures show three representations of its structure. If 18-annulene were aromatic (n=4), we might expect it to be planar. In fact, some programs will calculate a planar structure for it. Actual measurement shows that is very slightly distorted from planar, due to the repulsion of the hydrogens that are inside the ring. The following figure shows 3D representations of both "forms" of 18-annulene; they have been rotated so you view them "edge on". The top structure is the planar form that one might naively expect; the other is the slightly distorted structure, which closely corresponds to what is actually observed. The annulene name is used generically for cyclic molecules with alternating single and double bonds. The numeric prefix indicates the ring size. Benzene might be considered as 6-annulene, and cyclooctatetraene as 8-annulene. Note that all annulenes have the general formula C H (where x must be an even number), and that the term does not in itself imply aromatic character. Aromatic compounds are more stable than we might expect when we see a structure showing single and double bonds. We start our study of aromaticity with the classic case of benzene. But as we continue, we find examples of aromatic compounds that contain heteroatoms, charges, and rings of different sizes. As we explore these, we find that the key features of aromatic chemicals are dictacted by Hückel's rule: >Robert Bruner ( ) | 7,698 | 3,199 |
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In some molecules the can be very much more extensive than in ozone and benzene. This is particularly true of containing , i.e., long chains of alternating single and double bonds. An example is provided by vitamin A which has the structure Both the physical and chemical properties of this molecule indicate that all the bonds in the conjugated chain (shown in color) are intermediate in character between single and double bonds and that the pi electrons are free to move over the whole length of the conjugated chain. A very simple, though approximate, method of handling delocalized electrons mathematically is to treat them as though they were particles in a of the same length as the chain. In the case of vitamin A2 the distance from the carbon atom at one end of the conjugated chain to that on the other is about 1210 pm. If we feed this value and the mass of the electron (9.110 × 10 kg) into (Equation \(\ref{1}\)), we can obtain approximate values of the energy levels occupied by the delocalized electrons: \[E_{k}=\dfrac{1}{2m}\left(\dfrac{nh}{2d}\right)^{2}=\dfrac{1}{2\times9.110\times10^{-31} \text{ kg}}\left(\dfrac{n\times6.626\times10^{-34} \text{J s }}{2\times 1210\times 10^{-12} \text{ m}}\right) ^{2} \label{1} \] \[E_{k}=n^{2}(4.11\times10^{-20} \text{ J}) \label{2} \] The first seven energy levels derived from this formula are shown in Figure \(\Page {2}\). Note that since there are 12 pi electrons (one for each carbon atom on the conjugate chain), these will occupy the six lowest levels in accord with . This model allows us to calculate the wavelength of the main absorption band in the spectrum of vitamin A . If an electron from the highest occupied level ( = 6) is excited to the lowest unoccupied level ( = 7), the energy required can be calculated from Equation \(\ref{2}\): \[\Delta E = E_{7}-E_{6}=(7^{2}-6^{2})\times4.11\times10^{-20} \text{ J}=5.34\times10^{-19} \text{ J} \nonumber \] Thus \[ λ = \dfrac{hc}{\Delta E} = \dfrac{6.626\times10^{-34} \text{ Js}\times2.998\times0^{8} \text{ ms}^{-1}}{5.34\times10^{-19} \text{ J}} = 327 \text{ nm} \nonumber \] This is in reasonable agreement with the observed value of 350 nm, considering the approximate nature of the model. As a general rule, the longer a conjugated chain, the longer the wavelength at which it absorbs. Ethene (C H ), for example, has only one double bond and absorbs at 170 nm. Hexatriene (C H ) has three alternating double bonds and absorbs at 265 nm, while vitamin A , with six double bonds, absorbs at 350 nm. It is not difficult to explain this effect in terms of the particle-in-a-box model. According to Equation \(\ref{1}\) the energy of a level varies inversely with the square of the length of the box. Thus the longer the conjugated chain, the closer the energy levels will be to each other, and the less energy a photon need have to excite an electron. Naturally the lower the energy of a photon, the longer its wavelength will be. A similar effect is found for molecules containing several benzene rings. Since these correspond to Lewis formulas of alternating double and single bonds, they can also be regarded as conjugated systems containing delocalized electrons. Experimentally we find that the more benzene rings a molecule contains, the longer the wavelengths at which it absorbs: Thus increasing the extent of electron delocalization increases the wavelength at which a molecule will absorb light, whether the electron is delocalized over rings or chains. This behavior of delocalized electrons is important in the preparation of compounds which strongly absorb visible light, i.e., in the preparation of dyes. Very few compounds which are held together by sigma bonds alone are colored. The electrons are so tightly held that a very energetic photon is needed to excite them. In order for an organic molecule to absorb in the visible region of the spectrum, it must usually contain very delocalized pi electrons. Thus most dyes and most colored compounds occurring in living organisms turn out to be large molecules with extensive systems of conjugated double bonds. | 4,134 | 3,200 |
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It has been estimated that even a unicellular organism may contain as many as 5000 different substances, and the human body probably has well over 5 000 000. Only a few of these are exactly the same in both species, and so the total number of different compounds in the living portion of the earth (the ) is approximately (10 compounds/species) × 10 species = 10 compounds. If it were possible for chemists to synthesize one of these every second, 24 hours a day, 7 days a week, about 3000 years would be required to make all of them—obviously a hopeless task, even if we knew the composition and structure of each one. How then can we make sense out of the chemistry of living systems? Fortunately nearly all the substances found in living cells are —they are built up by different combinations of a limited number of relatively small molecules. For example, the basic structures of all proteins in all organisms consist of covalently linked chains containing 100 or more amino acid residues. Only 20 different amino acids are commonly incorporated in proteins, but the number of ways of arranging 100 of these in a chain taking any of the amino acids at random for each place in the chain is 20 \(\cong\) 10 , allowing an almost infinite variety of structures. Just as an understanding of the properties of atoms and their bonding characteristics was a significant aid in predicting the chemistry of molecules, a knowledge of the properties of simple molecular building blocks gives us a starting point for the study of biochemistry. Each of the building blocks and their polymeric forms has at least one major role to play in the chemistry of life. Most are quite versatile, serving several functions. Let us take a look at the building blocks of biochemistry: , or carbohydrates are molecules that follow the form C (H O) . A simple sugar can serve as an energy source for an organism. Simple sugars can dimerize into disaccharides or polymerize into Polysaccharides. Uses for polysacharides range from energy storage, such as glycogen stored in your liver and muscles, to structural support, such as cellulose that makes up cell walls for plants. polymerizes in ice formation. Much of this structure still remains in liquid form, and hydrogen bonding networks still remain. So not only does water provide the solvent medium for all chemical reactions in cells, its structure dictates the way proteins fold, and how membranes form. It also maintains rigidity of cell walls, and serves as a thermoregulator. By virtue of its large polarity water is a good solvent for ionic substances and therefore provides a means of transporting inorganic nutrients such as NH , NO , CO , PO , and monatomic ions throughout higher organisms. Its ability to dissolve a wide variety of substances also makes it useful for disposing of wastes. Many of the human body’s defense mechanisms against external toxic substances involve conversion into water-soluble forms and elimination via urine. , while not as prevalent as the chemical compounds discussed above, trace elements, such as zinc, copper, and iron are highly important to biological function, often serving key structural and chemical purposes in protein function. Vitamins are organic compounds that often augment proteins to help with biological functions. It is clear from this overview that both monomeric and polymeric forms of the chemical components of chemicals in living systems serve important functions. Also, those functions change depending on which building blocks are used for polymerization. However, the properties of these polymers can be understood in terms of the monomers and how they combine. | 3,679 | 3,201 |
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\( \newcommand{\phb}{\beta} % phase beta\)
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\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
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\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
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\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
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\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
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\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
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\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
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\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A is an intensive property that indicates the relative amount of a particular species or substance in a phase. We sometimes need to make a distinction between a species and a substance. A is any entity of definite elemental composition and charge and can be described by a chemical formula, such as H\(_2\)O, H\(_3\)O\(^+\), NaCl, or Na\(^+\). A is a species that can be prepared in a pure state (e.g., N\(_2\) and NaCl). Since we cannot prepare a macroscopic amount of a single kind of ion by itself, a charged species such as H\(_3\)O\(^+\) or Na\(^+\) is not a substance. Chap. 10 will discuss the special features of mixtures containing charged species. The of species \(i\) is defined by \begin{gather} \s{ x_i \defn \frac{n_i}{\sum_j n_j} \qquad \tx{or} \qquad y_i \defn \frac{n_i}{\sum_j n_j} } \tag{9.1.1} \cond{(\(P{=}1\))} \end{gather} where \(n_i\) is the amount of species \(i\) and the sum is taken over all species in the mixture. The symbol \(x_i\) is used for a mixture in general, and \(y_i\) is used when the mixture is a gas. The , or weight fraction, of species \(i\) is defined by \begin{gather} \s{ w_i \defn \frac{m(i)}{m} = \frac{n_i M_i}{\sum_j n_j M_j} } \tag{9.1.2} \cond{(\(P{=}1\))} \end{gather} where \(m(i)\) is the mass of species \(i\) and \(m\) is the total mass. The , or molarity, of species \(i\) in a mixture is defined by \begin{gather} \s{ c_i \defn \frac{n_i}{V} } \tag{9.1.3} \cond{(\(P{=}1\))} \end{gather} The symbol M is often used to stand for units of mol L\(^{-1}\), or mol dm\(^{-3}\). Thus, a concentration of \(0.5\units{M}\) is \(0.5\) moles per liter, or \(0.5\) molar. Concentration is sometimes called “amount concentration” or “molar concentration” to avoid confusion with number concentration (the number of per unit volume). An alternative notation for \(c\A\) is [A]. A is a mixture of substances. A , strictly speaking, is a mixture in which one substance, the , is treated in a special way. Each of the other species comprising the mixture is then a . The solvent is denoted by A and the solute species by B, C, and so on. (Some chemists denote the solvent by subscript \(1\) and use \(2\), \(3\), and so on for solutes.) Although in principle a solution can be a gas mixture, in this section we will consider only liquid and solid solutions. We can prepare a solution of varying composition by gradually mixing one or more solutes with the solvent so as to continuously increase the solute mole fractions. During this mixing process, the physical state (liquid or solid) of the solution remains the same as that of the pure solvent. When the sum of the solute mole fractions is small compared to \(x\A\) (i.e., \(x\A\) is close to unity), the solution is called . As the solute mole fractions increase, we say the solution becomes more . Mole fraction, mass fraction, and concentration can be used as composition variables for both solvent and solute, just as they are for mixtures in general. A fourth composition variable, molality, is often used for a solute. The of solute species B is defined by \begin{gather} \s{ m\B \defn \frac{n\B}{m(\tx{A})} } \tag{9.1.4} \cond{(solution)} \end{gather} where \(m(\tx{A})=n\A M\A\) is the mass of solvent. The symbol m is sometimes used to stand for units of mol kg\(^{-1}\), although this should be discouraged because m is also the symbol for meter. For example, a solute molality of \(0.6\units{m}\) is \(0.6\) moles of solute per kilogram of solvent, or \(0.6\) molal. We may write simplified equations for a binary solution of two substances, solvent A and solute B. Equations 9.1.1–9.1.4 become \begin{gather} \s{ x\B = \frac{n\B}{n\A + n\B} } \tag{9.1.5} \cond{(binary solution)} \end{gather} \begin{gather} \s{ w\B = \frac{n\B M\B}{n\A M\A + n\B M\B} } \tag{9.1.6} \cond{(binary solution)} \end{gather} \begin{gather} \s{ c\B = \frac{n\B}{V} = \frac{n\B \rho}{n\A M\A + n\B M\B} } \tag{9.1.7} \cond{(binary solution)} \end{gather} \begin{gather} \s{ m\B = \frac{n\B}{n\A M\A} } \tag{9.1.8} \cond{(binary solution)} \end{gather} The right sides of Eqs. 9.1.5–9.1.8 express the solute composition variables in terms of the amounts and molar masses of the solvent and solute and the density \(\rho\) of the solution. To be able to relate the values of these composition variables to one another, we solve each equation for \(n\B\) and divide by \(n\A\) to obtain an expression for the mole ratio \(n\B/n\A\): \begin{gather} \s{\tx{from Eq. 9.1.5}} \tag{9.1.9} \qquad \s{\frac{n\B}{n\A} = \frac{x\B}{1-x\B}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.6}} \tag{9.1.10} \qquad \s{\frac{n\B}{n\A} = \frac{M\A w\B}{M\B(1-w\B)}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.7}} \tag{9.1.11} \qquad \s{\frac{n\B}{n\A} = \frac{M\A c\B}{\rho - M\B c\B}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.8}} \tag{9.1.12} \qquad \s{\frac{n\B}{n\A} = M\A m\B} \cond{(binary solution)} \end{gather} These expressions for \(n\B/n\A\) allow us to find one composition variable as a function of another. For example, to find molality as a function of concentration, we equate the expressions for \(n\B/n\A\) on the right sides of Eqs. 9.1.11 and 9.1.12 and solve for \(m\B\) to obtain \begin{equation} m\B = \frac{c\B}{\rho - M\B c\B} \tag{9.1.13} \end{equation} A binary solution becomes more dilute as any of the solute composition variables becomes smaller. In the limit of infinite dilution, the expressions for \(n\B/n\A\) become: \begin{gather} \s{\begin{split} \frac{n\B}{n\A} & = x\B \cr & = \frac{M\A}{M\B}w\B \cr & = \frac{M\A}{\rho\A^*}c\B = V\mA^* c\B \cr & = \s{ M\A m\B } \end{split}} \tag{9.1.14} \cond{(binary solution at} \nextcond{infinite dilution)} \end{gather} where a superscript asterisk (\({}^*\)) denotes a pure phase. We see that, in the limit of infinite dilution, the composition variables \(x\B\), \(w\B\), \(c\B\), and \(m\B\) are proportional to one another. These expressions are also valid for solute B in a solute solution in which solute is very dilute; that is, in the limit \(x\A\ra 1\). The rule of thumb that the molarity and molality values of a dilute aqueous solution are approximately equal is explained by the relation \(M\A c\B/\rho\A^*=M\A m\B\) (from Eq. 9.1.14), or \(c\B/\rho\A^* = m\B\), and the fact that the density \(\rho\A^*\) of water is approximately \(1\units{kg L\(^{-1}\)}\). Hence, if the solvent is water and the solution is dilute, the numerical value of \(c\B\) expressed in mol L\(^{-1}\) is approximately equal to the numerical value of \(m\B\) expressed in mol kg\(^{-1}\). We can describe the composition of a phase with the amounts of each species, or with any of the composition variables defined earlier: mole fraction, mass fraction, concentration, or molality. If we use mole fractions or mass fractions to describe the composition, we need the values for all but one of the species, since the sum of all fractions is unity. Other composition variables are sometimes used, such as volume fraction, mole ratio, and mole percent. To describe the composition of a gas mixture, partial pressures can be used (Sec. 9.3.1). When the composition of a mixture is said to be or during changes of temperature, pressure, or volume, this means there is no change in the relative or of the various species. A mixture of fixed composition has fixed values of mole fractions, mass fractions, and molalities, but not necessarily of concentrations and partial pressures. Concentrations will change if the volume changes, and partial pressures in a gas mixture will change if the pressure changes. | 15,064 | 3,203 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.1%3A_Dynamic_Equilibrium |
In the , we discussed the principles of chemical kinetics, which deal with the , or how quickly a given chemical reaction occurs. We now turn our attention to the to which a reaction occurs and how reaction conditions affect the final concentrations of reactants and products. For most of the reactions that we have discussed so far, you may have assumed that once reactants are converted to products, they are likely to remain that way. In fact, however, virtually all chemical reactions are to some extent. That is, an opposing reaction occurs in which the products react, to a greater or lesser degree, to re-form the reactants. Eventually, the forward and reverse reaction rates become the same, and the system reaches the point at which the composition of the system no longer changes with time. Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (\(\ce{N_2O_4}\)) to nitrogen dioxide (\(\ce{NO_2}\)). You may recall that \(\ce{NO_2}\) is responsible for the brown color we associate with smog. When a sealed tube containing solid \(\ce{N_2O_4}\) (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of \(\ce{NO_2}\) appears (Figure \(\Page {1}\)). The reaction can be followed visually because the product (\(\ce{NO_2}\)) is colored, whereas the reactant (\(\ce{N_2O_4}\)) is colorless: \[\underset{colorless }{\ce{N2O4 (g)}} \ce{ <=>[k_f,k_r] } \underset{red-brown }{\ce{2NO2(g)}}\label{Eq1} \] The double arrow indicates that both the forward reaction \[\ce{N2O4 (g) ->[k_f] 2NO2(g)} \label{eq1B} \] and reverse reaction \[\ce{2NO2(g) ->[k_r] N2O4 (g) } \label{eq1C} \] occurring simultaneously (i.e, the reaction is reversible). However, this does not necessarily mean the system is equilibrium as the following chapter demonstrates. Figure \(\Page {2}\) shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of \(\ce{NO_2}\) were zero, then it increases as the concentration of \(\ce{N_2O_4}\) decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no \(\ce{N_2O_4}\) but an initial \(\ce{NO_2}\) concentration twice the initial concentration of \(\ce{N_2O_4}\) (Figure \(\Page {2a}\)), in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition (Figure \(\Page {2b}\)). Thus equilibrium can be approached from either direction in a chemical reaction. Figure \(\Page {3}\) shows the forward and reverse reaction rates for a sample that initially contains pure \(\ce{NO_2}\). Because the initial concentration of \(\ce{N_2O_4}\) is zero, the forward reaction rate (dissociation of \(\ce{N_2O_4}\)) is initially zero as well. In contrast, the reverse reaction rate (dimerization of \(\ce{NO_2}\)) is initially very high (\(2.0 \times 10^6\, M/s\)), but it decreases rapidly as the concentration of \(\ce{NO_2}\) decreases. As the concentration of \(\ce{N_2O_4}\) increases, the rate of dissociation of \(\ce{N_2O_4}\) increases—but more slowly than the dimerization of \(\ce{NO_2}\)—because the reaction is only first order in \(\ce{N_2O_4}\) (rate = \(k_f[N_2O_4]\), where \(k_f\) is the rate constant for the forward reaction in Equations \(\ref{Eq1}\) and \(\ref{eq1B}\)). Eventually, the forward and reverse reaction rates become identical, \(k_f = k_r\), and the system has reached chemical equilibrium. If the forward and reverse reactions occur at different rates, then the system is not at equilibrium. The rate of dimerization of \(\ce{NO_2}\) (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of \(\ce{N_2O_4}\) is zero, the rate of the dissociation reaction (forward reaction) at \(t = 0\) is also zero. As the dimerization reaction proceeds, the \(\ce{N_2O_4}\) concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of \(\ce{N_2O_4}\) and \(\ce{NO_2}\) no longer change. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation: \[2A \rightleftharpoons B \nonumber \] where the blue circles are \(A\) and the purple ovals are \(B\). Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium? : three reaction systems relative time to reach chemical equilibrium : Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium. : In systems 1 and 3, the concentration of A decreases from \(t_0\) through \(t_2\) but is the same at both \(t_2\) and \(t_3\). Thus systems 1 and 3 are at equilibrium by \(t_3\). In system 2, the concentrations of A and B are still changing between \(t_2\) and \(t_3\), so system 2 may not yet have reached equilibrium by \(t_3\). Thus system 2 took the longest to reach chemical equilibrium. In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium? system 2 A Introduction to Dynamic Equilibrium: At equilibrium, the forward and reverse reactions of a system proceed at equal rates. Chemical equilibrium is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached. | 6,401 | 3,204 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.06%3A_Ionization_Energies |
Figure \(\Page {1}\) plots the the ionization energies of the elements are plotted against atomic number. An obvious feature of this figure is that the elements with the highest ionization energies are the noble gases. Since the ionization energy measures the energy which must be to remove an electron, these high values mean that it is difficult to remove an electron from an atom of a noble gas. A second obvious feature is that the elements with the lowest ionization energies are the alkali metals. This means that it is easier to remove electrons from atoms of this group of elements than from any other group. Closer inspection also reveals the following two general tendencies: One can confirm these general trends by inspecting Figure \(\Page {1}\). As one moves from He to Ne to Ar one can see marked decreases in the ionization energy, confirming the trend of decreasing ionization energy as you move down a group. Moving from left to right across the periodic table produces an increase in the ionization energy, as can be observed by the upward trend as you go from Li to Ne. Ionization energies can be measured quite accurately for atoms, and the values obtained show some additional features which are less important than the two major trends mentioned above. For example, consider the data for elements in the second row of the periodic table. Numerical values for the relevant ionization energies are shown in Figure \(\Page {2}\) of ionization energies and electron affinities below. The general trend of increasing ionization energy across the table is broken at two points. Boron has a smaller value than beryllium, and oxygen has a smaller value than nitrogen. The first break occurs when the first electron is added to a subshell. As was mentioned several times in the previous chapter, a 2 electron is higher in energy and hence easier to remove than a 2 electron because it is more efficiently shielded from the nuclear charge. Thus the 2 electron in boron is easier to remove than a 2 electron in beryllium. The second exception to the general trend occurs in the case of oxygen, which has one more 2 electron than the half-filled subshell of nitrogen. The last electron in the oxygen atom is forced into an already occupied orbital where it is kept close to another electron. The repulsion between these two electrons makes one of them easier to remove, and so the ionization energy of oxygen is lower than might be expected. These discontinuities also appear periodically, as would be expected since they arise from the structure of the valence electrons. Sulfur and selenium, in the same group as oxygen, show the discontinuity in the trend of increasing ionization energy, which arises from the same half-filled . Aluminum and gallium, both in the same group as boron, similarly show a decrease in ionization energy compared to magnesium and calcium. While this trend does not seem to apply for indium, and thallium, it is important to remember that the chart is missing the transition metals. Looking at the graph of ionization energies, it is clear that indium(atomic number 49) have a lower ionization energy than cadmium (atomic number 48), and the same is true of mercury, the element preceding thallium (atomic number 81). Below is a full periodic table, with shading to demonstrate the periodic trend of ionization energy. The darker the shading, the higher the ionization energy. Notice the general trend of increasing darkness (or ionization energy) as one moves to the right and up. Take a moment to write down why this is so. | 3,589 | 3,205 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/08%3A_Gravimetric_Methods/8.03%3A_Volatilization_Gravimetry |
A second approach to gravimetry is to thermally or chemically decompose the sample and measure the resulting change in its mass. Alternatively, we can trap and weigh a volatile decomposition product. Because the release of a volatile species is an essential part of these methods, we classify them collectively as volatilization gravimetric methods of analysis. Whether an analysis is direct or indirect, volatilization gravimetry usually requires that we know the products of the decomposition reaction. This rarely is a problem for organic compounds, which typically decompose to form simple gases such as CO , H O, and N . For an inorganic compound, however, the products often depend on the decomposition temperature. One method for determining the products of a thermal decomposition is to monitor the sample’s mass as a function of temperature, a process called . Figure 8.3.1
shows a typical in which each change in mass—each “step” in the thermogram—represents the loss of a volatile product. As the following example illustrates, we can use a thermogram to identify a compound’s decomposition reactions. The thermogram in Figure 8.3.1
shows the mass of a sample of calcium oxalate monohydrate, CaC O •H O, as a function of temperature. The original sample of 17.61 mg was heated from room temperature to 1000 C at a rate of 20 C per minute. For each step in the thermogram, identify the volatilization product and the solid residue that remains. From 100–250 C the sample loses 17.61 mg – 15.44 mg, or 2.17 mg, which is \[\frac{2.17 \ \mathrm{mg}}{17.61 \ \mathrm{mg}} \times 100=12.3 \% \nonumber\] of the sample’s original mass. In terms of CaC O •H O, this corresponds to a decrease in the molar mass of \[0.123 \times 146.11 \ \mathrm{g} / \mathrm{mol}=18.0 \ \mathrm{g} / \mathrm{mol} \nonumber\] The product’s molar mass and the temperature range for the decomposition, suggest that this is a loss of H O , leaving a residue of CaC O . The loss of 3.38 mg from 350–550 C is a 19.2% decrease in the sample’s original mass, or a decrease in the molar mass of \[0.192 \times 146.11 \ \mathrm{g} / \mathrm{mol}=28.1 \ \mathrm{g} / \mathrm{mol} \nonumber\] which is consistent with the loss of CO and a residue of CaCO . Finally, the loss of 5.30 mg from 600-800 C is a 30.1% decrease in the sample’s original mass, or a decrease in molar mass of \[0.301 \times 146.11 \ \mathrm{g} / \mathrm{mol}=44.0 \ \mathrm{g} / \mathrm{mol} \nonumber\] This loss in molar mass is consistent with the release of CO , leaving a final residue of CaO. The three decomposition reactions are \[\begin{array}{c}{\mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}(s) \rightarrow \ \mathrm{CaC}_{2} \mathrm{O}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)} \\ {\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightarrow \ \mathrm{CaCO}_{3}(s)+\mathrm{CO}(g)} \\ {\mathrm{CaCO}_{3}(s) \rightarrow \ \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)}\end{array} \nonumber\] Identifying the products of a thermal decomposition provides information that we can use to develop an analytical procedure. For example, the thermogram in shows that we must heat a precipitate of CaC O •H O to a temperature between 250 and 400 C if we wish to isolate and weigh CaC O . Alternatively, heating the sample to 1000 C allows us to isolate and weigh CaO. Under the same conditions as , the thermogram for a 22.16 mg sample of MgC O •H O shows two steps: a loss of 3.06 mg from 100–250 C and a loss of 12.24 mg from 350–550 C. For each step, identify the volatilization product and the solid residue that remains. Using your results from this exercise and the results from Example 8.3.1
, explain how you can use thermogravimetry to analyze a mixture that contains CaC O •H O and MgC O •H O. You may assume that other components in the sample are inert and thermally stable below 1000 C. From 100–250 C the sample loses 13.8% of its mass, or a loss of \[0.138 \times 130.34 \ \mathrm{g} / \mathrm{mol}=18.0 \ \mathrm{g} / \mathrm{mol} \nonumber\] which is consistent with the loss of H O and a residue of MgC O . From 350–550 C the sample loses 55.23% of its original mass, or a loss of \[0.5523 \times 130.34 \ \mathrm{g} / \mathrm{mol}=71.99 \ \mathrm{g} / \mathrm{mol} \nonumber\] This weight loss is consistent with the simultaneous loss of CO and CO , leaving a residue of MgO. We can analyze the mixture by heating a portion of the sample to 300 C, 600 C, and 1000 C, recording the mass at each temperature. The loss of mass between 600 C and 1000 C, \(\Delta m_2\), is due to the loss of CO from the decomposition of CaCO to CaO, and is proportional to the mass of CaC O •H O in the sample. \[\mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\Delta m_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{CO}_{2}}{44.01 \ \mathrm{g} \ \mathrm{CO}_{2}} \times \frac{146.11 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{\mathrm{mol} \ \mathrm{CO}_{2}} \nonumber\] The change in mass between 300 C and 600 C, \(\Delta m_1\), is due to the loss of CO from CaC O •H O and the loss of CO and CO from MgC O •H O. Because we already know the amount of CaC O •H O in the sample, we can calculate its contribution to \(\Delta m_1\). \[\left(\Delta m_{1}\right)_{\mathrm{Ca}}=\mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\Delta m_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{CO}}{146.11 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}} \times \frac{28.01 \ \mathrm{g} \ \mathrm{CO}}{\mathrm{mol} \ \mathrm{CO}} \nonumber\] The change in mass between 300 C and 600 C due to the decomposition of MgC O •H O \[\left(m_{1}\right)_{\mathrm{Mg}}=\Delta m_{1}-\left(\Delta m_{1}\right)_{\mathrm{Ca}} \nonumber\] provides the mass of MgC O •H O in the sample. \[\mathrm{g} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}=\left(\Delta m_{1}\right)_{\mathrm{Mg}} \times \frac{1 \ \mathrm{mol}\left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)}{130.35 \ \mathrm{g} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}} \times \frac{78.02 \ \mathrm{g} \ \left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)}{\mathrm{mol}\ \left(\mathrm{CO} \ + \ \mathrm{CO}_{2}\right)} \nonumber\] Depending on the method of analysis, the equipment for volatilization gravimetry may be simple or complex. In the simplest experimental design, we place the sample in a crucible and decompose it at a fixed temperature using a Bunsen burner, a Meker burner, a laboratory oven, or a muffle furnace. The sample’s mass and the mass of the residue are measured using an analytical balance. Trapping and weighing the volatile products of a thermal decomposition requires specialized equipment. The sample is placed in a closed container and heated. As decomposition occurs, a stream of an inert purge-gas sweeps the volatile products through one or more selective absorbent traps. In a thermogravimetric analysis, the sample is placed on a small balance pan attached to one arm of an electromagnetic balance (Figure 8.3.2
). The sample is lowered into an electric furnace and the furnace’s temperature is increased at a fixed rate of few degrees per minute while monitoring continuously the sample’s weight. The instrument usually includes a gas line for purging the volatile decomposition products out of the furnace, and a heat exchanger to dissipate the heat emitted by the furnace. . (a) Instrumentation for conducting a thermogravimetric analysis. The balance sits on the top of the instrument with the sample suspended below. A gas line supplies an inert gas that sweeps the volatile decomposition products out of the furnace. The heat exchanger dissipates the heat from the furnace to a reservoir of water. (b) Close-up showing the balance pan, which sits on a moving platform, the thermocouple for monitoring temperature, a hook for lowering the sample pan into the furnace, and the opening to the furnace. After placing a small portion of the sample on the balance pan, the platform rotates over the furnace and transfers the balance pan to a hook that is suspended from the balance. Once the balance pan is in place, the platform rotates back to its initial position. The balance pan and the thermocouple are then lowered into the furnace. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical volatilization gravimetric method. Although each method is unique, the determination of Si in ores and alloys by forming volatile SiF provides an instructive example of a typical procedure. The description here is based on a procedure from Young, R. S. , Griffen: London, 1971, pp. 302–304. Silicon is determined by dissolving the sample in acid and dehydrating to precipitate SiO . Because a variety of other insoluble oxides also form, the precipitate’s mass is not a direct measure of the amount of silicon in the sample. Treating the solid residue with HF forms volatile SiF . The decrease in mass following the loss of SiF provides an indirect measure of the amount of silicon in the original sample. Transfer a sample of between 0.5 g and 5.0 g to a platinum crucible along with an excess of Na CO , and heat until a melt forms. After cooling, dissolve the residue in dilute HCl. Evaporate the solution to dryness on a steam bath and heat the residue, which contains SiO and other solids, for one hour at 110 C. Moisten the residue with HCl and repeat the dehydration. Remove any acid soluble materials from the residue by adding 50 mL of water and 5 mL of concentrated HCl. Bring the solution to a boil and filter through #40 filter paper ( ). Wash the residue with hot 2% v/v HCl followed by hot water. Evaporate the filtrate to dryness twice and, following the same procedure, treat to remove any acid-soluble materials. Combine the two precipitates and dry and ignite to a constant weight at 1200 C. After cooling, add 2 drops of 50% v/v H SO and 10 mL of HF. Remove the volatile SiF by evaporating to dryness on a hot plate. Finally, bring the residue to constant weight by igniting at 1200 C. 1. According to the procedure the sample should weigh between 0.5 g and 5.0 g. How should you decide upon the amount of sample to use? In this procedure the critical measurement is the decrease in mass following the volatilization of SiF . The reaction responsible for the loss of mass is \[\mathrm{SiO}_{2}(s)+4 \mathrm{HF}(a q) \rightarrow \mathrm{SiF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(l ) \nonumber\] Water and excess HF are removed during the final ignition, and do not contribute to the change in mass. The loss in mass, therefore, is equivalent to the mass of SiO present after the dehydration step. Every 0.1 g of Si in the original sample results in the loss of 0.21 g of SiO . How much sample we use depends on what is an acceptable uncertainty when we measure its mass. A 0.5-g sample that is 50% w/w in Si, for example, will lose 0.53 g. If we are using a balance that measures mass to the nearest ±0.1 mg, then the relative uncertainty in mass is approximately ±0.02%; this is a reasonable level of uncertainty for a gravimetric analysis. A 0.5-g sample that is only 5% w/w Si experiences a weight loss of only 0.053 g and has a relative uncertainty of ±0.2%. In this case a larger sample is needed. 2. Why are acid-soluble materials removed before we treat the dehydrated residue with HF? Any acid-soluble materials in the sample will react with HF or H SO . If the products of these reactions are volatile, or if they decompose at 1200 C, then the change in mass is not due solely to the volatilization of SiF . As a result, we will overestimate the amount of Si in our sample. 3. Why is H SO added with the HF? Many samples that contain silicon also contain aluminum and iron, which form Al O and Fe O when we dehydrate the sample. These oxides are potential interferents because they also form volatile fluorides. In the presence of H SO , however, aluminum and iron preferentially form non-volatile sulfates, which eventually decompose back to their respective oxides when we heat the residue to 1200 C. As a result, the change in weight after treating with HF and H SO is due only to the loss of SiF . Unlike precipitation gravimetry, which rarely is used as a standard method of analysis, volatilization gravimetric methods continue to play an important role in chemical analysis. Several important examples are discussed below. Determining the inorganic ash content of an organic material, such as a polymer, is an example of a direct volatilization gravimetric analysis. After weighing the sample, it is placed in an appropriate crucible and the organic material carefully removed by combustion, leaving behind the inorganic ash. The crucible that contains the residue is heated to a constant weight using either a burner or an oven before the mass of the inorganic ash is determined. Another example of volatilization gravimetry is the determination of dissolved solids in natural waters and wastewaters. In this method, a sample of water is transferred to a weighing dish and dried to a constant weight at either 103–105 C or at 180 C. Samples dried at the lower temperature retain some occluded water and lose some carbonate as CO ; the loss of organic material, however, is minimal at this temperature. At the higher temperature, the residue is free from occluded water, but the loss of carbonate is greater. In addition, some chloride, nitrate, and organic material is lost through thermal decomposition. In either case, the residue that remains after drying to a constant weight at 500 C is the amount of fixed solids in the sample, and the loss in mass provides an indirect measure of the sample’s volatile solids. Indirect analyses based on the weight of a residue that remains after volatilization are used to determine moisture in a variety of products and to determine silica in waters, wastewaters, and rocks. Moisture is determined by drying a preweighed sample with an infrared lamp or a low temperature oven. The difference between the original weight and the weight after drying equals the mass of water lost. The most important application of volatilization gravimetry is for the elemental analysis of organic materials. During combustion with pure O , many elements, such as carbon and hydrogen, are released as gaseous combustion products, such as CO and H O . Passing the combustion products through preweighed tubes that contain selective absorbents and measuring the increase in each tube’s mass provides a direct analysis for the mass of carbon and hydrogen in the sample. Instead of measuring mass, modern instruments for completing an elemental analysis use gas chromatography ( ) or infrared spectroscopy ( ) to monitor the gaseous decomposition products. Alkaline metals and earths in organic materials are determined by adding H SO to the sample before combustion. After combustion is complete, the metal remains behind as a solid residue of metal sulfate. Silver, gold, and platinum are determined by burning the organic sample, leaving a metallic residue of Ag, Au, or Pt. Other metals are determined by adding HNO before combustion, which leaves a residue of the metal oxide. Volatilization gravimetry also is used to determine biomass in waters and wastewaters. Biomass is a water quality index that provides an indication of the total mass of organisms contained within a sample of water. A known volume of the sample is passed through a preweighed 0.45-μm membrane filter or a glass-fiber filter and dried at 105 C for 24 h. The residue’s mass provides a direct measure of biomass. If samples are known to contain a substantial amount of dissolved inorganic solids, the residue is ignited at 500 C for one hour, which volatilizes the biomass. The resulting inorganic residue is wetted with distilled water to rehydrate any clay minerals and dried to a constant weight at 105 C. The difference in mass before and after ignition provides an indirect measure of biomass. For some applications, such as determining the amount of inorganic ash in a polymer, a quantitative calculation is straightforward and does not require a balanced chemical reaction. For other applications, however, the relationship between the analyte and the analytical signal depends upon the stoichiometry of any relevant reactions. Once again, a conservation of mass is useful when solving problems. A 101.3-mg sample of an organic compound that contains chlorine is combusted in pure O . The volatile gases are collected in absorbent traps with the trap for CO increasing in mass by 167.6 mg and the trap for H O increasing in mass by 13.7-mg. A second sample of 121.8 mg is treated with concentrated HNO , producing Cl that reacts with Ag to form 262.7 mg of AgCl. Determine the compound’s composition, as well as its empirical formula. A conservation of mass requires that all the carbon in the organic compound is in the CO produced during combustion; thus \[0.1676 \ \mathrm{g} \ \mathrm{CO}_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{C}}{44.010 \ \mathrm{g} \ \mathrm{CO}_{2}} \times \frac{12.011 \ \mathrm{g} \ \mathrm{C}}{\mathrm{mol} \ \mathrm{C}}=0.04574 \ \text{g C} \nonumber\] \[\frac{0.04574 \ \mathrm{g} \ \mathrm{C}}{0.1013 \ \mathrm{g} \text { sample }} \times 100=45.15 \% \mathrm{w} / \mathrm{w} \ \mathrm{C} \nonumber\] Using the same approach for hydrogen and chlorine, we find that \[0.0137 \ \mathrm{g} \ \mathrm{H}_{2} \mathrm{O} \times \frac{2 \ \mathrm{mol} \ \mathrm{H}}{18.015 \ \mathrm{g} \ \mathrm{H}_{2} \mathrm{O}} \times \frac{1.008 \ \mathrm{g} \ \mathrm{H}}{\mathrm{mol} \ \mathrm{H}}=1.533 \times 10^{-3} \mathrm{g} \ \mathrm{H} \nonumber\] \[\frac{1.533 \ \times 10^{-3} \mathrm{g} \ \mathrm{H}}{0.1003 \ \mathrm{g} \ \text { sample }} \times 100=1.53 \% \mathrm{w} / \mathrm{w} \ \mathrm{H} \nonumber\] \[0.2627 \ \mathrm{g} \ \mathrm{AgCl} \times \frac{1 \ \mathrm{mol} \ \mathrm{Cl}}{143.32 \ \mathrm{g} \ \mathrm{AgCl}} \times \frac{35.455 \ \text{g Cl}}{\mathrm{mol} \ \mathrm{Cl}}=0.06498 \ \mathrm{g} \ \mathrm{Cl} \nonumber\] \[\frac{0.06498 \ \mathrm{g} \ \mathrm{Cl}}{0.1218 \ \mathrm{g} \text { sample }} \times 100=53.35 \% \mathrm{w} / \mathrm{w} \ \mathrm{Cl} \nonumber\] Adding together the weight percents for C, H, and Cl gives a total of 100.03%; thus, the compound contains only these three elements. To determine the compound’s empirical formula we note that a gram of sample contains 0.4515 g of C, 0.0153 g of H and 0.5335 g of Cl. Expressing each element in moles gives 0.0376 moles C, 0.0152 moles H and 0.0150 moles Cl. Hydrogen and chlorine are present in a 1:1 molar ratio. The molar ratio of C to moles of H or Cl is \[\frac{\mathrm{mol} \ \mathrm{C}}{\mathrm{mol} \text{ H}} =\frac{\mathrm{mol} \ \mathrm{C}}{\mathrm{mol} \ \mathrm{Cl}}=\frac{0.0376}{0.0150}=2.51 \approx 2.5 \nonumber\] Thus, the simplest, or empirical formula for the compound is C H Cl . In an indirect volatilization gravimetric analysis, the change in the sample’s weight is proportional to the amount of analyte in the sample. Note that in the following example it is not necessary to apply a conservation of mass to relate the analytical signal to the analyte. A sample of slag from a blast furnace is analyzed for SiO by decomposing a 0.5003-g sample with HCl, leaving a residue with a mass of 0.1414 g. After treating with HF and H SO , and evaporating the volatile SiF , a residue with a mass of 0.0183 g remains. Determine the %w/w SiO in the sample. The difference in the residue’s mass before and after volatilizing SiF gives the mass of SiO in the sample; thus the sample contains \[0.1414 \ \mathrm{g}-0.0183 \ \mathrm{g}=0.1231 \ \mathrm{g} \ \mathrm{SiO}_{2} \nonumber\] and the %w/w SiO is \[\frac{0.1231 \ \mathrm{g} \ \mathrm{Si} \mathrm{O}_{2}}{0.5003 \ \mathrm{g} \text { sample }} \times 100=24.61 \% \mathrm{w} / \mathrm{w} \ \mathrm{SiO}_{2} \nonumber\] Heating a 0.3317-g mixture of CaC O and MgC O yields a residue of 0.1794 g at 600 C and a residue of 0.1294 g at 1000 C. Calculate the %w/w CaC O in the sample. You may wish to review your answer to as you consider this problem. In we developed an equation for the mass of CaC O •H O in a mixture of CaC O •H O, MgC O •H O, and inert materials. Adapting this equation to a sample that contains CaC O , MgC O , and inert materials is easy; thus \[\mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4}=(0.1794 \ \mathrm{g}-0.1294 \ \mathrm{g}) \times \frac{1 \ \mathrm{mol} \ \mathrm{CO}_{2}}{44.01 \ \mathrm{g} \ \mathrm{CO}_{2}} \times \frac{128.10 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4}}{\mathrm{mol} \ \mathrm{CO}_{2}}=0.1455 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4} \nonumber\] The %w/w CaC O in the sample is \[\frac{0.1455 \ \mathrm{g} \ \mathrm{CaC}_{2} \mathrm{O}_{4}}{0.3317 \ \mathrm{g} \text { sample }} \times 100=43.86 \% \mathrm{w} / \mathrm{w} \mathrm{CaC}_{2} \mathrm{O}_{4} \nonumber\] Finally, for some quantitative applications we can compare the result for a sample to a similar result obtained using a standard. A 26.23-mg sample of MgC O •H O and inert materials is heated to constant weight at 1200 C, leaving a residue that weighs 20.98 mg. A sample of pure MgC O •H O, when treated in the same fashion, undergoes a 69.08% change in its mass. Determine the %w/w MgC O •H O in the sample. The change in the sample’s mass is 5.25 mg, which corresponds to \[5.25 \ \mathrm{mg} \operatorname{lost} \times \frac{100.0 \ \mathrm{mg} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{69.08 \ \mathrm{mg} \text { lost }}=7.60 \ \mathrm{mg} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O} \nonumber\] The %w/w MgC O •H O in the sample is \[\frac{7.60 \ \mathrm{mg} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O}}{26.23 \ \mathrm{mg} \text { sample }} \times 100=29.0 \% \mathrm{w} / \mathrm{w} \ \mathrm{MgC}_{2} \mathrm{O}_{4} \cdot \mathrm{H}_{2} \mathrm{O} \nonumber\] The scale of operation, accuracy, and precision of a gravimetric volatilization method is similar to that described in the last section for precipitation gravimetry. The sensitivity of a direct analysis is fixed by the analyte’s chemical form following combustion or volatilization. We can improve the sensitivity of an indirect analysis by choosing conditions that give the largest possible change in mass. For example, the thermogram in shows us that an indirect analysis for CaC O •H O is more sensitive if we measure the change in mass following ignition at 1000 C than if we ignite the sample at 300 C. Selectivity is not a problem for a direct analysis if we trap the analyte using a selective absorbent trap. A direct analysis based on the residue’s weight following combustion or volatilization is possible if the residue contains only the analyte of interest. As noted earlier, an indirect analysis only is feasible when the change in mass results from the loss of a single volatile product that contains the analyte. Volatilization gravimetric methods are time and labor intensive. Equipment needs are few, except when combustion gases must be trapped, or for a thermogravimetric analysis, when specialized instrumentation is needed. | 23,309 | 3,206 |
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Earlier in the course we performed stoichiometric calculations with chemical reactions using quantities of moles and mass (typically in grams). These same principles can be applied to chemical reactions involving gases except that we first have to convert volumes of gases into moles. Hydrogen gas reacts with oxygen gas to produce water vapor via the following balanced chemical equation: 2H + O -> 2H O If the temperature is 320 K and pressure is 1.34 atm, what volume of oxygen is required to produce 65.0 g of water? Strategy: Since we are given the temperature and pressure, to find the volume of oxygen using the ideal gas law we need to first calculate the moles of oxygen. To find the moles of oxygen required, we can first calculate the moles of water in 65.0g. \[n_{water}=\frac{m_{water}}{mm_{water}}=\frac{\text{65.0 g}}{\text{18.0g/mol}}=\text{3.61 mol of water}\] From the balanced chemical equation, we can see that 1 equivalent of oxygen produces 2 equivalents of water. We can therefore write the following ratio: 1 mol O : 2 mol H O We can now solve for the amount of oxygen: \[(\text{3.61 mol } H_{2}O)\times(\frac{ \text{1 mol } O_{2}}{\text{2 mol } H_{2}O })= \text{1.81 mol } O_{2}\] Finally, now that we know how many moles of oxygen are required, we can calculate the volume of the oxygen using the ideal gas law and the temperature&pressure provided in question: \[PV=nRT\] \[V=\frac{nRT}{P}\] \[{V = \rm \frac{1.81 mol\ \cdot 0.08206 \frac{L atm}{mol K}\cdot 320K}{1.34 atm }}\] \[V = 35.5 \rm L\] | 1,591 | 3,207 |
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The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. For an ideal gas, a plot of \(PV/nRT\) versus \(P\) gives a horizontal line with an intercept of 1 on the \(PV/nRT\) axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure \(\Page {1a}\)). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure \(\Page {1b}\)). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure \(\Page {2}\) for \(N_2\). Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure \(\Page {3}\)). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of / is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is than expected (Figure \(\Page {4}\)). Thus as shown in Figure \(\Page {2}\), at low temperatures, the ratio of (PV/nRT\) is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the \(PV/nRT\) versus \(P\) plot for many gases. Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called equation, \[ \underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1} \] and are empirical constants that are different for each gas. The values of \(a\) and \(b\) are listed in Table \(\Page {1}\) for several common gases. The pressure term in Equation \(\ref{10.9.1}\) corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, \(n^2/V^2\) represents the concentration of the gas (\(n/V\)) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure \(\Page {4}\). The volume term corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on and , respectively. Because nonzero molecular volumes produce a measured volume that is than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is than that expected based on the ideal gas law, so the \(an^2/V^2\) term must be added to the measured pressure to correct for these effects. You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? volume of cylinder, mass of compound, pressure, and temperature safety Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. Obtain and values for Cl from Table \(\Page {1}\). Use the van der Waals equation (\(\ref{10.9.1}\)) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): \[\begin{align} n &=\dfrac{m}{M} \\[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \\[4pt] &=7.052\;mol\nonumber \end{align} \nonumber \] Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure: \[\begin{align} P &=\dfrac{nRT}{V} \\[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \\[4pt] &= 43.1\;atm \end{align} \nonumber \] If chlorine behaves like an ideal gas, you have a real problem! Now let’s use the van der Waals equation with the and values for Cl from Table \(\Page {1}\). Solving for \(P\) gives \[\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\\&=\rm28.2\;atm\end{align} \nonumber \] This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the 77 atm 67 atm Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals \(a\) coefficients are relatively easy to liquefy because large coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O , N , Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek , meaning “cold,” and , meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called , which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure \(\Page {5}\)). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of / versus at a given temperature; for an ideal gas, / versus = 1 under all conditions. At high pressures, most real gases exhibit larger / values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit / values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the , which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold . | 12,027 | 3,208 |
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is a class of separation techniques in which we separate analytes by their ability to move through a conductive medium—usually an aqueous buffer—in response to an applied electric field. In the absence of other effects, cations migrate toward the electric field’s negatively charged cathode. Cations with larger charge-to-size ratios—which favors ions of greater charge and of smaller size—migrate at a faster rate than larger cat- ions with smaller charges. Anions migrate toward the positively charged anode and neutral species do not experience the electrical field and remain stationary. As we will see shortly, under normal conditions even neutral species and anions migrate toward the cathode. There are several forms of electrophoresis. In slab gel electrophoresis the conducting buffer is retained within a porous gel of agarose or polyacrylamide. Slabs are formed by pouring the gel between two glass plates separated by spacers. Typical thicknesses are 0.25–1 mm. Gel electrophoresis is an important technique in biochemistry where it frequently is used to separate DNA fragments and proteins. Although it is a powerful tool for the qualitative analysis of complex mixtures, it is less useful for quantitative work. In the conducting buffer is retained within a capillary tube with an inner diameter that typically is 25–75 μm. The sample is injected into one end of the capillary tube, and as it migrates through the capillary the sample’s components separate and elute from the column at different times. The resulting looks similar to a GC or an HPLC chromatogram, and provides both qualitative and quantitative information. Only capillary electrophoretic methods receive further consideration in this section. In capillary electrophoresis we inject the sample into a buffered solution retained within a capillary tube. When an electric field is applied across the capillary tube, the sample’s components migrate as the result of two types of actions: electrophoretic mobility and electroosmotic mobility. is the solute’s response to the applied electrical field in which cations move toward the negatively charged cathode, anions move toward the positively charged anode, and neutral species remain stationary. The other contribution to a solute’s migration is , which occurs when the buffer moves through the capillary in response to the applied electrical field. Under normal conditions the buffer moves toward the cathode, sweeping most solutes, including the anions and neutral species, toward the negatively charged cathode. The velocity with which a solute moves in response to the applied electric field is called its , \(\nu_{ep}\); it is defined as \[\nu_{ep}=\mu_{ep} E \label{12.1}\] where \(\mu_{ep}\) is the solute’s electrophoretic mobility, and is the magnitude of the applied electrical field. A solute’s electrophoretic mobility is defined as \[\mu_{ep}=\frac{q}{6 \pi \eta r} \label{12.2}\] where is the solute’s charge, \(\eta\) is the buffer’s viscosity, and is the solute’s radius. Using Equation \ref{12.1} and Equation \ref{12.2} we can make several important conclusions about a solute’s electrophoretic velocity. Electrophoretic mobility and, therefore, electrophoretic velocity, increases for more highly charged solutes and for solutes of smaller size. Because is positive for a cation and negative for an anion, these species migrate in opposite directions. A neutral species, for which is zero, has an electrophoretic velocity of zero. When an electric field is applied to a capillary filled with an aqueous buffer we expect the buffer’s ions to migrate in response to their electrophoretic mobility. Because the solvent, H O, is neutral we might reasonably expect it to remain stationary. What we observe under normal conditions, however, is that the buffer moves toward the cathode. This phenomenon is called the electroosmotic flow. Electroosmotic flow occurs because the walls of the capillary tubing carry a charge. The surface of a silica capillary contains large numbers of silanol groups (–SiOH). At a pH level greater than approximately 2 or 3, the silanol groups ionize to form negatively charged silanate ions (–SiO ). Cations from the buffer are attracted to the silanate ions. As shown in Figure 12.7.1
, some of these cations bind tightly to the silanate ions, forming a fixed layer. Because the cations in the fixed layer only partially neutralize the negative charge on the capillary walls, the solution adjacent to the fixed layer—which is called the diffuse layer—contains more cations than anions. Together these two layers are known as the double layer. Cations in the diffuse layer migrate toward the cathode. Because these cations are solvated, the solution also is pulled along, producing the electroosmotic flow. The anions in the diffuse layer, which also are solvated, try to move toward the anode. Because there are more cations than anions, however, the cations win out and the electroosmotic flow moves in the direction of the cathode. The rate at which the buffer moves through the capillary, what we call its , \(\nu_{eof}\), is a function of the applied electric field, , and the buffer’s electroosmotic mobility, \(\mu_{eof}\). \[\nu_{eof}=\mu_{e o f} E \label{12.3}\] Electroosmotic mobility is defined as \[\mu_{eof}=\frac{\varepsilon \zeta}{4 \pi \eta} \label{12.4}\] where \(\epsilon\) is the buffer dielectric constant, \(\zeta\) is the zeta potential, and \(\eta\) is the buffer’s viscosity. The —the potential of the diffuse layer at a finite distance from the capillary wall—plays an important role in determining the electroosmotic flow velocity. Two factors determine the zeta potential’s value. First, the zeta potential is directly proportional to the charge on the capillary walls, with a greater density of silanate ions corresponding to a larger zeta potential. Below a pH of 2 there are few silanate ions and the zeta potential and the electroosmotic flow velocity approach zero. As the pH increases, both the zeta potential and the electroosmotic flow velocity increase. Second, the zeta potential is directly proportional to the thickness of the double layer. Increasing the buffer’s ionic strength provides a higher concentration of cations, which decreases the thickness of the double layer and decreases the electroosmotic flow. The definition of zeta potential given here admittedly is a bit fuzzy. For a more detailed explanation see Delgado, A. V.; González-Caballero, F.; Hunter, R. J.; Koopal, L. K.; Lyklema, J. “Measurement and Interpretation of Electrokinetic Phenomena,” , , 1753–1805. Although this is a very technical report, Sections 1.3–1.5 provide a good introduction to the difficulty of defining the zeta potential and of measuring its value. The electroosmotic flow profile is very different from that of a fluid moving under forced pressure. Figure 12.7.2
compares the electroosmotic flow profile with the hydrodynamic flow profile in gas chromatography and liquid chromatography. The uniform, flat profile for electroosmosis helps minimize band broadening in capillary electrophoresis, improving separation efficiency. A solute’s total velocity, \(\nu_{tot}\), as it moves through the capillary is the sum of its electrophoretic velocity and the electroosmotic flow velocity. \[\nu_{t o t}=\nu_{ep}+\nu_{eof} \nonumber\] As shown in Figure 12.7.3
, under normal conditions the following general relationships hold true. \[(\nu_{tot})_{cations} > \nu_{eof} \nonumber\] \[(\nu_{tot})_{neutrals} = \nu_{eof} \nonumber\] \[(\nu_{tot})_{anions} < \nu_{eof} \nonumber\] Cations elute first in an order that corresponds to their electrophoretic mobilities, with small, highly charged cations eluting before larger cations of lower charge. Neutral species elute as a single band with an elution rate equal to the electroosmotic flow velocity. Finally, anions are the last components to elute, with smaller, highly charged anions having the longest elution time. Another way to express a solute’s velocity is to divide the distance it travels by the elapsed time \[\nu_{tot}=\frac{l}{t_{m}} \label{12.5}\] where is the distance between the point of injection and the detector, and is the solute’s migration time. To understand the experimental variables that affect migration time, we begin by noting that \[\nu_{tot} = \mu_{tot}E = (\mu_{ep} + \mu_{eof})E \label{12.6}\] Combining Equation \ref{12.5} and Equation \ref{12.6} and solving for leaves us with \[t_{\mathrm{m}}=\frac{l}{\left(\mu_{ep}+\mu_{eof}\right) E} \label{12.7}\] The magnitude of the electrical field is \[E=\frac{V}{L} \label{12.8}\] where is the applied potential and is the length of the capillary tube. Finally, substituting Equation \ref{12.8} into Equation \ref{12.7} leaves us with the following equation for a solute’s migration time. \[t_{\mathrm{m}}=\frac{lL}{\left(\mu_{ep}+\mu_{eof}\right) V} \label{12.9}\] To decrease a solute’s migration time—which shortens the analysis time—we can apply a higher voltage or use a shorter capillary tube. We can also shorten the migration time by increasing the electroosmotic flow, although this decreases resolution. As we learned in , the efficiency of a separation is given by the number of theoretical plates, . In capillary electrophoresis the number of theoretic plates is \[N=\frac{l^{2}}{2 D t_{m}}=\frac{\left(\mu_{e p}+\mu_{eof}\right) E l}{2 D L} \label{12.10}\] where is the solute’s diffusion coefficient. From Equation \ref{12.10}, the efficiency of a capillary electrophoretic separation increases with higher voltages. Increasing the electroosmotic flow velocity improves efficiency, but at the expense of resolution. Two additional observations deserve comment. First, solutes with larger electrophoretic mobilities—in the same direction as the electroosmotic flow—have greater efficiencies; thus, smaller, more highly charged cations are not only the first solutes to elute, but do so with greater efficiency. Second, efficiency in capillary electrophoresis is independent of the capillary’s length. Theoretical plate counts of approximately 100 000–200 000 are not unusual. It is possible to design an electrophoretic experiment so that anions elute before cations—more about this later—in which smaller, more highly charged anions elute with greater efficiencies. In chromatography we defined the selectivity between two solutes as the ratio of their retention factors. In capillary electrophoresis the analogous expression for selectivity is \[\alpha=\frac{\mu_{ep, 1}}{\mu_{ep, 2}} \nonumber\] where \(\mu_{ep,1}\) and \(\mu_{ep,2}\) are the electrophoretic mobilities for the two solutes, chosen such that \(\alpha \ge 1\). We can often improve selectivity by adjusting the pH of the buffer solution. For example, \(\text{NH}_4^+\) is a weak acid with a p of 9.75. At a pH of 9.75 the concentrations of \(\text{NH}_4^+\) and NH are equal. Decreasing the pH below 9.75 increases its electrophoretic mobility because a greater fraction of the solute is present as the cation \(\text{NH}_4^+\). On the other hand, raising the pH above 9.75 increases the proportion of neutral NH , decreasing its electrophoretic mobility. The resolution between two solutes is \[R = \frac {0.177(\mu_{ep,2} - \mu_{ep,1})\sqrt{V}} {\sqrt{D(\mu_{avg} + \mu_{eof}}} \label{12.11}\] where \(\mu_{avg}\) is the average electrophoretic mobility for the two solutes. Increasing the applied voltage and decreasing the electroosmotic flow velocity improves resolution. The latter effect is particularly important. Although increasing electroosmotic flow improves analysis time and efficiency, it de- creases resolution. The basic instrumentation for capillary electrophoresis is shown in Figure 12.7.4
and includes a power supply for applying the electric field, anode and cathode compartments that contain reservoirs of the buffer solution, a sample vial that contains the sample, the capillary tube, and a detector. Each part of the instrument receives further consideration in this section. Figure 12.7.5
shows a cross-section of a typical capillary tube. Most capillary tubes are made from fused silica coated with a 15–35 μm layer of polyimide to give it mechanical strength. The inner diameter is typically 25–75 μm, which is smaller than the internal diameter of a capillary GC column, with an outer diameter of 200–375 μm. The capillary column’s narrow opening and the thickness of its walls are important. When an electric field is applied to the buffer solution, current flows through the capillary. This current leads to the release of heat, which we call . The amount of heat released is proportional to the capillary’s radius and to the magnitude of the electrical field. Joule heating is a problem because it changes the buffer’s viscosity, with the solution at the center of the capillary being less viscous than that near the capillary walls. Because a solute’s electrophoretic mobility depends on its viscosity (see Equation \ref{12.2}), solute species in the center of the capillary migrate at a faster rate than those near the capillary walls. The result is an additional source of band broadening that degrades the separation. Capillaries with smaller inner diameters generate less Joule heating, and capillaries with larger outer diameters are more effective at dissipating the heat. Placing the capillary tube inside a thermostated jacket is another method for minimizing the effect of Joule heating; in this case a smaller outer diameter allows for a more rapid dissipation of thermal energy. There are two common methods for injecting a sample into a capillary electrophoresis column: hydrodynamic injection and electrokinetic injection. In both methods the capillary tube is filled with the buffer solution. One end of the capillary tube is placed in the destination reservoir and the other end is placed in the sample vial. uses pressure to force a small portion of sample into the capillary tubing. A difference in pressure is applied across the capillary either by pressurizing the sample vial or by applying a vacuum to the destination reservoir. The volume of sample injected, in liters, is given by the following equation \[V_{\text {inj}}=\frac{\Delta P d^{4} \pi t}{128 \eta L} \times 10^{3} \label{12.12}\] where \(\Delta P\) is the difference in pressure across the capillary in pascals, is the capillary’s inner diameter in meters, is the amount of time the pressure is applied in seconds, \(\eta\) is the buffer’s viscosity in kg m s , and is the length of the capillary tubing in meters. The factor of 10 changes the units from cubic meters to liters. For a hydrodynamic injection we move the capillary from the source reservoir to the sample. The anode remains in the source reservoir. A hydrodynamic injection also is possible if we raise the sample vial above the destination reservoir and briefly insert the filled capillary. In a hydrodynamic injection we apply a pressure difference of \(2.5 \times 10^3\) Pa (a \(\Delta P \approx 0.02 \text{ atm}\)) for 2 s to a 75-cm long capillary tube with an internal diameter of 50 μm. Assuming the buffer’s viscosity is 10 kg m s , what volume and length of sample did we inject? Making appropriate substitutions into Equation \ref{12.12} gives the sample’s volume as \[V_{inj}=\frac{\left(2.5 \times 10^{3} \text{ kg} \text{ m}^{-1} \text{ s}^{-2}\right)\left(50 \times 10^{-6} \text{ m}\right)^{4}(3.14)(2 \text{ s})}{(128)\left(0.001 \text{ kg} \text{ m}^{-1} \text{ s}^{-1}\right)(0.75 \text{ m})} \times 10^{3} \mathrm{L} / \mathrm{m}^{3} \nonumber\] \[V_{inj} = 1 \times 10^{-9} \text{ L} = 1 \text{ nL} \nonumber\] Because the interior of the capillary is cylindrical, the length of the sample, , is easy to calculate using the equation for the volume of a cylinder; thus \[l=\frac{V_{\text {inj}}}{\pi r^{2}}=\frac{\left(1 \times 10^{-9} \text{ L}\right)\left(10^{-3} \text{ m}^{3} / \mathrm{L}\right)}{(3.14)\left(25 \times 10^{-6} \text{ m}\right)^{2}}=5 \times 10^{-4} \text{ m}=0.5 \text{ mm} \nonumber\] Suppose you need to limit your injection to less than 0.20% of the capillary’s length. Using the information from Example 12.7.1
, what is the maximum injection time for a hydrodynamic injection? The capillary is 75 cm long, which means that 0.20% of that sample’s maximum length is 0.15 cm. To convert this to the maximum volume of sample we use the equation for the volume of a cylinder. \[V_{i n j}=l \pi r^{2}=(0.15 \text{ cm})(3.14)\left(25 \times 10^{-4} \text{ cm}\right)^{2}=2.94 \times 10^{-6} \text{ cm}^{3} \nonumber\] Given that 1 cm is equivalent to 1 mL, the maximum volume is \(2.94 \times 10^{-6}\) mL or \(2.94 \times 10^{-9}\) L. To find the maximum injection time, we first solve Equation \ref{12.12} for \[t=\frac{128 V_{inj} \eta L}{P d^{4} \pi} \times 10^{-3} \text{ m}^{3} / \mathrm{L} \nonumber\] and then make appropriate substitutions. \[t=\frac{(128)\left(2.94 \times 10^{-9} \text{ L}\right)\left(0.001 \text{ kg } \text{ m}^{-1} \text{ s}^{-1}\right)(0.75 \text{ m})}{\left(2.5 \times 10^{3} \text{ kg } \mathrm{m}^{-1} \text{ s}^{-2}\right)\left(50 \times 10^{-6} \text{ m}\right)^{4}(3.14)} \times \frac{10^{-3} \text{ m}^{3}}{\mathrm{L}} = 5.8 \text{ s} \nonumber\] The maximum injection time, therefore, is 5.8 s. In an we place both the capillary and the anode into the sample and briefly apply an potential. The volume of injected sample is the product of the capillary’s cross sectional area and the length of the capillary occupied by the sample. In turn, this length is the product of the solute’s velocity (see Equation \ref{12.6}) and time; thus \[V_{inj} = \pi r^2 L = \pi r^2 (\mu_{ep} + \mu_{eof})E^{\prime}t \label{12.13}\] where is the capillary’s radius, is the capillary’s length, and \(E^{\prime}\) is the effective electric field in the sample. An important consequence of Equation \ref{12.13} is that an electrokinetic injection is biased toward solutes with larger electrophoretic mobilities. If two solutes have equal concentrations in a sample, we inject a larger volume—and thus more moles—of the solute with the larger \(\mu_{ep}\). The electric field in the sample is different that the electric field in the rest of the capillary because the sample and the buffer have different ionic compositions. In general, the sample’s ionic strength is smaller, which makes its conductivity smaller. The effective electric field is \[E^{\prime} = E \times \frac {\chi_\text{buffer}} {\chi_\text{sample}}\nonumber\] where \(\chi_\text{buffer}\) and \(\chi_{sample}\) are the conductivities of the buffer and the sample, respectively. When an analyte’s concentration is too small to detect reliably, it maybe possible to inject it in a manner that increases its concentration. This method of injection is called . Stacking is accomplished by placing the sample in a solution whose ionic strength is significantly less than that of the buffer in the capillary tube. Because the sample plug has a lower concentration of buffer ions, the effective field strength across the sample plug, \(E^{\prime}\), is larger than that in the rest of the capillary. We know from Equation \ref{12.1} that electrophoretic velocity is directly proportional to the electrical field. As a result, the cations in the sample plug migrate toward the cathode with a greater velocity, and the anions migrate more slowly—neutral species are unaffected and move with the electroosmotic flow. When the ions reach their respective boundaries between the sample plug and the buffer, the electrical field decreases and the electrophoretic velocity of the cations decreases and that for the anions increases. As shown in Figure 12.7.6
, the result is a stacking of cations and anions into separate, smaller sampling zones. Over time, the buffer within the capillary becomes more homogeneous and the separation proceeds without additional stacking. Migration in electrophoresis occurs in response to an applied electric field. The ability to apply a large electric field is important because higher voltages lead to shorter analysis times (Equation \ref{12.9}), more efficient separations (Equation \ref{12.10}), and better resolution (Equation \ref{12.11}). Because narrow bored capillary tubes dissipate Joule heating so efficiently, voltages of up to 40 kV are possible. Because of the high voltages, be sure to follow your instrument’s safety guidelines. Most of the detectors used in HPLC also find use in capillary electrophoresis. Among the more common detectors are those based on the absorption of UV/Vis radiation, fluorescence, conductivity, amperometry, and mass spectrometry. Whenever possible, detection is done “on-column” before the solutes elute from the capillary tube and additional band broadening occurs. UV/Vis detectors are among the most popular. Because absorbance is directly proportional to path length, the capillary tubing’s small diameter leads to signals that are smaller than those obtained in HPLC. Several approaches have been used to increase the pathlength, including a Z-shaped sample cell and multiple reflections (see Figure 12.7.7
). Detection limits are about 10 M. Better detection limits are obtained using fluorescence, particularly when using a laser as an excitation source. When using fluorescence detection a small portion of the capillary’s protective coating is removed and the laser beam is focused on the inner portion of the capillary tubing. Emission is measured at an angle of 90 to the laser. Because the laser provides an intense source of radiation that can be focused to a narrow spot, detection limits are as low as 10 M. Solutes that do not absorb UV/Vis radiation or that do not undergo fluorescence can be detected by other detectors. Table 12.7.1
provides a list of detectors for capillary electrophoresis along with some of their important characteristics. universal (total ion) selective (single ion) There are several different forms of capillary electrophoresis, each of which has its particular advantages. Four of these methods are described briefly in this section. The simplest form of capillary electrophoresis is . In CZE we fill the capillary tube with a buffer and, after loading the sample, place the ends of the capillary tube in reservoirs that contain additional buffer. Usually the end of the capillary containing the sample is the anode and solutes migrate toward the cathode at a velocity determined by their respective electrophoretic mobilities and the electroosmotic flow. Cations elute first, with smaller, more highly charged cations eluting before larger cations with smaller charges. Neutral species elute as a single band. Anions are the last species to elute, with smaller, more negatively charged anions being the last to elute. We can reverse the direction of electroosmotic flow by adding an alkylammonium salt to the buffer solution. As shown in Figure 12.7.8
, the positively charged end of the alkyl ammonium ions bind to the negatively charged silanate ions on the capillary’s walls. The tail of the alkyl ammonium ion is hydrophobic and associates with the tail of another alkyl ammonium ion. The result is a layer of positive charges that attract anions in the buffer. The migration of these solvated anions toward the anode reverses the electroosmotic flow’s direction. The order of elution is exactly opposite that observed under normal conditions. Coating the capillary’s walls with a nonionic reagent eliminates the electroosmotic flow. In this form of CZE the cations migrate from the anode to the cathode. Anions elute into the source reservoir and neutral species remain stationary. Capillary zone electrophoresis provides effective separations of charged species, including inorganic anions and cations, organic acids and amines, and large biomolecules such as proteins. For example, CZE was used to separate a mixture of 36 inorganic and organic ions in less than three minutes [Jones, W. R.; Jandik, P. . , , 385–393]. A mixture of neutral species, of course, can not be resolved. One limitation to CZE is its inability to separate neutral species. overcomes this limitation by adding a surfactant, such as sodium dodecylsulfate (Figure 12.7.9
a) to the buffer solution. Sodium dodecylsulfate, or SDS, consists of a long-chain hydrophobic tail and a negatively charged ionic functional group at its head. When the concentration of SDS is sufficiently large a micelle forms. A micelle consists of a spherical agglomeration of 40–100 surfactant molecules in which the hydrocarbon tails point inward and the negatively charged heads point outward (Figure 12.7.9
b). Because micelles have a negative charge, they migrate toward the cathode with a velocity less than the electroosmotic flow velocity. Neutral species partition themselves between the micelles and the buffer solution in a manner similar to the partitioning of solutes between the two liquid phases in HPLC. Because there is a partitioning between two phases, we include the descriptive term chromatography in the techniques name. Note that in MEKC both phases are mobile. The elution order for neutral species in MEKC depends on the extent to which each species partitions into the micelles. Hydrophilic neutrals are insoluble in the micelle’s hydrophobic inner environment and elute as a single band, as they would in CZE. Neutral solutes that are extremely hy- drophobic are completely soluble in the micelle, eluting with the micelles as a single band. Those neutral species that exist in a partition equilibrium between the buffer and the micelles elute between the completely hydro- philic and completely hydrophobic neutral species. Those neutral species that favor the buffer elute before those favoring the micelles. Micellar electrokinetic chromatography is used to separate a wide variety of samples, including mixtures of pharmaceutical compounds, vitamins, and explosives. In the capillary tubing is filled with a polymeric gel. Because the gel is porous, a solute migrates through the gel with a velocity determined both by its electrophoretic mobility and by its size. The ability to effect a separation using size is helpful when the solutes have similar electrophoretic mobilities. For example, fragments of DNA of varying length have similar charge-to-size ratios, making their separation by CZE difficult. Because the DNA fragments are of different size, a CGE separation is possible. The capillary used for CGE usually is treated to eliminate electroosmotic flow to prevent the gel from extruding from the capillary tubing. Samples are injected electrokinetically because the gel provides too much resistance for hydrodynamic sampling. The primary application of CGE is the separation of large biomolecules, including DNA fragments, proteins, and oligonucleotides. Another approach to separating neutral species is . In CEC the capillary tubing is packed with 1.5–3 μm particles coated with a bonded stationary phase. Neutral species separate based on their ability to partition between the stationary phase and the buffer, which is moving as a result of the electroosmotic flow; Figure 12.7.10
provides a representative example for the separation of a mixture of hydrocarbons. A CEC separation is similar to the analogous HPLC separation, but without the need for high pressure pumps. Efficiency in CEC is better than in HPLC, and analysis times are shorter. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of a vitamin B complex by capillary zone electrophoresis or by micellar electrokinetic capillary chromatography provides an instructive example of a typical procedure. The description here is based on Smyth, W. F. , Wiley Teubner: Chichester, England, 1996, pp. 154–156. The water soluble vitamins B (thiamine hydrochloride), B (riboflavin), B (niacinamide), and B (pyridoxine hydrochloride) are determined by CZE using a pH 9 sodium tetraborate-sodium dihydrogen phosphate buffer, or by MEKC using the same buffer with the addition of sodium dodecyl sulfate. Detection is by UV absorption at 200 nm. An internal standard of -ethoxybenzamide is used to standardize the method. Crush a vitamin B complex tablet and place it in a beaker with 20.00 mL of a 50 % v/v methanol solution that is 20 mM in sodium tetraborate and 100.0 ppm in -ethoxybenzamide. After mixing for 2 min to ensure that the B vitamins are dissolved, pass a 5.00-mL portion through a 0.45-μm filter to remove insoluble binders. Load an approximately 4 nL sample into a capillary column with an inner diameter of a 50 μm. For CZE the capillary column contains a 20 mM pH 9 sodium tetraborate-sodium dihydrogen phosphate buffer. For MEKC the buffer is also 150 mM in sodium dodecyl sulfate. Apply a 40 kV/m electrical field to effect both the CZE and MEKC separations. 1. Methanol, which elutes at 4.69 min, is included as a neutral species to indicate the electroosmotic flow. When using standard solutions of each vitamin, CZE peaks are found at 3.41 min, 4.69 min, 6.31 min, and 8.31 min. Examine the structures and p information in Figure 12.7.11
and identify the order in which the four B vitamins elute. At a pH of 9, vitamin B is a cation and elutes before the neutral species methanol; thus it is the compound that elutes at 3.41 min. Vitamin B is a neutral species at a pH of 9 and elutes with methanol at 4.69 min. The remaining two B vitamins are weak acids that partially ionize to weak base anions at a pH of 9. Of the two, vitamin B is the stronger acid (a p of 9.0 versus a p of 9.7) and is present to a greater extent in its anionic form. Vitamin B , therefore, is the last of the vitamins to elute. 2. The order of elution when using MEKC is vitamin B (5.58 min), vitamin B (6.59 min), vitamin B (8.81 min), and vitamin B (11.21 min). What conclusions can you make about the solubility of the B vitamins in the sodium dodecylsulfate micelles? The micelles elute at 17.7 min. The elution time for vitamin B shows the greatest change, increasing from 3.41 min to 11.21 minutes. Clearly vitamin B has the greatest solubility in the micelles. Vitamin B and vitamin B3 have a more limited solubility in the micelles, and show only slightly longer elution times in the presence of the micelles. Interestingly, the elution time for vitamin B decreases in the presence of the micelles. 3. For quantitative work an internal standard of -ethoxybenzamide is added to all samples and standards. Why is an internal standard necessary? Although the method of injection is not specified, neither a hydrodynamic injection nor an electrokinetic injection is particularly reproducible. The use of an internal standard compensates for this limitation. When compared to GC and HPLC, capillary electrophoresis provides similar levels of accuracy, precision, and sensitivity, and it provides a comparable degree of selectivity. The amount of material injected into a capillary electrophoretic column is significantly smaller than that for GC and HPLC—typically 1 nL versus 0.1 μL for capillary GC and 1–100 μL for HPLC. Detection limits for capillary electrophoresis, however, are 100–1000 times poorer than that for GC and HPLC. The most significant advantages of capillary electrophoresis are improvements in separation efficiency, time, and cost. Capillary electrophoretic columns contain substantially more theoretical plates (\(\approx 10^6\) plates/m) than that found in HPLC (\(\approx 10^5\) plates/m) and capillary GC columns (\(\approx 10^3\) plates/m), providing unparalleled resolution and peak capacity. Separations in capillary electrophoresis are fast and efficient. Furthermore, the capillary column’s small volume means that a capillary electrophoresis separation requires only a few microliters of buffer, compared to 20–30 mL of mobile phase for a typical HPLC separation. | 32,062 | 3,210 |
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Although the calculations in this chapter are relatively straightforward, it can be tedious to work problems using nothing more than a calculator. Both Excel and R include functions for many common statistical calculations. In addition, R provides useful functions for visualizing your data. Excel has built-in functions that we can use to complete many of the statistical calculations covered in this chapter, including reporting descriptive statistics, such as means and variances, predicting the probability of obtaining a given outcome from a binomial distribution or a normal distribution, and carrying out significance tests. Table 4.8.1
provides the syntax for many of these functions; you can information on functions not included here by using Excel’s Help menu. Let’s use Excel to provide a statistical summary of the data in . Enter the data into a spreadsheet, as shown in Figure 4.8.1
. To calculate the sample’s mean, for example, click on any empty cell, enter the formula = (b2:b8) and press Return or Enter to replace the cell’s content with Excel’s calculation of the mean (3.117285714), which we round to 3.117. Excel does not have a function for the range, but we can use the functions that report the maximum value and the minimum value to calculate the range; thus = (b2:b8) – (b2:b8) returns 0.142 as an answer. In we showed that 91.10% of a manufacturer’s analgesic tablets contained between 243 and 262 mg of aspirin. We arrived at this result by calculating the deviation, , of each limit from the population’s expected mean, \(\mu\), of 250 mg in terms of the population’s expected standard deviation, \(\sigma\), of 5 mg. After we calculated values for , we used the table in to find the area under the normal distribution curve between these two limits. We can complete this calculation in Excel using the function As shown in Figure 4.8.2
, the function calculates the probability of obtaining a result less than from a normal distribution with a mean of \(\mu\) and a standard deviation of \(\sigma\). To solve using Excel enter the following formulas into separate cells = (243, 250, 5, TRUE) = (262, 250, 5, TRUE) obtaining results of 0.080756659 and 0.991802464. Subtracting the smaller value from the larger value and adjusting to the correct number of significant figures gives the probability as 0.9910, or 99.10%. Excel also includes a function for working with binomial distributions. The function’s syntax is = ( , , , TRUE or FALSE) where is the number of times a particular outcome occurs in trials, and is the probability that occurs in a single trial. Setting the function’s last term to TRUE gives the total probability for any result up to and setting it to FALSE gives the probability for . Using to test this function, we use the formula = (0, 27, 0.0111, FALSE) to find the probability of finding no atoms of C atoms in a molecule of cholesterol, C H O, which returns a value of 0.740 after adjusting for significant figures. Using the formula = (2, 27, 0.0111, TRUE) we find that 99.7% of cholesterol molecules contain two or fewer atoms of C. As shown in Table 4.8.1
, Excel includes functions for the following significance tests covered in this chapter: Let’s use these functions to complete a -test on the data in , which contains results for two experiments to determine the mass of a circulating U. S. penny. Enter the data from into a spreadsheet as shown in Figure 4.8.3
. Because the data in this case are unpaired, we will use Excel to complete an unpaired -test. Before we can complete the -test, we use an -test to determine whether the variances for the two data sets are equal or unequal. To complete the -test, we click on any empty cell, enter the formula = (b2:b8, c2:c6) and press Return or Enter, which replaces the cell’s content with the value of \(\alpha\) for which we can reject the null hypothesis of equal variances. In this case, Excel returns an \(\alpha\) of 0.566 105 03; because this value is not less than 0.05, we retain the null hypothesis that the variances are equal. Excel’s -test is two-tailed; for a one-tailed -test, we use the same function, but divide the result by two; thus = (b2:b8, c2:c6)/2 Having found no evidence to suggest unequal variances, we next complete an unpaired -test assuming equal variances, entering into any empty cell the formula = (b2:b8, c2:c6, 2, 2) where the first 2 indicates that this is a two-tailed -test, and the second 2 indicates that this is an unpaired -test with equal variances. Pressing Return or Enter replaces the cell’s content with the value of \(\alpha\) for which we can reject the null hypothesis of equal means. In this case, Excel returns an \(\alpha\) of 0.211 627 646; because this value is not less than 0.05, we retain the null hypothesis that the means are equal. See and for our earlier solutions to this problem. The other significance tests in Excel work in the same format. The following practice exercise provides you with an opportunity to test yourself. Rework and using Excel. You will find small differences between the values you obtain using Excel’s built in functions and the worked solutions in the chapter. These differences arise because Excel does not round off the results of intermediate calculations. R is a programming environment that provides powerful capabilities for analyzing data. There are many functions built into R’s standard installation and additional packages of functions are available from the R web site ( ). Commands in R are not available from pull down menus. Instead, you interact with R by typing in commands. You can download the current version of R from . on the link for Download: CRAN and find a local mirror site. on the link for the mirror site and then use the link for Linux, MacOS X, or Windows under the heading “Download and Install R.” Let’s use R to provide a statistical summary of the data in . To do this we first need to create an object that contains the data, which we do by typing in the following command. > penny1 = c(3.080, 3.094, 3.107, 3.056, 3.112, 3.174, 3.198) In R, the symbol ‘>’ is a prompt, which indicates that the program is waiting for you to enter a command. When you press ‘Return’ or ‘Enter,’ R executes the command, displays the result (if there is a result to return), and returns the > prompt. Table 4.8.2
lists some of the commands in R for calculating basic descriptive statistics. As is the case for Excel, R does not include stand alone commands for all descriptive statistics of interest to us, but we can calculate them using other commands. Using a command is easy—simply enter the appropriate code at the prompt; for example, to find the sample’s variance we enter > (penny1) [1] 0.002221918 In we showed that 91.10% of a manufacturer’s analgesic tablets contained between 243 and 262 mg of aspirin. We arrived at this result by calculating the deviation, , of each limit from the population’s expected mean, \(\mu\), of 250 mg in terms of the population’s expected standard deviation, \(\sigma\), of 5 mg. After we calculated values for , we used the table in to find the area under the normal distribution curve between these two limits. We can complete this calculation in R using the function . The function’s general format is (\(x, \mu, \sigma\)) where is the limit of interest, \(\mu\) is the distribution’s expected mean, and \(\sigma\) is the distribution’s expected standard deviation. The function returns the probability of obtaining a result of less than (Figure 4.8.4
).
Shown in blue is the area returned by the function (\(x, \mu, \sigma\)). Here is the output of an R session for solving . > pnorm(243, 250, 5) [1] 0.08075666 > pnorm(262, 250, 5) [1] 0.9918025 Subtracting the smaller value from the larger value and adjusting to the correct number of significant figures gives the probability as 0.9910, or 99.10%. R also includes functions for binomial distributions. To find the probability of obtaining a particular outcome, , in trials we use the function. ( , , ) where is the number of times a particular outcome occurs in trials, and is the probability that occurs in a single trial. Using to test this function, we find that the probability of finding no atoms of C atoms in a molecule of cholesterol, C H O is > dbinom(0, 27, 0.0111) [1] 0.7397997 0.740 after adjusting the significant figures. To find the probability of obtaining any outcome up to a maximum value of , we use the function. ( , , ) To find the percentage of cholesterol molecules that contain 0, 1, or 2 atoms of C, we enter > pbinom(2, 27, 0.0111) [1] 0.9967226 and find that the answer is 99.7% of cholesterol molecules. R includes commands for the following significance tests covered in this chapter: Let’s use these functions to complete a -test on the data in , which contains results for two experiments to determine the mass of a circulating U. S. penny. First, enter the data from into two objects. > penny1 = c(3.080, 3.094, 3.107, 3.056, 3.112, 3.174, 3.198) > penny2 = c(3.052, 3.141, 3.083, 3.083, 3.048) Because the data in this case are unpaired, we will use R to complete an unpaired -test. Before we can complete a -test we use an -test to determine whether the variances for the two data sets are equal or unequal. To complete a two-tailed -test in R we use the command ( , ) where and are the objects that contain the two data sets. Figure 4.8.5
shows the output from an R session to solve this problem. Note that R does not provide the critical value for (0.05, 6, 4); instead it reports the 95% confidence interval for . Because this confidence interval of 0.204 to 11.661 includes the expected value for of 1.00, we retain the null hypothesis and have no evidence for a difference between the variances. R also provides the probability of incorrectly rejecting the null hypothesis, which in this case is 0.5561. For a one-tailed -test the command is one of the following var.test( , alternative = “greater”) var.test( , alternative = “less”) where “greater” is used when the alternative hypothesis is \(s_X^2 > s_Y^2\), and “less” is used when the alternative hypothesis is \(s_X^2 < s_Y^2\). Having found no evidence suggesting unequal variances, we now complete an unpaired -test assuming equal variances. The basic syntax for a two-tailed -test is t.test( , , ) where and are the objects that contain the data sets. You can change the underlined terms to alter the nature of the -test. Replacing “var.equal = FALSE” with “var.equal = TRUE” makes this a two-tailed -test with equal variances, and replacing “paired = FALSE” with “paired = TRUE” makes this a paired -test. The term “mu = 0” is the expected difference between the means, which for this problem is 0. You can, of course, change this to suit your needs. The underlined terms are default values; if you omit them, then R assumes you intend an unpaired two-tailed -test of the null hypothesis that = with unequal variances. Figure 4.8.6
shows the output of an R session for this problem. We can interpret the results of this -test in two ways. First, the -value of 0.2116 means there is a 21.16% probability of incorrectly rejecting the null hypothesis. Second, the 95% confidence interval of –0.024 to 0.0958 for the difference between the sample means includes the expected value of zero. Both ways of looking at the results provide no evidence for rejecting the null hypothesis; thus, we retain the null hypothesis and find no evidence for a difference between the two samples. The other significance tests in R work in the same format. The following practice exercise provides you with an opportunity to test yourself. Rework and using R. Shown here are copies of R sessions for each problem. You will find small differences between the values given here for and for and those values shown with the worked solutions in the chapter. These differences arise because R does not round off the results of intermediate calculations. > AnalystA = c(86.82, 87.04, 86.93, 87.01, 86.20, 87.00) > AnalystB = c(81.01, 86.15, 81.73, 83.19, 80.27, 83.94) > var.test(AnalystB, AnalystA) F test to compare two variances data: AnalystB and AnalystA F = 45.6358, num df = 5, denom df = 5, p-value = 0.0007148 alternative hypothesis: true ratio of variances is not equal to 1 95 percent confidence interval: 6.385863 326.130970 sample estimates: ratio of variances 45.63582 > t.test(AnalystA, AnalystB, var.equal=FALSE) Welch Two Sample t-test data: AnalystA and AnalystB t = 4.6147, df = 5.219, p-value = 0.005177 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 1.852919 6.383748 sample estimates: mean of x mean of y 86.83333 82.71500 > micro = c(129.5, 89.6, 76.6, 52.2, 110.8, 50.4, 72.4, 141.4, 75.0, 34.1, 60.3)
> elect = c(132.3, 91.0, 73.6, 58.2, 104.2, 49.9, 82.1, 154.1, 73.4, 38.1, 60.1) > t.test(micro,elect,paired=TRUE) Paired t-test data: micro and elect t = -1.3225, df = 10, p-value = 0.2155 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -6.028684 1.537775 sample estimates: mean of the differences –2.245455 Unlike Excel, R also includes functions for evaluating outliers. These functions are not part of R’s standard installation. To install them enter the following command within R ( ). > install.packages(“outliers”) After you install the package, you must load the functions into R by using the following command ( ). > library(“outliers”) You need to install a package once, but you need to load the package each time you plan to use it. There are ways to configure R so that it automatically loads certain packages; see for more information (click to view a PDF version of this document). Let’s use this package to find the outlier in using both Dixon’s -test and Grubb’s test. The commands for these tests are ( , type = 10, two.sided = TRUE) ( , type = 10, two.sided = TRUE) where is the object that contains the data, “type = 10” specifies that we are looking for one outlier, and “two.sided = TRUE” indicates that we are using the more conservative two-tailed test. Both tests have other variants that allow for the testing of outliers on both ends of the data set (“type = 11”) or for more than one outlier (“type = 20”), but we will not consider these here. Figure 4.8.7
shows the output of a session for this problem. For both tests the very small -value indicates that we can treat as an outlier the penny with a mass of 2.514 g. One of R’s more useful features is the ability to visualize data. Visualizing data is important because it provides us with an intuitive feel for our data that can help us in applying and evaluating statistical tests. It is tempting to believe that a statistical analysis is foolproof, particularly if the probability for incorrectly rejecting the null hypothesis is small. Looking at a visual display of our data, however, can help us determine whether our data is normally distributed—a requirement for most of the significance tests in this chapter—and can help us identify potential outliers. There are many useful ways to look at data, four of which we consider here. Visualizing data is important, a point we will return to in when we consider the mathematical modeling of data. To plot data in R, we will use the package “lattice,” which you will need to load using the following command. > library(“lattice”) To demonstrate the types of plots we can generate, we will use the object “penny,” which contains the masses of the 100 pennies in . You do not need to use the command install.package this time because lattice was automatically installed on your computer when you downloaded R. Our first visualization is a histogram. To construct the histogram we use mass to divide the pennies into bins and plot the number of pennies or the percent of pennies in each bin on the -axis as a function of mass on the -axis. Figure 4.8.8
shows the result of entering the command > (penny, type = “percent”, xlab = “Mass (g)”, ylab = “Percent of Pennies”, main = “Histogram of Data in Table 4.4.3”) A histogram allows us to visualize the data’s distribution. In this example the data appear to follow a normal distribution, although the largest bin does not include the mean of 3.095 g and the distribution is not perfectly symmetric. One limitation of a histogram is that its appearance depends on how we choose to bin the data. Increasing the number of bins and centering the bins around the data’s mean gives a histogram that more closely approximates a normal distribution ( ). An alternative to the histogram is a kernel density plot, which basically is a smoothed histogram. In this plot each value in the data set is replaced with a normal distribution curve whose width is a function of the data set’s standard deviation and size. The resulting curve is a summation of the individual distributions. Figure 4.8.9
shows the result of entering the command > (penny, xlab = “Mass of Pennies (g)”, main = “Kernel Density Plot of Data in Table 4.4.3”) The circles at the bottom of the plot show the mass of each penny in the data set. This display provides a more convincing picture that the data in are normally distributed, although we see evidence of a small clustering of pennies with a mass of approximately 3.06 g. We analyze samples to characterize the parent population. To reach a meaningful conclusion about a population, the samples must be representative of the population. One important requirement is that the samples are random. A dot chart provides a simple visual display that allows us to examine the data for non-random trends. Figure 4.8.10
shows the result of entering > (penny, xlab = “Mass of Pennies (g)”, ylab = “Penny Number”, main = “Dotchart of Data in Table 4.4.3”) In this plot the masses of the 100 pennies are arranged along the -axis in the order in which they were sampled. If we see a pattern in the data along the -axis, such as a trend toward smaller masses as we move from the first penny to the last penny, then we have clear evidence of non-random sampling. Because our data do not show a pattern, we have more confidence in the quality of our data. The last plot we will consider is a box plot, which is a useful way to identify potential outliers without making any assumptions about the data’s distribution. A box plot contains four pieces of information about a data set: the median, the middle 50% of the data, the smallest value and the largest value within a set distance of the middle 50% of the data, and possible outliers. Figure 4.8.11
shows the result of entering > (penny, xlab = “Mass of Pennies (g)”, main = “Boxplot of Data in Table 4.4.3)” The black dot (•) is the data set’s median. The rectangular box shows the range of masses spanning the middle 50% of the pennies. This also is known as the interquartile range, or IQR. The dashed lines, which are called “whiskers,” extend to the smallest value and the largest value that are within \(\pm 1.5 \times \text{IQR}\) of the rectangular box. Potential outliers are shown as open circles (o). For normally distributed data the median is near the center of the box and the whiskers will be equidistant from the box. As is often the case in statistics, the converse is not true—finding that a boxplot is perfectly symmetric does not prove that the data are normally distributed. To find the interquartile range you first find the median, which divides the data in half. The median of each half provides the limits for the box. The IQR is the median of the upper half of the data minus the median for the lower half of the data. For the data in the median is 3.098. The median for the lower half of the data is 3.068 and the median for the upper half of the data is 3.115. The IQR is 3.115 – 3.068 = 0.047. You can use the command “summary(penny)” in R to obtain these values. The lower “whisker” extend to the first data point with a mass larger than 3.068 – 1.5 \(\times\) IQR = 3.068 – 1.5 \(\times\) 0.047 = 2.9975 which for this data is 2.998 g. The upper “whisker” extends to the last data point with a mass smaller than 3.115 + 1.5 \(\times\) IQR = 3.115 + 1.5 \(\times\) 0.047 = 3.1855 which for this data is 3.181 g. The box plot in Figure 4.8.11
is consistent with the histogram ( ) and the kernel density plot ( ). Together, the three plots provide evidence that the data in are normally distributed. The potential outlier, whose mass of 3.198 g, is not sufficiently far away from the upper whisker to be of concern, particularly as the size of the data set ( = 100) is so large. A Grubb’s test on the potential outlier does not provide evidence for treating it as an outlier. Use R to create a data set consisting of 100 values from a uniform distribution by entering the command > data = runif(100, min = 0, max = 100) A uniform distribution is one in which every value between the minimum and the maximum is equally probable. Examine the data set by creating a histogram, a kernel density plot, a dot chart, and a box plot. Briefly comment on what the plots tell you about the your sample and its parent population. Because we are selecting a random sample of 100 members from a uniform distribution, you will see subtle differences between your plots and the plots shown as part of this answer. Here is a record of my R session and the resulting plots. > data = runif(100, min = 0, max = 0) > data [1] 18.928795 80.423589 39.399693 23.757624 30.088554 [6] 76.622174 36.487084 62.186771 81.115515 15.726404 [11] 85.765317 53.994179 7.919424 10.125832 93.153308 [16] 38.079322 70.268597 49.879331 73.115203 99.329723 [21] 48.203305 33.093579 73.410984 75.128703 98.682127 [26] 11.433861 53.337359 81.705906 95.444703 96.843476 [31] 68.251721 40.567993 32.761695 74.635385 70.914957 [36] 96.054750 28.448719 88.580214 95.059215 20.316015 [41] 9.828515 44.172774 99.648405 85.593858 82.745774 [46] 54.963426 65.563743 87.820985 17.791443 26.417481 [51] 72.832037 5.518637 58.231329 10.213343 40.581266 [56] 6.584000 81.261052 48.534478 51.830513 17.214508 [61] 31.232099 60.545307 19.197450 60.485374 50.414960 [66] 88.908862 68.939084 92.515781 72.414388 83.195206 [71] 74.783176 10.643619 41.775788 20.464247 14.547841 [76] 89.887518 56.217573 77.606742 26.956787 29.641171 [81] 97.624246 46.406271 15.906540 23.007485 17.715668 [86] 84.652814 29.379712 4.093279 46.213753 57.963604 [91] 91.160366 34.278918 88.352789 93.004412 31.055807 [96] 47.822329 24.052306 95.498610 21.089686 2.629948 > histogram(data, type = “percent”) > densityplot(data) > dotchart(data) > bwplot(data) The histogram (far left) divides the data into eight bins, each of which contains between 10 and 15 members. As we expect for a uniform distribution, the histogram’s overall pattern suggests that each outcome is equally probable. In interpreting the kernel density plot (second from left), it is important to remember that it treats each data point as if it is from a normally distributed population (even though, in this case, the underlying population is uniform). Although the plot appears to suggest that there are two normally distributed populations, the individual results shown at the bottom of the plot provide further evidence for a uniform distribution. The dot chart (second from right) shows no trend along the -axis, which indicates that the individual members of this sample were drawn at random from the population. The distribution along the -axis also shows no pattern, as expected for a uniform distribution, Finally, the box plot (far right) shows no evidence of outliers. | 23,742 | 3,211 |
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To standardize an analytical method we also must determine the analyte’s sensitivity, , in Equation 5.3.1
or Equation 5.3.2
. \[S_{total} = k_A n_A + S_{reag} \label{5.1}\] \[S_{total} = k_A C_A + S_{reag} \label{5.2}\] In principle, it is possible to derive the value of for any analytical method if we understand fully all the chemical reactions and physical processes responsible for the signal. Unfortunately, such calculations are not feasible if we lack a sufficiently developed theoretical model of the physical processes or if the chemical reaction’s evince non-ideal behavior. In such situations we must determine the value of by analyzing one or more standard solutions, each of which contains a known amount of analyte. In this section we consider several approaches for determining the value of . For simplicity we assume that is accounted for by a proper reagent blank, allowing us to replace in Equation \ref{5.1} and Equation \ref{5.2} with the analyte’s signal, . \[S_A = k_A n_A \label{5.3}\] \[S_A = k_A C_A \label{5.4}\] Equation \ref{5.3} and Equation \ref{5.4} essentially are identical, differing only in whether we choose to express the amount of analyte in moles or as a concentration. For the remainder of this chapter we will limit our treatment to Equation \ref{5.4}. You can extend this treatment to Equation \ref{5.3} by replacing with . The simplest way to determine the value of in Equation \ref{5.4} is to use a single-point standardization in which we measure the signal for a standard, , that contains a known concentration of analyte, . Substituting these values into Equation \ref{5.4} \[k_A = \frac {S_{std}} {C_{std}} \label{5.5}\] gives us the value for . Having determined , we can calculate the concentration of analyte in a sample by measuring its signal, , and calculating using Equation \ref{5.6}. \[C_A = \frac {S_{samp}} {k_A} \label{5.6}\] A single-point standardization is the least desirable method for standardizing a method. There are two reasons for this. First, any error in our determination of carries over into our calculation of . Second, our experimental value for is based on a single concentration of analyte. To extend this value of to other concentrations of analyte requires that we assume a linear relationship between the signal and the analyte’s concentration, an assumption that often is not true [Cardone, M. J.; Palmero, P. J.; Sybrandt, L. B. , , 1187–1191]. Figure 5.3.1
shows how assuming a constant value of leads to a determinate error in if becomes smaller at higher concentrations of analyte. Despite these limitations, single-point standardizations find routine use when the expected range for the analyte’s concentrations is small. Under these conditions it often is safe to assume that is constant (although you should verify this assumption experimentally). This is the case, for example, in clinical labs where many automated analyzers use only a single standard. The better way to standardize a method is to prepare a series of standards, each of which contains a different concentration of analyte. Standards are chosen such that they bracket the expected range for the analyte’s concentration. A should include at least three standards, although more are preferable. A plot of versus is called a . The exact standardization, or calibration relationship, is determined by an appropriate curve-fitting algorithm. Linear regression, which also is known as the method of least squares, is one such algorithm. Its use is covered in Section 5.4. There are two advantages to a multiple-point standardization. First, although a determinate error in one standard introduces a determinate error, its effect is minimized by the remaining standards. Second, because we measure the signal for several concentrations of analyte, we no longer must assume is independent of the analyte’s concentration. Instead, we can construct a calibration curve similar to the “actual relationship” in Figure 5.3.1
. The most common method of standardization uses one or more , each of which contains a known concentration of analyte. We call these standards “external” because they are prepared and analyzed separate from the samples. Appending the adjective “external” to the noun “standard” might strike you as odd at this point, as it seems reasonable to assume that standards and samples are analyzed separately. As we will soon learn, however, we can add standards to our samples and analyze both simultaneously. With a single external standard we determine using EEquation \ref{5.5} and then calculate the concentration of analyte, , using Equation \ref{5.6}. A spectrophotometric method for the quantitative analysis of Pb in blood yields an of 0.474 for a single standard for which the concentration of lead is 1.75 ppb. What is the concentration of Pb in a sample of blood for which is 0.361? Equation \ref{5.5} allows us to calculate the value of using the data for the single external standard. \[k_A = \frac {S_{std}} {C_{std}} = \frac {0.474} {1.75 \text{ ppb}} = 0.2709 \text{ ppb}^{-1} \nonumber\] Having determined the value of , we calculate the concentration of Pb in the sample of blood is calculated using Equation \ref{5.6}. \[C_A = \frac {S_{samp}} {k_A} = \frac {0.361} {0.2709 \text{ ppb}^{-1}} = 1.33 \text{ ppb} \nonumber\] Figure 5.3.2
shows a typical multiple-point external standardization. The volumetric flask on the left contains a reagent blank and the remaining volumetric flasks contain increasing concentrations of Cu . Shown below the volumetric flasks is the resulting calibration curve. Because this is the most common method of standardization, the resulting relationship is called a . When a calibration curve is a straight-line, as it is in Figure 5.3.2
, the slope of the line gives the value of . This is the most desirable situation because the method’s sensitivity remains constant throughout the analyte’s concentration range. When the calibration curve is not a straight-line, the method’s sensitivity is a function of the analyte’s concentration. In , for example, the value of is greatest when the analyte’s concentration is small and it decreases continuously for higher concentrations of analyte. The value of at any point along the calibration curve in is the slope at that point. In either case, a calibration curve allows to relate to the analyte’s concentration. A second spectrophotometric method for the quantitative analysis of Pb in blood has a normal calibration curve for which \[S_{std} = (0.296 \text{ ppb}^{-1} \times C_{std}) + 0.003 \nonumber\] What is the concentration of Pb in a sample of blood if is 0.397? To determine the concentration of Pb in the sample of blood, we replace in the calibration equation with and solve for . \[C_A = \frac {S_{samp} - 0.003} {0.296 \text{ ppb}^{-1}} = \frac {0.397 - 0.003} {0.296 \text{ ppb}^{-1}} = 1.33 \text{ ppb} \nonumber\] It is worth noting that the calibration equation in this problem includes an extra term that does not appear in Equation \ref{5.6}. Ideally we expect our calibration curve to have a signal of zero when is zero. This is the purpose of using a reagent blank to correct the measured signal. The extra term of +0.003 in our calibration equation results from the uncertainty in measuring the signal for the reagent blank and the standards. shows a normal calibration curve for the quantitative analysis of Cu . The equation for the calibration curve is \[S_{std} = 29.59 \text{ M}^{-1} \times C_{std} + 0.015 \nonumber\] What is the concentration of Cu in a sample whose absorbance, , is 0.114? Compare your answer to a one-point standardization where a standard of \(3.16 \times 10^{-3} \text{ M}\) Cu gives a signal of 0.0931. Substituting the sample’s absorbance into the calibration equation and solving for give \[S_{samp} = 0.114 = 29.59 \text{ M}^{-1} \times C_{A} + 0.015 \nonumber\] \[C_A = 3.35 \times 10^{-3} \text{ M} \nonumber\] For the one-point standardization, we first solve for \[k_A = \frac {S_{std}} {C_{std}} = \frac {0.0931} {3.16 \times 10^{-3} \text{ M}} = 29.46 \text{ M}^{-1} \nonumber\] and then use this value of to solve for . \[C_A = \frac {S_{samp}} {k_A} = \frac {0.114} {29.46 \text{ M}^{-1}} = 3.87 \times 10^{-3} \text{ M} \nonumber\] When using multiple standards, the indeterminate errors that affect the signal for one standard are partially compensated for by the indeterminate errors that affect the other standards. The standard selected for the one-point standardization has a signal that is smaller than that predicted by the regression equation, which underestimates and overestimates . An external standardization allows us to analyze a series of samples using a single calibration curve. This is an important advantage when we have many samples to analyze. Not surprisingly, many of the most common quantitative analytical methods use an external standardization. There is a serious limitation, however, to an external standardization. When we determine the value of using Equation \ref{5.5}, the analyte is present in the external standard’s matrix, which usually is a much simpler matrix than that of our samples. When we use an external standardization we assume the matrix does not affect the value of . If this is not true, then we introduce a proportional determinate error into our analysis. This is not the case in Figure 5.3.3
, for instance, where we show calibration curves for an analyte in the sample’s matrix and in the standard’s matrix. In this case, using the calibration curve for the external standards leads to a negative determinate error in analyte’s reported concentration. If we expect that matrix effects are important, then we try to match the standard’s matrix to that of the sample, a process known as . If we are unsure of the sample’s matrix, then we must show that matrix effects are negligible or use an alternative method of standardization. Both approaches are discussed in the following section. The matrix for the external standards in , for example, is dilute ammonia. Because the \(\ce{Cu(NH3)4^{2+}}\) complex absorbs more strongly than Cu , adding ammonia increases the signal’s magnitude. If we fail to add the same amount of ammonia to our samples, then we will introduce a proportional determinate error into our analysis. We can avoid the complication of matching the matrix of the standards to the matrix of the sample if we carry out the standardization in the sample. This is known as the . The simplest version of a standard addition is shown in Figure 5.3.4
. First we add a portion of the sample, , to a volumetric flask, dilute it to volume, , and measure its signal, . Next, we add a second identical portion of sample to an equivalent volumetric flask along with a spike, , of an external standard whose concentration is . After we dilute the spiked sample to the same final volume, we measure its signal, . The following two equations relate and to the concentration of analyte, , in the original sample. \[S_{samp} = k_A C_A \frac {V_o} {V_f} \label{5.7}\] \[S_{spike} = k_A \left( C_A \frac {V_o} {V_f} + C_{std} \frac {V_{std}} {V_f} \right) \label{5.8}\] As long as is small relative to , the effect of the standard’s matrix on the sample’s matrix is insignificant. Under these conditions the value of is the same in Equation \ref{5.7} and Equation \ref{5.8}. Solving both equations for and equating gives \[\frac {S_{samp}} {C_A \frac {V_o} {V_f}} = \frac {S_{spike}} {C_A \frac {V_o} {V_f} + C_{std} \frac {V_{std}} {V_f}} \label{5.9}\] which we can solve for the concentration of analyte, , in the original sample. A third spectrophotometric method for the quantitative analysis of Pb in blood yields an of 0.193 when a 1.00 mL sample of blood is diluted to 5.00 mL. A second 1.00 mL sample of blood is spiked with 1.00 mL of a 1560-ppb Pb external standard and diluted to 5.00 mL, yielding an of 0.419. What is the concentration of Pb in the original sample of blood? We begin by making appropriate substitutions into Equation \ref{5.9} and solving for . Note that all volumes must be in the same units; thus, we first convert from 1.00 mL to \(1.00 \times 10^{-3} \text{ mL}\). \[\frac {0.193} {C_A \frac {1.00 \text{ mL}} {5.00 \text{ mL}}} = \frac {0.419} {C_A \frac {1.00 \text{ mL}} {5.00 \text{ mL}} + 1560 \text{ ppb} \frac {1.00 \times 10^{-3} \text{ mL}} {5.00 \text{ mL}}} \nonumber\] \[\frac {0.193} {0.200C_A} = \frac {0.419} {0.200C_A + 0.3120 \text{ ppb}} \nonumber\] \[0.0386C_A + 0.0602 \text{ ppb} = 0.0838 C_A \nonumber\] \[0.0452 C_A = 0.0602 \text{ ppb} \nonumber\] \[C_A = 1.33 \text{ ppb} \nonumber\] The concentration of Pb in the original sample of blood is 1.33 ppb. It also is possible to add the standard addition directly to the sample, measuring the signal both before and after the spike (Figure 5.3.5
). In this case the final volume after the standard addition is + and Equation \ref{5.7}, Equation \ref{5.8}, and Equation \ref{5.9} become \[S_{samp} = k_A C_A \nonumber\] \[S_{spike} = k_A \left( C_A \frac {V_o} {V_o + V_{std}} + C_{std} \frac {V_{std}} {V_o + V_{std}} \right) \label{5.10}\] \[\frac {S_{samp}} {C_A} = \frac {S_{spike}} {C_A \frac {V_o} {V_o + V_{std}} + C_{std} \frac {V_{std}} {V_o + V_{std}}} \label{5.11}\] A fourth spectrophotometric method for the quantitative analysis of Pb in blood yields an of 0.712 for a 5.00 mL sample of blood. After spiking the blood sample with 5.00 mL of a 1560-ppb Pb external standard, an of 1.546 is measured. What is the concentration of Pb in the original sample of blood? \[\frac {0.712} {C_A} = \frac {1.546} {C_A \frac {5.00 \text{ mL}} {5.005 \text{ mL}} + 1560 \text{ ppb} \frac {5.00 \times 10^{-3} \text{ mL}} {5.005 \text{ mL}}} \nonumber\] \[\frac {0.712} {C_A} = \frac {1.546} {0.9990C_A + 1.558 \text{ ppb}} \nonumber\] \[0.7113C_A + 1.109 \text{ ppb} = 1.546C_A \nonumber\] \[C_A = 1.33 \text{ ppb} \nonumber\] The concentration of Pb in the original sample of blood is 1.33 ppb. We can adapt a single-point standard addition into a multiple-point standard addition by preparing a series of samples that contain increasing amounts of the external standard. Figure 5.3.6
shows two ways to plot a standard addition calibration curve based on Equation \ref{5.8}. In Figure 5.3.6
a we plot against the volume of the spikes, . If is constant, then the calibration curve is a straight-line. It is easy to show that the -intercept is equivalent to – / . Beginning with Equation \ref{5.8} show that the equations in Figure 5.3.6
a for the slope, the -intercept, and the -intercept are correct. We begin by rewriting Equation \ref{5.8} as \[S_{spike} = \frac {k_A C_A V_o} {V_f} + \frac {k_A C_{std}} {V_f} \times V_{std} \nonumber\] which is in the form of the equation for a straight-line \[y = y\text{-intercept} + \text{slope} \times x\text{-intercept} \nonumber\] where is and is . The slope of the line, therefore, is / and the -intercept is / . The -intercept is the value of when is zero, or \[0 = \frac {k_A C_A V_o} {V_f} + \frac {k_A C_{std}} {V_f} \times x\text{-intercept} \nonumber\] \[x\text{-intercept} = - \frac {k_A C_A V_o / V_f} {K_A C_{std} / V_f} = - \frac {C_A V_o} {C_{std}} \nonumber\] Beginning with Equation \ref{5.8} show that the Equations in Figure 5.3.6
b for the slope, the -intercept, and the -intercept are correct. We begin with Equation \ref{5.8} \[S_{spike} = k_A \left( C_A \frac {V_o} {V_f} + C_{std} \frac {V_{std}} {V_f} \right) \nonumber\] rewriting it as \[S_{spike} = \frac {k_A C_A V_o} {V_f} + k_A \left( C_{std} \frac {V_{std}} {V_f} \right) \nonumber\] which is in the form of the linear equation \[y = y\text{-intercept} + \text{slope} \times x\text{-intercept} \nonumber\] where is and is \(\times\) / . The slope of the line, therefore, is , and the -intercept is / . The -intercept is the value of when is zero, or \[x\text{-intercept} = - \frac {k_A C_A V_o/V_F} {k_A} = - \frac {C_A V_o} {V_f} \nonumber\] Because we know the volume of the original sample, , and the concentration of the external standard, , we can calculate the analyte’s concentrations from the -intercept of a multiple-point standard additions. A fifth spectrophotometric method for the quantitative analysis of Pb in blood uses a multiple-point standard addition based on Equation \ref{5.8}. The original blood sample has a volume of 1.00 mL and the standard used for spiking the sample has a concentration of 1560 ppb Pb . All samples were diluted to 5.00 mL before measuring the signal. A calibration curve of versus has the following equation \[S_{spike} = 0.266 + 312 \text{ mL}^{-1} \times V_{std} \nonumber\] What is the concentration of Pb in the original sample of blood? To find the -intercept we set equal to zero. \[S_{spike} = 0.266 + 312 \text{ mL}^{-1} \times V_{std} \nonumber\] Solving for , we obtain a value of \(-8.526 \times 10^{-4} \text{ mL}\) for the -intercept. Substituting the -intercept’s value into the equation from Figure 5.3.6
a \[-8.526 \times 10^{-4} \text{ mL} = - \frac {C_A V_o} {C_{std}} = - \frac {C_A \times 1.00 \text{ mL}} {1560 \text{ ppb}} \nonumber\] and solving for gives the concentration of Pb in the blood sample as 1.33 ppb. shows a standard additions calibration curve for the quantitative analysis of Mn . Each solution contains 25.00 mL of the original sample and either 0, 1.00, 2.00, 3.00, 4.00, or 5.00 mL of a 100.6 mg/L external standard of Mn . All standard addition samples were diluted to 50.00 mL with water before reading the absorbance. The equation for the calibration curve in is \[S_{std} = 0.0854 \times V_{std} + 0.1478 \nonumber\] What is the concentration of Mn in this sample? Compare your answer to the data in , for which the calibration curve is \[S_{std} = 0.425 \times C_{std}(V_{std}/V_f) + 0.1478 \nonumber\] Using the calibration equation from , we find that the -intercept is \[x\text{-intercept} = - \frac {0.1478} {0.0854 \text{ mL}^{-1}} = - 1.731 \text{ mL} \nonumber\] If we plug this result into the equation for the -intercept and solve for , we find that the concentration of Mn is \[C_A = - \frac {x\text{-intercept} \times C_{std}} {V_o} = - \frac {-1.731 \text{ mL} \times 100.6 \text{ mg/L}} {25.00 \text{ mL}} = 6.96 \text{ mg/L} \nonumber\] For Figure 5.3.6
b, the -intercept is \[x\text{-intercept} = - \frac {0.1478} {0.0425 \text{ mL/mg}} = - 3.478 \text{ mg/mL} \nonumber\] and the concentration of Mn is \[C_A = - \frac {x\text{-intercept} \times V_f} {V_o} = - \frac {-3.478 \text{ mg/mL} \times 50.00 \text{ mL}} {25.00 \text{ mL}} = 6.96 \text{ mg/L} \nonumber\] Since we construct a standard additions calibration curve in the sample, we can not use the calibration equation for other samples. Each sample, therefore, requires its own standard additions calibration curve. This is a serious drawback if you have many samples. For example, suppose you need to analyze 10 samples using a five-point calibration curve. For a normal calibration curve you need to analyze only 15 solutions (five standards and ten samples). If you use the method of standard additions, however, you must analyze 50 solutions (each of the ten samples is analyzed five times, once before spiking and after each of four spikes). We can use the method of standard additions to validate an external standardization when matrix matching is not feasible. First, we prepare a normal calibration curve of versus and determine the value of from its slope. Next, we prepare a standard additions calibration curve using Equation \ref{5.8}, plotting the data as shown in . The slope of this standard additions calibration curve provides an independent determination of . If there is no significant difference between the two values of , then we can ignore the difference between the sample’s matrix and that of the external standards. When the values of are significantly different, then using a normal calibration curve introduces a proportional determinate error. To use an external standardization or the method of standard additions, we must be able to treat identically all samples and standards. When this is not possible, the accuracy and precision of our standardization may suffer. For example, if our analyte is in a volatile solvent, then its concentration will increase if we lose solvent to evaporation. Suppose we have a sample and a standard with identical concentrations of analyte and identical signals. If both experience the same proportional loss of solvent, then their respective concentrations of analyte and signals remain identical. In effect, we can ignore evaporation if the samples and the standards experience an equivalent loss of solvent. If an identical standard and sample lose different amounts of solvent, however, then their respective concentrations and signals are no longer equal. In this case a simple external standardization or standard addition is not possible. We can still complete a standardization if we reference the analyte’s signal to a signal from another species that we add to all samples and standards. The species, which we call an , must be different than the analyte. Because the analyte and the internal standard receive the same treatment, the ratio of their signals is unaffected by any lack of reproducibility in the procedure. If a solution contains an analyte of concentration and an internal standard of concentration , then the signals due to the analyte, , and the internal standard, , are \[S_A = k_A C_A \nonumber\] \[S_{IS} = k_{SI} C_{IS} \nonumber\] where \(k_A\) and \(k_{IS}\) are the sensitivities for the analyte and the internal standard, respectively. Taking the ratio of the two signals gives the fundamental equation for an internal standardization. \[\frac {S_A} {S_{IS}} = \frac {k_A C_A} {k_{IS} C_{IS}} = K \times \frac {C_A} {C_{IS}} \label{5.12}\] Because is a ratio of the analyte’s sensitivity and the internal standard’s sensitivity, it is not necessary to determine independently values for either or . In a single-point internal standardization, we prepare a single standard that contains the analyte and the internal standard, and use it to determine the value of in Equation \ref{5.12}. \[K = \left( \frac {C_{IS}} {C_A} \right)_{std} \times \left( \frac {S_A} {S_{IS}} \right)_{std} \label{5.13}\] Having standardized the method, the analyte’s concentration is given by \[C_A = \frac {C_{IS}} {K} \times \left( \frac {S_A} {S_{IS}} \right)_{samp} \nonumber\] A sixth spectrophotometric method for the quantitative analysis of Pb in blood uses Cu as an internal standard. A standard that is 1.75 ppb Pb and 2.25 ppb Cu yields a ratio of ( / ) of 2.37. A sample of blood spiked with the same concentration of Cu gives a signal ratio, ( / ) , of 1.80. What is the concentration of Pb in the sample of blood? Equation \ref{5.13} allows us to calculate the value of using the data for the standard \[K = \left( \frac {C_{IS}} {C_A} \right)_{std} \times \left( \frac {S_A} {S_{IS}} \right)_{std} = \frac {2.25 \text{ ppb } \ce{Cu^{2+}}} {1.75 \text{ ppb } \ce{Pb^{2+}}} \times 2.37 = 3.05 \frac {\text{ppb } \ce{Cu^{2+}}} {\text{ppb } \ce{Pb^{2+}}} \nonumber\] The concentration of Pb , therefore, is \[C_A = \frac {C_{IS}} {K} \times \left( \frac {S_A} {S_{IS}} \right)_{samp} = \frac {2.25 \text{ ppb } \ce{Cu^{2+}}} {3.05 \frac {\text{ppb } \ce{Cu^{2+}}} {\text{ppb } \ce{Pb^{2+}}}} \times 1.80 = 1.33 \text{ ppb } \ce{Pb^{2+}} \nonumber\] A single-point internal standardization has the same limitations as a single-point normal calibration. To construct an internal standard calibration curve we prepare a series of standards, each of which contains the same concentration of internal standard and a different concentrations of analyte. Under these conditions a calibration curve of ( / ) versus is linear with a slope of / . Although the usual practice is to prepare the standards so that each contains an identical amount of the internal standard, this is not a requirement. A seventh spectrophotometric method for the quantitative analysis of Pb in blood gives a linear internal standards calibration curve for which \[\left( \frac {S_A} {S_{IS}} \right)_{std} = (2.11 \text{ ppb}^{-1} \times C_A) - 0.006 \nonumber\] What is the ppb Pb in a sample of blood if ( / )samp is 2.80? To determine the concentration of Pb in the sample of blood we replace ( / ) in the calibration equation with ( / ) and solve for . \[C_A = \frac {\left( \frac {S_A} {S_{IS}} \right)_{samp} + 0.006} {2.11 \text{ ppb}^{-1}} = \frac {2.80 + 0.006} {2.11 \text{ ppb}^{-1}} = 1.33 \text{ ppb } \ce{Pb^{2+}} \nonumber\] The concentration of Pb in the sample of blood is 1.33 ppb. In some circumstances it is not possible to prepare the standards so that each contains the same concentration of internal standard. This is the case, for example, when we prepare samples by mass instead of volume. We can still prepare a calibration curve, however, by plotting \((S_A / S_{IS})_{std}\) versus / , giving a linear calibration curve with a slope of . You might wonder if it is possible to include an internal standard in the method of standard additions to correct for both matrix effects and uncontrolled variations between samples; well, the answer is yes as described in the paper “Standard Dilution Analysis,” the full reference for which is Jones, W. B.; Donati, G. L.; Calloway, C. P.; Jones, B. T. , , 2321-2327. | 25,573 | 3,212 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.08%3A_Electron_Affinities |
Electron affinities are more difficult to measure experimentally than are ionization energies, and far fewer values are available. The relationship of the periodic table with those electron affinities that have been measured or estimated from calculations can be seen on the table of ionization energies and electron affinities (in the red), seen below. \(\Page {1}\) Ionization Energies and Electron Affinities. The electron affinities are the red values above the atomic symbol. It is not easy to discern many obvious regularities in this table, especially since some of the electron-affinity values quoted are negative, indicating that energy is sometimes to force an extra electron onto an atom. Nevertheless, it is quite obvious which of the periodic groups correspond to the highest electron affinities. All the halogens have values of about 300 kJ mol while the group VI nonmetals have somewhat lower values, in the region of 200 kJ mol or less. The high electron affinities of the halogens are a result of their having an almost complete outer shell of electrons. The element fluorine, for example, has the structure 1 2 2 , in which one of the 2 orbitals contains but one electron. If an extra electron (in the red in both diagrams below) is added to this atom to form a fluoride ion, the electron can pair with the electron in the half-filled 2 orbital (as seen in the orbital diagram below). The added electron will be shielded from the nucleus by the 1 electrons, but the 2 and 2 electrons are in the same shell and will shield it rather poorly. There will thus be quite a large effective nuclear charge (a rough estimate is +5) attracting the added electron. Because of this overall attraction, energy will be when the electron is captured by the fluorine atom. Similar reasoning also explains why oxygen also has a high electron affinity. Here, though, the nuclear charge is smaller, and the attraction for the added electron distinctly less. | 1,982 | 3,213 |
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of matter mean the state of a material, like solid, liquid or gas. How can we predict whether a material will be a solid, liquid or gas under certain conditions? We have to know about the forces that hold the material together. In solids, the separate molecules or ions are held tightly in their positions by some type of force. They might vibrate a little bit in place, but they don't move around. For this reason, solids don't easily change shape. You might be able to bend or break them, but usually they don't change shape by themselves. In liquids, the separate molecules or ions are held close together by some type of force, but they can move around while staying close together. For this reason, a liquid can change shape to fit whatever is holding it. But like solids, liquids have approximately constant volume. Even if the liquid flows into a new shape, the distance between the molecules doesn't change, so the total volume stays the same. In a gas, usually the molecules bounce around as though there are no forces between them. (See .) At very high pressure, when they are forced to be close together, we might start to notice that there are some forces between the molecules (because the pressure is less than we expect from the Ideal Gas Law) but usually they move around separately. Because they aren't really attracted to each other and have a lot of kinetic energy (at least at normal temperatures), they fill the whole space they have. So gases can take any shape, and also can change volume a lot. The higher the temperature, the more kinetic energy the molecules or ions have. With more kinetic energy, it's harder for them to stay in their place in a solid, or not to bounce right out of a liquid and become a gas. So as we increase the temperature, we might see from solid to liquid to gas. We can think about these transitions using equilibrium, like when we think about and equilibria. Think about a liquid in a closed container, like a bottle half full of water. Some of the molecules have bigger kinetic energy and some have smaller kinetic energy. If a molecule with big kinetic energy is on the surface of the liquid, it might fly off and enter the gas state. At the same time, other molecules in the gas state might bump the surface of the liquid, and if they don't have very much kinetic energy, they might stay there and join the liquid. This is a dynamic equilibrium: the molecules go back and forth between the 2 states. If we increase the temperature, the average kinetic energy increases, and that means the molecules are more likely to have enough kinetic energy to go into or stay in the gas phase. Liquid molecules will become gas molecules more often, and gas molecules will become liquid molecules less often. Then the equilibrium will move, so a bigger % of the total molecules are gas. In many cases, there is a specific temperature above which all of a material goes from solid to liquid (a melting point) or from liquid to gas (a boiling point). What temperature that is depends on the strength and type of forces between the molecules and ions. If the forces are strong, then more kinetic energy is needed to make the molecules move around or separate and become gas, which means the melt point and boiling point are higher. If the forces between molecules are very weak, then the material may be a gas, and it may be hard to cool it enough to make a liquid. In the next sections, we will talk about the forces between molecules that determine boiling points and melting points and other important properties. (Thus, they are called , to separate them from the forces inside molecules that hold the molecules together.) These forces are often called Van der Waals forces after Johannes van der Waals, who wrote the . Van der Waals figured out that the reason gas pressures are often lower than we expect (at high pressure) is because there are attractions between the molecules. Actually Van der Waals was the son of a carpenter who wasn't allowed to enter university because he didn't have the right expensive primary education. But he took classes anyway, became a teacher, and eventually they changed the rules for university admission, so he got a doctorate, became famous, became a professor, and won the Nobel Prize. | 4,299 | 3,214 |
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The International Union of Pure and Applied Chemistry (IUPAC) maintains a web-based compendium of analytical terminology. You can find it at the following web site. The following papers provide alternative schemes for classifying analytical methods. Further details on criteria for evaluating analytical methods are found in the following series of papers. For a point/counterpoint debate on the meaning of sensitivity consult the following two papers and two letters of response. Several texts provide analytical procedures for specific analytes in well-defined matrices. For a review of the importance of analytical methodology in today’s regulatory environment, consult the following text. | 704 | 3,215 |
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Organic chemistry is an art-form and a craft. In research, we often talk about candidates that “have it”, the “magic touch” to get a compound analytically pure, or solve a very difficult synthetic challenge. In this chapter we will go through some common techniques that you will encounter in the organic chemistry labs. The aim of this chapter is not to give you a comprehensive introduction in the theory behind these techniques, but to highlight technical aspects to make sure you also develop this “magic touch.” We will discuss recrystallization, distillation, liquid-liquid extraction, TLC, chromatography and lastly sublimation.
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\(\Page {1}\) Molar Perhaps the most obvious feature of the table of molecular entropies is a general increase in the molar entropy as we move from solids to liquids to gases. In a solid, the molecules are only capable of restricted vibrations around a fixed point, but when a solid melts, the molecules, though still hampered by their mutual attraction, are much freer to move around. Thus when a solid melts, the molar entropy of the substance increases. When a liquid vaporizes, the restrictions on the molecules’ ability to move around are relaxed almost completely and a further and larger increase in the entropy occurs. When 1 mol of ice melts, for example, its entropy increases by 22 J K , while on boiling the entropy increase is 110 J K . The more atoms there are in a molecule, the more ways the molecule can change its shape by vibrating. In consequence there are more ways in which the energy can be distributed among the molecules. which are isoelectronic with sodium fluoride. Since there is very little difference in the molar masses, the entropy decrease can only be attributed to the increase in the coulombic attraction between the ions as we move from the singly charged ions Na and F through the doubly charged ions Mg and O , to the triply charged ions A1 and N . (While it is true that there is a fair degree of covalent character to the bonding in AIN, the effect of this will be to increase the strength of the bonding.) From each of the following pairs of compounds choose the one with the higher standard molar entropy at 25°C. Give brief reasons for your choice. | 1,614 | 3,218 |
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\[ C_6H_{12}O_6 (s) + 6 O_2 (g) \rightarrow 6 CO_2 (g) + 6 H_2O (l) \label{4.2.1}\] Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain does not run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room? The initial step in solving a problem of this type is to write the balanced chemical equation for the reaction. Inspection shows that it is balanced as written, so the strategy outlined in , can be adapted as follows: 1. Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar: \[ moles \, glucose = 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose } = 0.251 \, mol \, glucose \] 2. According to the balanced chemical equation, 6 mol of CO is produced per mole of glucose; the mole ratio of CO to glucose is therefore 6:1. The number of moles of CO produced is thus \[ moles \, CO_2 = mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \] \[ = 0.251 \, mol \, glucose \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose } \] \[ = 1.51 \, mol \, CO_2 \] 3. Use the molar mass of CO (44.010 g/mol) to calculate the mass of CO corresponding to 1.51 mol of CO : \[ 45.3 \, g \, glucose \times {1 \, mol \, glucose \over 180.2 \, g \, glucose} \times {6 \, mol \, CO_2 \over 1 \, mol \, glucose} \times {44.010 \, g \, CO_2 \over 1 \, mol \, CO_2} = 66.4 \, g \, CO_2 \] Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in , .) This amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. Similar methods can be used to calculate the amount of oxygen consumed or the amount of water produced. The balanced chemical equation was used to calculate the mass of product that is formed from a certain amount of reactant. It can also be used to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant. Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): | 2,709 | 3,220 |
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. In addition to these individual basis; please contact A liquid, unlike a gas, is virtually . Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases? Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain. Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases? What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions? Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers. Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds? Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions? Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases? Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions? In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why? The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend. Identify the most important intermolecular interaction in each of the following. Identify the most important intermolecular interaction in each of the following. Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.) Arrange Kr, Cl , H , N , Ne, and O in order of increasing polarizability. Explain your reasoning. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds. The structures of ethanol, ethylene glycol, and glycerin are as follows: Arrange these compounds in order of increasing boiling point. Explain your rationale. Do you expect the boiling point of H S to be higher or lower than that of H O? Justify your answer. Ammonia (NH ), methylamine (CH NH ), and ethylamine (CH CH NH ) are gases at room temperature, while propylamine (CH CH CH NH ) is a liquid at room temperature. Explain these observations. Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not? Which compound in the following pairs will have the higher boiling point? Explain your reasoning. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input? Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why? How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H O molecule? What effect does this have on the structure and density of ice? Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular H⋅⋅⋅I interactions in liquid HI. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of hydrogen bonds per H O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering. Why is a water droplet round? How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension? Explain the role of intermolecular and intramolecular forces in surface tension. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why? Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram. Of CH Cl , hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m , whereas the surface tension of cyclohexanone, which is very similar chemically, is only 25.45 mN/m . Why is the surface tension of cyclohexanone so much less than that of cyclohexanol? What is the relationship between Explain your answers in terms of a microscopic picture. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus? Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning. How does viscosity depend on molecular shape? What molecular features make liquids highly viscous? Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid. The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity. The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning. Surface tension data (in dyn/cm) for propanoic acid (C H O ), and 2-propanol (C H O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar? 3. The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol. In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored? Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water? When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? Why do substances with high enthalpies of fusion tend to have high melting points? Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion? What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added? If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change? Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid? When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly. 3 The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are: \[ \text{solid} \rightarrow \text{liquid}: \Delta H_{fus} \] \[ \text{liquid} \rightarrow \text{gas}: \Delta H_{vap} \] \[ \text{solid} \rightarrow \text{gas}: \Delta{H_{sub}}= \Delta{H_{fus}} + \Delta{H_{vap}}\] The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below: The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to Δ , which is a positive number: Δ = Δ + Δ . The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely. The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point. A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor. Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid. The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen. The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane. Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C. Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C? A 0.542 g sample of I requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I ? A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound? One fuel used for jet engines and rockets is aluminum borohydride [Al(BH ) ], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a Δ of 30 kJ/mol and a molar heat capacity ( ) of 194.6 J/(mol•K)? How much energy is released when freezing 100.0 g of dimethyl disulfide (C H S ) initially at 20°C? Use the following information: melting point = −84.7°C, Δ = 9.19 kJ/mol, = 118.1 J/(mol•K). Δ Δ How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm? How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C? How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C? How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C? The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K). 45.0 kJ/mol 488 kJ 32.6 kJ 57 g What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure? What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility? An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol? Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away? If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer. What is the relationship between the vapor pressure of a liquid and At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure. Acetylene (C H ), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of Δ for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K? The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of Δ for water. The Δ of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm? The normal boiling point of sodium is 883°C. If Δ is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C? An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in Table 11.6 in Section 11.5 to identify the liquid. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in Table 11.6 in Section 11.5 to identify the liquid. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid? If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid? The vapor pressure of liquid SO is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K. The vapor pressure of CO at various temperatures is given in the following table: vapor pressure at 273 K is 3050 mmHg; Δ = 18.7 kJ/mol, 1.44 kJ 12.5°C Δ = 28.9 kJ/mol, -hexane Δ = 7.81 kJ/mol, 36°C The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the , the liquid is less dense than the solid, while if it slopes up and to the , the liquid is denser than the solid. 1. Compare the solid and liquid states in terms of a. rigidity of structure. b. long-range order. c. short-range order. 2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid? a. rigidity of structure b. long-range order c. short-range order 3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas? 4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion? 5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous. 6. Explain why each characteristic would or would not favor the formation of an amorphous solid. a. slow cooling of pure molten material b. impurities in the liquid from which the solid is formed c. weak intermolecular attractive forces 7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point? 3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases. 7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature. 1. four 3. fcc 5. molybdenum 7. sodium, unit cell edge = 428 pm, r = 185 pm 9. d = 0.5335 g/cm , r =151.9 pm 1. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents: Match each vial with its contents. 2. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning. 3. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why? 4. Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials? 5. Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy? 6. How are intermetallic compounds different from interstitial alloys or substitutional alloys? 1. a. NaCl, ionic solid b. quartz, covalent solid c. zinc, metal d. sucrose, molecular solid 5. In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy. 1. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La , 104 pm; O , 132 pm; Fe , 83 pm; and Br , 196 pm. Explain your reasoning. 2. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature. 3. Which has the higher melting point? Explain your reasoning in each case. a. Os or Hf b. SnO or ZrO c. Al O or SiO 4. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature. 3. a. Osmium has a higher melting point, due to more valence electrons for metallic bonding. b. Zirconium oxide has a higher melting point, because it has greater ionic character. c. Aluminum oxide has a higher melting point, again because it has greater ionic character. | 27,217 | 3,222 |
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Having identified basic sites as areas of high electron density and acidic protons as hydrogen atoms of low electron density, we can now establish how curved arrows are used to indicate the movement of electrons in acid-base reactions. The rules are basically the same as for resonance structures. (a) Full headed arrows indicate electron pair movement. (b) The arrow always originates at the basic site, i.e. the source of electrons (typically nonbonding electrons, but can also be a p-bond). (c) The arrow indicates a newly formed σ-bond between the basic site of one molecule and the acidic proton of another molecule. As this new bond is being formed, the atom (or group of atoms) to which the acidic proton is attached leaves as the conjugate base. Sometimes this atom or group of atoms is referred to as a SINCE STRONG ACIDS HAVE WEAK CONJUGATE BASES, THE BEST LEAVING GROUPS ARE WEAK BASES. IN OTHER WORDS, EQUILIBRIUM FAVORS DISPLACEMENT OF THE WEAKER BASE BY THE STRONGER BASE. The red arrow originates at the electron source and moves towards the acidic proton. It indicates a new bond that forms between the oxygen and the acidic proton in HCl. As oxygen bonds to the acidic proton, chlorine leaves with the electrons as chloride ion, the conjugate base of HCl. It can therefore be referred to as the leaving group. In this case, water acts as the leaving group. | 1,386 | 3,224 |
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Ligand Field Theory can be considered an extension of such that all levels of covalent interactions can be incorporated into the model. Treatment of the bonding in LFT is generally done using . A qualitative approach that can be used for octahedral metal complexes is given in the following 3 diagrams. In the first diagram, the 3d, 4s and 4p metal ion atomic orbitals are shown together with the ligand group orbitals that would have the correct symmetry to be able to overlap with them. The of ligand orbitals are generated by taking 6 sigma orbitals from the ligands, designated as σ , σ , σ , σ , σ , σ and then combining them to make 6 ligand group orbitals. (labelled e , a , t ) In the second diagram only sigma bonding is considered and it shows the combination of the metal 3d, 4s and 4p orbitals with OCCUPIED ligand group orbitals (using 1 orbital from each ligand). The result is that that the metal electrons would be fed into t2g and eg* molecular orbitals which is similar to the CFT model except that the eg orbital is now eg*. In the third diagram, π (pi) bonding is considered. In general π bonds are weaker than σ (sigma) bonds and so the effect is to modify rather than dramatically alter the description. 2 orbitals from each ligand are combined to give a total of 12 which are subdivided into four sets with three ligand group orbitals in each set. These are labelled t , t , t and t . The metal t orbital is the most suitable for interaction and this is shown in the 2 cases above.
Case A is the same as above, ignoring π interactions. Returning to the problem of correctly placing ligands in the Spectrochemical series, the halides are examples of case A and groups like CN- and CO are examples of case B. It is possible then to explain the Spectrochemical series once covalent effects are considered. Some convincing arguments for covalency and effects on Δ come from a study of the IR spectra recorded for simple carbonyl compounds e.g. M(CO) . Note that the IR values we are dealing with relate to the CO bond and not the M-C so when the CO frequency gets less then it is losing triple bond character and becoming more like a double bond. This is expected if electrons are pushed back from the metal into what were empty π antibonding orbitals. | 2,301 | 3,225 |
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Quinones are not aromatic compounds but are conjugated cyclic diketones. Yet it is convenient to discuss their chemistry at this point because quinones and the related aromatic arenols are readily interconverted, and their chemistry is largely interdependent. A variety of quinonelike structures have been prepared, the most common of which are the 1,2 - and 1,4-quinones as exemplified by 1,2- and 1,4-benzenediones. Usually the 1,2-quinones are more difficult to make and are more reactive than the 1,4-quinones. A few 1,6- and 1,8-quinones also are known. No 1,3-quinones are known, possibly because they would have nonplanar, highly strained structures and therefore would be unstable: A number of quinones are known in which the quinone arrangement extends over more than one ring. Examples are: A characteristic and important reaction of quinones is reduction to the corresponding arenediols. The reduction products of 1,4-quinones are called : \(\tag{26-2}\) Reduction can be achieved electrochemically and with a variety of reducing agents (metals in aqueous acid, catalytic hydrogenation). Such reductions are unusual among organic reactions in being sufficiently rapid and reversible to give easily reproducible electrode potentials in an electrolytic cell. The position of the 1,4-benzenediol-1,4,-benzenedione equilibrium (Equation 26-2) is proportional to the square of the hydrogen-ion concentration. Therefore the electrode potential is sensitive to pH; a change of one unit of pH in water solution changes the potential of the electrode by \(0.059 \: \text{V}\). Before the invention of the glass-electrode pH meter, the half-cell potential developed by this equilibrium was used widely to determine pH values of aqueous solutions. The method is not very useful above pH 9 because the quinone reacts irreversibly with alkali. Numerous studies have been made of the relationship between half-cell reduction potentials and the structures of quinones. As might be expected, the potentials are greatest when the resonance stabilization associated with formation of the aromatic ring is greatest. When alcohols solutions of hydroquinone and quinone are mixed, a brown-red color develops and a green-black 1:1 complex crystallizes that is known as . This substance is a charge-transfer complex ( ), with the diol acting as the electron donor and the dione as the electron acceptor. Quinhydrone is not very soluble and dissociates considerably to its components in solution. The reduction of a quinone requires two electrons, and it is possible that these electrons could be transferred either together or one at a time. The product of a single-electron transfer leads to what appropriately is called a , \(1\), with both a negative charge and an odd electron (a radical anion): The formation of relatively stable semiquinone radicals by electrolytic reduction of quinones has been established by a variety of methods. Some semiquinone radicals undergo reversible dimerization reactions to form peroxides. A particularly stable cation-radical of the semiquinone type is formed by mild oxidation of \(\ce{N}\),\(\ce{N}\),\(\ce{N'}\),\(\ce{N'}\)-tetramethyl-1,4-benzenediamine. The cation, which is isolable as a brilliant-blue perchlorate salt, \(2\), is called "Wurster's Blue": Oxidation and reduction in biochemical systems involve many reactions that are similar to the arenediol-arenediol couple. We have mentioned several previously: \(\ce{NADP}^\oplus \rightleftharpoons \ce{NADPH}\) ( ), and \(\ce{FADH_2} \rightleftharpoons \ce{FAD}\) ( ). An important question in metabolic oxidation is just how reduction of oxygen \(\left( \frac{1}{2} \ce{O_2} + 2 \ce{H}^\oplus + 2 \ce{e} \rightarrow \ce{H_2O} \right)\) is linked to the oxidation of \(\ce{NADH}\) \(\left( \ce{NADH} \rightarrow \ce{NAD}^\oplus + \ce{H}^\oplus + 2 \ce{e} \right)\). The route for transfer of electrons from \(\ce{NADH}\) to oxygen (oxidation plus phosphorylation; ) is indirect, complicated, and involves, in an early stage, oxidation of \(\ce{NADH}\) by flavin mononucleotide \(\left( \ce{FMN} \right)\) by the reaction \(\ce{FMN} + \ce{NADH} + \ce{H}^\oplus \rightarrow \ce{FMNH_2} + \ce{NAD}^\oplus\). But the reduced form of \(\ce{FMN}\), \(\ce{FMNH_2}\), does not react directly with oxygen. Instead, it reduces a quinone called to the corresponding arenediol: The effect of this step is to form a slightly polar reductant ( \(\ce{H_2}\)) from a strongly polar reductant \(\left( \ce{FMNH_2} \right)\), and this permits the reduced material to penetrate into a less polar region of the oxidative apparatus. The reduced does not react directly with oxygen but is a participant in a chain of oxidation-reduction reactions involving electron transfer between a number of iron-containing proteins known as . At the end of this chain of reaction, the reduced form of a copper-containing cytochrome actually reacts with oxygen. The sequence of electron-carriers may be summarized as \[\ce{NADH} \rightarrow \ce{FMNH_2} \rightarrow \textbf{CoQ}\ce{H_2} \rightarrow \text{cytochromes} \rightarrow \ce{O_2}\] A related process occurs in photosynthesis ( ). You will recall that a critical part of photosynthesis involves the transfer of electrons from the photosystem that oxidizes water \(\left( \ce{H_2O} \rightarrow \frac{1}{2} \ce{O_2} + 2 \ce{H}^\oplus + 2 \ce{e} \right)\) to the photosystem that reduces \(\ce{NADP}^\oplus\) \(\left( \ce{NADP}^\oplus + 2 \ce{H}^\oplus + 2 \ce{e} \rightarrow \ce{NADPH} + \ce{H}^\oplus \right)\). As in oxidative phosphorylation, the electron-transfer route is complex. However, one of the electron carriers is a quinone called that closely resembles coenzyme Q found in animals. Plastoquinone, like coenzyme Q, is reduced to the hydroquinone form, which is part of an electron-transport chain involving iron- and copper-containing proteins: Among other naturally occurring substances having quinone-type structures, one of the most important is the blood antihemorrhagic factor, vitamin K\(_1\), which occurs in green plants and is a substituted 1,4-naphthalenedione: The structure of vitamin K\(_1\) has been established by degradation and by synthesis. Surprisingly, the long alkyl side chain of vitamin K\(_1\) is not necessary for its action in aiding blood clotting because 2-methyl-1,4-naphthoquinone is almost equally active on a molar basis. Besides playing a vital role in the oxidation-reduction processes of living organisms, quinones occur widely as natural pigments found mainly in plants, fungi, lichens, marine organisms, and insects (see alizarin, , as representative of a natural anthraquinone-type dye). Photography makes important practical use of the arenediol-arenediol oxidation-reduction system. Exposure of the minute grains of silver bromide in a photographic emulsion to blue light (or any visible light in the presence of suitably sensitizing dyes) produces a stable activated form of silver bromide, the activation involving generation of some sort of crystal defect. Subsequently, when the emulsion is brought into contact with a developer, which may be an alkaline aqueous solution of 1,4-benzenediol (hydroquinone) and sodium sulfite, the particles of activated silver bromide are reduced to silver metal much more rapidly than the ordinary silver bromide. Removal of the unreduced silver bromide with sodium thiosulfate ("fixing") leaves a suspension of finely divided silver in the emulsion in the form of the familiar photographic negative. A variety of compounds are used as photographic developing agents. They are not all arenediols. In fact, most are aromatic aminoalcohols or diamines, but irrespective of their structural differences, they all possess the ability to undergo redox reactions of the type described for 1,4-benzenediol. Structural formulas and commercial names for several important developers are 1,4-Benzenediamine also is an effective developing agent, but it may cause dermatitis in sensitive individuals. \(\ce{N}\),\(\ce{N}\)-Diethyl-1,4-benzenediamine is used as a developer in color photography. These substances react with silver bromide to produce benzenediamine derivatives: Being \(\alpha\),\(\beta\)-unsaturated ketones, quinones have the potential of forming 1,4-addition products in the same way as their open-chain analogs ( ). 1,4-Benzenedione undergoes such additions rather readily. The products are unstable and undergo enolization to give substituted 1,4-benzenediols. Two examples are the addition of hydrogen chloride and the acid-catalyzed addition of ethanoic anhydride. In the latter reaction, the hydroxyl groups of the adduct are acylated by the anhydride. Hydrolysis of the product yields 1,2,4-benzenetriol: Quinones usually undergo Diels-Alder additions readily, provided that they have at least one double bond that is not part of an aromatic ring. With 1,4-benzenedione and 1,3-butadiene, either the mono- or diadduct can be obtained. The monoadduct enolizes under the influence of acid or base to a 1,4-benzenediol derivative: Benzoquinones owe their unusual properties as \(\alpha\),\(\beta\)-unsaturated ketones to the ease by which they are transformed to stable aromatic systems. How would these properties change if the quinone were derived from nonaromatic structures, such as cyclobutadiene, cyclooctatetraene, or pentalene? There is no final answer to this question because few such substances have been prepared, the best known so far being the mono- and diphenylcyclobutenediones: For example, \(3\) can be prepared from sulfuric acid hydrolysis of the cycloaddition product of ethynylbenzene and trifluorochloroethene ( ): The dione, \(3\), is a yellow crystalline solid that, despite its strained four-membered ring, is much less reactive than 1,2-benzenedione ( -benzoquinone). It cannot be reduced to a cyclobutenediol, does not undergo Diels-Alder reactions, and with bromine gives a substitution product rather than addition. The bromo compound so formed hydrolyzes rapidly to a hydroxy compound, \(4\), which is an extraordinarily strong acid having an ionization constant about \(10^9\) times that of benzenol: A related compound, 3,4-dihydroxy-1,2-cyclobutenedione, \(5\), also has been prepared and is a very strong dibasic acid. It is sometimes called "squaric acid": From the data so far available, it appears that the quinones corresponding to cyclobutadiene have more aromatic character than do the cyclobutadienes themselves. and (1977) | 10,506 | 3,226 |
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What is the total mass (amu) of carbon in each of the following molecules? (a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu What is the total mass of hydrogen in each of the molecules? Calculate the molecular or formula mass of each of the following: (a) P (b) H O (c) Ca(NO ) (d) CH CO H (acetic acid) (e) C H O (sucrose, cane sugar). (a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu Determine the molecular mass of the following compounds: (a) (b) (c) (d) Determine the molecular mass of the following compounds: (a) (b) (c) (d) Which molecule has a molecular mass of 28.05 amu? (a) (b) (c) Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula. Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams. Compare 1 mole of H , 1 mole of O , and 1 mole of F . Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C H OH), 0.60 mol of formic acid (HCO H), or 1.0 mol of water (H O)? Explain why. Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom. Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C H OH), 1 mol of formic acid (HCO H), or 1 mol of water (H O)? Explain why. How are the molecular mass and the molar mass of a compound similar and how are they different? The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 10 molecules. Calculate the molar mass of each of the following compounds: Calculate the molar mass of each of the following: (a) 256.528 g/mol; (b) 72.150 g mol ; (c) 378.103 g mol ; (d) 58.080 g mol ; (e) 180.158 g mol Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals: Calculate the molar mass of each of the following: (a) 197.382 g mol ; (b) 257.163 g mol ; (c) 194.193 g mol ; (d) 60.056 g mol ; (e) 306.464 g mol Determine the number of moles of compound and the number of moles of each type of atom in each of the following: Determine the mass of each of the following: Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following: Determine the mass of each of the following: The approximate minimum daily dietary requirement of the amino acid leucine, C H NO , is 1.1 g. What is this requirement in moles? Determine the mass in grams of each of the following: (a) 9.60 g; (b) 19.2 g; (c) 28.8 g A 55-kg woman has 7.5 × 10 mol of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams? Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO , a semiprecious stone. zirconium: 2.038 × 10 atoms; 30.87 g; silicon: 2.038 × 10 atoms; 9.504 g; oxygen: 8.151 × 10 atoms; 21.66 g Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH , 0.6 mol of C H , or 0.4 mol of C H . Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO , 266 g of Al Cl , or 225 g of Al S . AlPO : 1.000 mol Al Cl : 1.994 mol Al S : 3.00 mol Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond? The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone? 3.113 × 10 C atoms One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance? A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C H O ) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar? 0.865 servings, or about 1 serving. A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na PO F) in 100 mL. Which of the following represents the least number of molecules? 20.0 g H O represents the least number of molecules since it has the least number of moles. What information do we need to determine the molecular formula of a compound from the empirical formula? Calculate the following to four significant figures: (a) % N = 82.24% % H = 17.76%; (b) % Na = 29.08% % S = 40.56% % O = 30.36%; (c) % Ca = 38.76% Determine the following to four significant figures: Determine the percent ammonia, NH , in Co(NH ) Cl , to three significant figures. % NH = 38.2% Determine the percent water in CuSO ∙5H O to three significant figures. Determine the empirical formulas for compounds with the following percent compositions: (a) 15.8% carbon and 84.2% sulfur (b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen (a) CS (b) CH O Determine the empirical formulas for compounds with the following percent compositions: (a) 43.6% phosphorus and 56.4% oxygen (b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula? C H Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula? Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol. Mg Si H O (empirical formula), Mg Si H O (molecular formula) Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers: A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye. C H N Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L. What information do we need to calculate the molarity of a sulfuric acid solution? We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution. What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different? Determine the molarity for each of the following solutions: Determine the molarity of each of the following solutions: Answers: a.) 0.9713 M b.) 5.25x10 M c.) 6.122x10 M d.) 7.62 M e.) 0.5653 M f.) 1.13x10 M Consider this question: What is the mass of the solute in 0.500 L of 0.30 glucose, C H O , used for intravenous injection? (a) Outline the steps necessary to answer the question. (a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g Consider this question: What is the mass of solute in 200.0 L of a 1.556- solution of KBr? Answer: (a) (b) \(\dfrac{1.556\:moles\:\ce{KBr}}{1\:\cancel{L}}\times 200.0\:\cancel{L}=311.2\:moles\:\ce{KBr}\) \(311.2\:\cancel{moles}\:\ce{KBr}\times\dfrac{119.0\:g\:\ce{KBr}}{1\:\cancel{mole}\:\ce{KBr}}=37,030\:g\) 37,030g; 37.03 kg Calculate the number of moles and the mass of the solute in each of the following solutions: (a) 37.0 mol H SO ; 3.63 × 10 g H SO ; (b) 3.8 × 10 mol NaCN; 1.9 × 10 g NaCN; (c) 73.2 mol H CO; 2.20 kg H CO; (d) 5.9 × 10 mol FeSO ; 8.9 × 10 g FeSO Calculate the number of moles and the mass of the solute in each of the following solutions: Answers: a. 2.67x10 moles KI; 4.44x10 g KI b. 1.7x10 moles H SO ; 1.6x10 g H SO c. 2.838x10 moles K CrO ; 5.510g K CrO d. 39.0 moles (NH ) SO ; 5,160 g (NH ) SO Consider this question: What is the molarity of KMnO in a solution of 0.0908 g of KMnO in 0.500 L of solution? Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl? Answer: (a) (b) \(0.3366\:\cancel{g}\:\ce{HCl}\times\dfrac{1\:mole\:\ce{HCl}}{36.46\:\cancel{g}\:\ce{HCl}}=9.232\times10^{-3}\:moles\:\ce{HCl}\) \(35.23\:mL = 0.03523\:L\) \(\dfrac{9.232\times10^{-3}\:moles\:\ce{HCl}}{0.03523\:L}=0.2621\:M\:\ce{HCl}\) 0.2621 M ; Calculate the molarity of each of the following solutions: (a) 0.195 g of cholesterol, C H O, in 0.100 L of serum, the average concentration of cholesterol in human serum (b) 4.25 g of NH in 0.500 L of solution, the concentration of NH in household ammonia (c) 1.49 kg of isopropyl alcohol, C H OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol (d) 0.029 g of I in 0.100 L of solution, the solubility of I in water at 20 °C (a) 5.04 × 10 ; (b) 0.499 ; (c) 9.92 ; (d) 1.1 × 10 Calculate the molarity of each of the following solutions: There is about 1.0 g of calcium, as Ca , in 1.0 L of milk. What is the molarity of Ca in milk? 0.025 What volume of a 1.00- Fe(NO ) solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 ? If 0.1718 L of a 0.3556- C H OH solution is diluted to a concentration of 0.1222 , what is the volume of the resulting solution? 0.5000 L If 4.12 L of a 0.850 -H PO solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution? What volume of a 0.33- C H O solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 ? 1.9 mL What is the concentration of the NaCl solution that results when 0.150 L of a 0.556- solution is allowed to evaporate until the volume is reduced to 0.105 L? What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume? What is the final concentration of the solution produced when 225.5 mL of a 0.09988- solution of Na CO is allowed to evaporate until the solution volume is reduced to 45.00 mL? A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution? 11.9 An experiment in a general chemistry laboratory calls for a 2.00- solution of HCl. How many mL of 11.9 HCl would be required to make 250 mL of 2.00 HCl? What volume of a 0.20- K SO solution contains 57 g of K SO ? 1.6 L The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K Cr O ), what is the maximum permissible molarity of that substance? 1 | 11,314 | 3,227 |
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Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,229 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Halogenation_of_Benzene_and_Methylbenzene |
This page looks at the reactions of benzene and methylbenzene (toluene) with chlorine and bromine under various conditions. Benzene reacts with chlorine or bromine in the presence of a catalyst, replacing one of the hydrogen atoms on the ring by a chlorine or bromine atom. The reactions happen at room temperature. The catalyst is either aluminum chloride (or aluminum bromide if you are reacting benzene with bromine) or iron. Strictly speaking iron is not a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, \(FeCl_3\), or iron(III) bromide, \(FeBr_3\). \[ 2Fe + 3Cl_2 \rightarrow 2FeCl_3\] \[ 2Fe + 3Br_2 \rightarrow 2FeBr_3\] These compounds act as the catalyst and behave exactly like aluminum chloride, \(AlCl_3\), or aluminum bromide, \(AlBr_3\), in these reactions. The reaction between benzene and chlorine in the presence of either aluminum chloride or iron gives chlorobenzene. or, written more compactly: \[ C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl\] The reaction between benzene and bromine in the presence of either aluminum bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and is more readily available. or \[C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr\] In the presence of ultraviolet light (but without a catalyst present), hot benzene will also undergo an addition reaction with chlorine or bromine. The ring delocalization is permanently broken and a chlorine or bromine atom adds on to each carbon atom. For example, if you bubble chlorine gas through hot benzene exposed to UV light for an hour, you get 1,2,3,4,5,6-hexachlorocyclohexane. Bromine would behave similarly. The chlorines and hydrogens can stick up and down at random above and below the ring and this leads to a number of geometric isomers. Although there aren't any carbon-carbon double bonds, the bonds are still "locked" and unable to rotate. One of these isomers was once commonly used as an insecticide known variously as BHC, HCH and Gammexane. This is one of the "chlorinated hydrocarbons" which caused so much environmental harm. It is possible to get two quite different substitution reactions between methylbenzene and chlorine or bromine depending on the conditions used. The chlorine or bromine can substitute into the ring or into the methyl group. Substitution in the ring happens in the presence of aluminum chloride (or aluminum bromide if you are using bromine) or iron, and in the absence of UV light. The reactions happen at room temperature. This is exactly the same as the reaction with benzene, except that you have to worry about where the halogen atom attaches to the ring relative to the position of the methyl group. Methyl groups are 2,4-directing, which means that incoming groups will tend to go into the 2 or 4 positions on the ring - assuming the methyl group is in the 1 position. In other words, the new group will attach to the ring next door to the methyl group or opposite it. With chlorine, substitution into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene. With bromine, you would get the equivalent bromine compounds. If chlorine or bromine react with boiling methylbenzene in the absence of a catalyst but in the presence of UV light, substitution happens in the methyl group rather than the ring. For example, with chlorine (bromine would be similar): The organic product is (chloromethyl)benzene. The brackets in the name emphasize that the chlorine is part of the attached methyl group, and isn't on the ring. One of the hydrogen atoms in the methyl group has been replaced by a chlorine atom. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms. That means that you could also get (dichloromethyl)benzene and (trichloromethyl)benzene as the other hydrogen atoms in the methyl group are replaced one at a time. If you use enough chlorine you will eventually get (trichloromethyl)benzene, but any other proportions will always lead to a mixture of products. I haven't been able to track down anything similar to the reaction between benzene and chlorine in which six chlorine atoms add around the ring. That perhaps isn't surprising. Chlorine adds to benzene in the presence of ultraviolet light. With methylbenzene under those conditions, you get substitution in the methyl group. That is energetically easier because it doesn't involve breaking the delocalized electron system. Whether you would get addition to the ring if you used a large excess of chlorine and did the reaction for a long time, I don't know. Once all the hydrogens in the methyl group had been substituted, perhaps you might then get addition to the ring as well. Jim Clark ( ) | 4,812 | 3,230 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Metabolic_Pathways/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,231 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/01%3A_The_Basics/1.02%3A_Pressure_and_Molar_Volume |
Italian physicist Evangelista Torricelli (1608 – 1647) (Evangelista Torricelli) was the inventor of an ingenious device that could be used to measure air . Basically, he took a glass tube closed at one end, and filled it with mercury. He then inverted it, submerging the open end below the surface level in a pool of mercury. The mercury in the glass tube was then allowed to drain, leaving a vacuum (known as a “Torrocellian vacuum”) in the open space at the closed end of the tube. Remarkably, the tube did not drain completely! Torricelli, was able to use the residual column height to measure the pressure of the air pushing down on the surface of the pool of mercury. The larger the pressure pushing down on the exposed surface, the larger the column height is observed to be. The ambient air pressure can be computed by equating the force generated by the mass of the mercury in the column to the force generated by ambient air pressure (after normalizing for surface area). The resulting relationship is where \(\rho\) is the density of the mercury (13.1 g/cm ), \(g\) is the acceleration due to gravity, and \(h\) is the height of the column. Torricelli found that at sea level, the height of the column was 76 cm. \[ p = (13.1\, g/cm^3 ) (9.8 \,m/s) ( 76 \, cm) \left(\dfrac{100^2 cm^2}{m^2} \right) \left( \dfrac{1\, kg}{1000\, g} \right) \left( \dfrac{1 kg \, m/s^2}{N} \right) \nonumber \] \[ 100,000 N/m^2= 1 \times 10 ^5 Pa \nonumber \] A force of 1 N acting on an area of 1 m is a Pascal (Pa). A standard atmosphere is 101,325 Pa (101.325 kPa), or 76.0 cm Hg (760 mm Hg.) Another commonly used unit of pressure is the bar: \[1\, bar = 100,000 \,Pa \nonumber \] | 1,690 | 3,232 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/14%3A_Hydrodynamics/14.01%3A_Newtonian_Fluids |
Viscosity measures the resistance to shear forces. A fluid is placed between two plates of area a separated along z, and one plate is moved relative to the other by applying a shear force along x. At contact, the velocity of the fluid at the interface with either plate is equal to the velocity of the plate as a result of intermolecular interactions: \(\vec{v}_x(z=0)=0\). This is known as the no-slip boundary condition. The movement of one plate with respect to the other sets up a velocity gradient along z. This velocity gradient is equal to the strain rate. The relationship between the shear velocity gradient and the force is \[ \vec{f}_x = a \eta \dfrac{d \vec{v}_x}{dz} \nonumber \] where η, the dynamic viscosity (kg m s ), is the proportionality factor. For water at 25°C, the dynamic viscosity is η = 8.9×10 Pa s. A normal stress is a pressure (force per unit area), and these forces are transmitted through a fluid as a result of the conservation of momentum in an incompressible medium. This force transduction also means that a stress applied in one direction can induce a strain in another, i.e. a stress tensor is needed to describe the proportionality between the stress and strain vectors. In an anisotropic particulate system, force transmission from one region of the fluid to another results from “force chains” involving steaming motion of particles that repel each other. These force chains are not simply unidirectional, but also branch into networks that bypass unaffected regions of the system. | 1,536 | 3,233 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Potentials/Reduction_Potential_Intuition |
Reduction potentials are relative and are reported relative to the reduction of protons in a standard hydrogen electrode (SHE) in Tables P1 and P2. However, what would we see if we used some other sort of electrode for comparison? For example, instead of a hydrogen electrode, we might use a fluorine electrode, in which we have fluoride salts and fluorine gas in solution with a platinum electrode. The "reduction potentials" we measure would all be relative to this reaction now. They would tell us: how much more motivated is this ion to gain an electron than fluorine? The resulting table would look something like this: The trouble is, there is not much out there that would be more motivated than fluorine to gain an electron. All of the potentials in the table are negative because none of the species on the left could take an electron away from fluoride ion. However, if we turned each of these reactions around, the potentials would all become positive. That means gold, for instance, could give an electron to fluorine to become \(\ce{Au^{+}}\). This electron transfer would be spontaneous, and a voltmeter in the circuit between the gold electrode and the fluorine electrode would measure a voltage of -1.04 V. That potential is generated by the reaction: \[ \ce{ 2 Au (s) + F_2 (g) -> 2 Au^{+} + 2 F^{-}} \nonumber \] Notice that, because 2 electrons are needed to reduce the fluorine to fluoride, and because gold only supplies one electron, two atoms of gold would be needed to supply enough electrons. The reduction potentials in Table \(\Page {1}\) are, indirectly, an index of differences in electronic energy levels. The electron on gold is at a higher energy level than if it were on fluoride. It is thus motivated to spontaneously transfer to the fluorine atom, generating a potential in the circuit of -1.04V. Silver metal is even more motivated to donate an electron to fluorine. An electron from silver can "fall" even further than an electron from gold, to a lower energy level on fluoride. The potential in that case would be - 2.074 V. There are a couple of things to note here. The first is that, if potential is an index of the relative energy level of an electron, it does not matter whether one electron or two is transferred. They are transferred from the same, first energy level to the same, second energy level. The distance that the electron falls is the same regardless of the number of electrons that fall. A reduction potential reflects an inherent property of the material and does not depend on how many electrons are being transferred. Another important note is that, if reduction potentials provide a glimpse of electronic energy levels, we may be able to deduce new relationships from previous information. For example, if an electron on gold is 1.04 V above an electron on fluoride, and an electron on silver is 1.04 V above an electron on fluoride, what can we deduce about the relative energy levels of an electron on silver vs. gold? The answer is that the electron on silver is 1.034 V above the electron on gold. We know this because reduction potentials are "state functions", reflecting an intrinsic property of a material. It does not matter how we get from one place to another; the answer will always be the same. That means that if we transfer an electron from silver to gold indirectly, via fluorine, the overall potential will be the same as if we transfer the electron directly from silver to gold. So, the electron drops from silver to fluorine (a drop of 2.074 V). The electron then hops (under duress) up to gold (a climb of 1.04 V). The net drop is only 1.034 V. That's the same value we would expect to measure if we took a standard solution of gold salts and a gold electrode and connected it, via a circuit, to a standard solution of silver salts and a silver electrode. In fact, the table above does not reflect any experimental measurements; it's simply the table of standard reduction potentials from the previous page, with the reduction potential of fluorine subtracted from all the other values. In other words, mapping out the distance from iron to SHE (in terms of the reduction potential for \(Fe^{2+} + 2 e^- \rightarrow Fe(s)\)), together with the distance from SHE to fluorine gives the potential relative to IFE (imaginary fluorine electrode). Frequently in biology, electron transfers are made more efficient through a series of smaller drops rather than one big jump. To think about this, consider the transfer of an electron from lithium to fluorine. (Neither of these species is likely to be found in an organism, but this transfer is a good illustration of a big energy difference.) An "activity series" is a ranking of elements in terms of their "activity" or their ability to provide electrons. The series is normally written in a column, with the strongest reducing metals at the top. Beside these elements, we write the ion produced when the metal loses its electron(s). Looking at the table of reduction potentials relative to fluorine on this page, construct an activity series for the available elements. Construct an activity series for the alkali metals using the following standard reduction potentials (relative to SHE): Fr, -2.9 V; Cs, -3.026 V; Rb, -2.98 V; K, -2.931 V; Na, -2.71 V; Li, -3.04 V. In reality, the energy gap that leads to a reduction potential is sometimes more complicated than following an electron as it moves from one level to another. Use the activity series you have constructed for the alkali metals to compare and contrast the redox potential with your expectations of energy level / ease of electron donation based on standard periodic trends. , | 5,680 | 3,234 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.01%3A_Prelude_to_Covalent_Bonding |
While other sections concentrate on and Lewis diagrams for simple covalent molecules, there are numerous examples of molecules which are quite stable, but contain one or more atoms which do not have a noble-gas electron configuration. Furthermore, structural formulas like those from sections under only show which atoms are connected to which. They do not tell us how the atoms are arranged in three-dimensional space. In other words a Lewis diagram does not show the shape of a molecule. It may actually be misleading because most molecules don't have the 90 or 180 bond angles that are conventionally used in the Lewis Diagram. In the following sections, we will develop a more detailed picture of molecules—including some which do not obey the octet rule. You will learn how both the shapes and bonding of molecules may be described in terms of orbitals. In addition it will become apparent that the distinction between covalent and ionic bonding is not so sharp as it may have seemed. You will find that many covalent molecules are electrically unbalanced, causing their properties to tend toward those of ion pairs. Rules will be developed so that you can predict which combinations of atoms will exhibit this kind of behavior. are used as the first step in predicting the structures for covalently bonded molecules. We will see what supplemental information is required to describe covalent molecules. We will describe some molecules which and some which contain , as well as simple molecules. The structure of a molecule is determined by the positions of atomic nuclei in three-dimensional space, but it is repulsions among electron pairs—bonding pairs and lone pairs—which determine molecular geometry. When dealing with molecular shapes, it is often convenient to think in terms of sp, , , or other . These correspond to the same overall electron density and to the same physical reality as do the and atomic orbitals considered earlier, but hybrid orbitals emphasize the directions in which electron density is concentrated. Hybrid orbitals may also be used to describe multiple bonding, in which case bonds must be bent so that two or three of them can link the same pair of atoms. A second equivalent approach to multiple bonding involves sigma and pi bonds. Electron density in a sigma bond is concentrated directly between the bonded nuclei, while that of the pi bond is divided in two—half on one side and half on the other side of the sigma bond. No chemical bond can be 100 percent ionic, and, except for those between identical atoms, 100 percent covalent bonds do not exist either. Electron clouds—especially large, diffuse ones—are easily polarized, affecting the electrical balance of atoms, ions, or molecules. Large negative ions are readily polarized by small positive ions, increasing the covalent character of the bond between them. Electron density in covalent bonds shifts toward the more electronegative atom, producing partial charges on each atom and hence a dipole. In a polyatomic molecule, bond dipoles must be added as vectors to obtain a resultant which indicates molecular polarity. In the case of symmetric molecules, the effects of individual bond dipoles cancel and a nonpolar molecule results. Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such may be of three types. All molecules are attracted together by weak . These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, occur. This results in higher melting and boiling points than for nonpolar substances. The third type of intermolecular force, the , occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the . This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do. Oxidation numbers are used by chemists to keep track of electrons during the course of a chemical reaction. They may be obtained by arbitrarily assigning valence electrons to the more electronegative of two bonded atoms and calculating the resulting charge as if the bond were 100 percent ionic. Alternatively, some simple rules are available to predict the oxidation number of each atom in a formula. Oxidation numbers are used in the names of compounds and are often helpful in predicting formulas and writing Lewis diagrams. In addition to deficiency of electrons and expansion of the valence shell, the octet rule is violated by species which have one or more unpaired electrons. Such free radicals are usually quite reactive. A difficulty of another sort occurs in benzene and other molecules for which more than one Lewis diagram can be drawn. Rearranging electrons (but not atomic nuclei) results in several structures which are referred to collectively as a resonance hybrid. Like an hybrid, a resonance hybrid is a combination of the contributing structures and has properties intermediate between them. | 5,569 | 3,236 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photosynthesis/Photosystem_II/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,237 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/11%3A_Fluids/11.01%3A_Introduction |
The kinetic molecular theory of gases described in gives a reasonably accurate description of the behavior of gases. A similar model can be applied to liquids, but it must take into account the nonzero volumes of particles and the presence of strong intermolecular attractive forces. In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. In contrast, the molecules in liquids are very close together, with essentially no empty space between them. As in gases, however, the molecules in liquids are in constant motion, and their kinetic energy (and hence their speed) depends on their temperature. We begin our discussion by examining some of the characteristic properties of liquids to see how each is consistent with a modified kinetic molecular description. The molecules of a liquid are packed relatively close together. Consequently, liquids are much denser than gases. The density of a liquid is typically about the same as the density of the solid state of the substance. Densities of liquids are therefore more commonly measured in units of grams per cubic centimeter (g/cm ) or grams per milliliter (g/mL) than in grams per liter (g/L), the unit commonly used for gases. Liquids exhibit short-range order because strong intermolecular attractive forces cause the molecules to pack together rather tightly. Because of their higher kinetic energy compared to the molecules in a solid, however, the molecules in a liquid move rapidly with respect to one another. Thus unlike the ions in the ionic solids discussed in , the molecules in liquids are not arranged in a repeating three-dimensional array. Unlike the molecules in gases, however, the arrangement of the molecules in a liquid is not completely random. Liquids have so little empty space between their component molecules that they cannot be readily compressed. Compression would force the atoms on adjacent molecules to occupy the same region of space. The intermolecular forces in liquids are strong enough to keep them from expanding significantly when heated (typically only a few percent over a 100°C temperature range). Thus the volumes of liquids are somewhat fixed. Notice from that the density of water, for example, changes by only about 3% over a 90-degree temperature range. Molecules in fluids diffuse because they are in constant motion ( ). A molecule in a liquid cannot move far before colliding with another molecule, however, so the mean free path in liquids is very short, and the rate of diffusion is much slower than in gases. A drop of an aqueous solution containing a marker dye is added to a larger volume of water. As it diffuses, the color of the dye becomes fainter at the edges. Liquids can flow, adjusting to the shape of their containers, because their molecules are free to move. This freedom of motion and their close spacing allow the molecules in a liquid to move rapidly into the openings left by other molecules, in turn generating more openings, and so forth ( ). The properties of liquids can be explained using a modified version of the kinetic molecular theory of gases described in . This model explains the higher density, greater order, and lower compressibility of liquids versus gases; the thermal expansion of liquids; why they diffuse; and why they adopt the shape (but not the volume) of their containers. A liquid, unlike a gas, is virtually . Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases? Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain. Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases? | 4,037 | 3,239 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure |
Suppose we seal a pure liquid and its vapor in a strong glass tube and heat it to a very high temperature. As we increase the temperature, the vapor pressure will rise. (It is not a good idea to heat a liquid this way unless you are sure the container can withstand the increased pressure.) The rising vapor pressure corresponds to a greater number of molecules in the limited volume of the vapor phase. In other words, the vapor becomes considerably denser. Eventually we reach a temperature at which the density of the vapor becomes the same as that of the liquid. Since liquids are usually distinguished from gases on the basis of density, at this point both have become identical. The temperature at which this occurs is called the , and the pressure is called the . The accompanying videos illustrate what happens experimentally in the case of Chlorine. In the first video, as the temperature nears the critical temperature, liquid and vapor become very similar in appearance and the meniscus between them becomes difficult to distinguish. Finally, at the critical temperature the meniscus disappears completely. Above the critical temperature the sample is quite uniform and it is difficult to know whether to call it a liquid or a gas. In the second half of the video, the flask is brought back below the critical temperature. The speed of gas molecules decreases to a point where intermolecular forces can cause a liquid phase to condense out. The meniscus reappears, and the Chlorine separates back again into a liquid and vapor phase. Once a gas is above its critical temperature, it is to get it to separate into a liquid layer below and a vapor layer above no matter how great a pressure is applied, as can be seen in the graph below. On the graph, once the temperature is higher than 300 K, it is not possible to revert to liquid form. Increasing the pressure only leads to the transition from gas to supercritical fluid. Oxygen, for instance, is well above its critical temperature at room temperature. If we increase the pressure on it to a few thousand atmospheres, its density becomes so high that we are forced to classify it as a liquid. Nevertheless as we increase the pressure, there is no point at which drops of liquid suddenly appear in the gas. Instead the oxygen gradually changes from something which is obviously a gas to something which is obviously a liquid (a supercritical fluid). Conversely, if we gradually relax the pressure, there is no point at which the oxygen will start to boil. The following table lists the critical temperatures and critical pressures for some well-known gases and liquids. Such data are often quite useful. Many gases are sold commercially in strong steel cylinders at high pressures. The behavior of the gas in such a cylinder depends on whether it is above or below the critical temperature. The critical temperature of propane, for instance, is 97°C, well above room temperature. Thus propane in a high-pressure cylinder consists of a mixture of liquid and vapor, and you can sometimes hear the liquid sloshing about inside.
\(\Page {1}\) The pressure of the gas in such a cylinder will be the vapor pressure of propane at namely, 9.53 atm (965.4 kPa).As long as there is some liquid left in the cylinder, the pressure will remain at 9.53 atm. Only when all the liquid has evaporated the pressure begin to drop. At that point the cylinder will be virtually empty. A very different behavior is found in the case of a cylinder of oxygen. Since oxygen is above its critical temperature at 20°C, the cylinder will contain a uniform fluid rather than a liquid-vapor mixture. As we use up the oxygen, the pressure will gradually decrease to 1 atm, at which point no more O will escape from the cylinder. The principles discussed in the preceding paragraph apply to the aerosol sprays most of us encounter every day. Such spray cans contain a small quantity of the active ingredient—hair conditioner, deodorant, shaving cream, and the like—and a large quantity of propellant. The propellant is a substance, such as propane, whose critical temperature is well above room temperature. Therefore it can be liquefied at the high pressure in the spray can. When the valve is opened, the vapor pressure of the liquid propellant causes the active ingredient and the propellant to spray out of the can. As long as liquid propellant remains, the pressure inside the can will be constant (it will be the vapor pressure), and the spray will be reproducible. It should be obvious why such cans always bear a warning against throwing them in a fire--vapor pressure increases more rapidly at , and so heating an enclosed liquid is far more likely to produce an explosion than heating a gas alone. (The latter case was described in an .) | 4,800 | 3,240 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/07_Molecular_Geometry_and_Electron_Domain_Theory |
We begin by assuming a for chemical bonding based on valence shell electron pair sharing and the octet rule. We thus assume the nuclear structure of the atom, and we further assume the existence of a valence shell of electrons in each atom which dominates the chemical behavior of that atom. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. A number of atoms, including \(\ce{C}\), \(\ce{N}\), \(\ce{O}\), \(\ce{P}\), and \(\ce{S}\), can form double or triple bonds as needed to complete an octet. We know that double bonds are generally stronger and have shorter lengths than single bonds, and triple bonds are stronger and shorter than double bonds. We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. The relationship between bonding, structure, and properties is comparatively simple in molecules, which contain two atoms only, e.g. \(\ce{HCl}\) or \(\ce{O_2}\). A molecule contains more than two atoms. An example of the complexities which arise with polyatomic molecules is molecular geometry: how are the atoms in the molecule arranged with respect to one another? In a diatomic molecule, only a single molecular geometry is possible since the two atoms must lie on a line. However, with a triatomic molecule (three atoms), there are two possible geometries: the atoms may lie on a line, producing a linear molecule, or not, producing a bent molecule. In molecules with more than three atoms, there are many more possible geometries. What geometries are actually observed? What determines which geometry will be observed in a particular molecule? We seek a model which allows us to understand the observed geometries of molecules and thus to predict these geometries. Once we have developed an understanding of the relationship between molecular structure and chemical bonding, we can attempt an understanding of the relationship of the structure and bonding in a polyatomic molecule to the physical and chemical properties we observe for those molecules. The geometry of a molecule includes a description of the arrangements of the atoms in the molecule. At a simple level, the molecular structure tell us which atoms are bonded to which. At a more detailed level, the geometry includes the lengths of all of these bonds, that is, the distances between the atoms which are bonded together, and the angles between pairs of bonds. For example, we find that in water, \(\ce{H_2O}\), the two hydrogens are bonded to the oxygen and each \(\ce{O-H}\) bond length is \(95.72 \: \text{pm}\) (where \(1 \: \text{pm} = 10^{-12} \: \text{m}\)). Furthermore, \(\ce{H_2O}\) is a bent molecule, with the \(\ce{H-O-H}\) angle equal to \(104.5^\text{o}\). (The measurement of these geometric properties is difficult, involving the measurement of the frequencies at which the molecule rotates in the gas phase. In molecules in crystalline form, the geometry of the molecule is revealed by irradiating the crystal with x-rays and analyzing the patterns formed as the x-rays diffract off of the crystal.) Not all triatomic molecules are bent, however. As a common example, \(\ce{CO_2}\) is a linear molecule. Larger polyatomics can have a variety of shapes, as illustrated in Figure 7.1. Ammonia, \(\ce{NH_3}\), is a pyramid-shaped molecule, with the hydrogens in an equilateral triangle, the nitrogen above the plane of the triangle, and a \(\ce{H-N-H}\) angle equal to \(107^\text{o}\). The geometry of \(\ce{CH_4}\) is that of a tetrahedron, with all \(\ce{H-C-H}\) angles equal to \(109.5^\text{o}\). (See also Figure 7.1.) Ethane, \(\ce{C_2H_6}\), has a geometry related to that of methane. The two carbons are bonded together, and each is bonded to three hydrogens. Each \(\ce{H-C-H}\) angle is \(109.5^\text{o}\) and each \(\ce{H-C-C}\) angle is \(109.5^\text{o}\). By contrast, in ethene, \(\ce{C_2H_4}\), each \(\ce{H-C-H}\) bond angle is \(116.6^\text{o}\), and each \(\ce{H-C-C}\) bond angle is \(121.7^\text{o}\). All six atoms of ethene lie in the same plane. Thus, ethene and ethane have very different geometries, despite the similarities in their molecular formulae. We begin our analysis of these geometries by noting that, in the molecules listed above which do contain double or triple bonds (\(\ce{H_2O}\), \(\ce{NH_3}\), \(\ce{CH_4}\), and \(\ce{C_2H_6}\)), the bond angles are very similar, each equal to or very close to the tetrahedral angle \(109.5^\text{o}\). To account for the observed angle, we begin with our valence shell electron pair sharing model, and we note that, in the Lewis structures of these molecules, the central atom in each bond angle of these molecules contains four pairs of valence shell electrons. For methane and ethane, these four electron pairs are all shared with adjacent bonded atoms, whereas in ammonia or water, one or two (respectively) of the electron pairs are not shared with any other atom. These unshared electron pairs are called . Notice that, in the two molecules with no lone pairs, all bond angles are equal to the tetrahedral angle, whereas the bond angles are only close in the molecules with lone pairs. One way to understand this result is based on the mutual repulsion of the negative charges on the valence shell electrons. Although the two electrons in each bonding pair must remain relatively close together in order to form the bond, different pairs of electrons should arrange themselves in such a way that the distances between the pairs are as large as possible. Focusing for the moment on methane, the four pairs of electrons must be equivalent to one another, since the four \(\ce{C-H}\) bonds are equivalent, so we can assume that the electron pairs are all the same distance from the central carbon atom. How can we position four electron pairs at a fixed distance from the central atom but as far apart from one another as possible? A little reflection reveals that this question is equivalent to asking how to place four points on the surface of a sphere spread out from each other as far apart as possible. A bit of experimentation reveals that these four points must sit at the corners of a tetrahedron, an equilateral triangular pyramid, as may be seen in Figure 7.2a. If the carbon atom is at the center of this tetrahedron and the four electron pairs placed at the corners, then the hydrogen atoms also form a tetrahedron about the carbon. This is, as illustrated in Figure 7.2b, the correct geometry of a methane molecule. The angle formed by any two corners of a tetrahedron and the central atom is \(109.5^\text{o}\), exactly in agreement with the observed angle in methane. This model also works well in predicting the bond angles in ethane. a. b. We conclude that molecular geometry is determined by minimizing the mutual repulsion of the valence shell electron pairs. As such, this model of molecular geometry is often referred to as the . For reasons that will become clear, extension of this model implies that a better name is the . This model also accounts, at least approximately, for the bond angles of \(\ce{H_2O}\) and \(\ce{NH_3}\). These molecules are clearly not tetrahedral, like \(\ce{CH_4}\), since neither contains the requisite five atoms to form the tetrahedron. However, each molecule does contain a central atom surrounded by four pairs of valence shell electrons. We expect from our Electron Domain model that those four pairs should be arrayed in a tetrahedron, without regard to whether they are bonding or lone-pair electrons. Then attaching the hydrogens (two for oxygen, three for nitrogen) produces a prediction of bond angles of \(109.5^\text{o}\), very close indeed to the observed angles of \(104.5^\text{o}\) in \(\ce{H_2O}\) and \(107^\text{o}\) in \(\ce{NH_3}\). Note, however, that we do not describe the geometries of \(\ce{H_2O}\) and \(\ce{NH_3}\) as "tetrahedral", since the of the molecules do not form tetrahedrons, even if the valence shell electron pairs do. (It is worth noting that these angles are not exactly equal to \(109.5^\text{o}\), as in methane. These deviations will be discussed later.) We have developed the Electron Domain model to this point only for geometries of molecules with four pairs of valence shell electrons. However, there are a great variety of molecules in which atoms from Period 3 and beyond can have more than an octet of valence electrons. We consider two such molecules illustrated in Figure 7.3. First, \(\ce{PCl_5}\) is a stable gaseous compound in which the five chlorine atoms are each bonded to the phosphorus atom. Experiments reveal that the geometry of \(\ce{PCl_5}\) is that of a : three of the chlorine atoms form an equilateral triangle with the \(\ce{P}\) atom in the center, and the other two chlorine atoms are on top of and below the \(\ce{P}\) atom. Thus there must be 10 valence shell electrons around the phosphorus atom. Hence, phosphorus exhibits what is called an in \(\ce{PCl_5}\). Applying our Electron Domain model, we expect the five valence shell electron pairs to spread out optimally to minimize their repulsions. The required geometry can again be found by trying to place five points on the surface of a sphere with maximum distances amongst these points. A little experimentation reveals that this can be achieved by placing the five points to form a trigonal bipyramid. Hence, Electron Domain theory accounts for the geometry of \(\ce{PCl_5}\). Second, \(\ce{SF_6}\) is a fairly unreactive gaseous compound in which all six fluorine atoms are bonded to the central sulfur atom. Again, it is clear that the octet rule is violated by the sulfur atom, which must therefore have an expanded valence. The observed geometry of \(\ce{SF_6}\), as shown in Figure 7.2, is highly symmetric: all bond lengths are identical and all bond angles are \(90^\text{o}\). The \(\ce{F}\) atoms form an about the central \(\ce{S}\) atom: four of the \(\ce{F}\) atoms form a square with the \(\ce{S}\) atom at the center, and the other two \(\ce{F}\) atoms are above and below the \(\ce{S}\) atom. To apply our Electron Domain model to understand this geometry, we must place six points, representing the six electron pairs about the central \(\ce{S}\) atom, on the surface of a sphere with maximum distances between the points. The requisite geometry is found, in fact, to be that of an octahedron, in agreement with the observed geometry. As an example of a molecule with an atom with less than an octet of valence shell electrons, we consider boron trichloride, \(\ce{BCl_3}\). The geometry of \(\ce{BCl_3}\) is also given in Figure 7.2: it is , with all four atoms lying in the same plane, and all \(\ce{Cl-B-Cl}\) bond angles equal to \(120^\text{o}\). The three \(\ce{Cl}\) atoms form an equilateral triangle. The Boron atom has only three pairs of valence shell electrons in \(\ce{BCl_3}\). In applying Electron Domain theory to understand this geometry, we must place three points on the surface of a sphere with maximum distance between the points. We find that the three points form an equilateral triangle in a plane with the center of the sphere, so Electron Domain is again in accord with the observed geometry. We conclude from these predictions and observations that the Electron Domain model is a reasonably accurate way to understand molecular geometries, even in molecules which violate the octet rule. In each of the molecules considered up to this point, the electron pairs are either in single bond or in lone pairs. In current form, the Electron Domain model does account for the observed geometry of \(\ce{C_2H_4}\), in which each \(\ce{H-C-H}\) bond angle is \(116.6^\text{o}\) and each \(\ce{H-C-C}\) bond angle is \(121.7^\text{o}\) and all six atoms lie in the same plane. Each carbon atom in this molecule is surrounded by four pairs of electrons, all of which are involved in bonding, i.e. there are no lone pairs. However, the arrangement of these electron pairs, and thus the bonded atoms, about each carbon is not even approximately tetrahedral. Rather, the \(\ce{H-C-H}\) and \(\ce{H-C-C}\) bond angles are much closer to \(120^\text{o}\), the angle which would be expected if electron pairs were separated in the optimal arrangement, as just discussed for \(\ce{BCl_3}\). This observed geometry can be understood by re-examining the Lewis structure. Recall that, although there are four electron pairs about each carbon atom, two of these pairs form a double bond between the carbon atoms. It is tempting to assume that these four electron pairs are forced apart to form a tetrahedron as in previous molecules. However, if this were the case, the two pairs involved in the double bond would be separated by an angle of \(109.5^\text{o}\) which would make it impossible for both pairs to be localized between the carbon atoms. To preserve the double bond, we must assume that the two electron pairs in the double bond remain in the same vicinity. Given this assumption, separating the three groups of electron pairs about a carbon atom produces an expectation that all three pairs should lie in the same plane as the carbon atom, separated by \(120^\text{o}\) angles. This agrees very closely with the observed bond angles. We conclude that our model can be extended to understanding the geometries of molecules with double (or triple) bonds by treating the multiple bond as two electron pairs confined to a single . It is for this reason that we refer to the model as Electron Domain theory. Applied in this form, Electron Domain theory can help us understand the linear geometry of \(\ce{CO_2}\). Again, there are four electron pairs in the valence shell of the carbon atom, but these are grouped into only two domains of two electron pairs each, corresponding to the two \(\ce{C=O}\) double bonds. Minimizing the repulsion between these two domains forces the oxygen atoms to directly opposite sides of the carbon, producing a linear molecule. Similar reasoning using Electron Domain theory as applied to triple bonds correctly predicts that acetylene, \(\ce{HCCH}\), is a linear molecule. If the electron pairs in the triple bond are treated as a single domain, then each carbon atom has only two domains each. Forcing these domains to opposite sides from one another accurately predicts \(180^\text{o}\) \(\ce{H-C-C}\) bond angles. It is interesting to note that some molecular geometries (\(\ce{CH_4}\), \(\ce{CO_2}\), \(\ce{HCCH}\)) are exactly predicted by the Electron Domain model, whereas in other molecules, the model predictions are only approximately correct. For example, the observed angles in ammonia and water each differ slightly from the tetrahedral angle. Here again, there are four pairs of valence shell electrons about the central atoms. As such, it is reasonable to conclude that the bond angles are determined by the mutual repulsion of these electron pairs, and are thus expected to be \(109.5^\text{o}\), which is close but not exact. One clue as to a possible reason for the discrepancy is that the bond angles in ammonia and water are both than \(109.5^\text{o}\). Another is that both ammonia and water molecules have lone pair electrons, whereas there are no lone pairs in a methane molecule, for which the Electron Domain prediction is exact. Moreover, the bond angle in water, with two lone pairs, is less than the bond angles in ammonia, with a single lone pair. We can straightforwardly conclude from these observations that the lone pairs of electrons must produce a greater repulsive effect than do the bonded pairs. Thus, in ammonia, the three bonded pairs of electrons are forced together slightly compared to those in methane, due to the greater repulsive effect of the lone pair. Likewise, in water, the two bonded pairs of electrons are even further forced together by the two lone pairs of electrons. This model accounts for the comparative bond angles observed experimentally in these molecules. The valence shell electron pairs repel one another, establishing the geometry in which the energy of their interaction is minimized. Lone pair electrons apparently generate a greater repulsion, thus slightly reducing the angles between the bonded pairs of electrons. Although this model accounts for the observed geometries, why should lone pair electrons generate a greater repulsive effect? We must guess at a qualitative answer to this question, since we have no description at this point for where the valence shell electron pairs actually are or what it means to share an electron pair. We can assume, however, that a pair of electrons shared by two atoms must be located somewhere between the two nuclei, otherwise our concept of "sharing" is quite meaningless. Therefore, the powerful tendency of the two electrons in the pair to repel one another must be significantly offset by the localization of these electrons between the two nuclei which share them. By contrast, a lone pair of electrons need not be so localized, since there is no second nucleus to draw them into the same vicinity. Thus more free to move about the central atom, these lone pair electrons must have a more significant repulsive effect on the other pairs of electrons. These ideas can be extended by more closely examining the geometry of ethene, \(\ce{C_2H_4}\). Recall that each \(\ce{H-C-H}\) bond angle is \(116.6^\text{o}\) and each \(\ce{H-C-C}\) bond angle is \(121.7^\text{o}\), whereas the Electron Domain theory prediction is for bond angles exactly equal to \(120^\text{o}\). We can understand why the \(\ce{H-C-H}\) bond angle is slightly less than \(120^\text{o}\) by assuming that the two pairs of electrons in the \(\ce{C=C}\) double bond produce a greater repulsive effect than do either of the single pairs of electrons in the \(\ce{C-H}\) single bonds. The result of this greater repulsion is a slight "pinching" of the \(\ce{H-C-H}\) bond angle to less than \(120^\text{o}\). The concept that lone pair electrons produce a greater repulsive effect than do bonded pairs can be used to understand other interesting molecular geometries. Sulfur tetrafluoride, \(\ce{SF_4}\), is a particularly interesting example, shown in Figure 7.4. Note that two of the fluorines form close to a straight line with the central sulfur atom, but the other two are approximately perpendicular to the first two and at an angle of \(101.5^\text{o}\) to each other. Viewed sideways, this structure looks something like a seesaw. To account for this structure, we first prepare a Lewis structure. We find that each fluorine atom is singly bonded to the sulfur atom, and that there is a lone pair of electrons on the sulfur. Thus, with five electron pairs around the central atom, we expect the electrons to arrange themselves in a trigonal bipyramid, similar to the arrangement in \(\ce{PCl_5}\) in Figure 7.3. In this case, however, the fluorine atoms and the lone pair could be arranged in two different ways with two different resultant molecular structures. The lone pair can either go on the axis of the trigonal bipyramid (i.e. "above" the sulfur) or on the equator of the bipyramid (i.e. "beside" the sulfur). The actual molecular structure in Figure 7.4 shows clearly that the lone pair goes on the equatorial position. This can be understood if we assume that the lone pair produces a greater repulsive effect than do the bonded pairs. With this assumption, we can deduce that the lone pair should be placed in the trigonal bipyramidal arrangement as far as possible from the bonded pairs. The equatorial position does a better job of this, since only two bonding pairs of electrons are at approximately \(90^\text{o}\) away from three bonding pairs. Therefore, our Electron Domain model assumptions are consistent with the observed geometry of \(\ce{SF_4}\). Note that these assumptions also correctly predict the observed distortions away from the \(180^\text{o}\) and \(120^\text{o}\) angles which would be predicted by a trigonal bipyramidal arrangement of the five electron pairs. Using a styrofoam or rubber ball, prove to yourself that a tetrahedral arrangement provides the maximum separation of four points on the surface of the ball. Repeat this argument to find the expected arrangements for two, three, five, and six points on the surface of the ball. Explain why arranging points on the surface of a sphere can be considered equivalent to arranging electron pairs about a central atom. The valence shell electron pairs about the central atom in each of the molecules \(\ce{H_2O}\), \(\ce{NH_3}\), and \(\ce{CH_4}\) are arranged approximately in a tetrahedron. However, only \(\ce{CH_4}\) is considered a tetrahedral molecule. Explain why these statements are not inconsistent. Explain how a comparison of the geometries of \(\ce{H_2O}\) and \(\ce{CH_4}\) leads to a conclusion that lone pair electrons produce a greater repulsive effect than do bonded pairs of electrons. Give a physical reason why this might be expected. Explain why the octet of electrons about each carbon atom in ethene, \(\ce{C_2H_4}\), are not arranged even approximately in a tetrahedron. Assess the accuracy of the following reasoning and conclusions:
A trigonal bipyramid forms when there are five electron domains. If one ED is a lone pair, then the lone pair takes an equatorial position and the molecule has a seesaw geometry. If two EDs are lone pairs, we have to decide among the following options: both axial, both equatorial, or one axial and one equatorial. By placing both lone pairs in the axial positions, the lone pairs are as far apart as possible, so the trigonal planar structure is favored. Assess the accuracy of the following reasoning and conclusions:
The \(\ce{Cl-X-Cl}\) bond angles in the two molecules shown in Figure 7.5 are identical, because the bond angle is determined by the repulsion of the two \(\ce{Cl}\) atoms, which is identical in the two molecules. ; Chemistry) | 22,174 | 3,241 |
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it was mentioned that many of the properties of , , and gases could be accounted for if we assumed that substances are made of atoms or molecules which are constantly in motion. and the other have now given us much more quantitative information about gases, and it is worth asking whether with the previous model we can make quantitative predictions in agreement with these laws. In answering this question, we will also gain important insights into the nature of temperature and of heat energy. The microscopic theory of gas behavior based on molecular motion is called the . Its basic postulates are listed in Table 1: \(\Page {1}\) Postulates of the Kinetic Theory of Gases. The molecules in a gas are small and very far apart. Most of the volume which a gas occupies is empty space. Gas molecules are in constant random motion. Just as many molecules are moving in one direction as in any other. Molecules can collide with each other and with the walls of the container. Collisions with the walls account for the pressure of the gas. When collisions occur, the molecules lose no kinetic energy; that is, the collisions are said to be . The total kinetic energy of all the molecules remains constant unless there is some outside interference with the The molecules exert no attractive or repulsive forces on one another except during the process of collision. Between collisions, they move in straight lines. From them it is possible to derive the following expression for the pressure of a gas in terms of the properties of its molecules: \[P=\frac{\text{1}N}{\text{3}V}m\text{(}u^{\text{2}}\text{)}_{\text{ave}} \quad \label{1} \] Where = pressure and volume of the gas = number of molecules = mass of each molecule ( ) = (or mean) of the of all individual molecular velocities. This mean square velocity must be used because pressure is proportional to the square of molecular velocity, and molecular collisions cause different molecules to have quite different velocities. Rather than concerning ourselves with the procedure for deriving Eq. \(\ref{1}\), let us inspect the equation and see that its general features are much as we would expect. In some ways, the ability to do this with a formula is more useful than the ability to derive it. Figure \(\Page {1}\) First of all, the equation tells us that the pressure of a gas is proportional to the number of molecules divided by the volume. This is shown graphically in Figure \(\Page {1}\), where a computer has drawn the same number of gas molecules occupying each of three different volumes. The “tail” on each molecule shows the exact path followed by that molecule in the previous microsecond—the longer the tail, the faster the molecule was going. The average of the squares of the tail lengths is proportional to ( ) and is the same in all three diagrams. It is also assumed that all the molecules have equal masses. As you can see, reducing the volume of the gas increases the number of collisions per unit area on the walls of the container. Each collision exerts force on the wall; force per unit area is pressure, and so the number of collisions per unit area is proportional to pressure. Halving the volume doubles the pressure, a prediction which agrees with the experimental facts summarized in Boyle’s law. Equation \(\ref{1}\) also says that the pressure is proportional to the mass of each gas molecule. Again, this is what we would expect. Heavy molecules give a bigger “push“(the technical term for this is ) against the wall than do light ones with the same velocity. Finally, the equation tells us that pressure is proportional to the average of the squares of the molecular velocities. This dependence on the square of velocity is reasonable if we realize that doubling the velocity of a molecule has effects. First, the molecule can move farther in a given length of time, doubling the number of collisions with the walls. This would double the pressure. Second, doubling the velocity of a molecule doubles the push or impulse of each collision. This doubles the pressure again. Therefore doubling a molecule’s velocity quadruples the pressure, and for a large number of molecules, is proportional to the mean velocity. | 4,249 | 3,242 |
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These are reactions in which one atom in a molecule is replaced by another atom or group of atoms. Free radical substitution often involves breaking a carbon-hydrogen bond in alkanes such as A new bond is then formed to something else. It also happens in alkyl groups like methyl, ethyl (and so on) wherever these appear in more complicated molecules. For example, ethanoic acid is CH COOH and contains a methyl group. The carbon-hydrogen bonds in the methyl group behave just like those in methane, and can be broken and replaced by something else in the same way. A simple example of substitution is the reaction between methane and chlorine in the presence of UV light (or sunlight). \[ CH_4 + Cl_2\rightarrow CH_3Cl + HCl\] Notice that one of the hydrogen atoms in the methane has been replaced by a chlorine atom. That's substitution. Free radicals are atoms or groups of atoms which have a single unpaired electron. A free radical substitution reaction is one involving these radicals. Free radicals are formed if a bond splits evenly - each atom getting one of the two electrons. The name given to this is homolytic fission. Jim Clark ( ) | 1,157 | 3,243 |
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. In addition to these individual basis; please contact What is the relationship between mechanical work and energy? Does a person with a mass of 50 kg climbing a height of 15 m do work? Explain your answer. Does that same person do work while descending a mountain? If a person exerts a force on an immovable object, does that person do work? Explain your answer. Explain the differences between electrical energy, nuclear energy, and chemical energy. The chapter describes thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy. Which form(s) of energy are represented by each of the following? Describe the various forms of energy that are interconverted when a flashlight is switched on. Describe the forms of energy that are interconverted when the space shuttle lifts off. Categorize each of the following as representing kinetic energy or potential energy. Are the units for potential energy the same as the units for kinetic energy? Can an absolute value for potential energy be obtained? Explain your answer. Categorize each of the following as representing kinetic energy or potential energy. Why does hammering a piece of sheet metal cause the metal to heat up? 3. Technically, the person is not doing any work, since the object does not move. 11. The kinetic energy of the hammer is transferred to the metal. How much energy (in kilojoules) is released or stored when each of the following occurs? Calculate how much energy (in kilojoules) is released or stored when each of the following occurs: A car weighing 1438 kg falls off a bridge that is 211 ft high. Ignoring air resistance, how much energy is released when the car hits the water? A 1 tn roller coaster filled with passengers reaches a height of 28 m before accelerating downhill. How much energy is released when the roller coaster reaches the bottom of the hill? Assume no energy is lost due to friction. 1. 5. 250 kJ released In some simple case, when a reaction only involves solids, liquids or liquid solution, we can say that \[Zn_{(s)} + 2HCl_{(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Cl^−_{(aq)} + H_2(g)\] When 3.00 g of zinc metal is added to a dilute HCl solution at 1.00 atm and 25°C, and this reaction is allowed to go to completion at constant pressure, 6.99 kJ of heat must be removed to return the final solution to its original temperature. What are the values of q and w, and what is the change in internal energy? Calculate the amount of work done against a pressure of 1.0 atm when 4.0 mol of acetylene are allowed to react with 10 mol of O at 1.0 atm at 20°C. What is the change in internal energy for the reaction? Heat implies the flow of energy from one object to another. Describe the energy flow in an a. exothermic reaction. b. endothermic reaction. When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic. Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer. Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound? Determine whether energy is consumed or released in each scenario. Explain your reasoning. The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact. The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why. Is the distance a person travels on a trip a state function? Why or why not? Using erive a mathematical relationship between and . Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C. Using Table 9.6.1, how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K). In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change? A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 9.5.1 and solubility rules to identify the salt. Using the data in Table 9.8.2, calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)? A 3.00 g sample of TNT (trinitrotoluene, C H N O ) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the Δ of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the Δ of TNT? = × (molar mass) For Cu: = 58 J; For Al: = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: = 0.385 and 0.897 (g•K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al. 4.7 kJ Δ = −0.56 kJ/g; based on reaction with AgNO , salt contains halide; dividing Δ values mass of salts gives lithium bromide as best match, with −0.56 kJ/g. = 43.1°C; the combustion reaction is \[4C_7H_5N_3O_{6(s)} + 21O_{2(g)} \rightarrow 28CO_{2(g)} + 10H_2O_{(g)} + 6N_{2(g)}\] with \[Δ_f^οH (TNT) = −65.5\; kJ/mol\] Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b. identify each step as being exothermic or endothermic. Based on the following energy diagram, a. write an equation showing how the value of Δ could be determined if the values of Δ and Δ are known. b. identify each step as being exothermic or endothermic. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? In their elemental form, A and B exist as diatomic molecules. Given the following reactions, each with an associated Δ °, describe how you would calculate ΔH for the compound AB . \( \begin{matrix}
2AB & \rightarrow & A_{2} + B _{2} & \Delta H_{1}^{o}\\
3AB & \rightarrow & AB_{2} + A _{2}B & \Delta H_{2}^{o} \\
2A_{2}B &\rightarrow & 2A_{2} + B _{2} & \Delta H_{3}^{o}
\end{matrix} \) Methanol is used as a fuel in Indianapolis 500 race cars. Use the following table to determine whether methanol or 2,2,4-trimethylpentane (isooctane) releases more energy per liter during combustion. a. Use the enthalpies of combustion given in the following table to determine which organic compound releases the greatest amount of energy per gram during combustion. b. Calculate the standard enthalpy of formation of 1-ethyl-2-methylbenzene. Given the enthalpies of combustion, which organic compound is the best fuel per gram? 1. 2. a. To one decimal place Octane provides the largest amount of heat per gram upon combustion. b, Using , calculate for each chemical reaction. a. 2Mg(s) + O (g) → 2MgO(s) b. CaCO (s, calcite) → CaO(s) + CO (g) c. AgNO (s) + NaCl(s) → AgCl(s) + NaNO (s) Using , determine for each chemical reaction. a. 2Na(s) + Pb(NO ) (s) → 2NaNO (s) + Pb(s) b. Na CO (s) + H SO (l) → Na SO (s) + CO (g) + H O(l) c. 2KClO (s) → 2KCl(s) + 3O (g) Calculate for each chemical equation. If necessary, balance the chemical equations. a. Fe(s) + CuCl (s) → FeCl (s) + Cu(s) b. (NH ) SO (s) + Ca(OH) (s) → CaSO (s) + NH (g) + H O(l) c. Pb(s) + PbO (s) + H SO (l) → PbSO (s) + H O(l) Calculate for each reaction. If necessary, balance the chemical equations. a. 4HBr(g) + O (g) → 2H O(l) + 2Br (l) b. 2KBr(s) + H SO (l) → K SO (s) + 2HBr(g) c. 4Zn(s) + 9HNO (l) → 4Zn(NO ) (s) + NH (g) + 3H O(l) Use the data in to calculate for the reaction Sn(s, white) + 4HNO (l) → SnO (s) + 4NO (g) + 2H O(l). Use the data in to calculate for the reaction P O (s) + 6H O(l) → 4H PO (l). How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas? How much energy is released or consumed if 10.0 g of N O is completely decomposed to produce gaseous nitrogen dioxide and oxygen? In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction: Calculate ΔH for this reaction. Is the reaction exothermic or endothermic? Would you expect heat to be evolved during each reaction? How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate. a. −1203 kJ/mol O b. 179.2 kJ c. −59.3 kJ −174.1 kJ/mol −20.3 kJ −34.3 kJ/mol Cl ; exothermic Δ = −2.86 kJ CaCl : −4.6 kJ; K CO , −0.65 kJ; mixing, −0.28 kJ Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, Δ = −3581.7 kJ/mol). Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories? While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From how many grams of glucose would be needed to provide the energy to accomplish this task? A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C. | 11,499 | 3,244 |
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The compounds discussed in this chapter are all considered to be derived from carboxylic acids, and include acid halides, acid anhydrides, esters and amides (thioesters and acyl phosphates are also briefly mentioned). As you proceed through the chapter, you should be looking for similarities in behaviour among the various classes of compounds. These similarities can be readily understood once you appreciate the fact that, in most of their reactions, carboxylic acid derivatives react via the nucleophilic acyl substitution mechanism. In this chapter, we describe the nomenclature of the various types of carboxylic acid derivatives, and explain the relative reactivity of these compounds in terms of resonance contributions to the ground state of each type of compound. We describe the reactions of carboxylic acids, acid halides, acid anhydrides, esters, amides, polyamides and polyesters in detail, and discuss the biological importance of thiol esters briefly. The chapter concludes with a look at how infrared spectroscopy and NMR spectroscopy can be used in the identification of unknown carboxylic acid derivatives.
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In we point out that when an ionic compound dissolves in water, the positive and negative ions originally present in the crystal lattice persist in solution. Their ability to move nearly independently through the solution permits them to carry positive or negative electrical charges from one place to another. Hence the solution conducts an electrical current. Substances whose solutions conduct electricity are called electrolytes. All soluble ionic compounds are strong electrolytes. They conduct very well because they provide a plentiful supply of ions in solution. Some polar covalent compounds are also strong electrolytes. Common examples are HCl, HBr, HI and H SO , all of which react with H O to form large concentrations of ions. A solution of HCl, for example, conducts even better than one of NaCl having the same concentration. The effect of the concentration of ions on the electrical current flowing through a solution is illustrated in Figure \(\Page {1}\). Part of the figure shows what happens when a battery is connected through an electrical meter to two inert metal strips ( ) dipping in ethanol. Each cubic decimeter of such a solution contains 0.10 mol NaCl (that is, 0.10 mol Na and 0.10 mol Cl ). An electrical current is carried through the solution both by the Na ions moving toward the negative electrode and by the Cl ions which are attracted toward the positive electrode. The dial on the meter indicates the quantity of current. Figure 1 shows that if we replace the 0.10- NaCl solution with a 0.05- NaCl solution, the meter reading falls to about one-half its former value. Halving the concentration of NaCl halves the number of ions between the electrodes, and half as many ions can only carry half as much electrical charge. Therefore the current is half as great. Because it responds in such a direct way to the concentration of ions, conductivity of electrical current is a useful tool in the study of solutions. Conductivity measurements reveal that most covalent compounds, if they dissolve in water at all, retain their original molecular structures. Neutral molecules cannot carry electrical charges through the solution, and so no current flows. A substance whose aqueous solution conducts no better than water itself is called a . Some examples are oxygen, O , ethanol, C H OH, and sugar, C H O . Some covalent substances behave as —their solutions allow only a small current flow, but it is greater than that of the pure solvent. An example is mercury(II) chloride (seen in the Figure above). For a 100- HgCl solution the meter reading shows only about 0.2 percent as much current as for 0.10 NaCl. A crystal of HgCl consists of discrete molecules, like those shown for HgBr in Figure \(\Page {2}\). When the solid dissolves, most of these molecules remain intact, but a few dissociate into ions according to the equation \[ \underbrace{HgCl_2}_{99.8\%} \rightleftharpoons \underbrace{HgCl^+}_{0.2\%} + Cl^- \nonumber \] (The double arrows indicate that the ionization proceeds only to a limited extent and an equilibrium state is attained.) Since only 0.2 percent of the HgCl forms ions, the 0.10 solution can conduct only about 0.2 percent as much current as 0.10 NaCl. Conductivity measurements can tell us more than whether a substance is a strong, a weak, or a nonelectrolyte. Consider, for instance, the data in Table \(\Page {1}\) which shows the electrical current conducted through various aqueous solutions under identical conditions. At the rather low concentration of 0.001 , the strong electrolyte solutions conduct between 2500 and 10 000 times as much current as pure H O and about 10 times as much as the weak electrolytes HC H O (acetic acid) and NH (ammonia). Closer examination of the data for strong electrolytes reveals that some compounds which contain H or OH groups [such as HCl or Ba(OH) ] conduct unusually well. If these compounds are excluded, we find that 1:1 electrolytes (compounds which consist of equal numbers of +1 ions and –1 ions) usually conduct about half as much current as 2:2 electrolytes (+2 and -2 ions), 1:2 electrolytes (+1 and -2 ions), or 2:1 electrolytes (+2 and -1 ions). * All measurements refer to a cel1 in which the distance between the electrodes is 1.0 mm and the area of each electrode is 1.0 cm². A potential difference of 1.0 V is applied to produce the tabulated currents. There is a simple reason for this behavior. Under similar conditions, most ions move through water at comparable speeds. This means that ions like Mg or SO , which are doubly charged, will carry twice as much current through the solution as will singly charged ions like Na or Cl . Consequently, a 0.001 solution of a 2:2 electrolyte like MgSO will conduct about twice as well as a 0.001 M solution of a 1:1 electrolyte like NaCl. A similar argument applies to solutions of 1:2 and 2:1 electrolytes. A solution like 0.001 Na SO conducts about twice as well as 0.001 NaCl partly because there are twice as many Na ions available to move when a battery is connected, but also because SO ions carry twice as much charge as Cl ions when moving at the same speed. These differences in conductivity between different types of strong electrolytes can sometimes be very useful in deciding what ions are actually present in a given electrolyte solution as the following example makes clear. A second, slightly more subtle, conclusion can be drawn from the data in Table \(\Page {1}\). When an electrolyte dissolves, each type of ion makes an independent contribution to the current the solution conducts. This can be seen by comparing NaCl with KCl, and NaI with KI. In each case the compound containing K conducts about 0.2 mA more than the one containing Na . If we apply this observation to Na CO and K CO , each of which produces twice as many Na or K ions in solution, we find that the difference in current is also twice as great—about 0.4 mA. Thus conductivity measurements confirm our statement that each ion exhibits its own characteristic properties in aqueous solutions, independent of the presence of other ions. One such characteristic property is the quantity of electrical current that a given concentration of a certain type of ion can carry. At 18°C a 0.001- aqueous solution of potassium hydrogen carbonate, KHCO , conducts a current of 1.10 mA in a cell of the same design as that used to obtain the data in Table 11.1. What ions are present in solution? Referring to Table 6.2 which lists possible polyatomic ions, we can arrive at three possibilities for the ions from which KHCO is made: Since the current conducted by the solution falls in the range of 1.0 to 1.3 mA characteristic of 1:1 electrolytes, possibility is the only reasonable choice. | 6,794 | 3,248 |
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have their own classes of , different and , and various applications (photography, cancer treatment, etc), so it makes sense that they would have a naming system as well. Consisting of a metal and ligands, their formulas follow the pattern [Metal ligands] , while names are written Prefix Ligands Metal (Oxidation State). According to the , ligands are Lewis bases since they can donate electrons to the central metal atom. The metals, in turn, are Lewis acids since they accept electrons. Coordination complexes consist of a ligand and a metal center cation. The overall charge can be positive, negative, or neutral. Coordination compounds are complex or contain complex ions, for example: A ligand can be an anion or a neutral molecule that to the complex (NH , H O, Cl ). The number of ligands that attach to a metal depends on whether the ligand is monodentate or polydentate. To begin naming coordination complexes, here are some things to keep in mind. Ligands that act as anions which end in "-ide" are replaced with an ending "-o" (e.g., Chloride → Chloro). Anions ending with "-ite" and "-ate" are replaced with endings "-ito" and "-ato" respectively (e.g., Nitrite → Nitrito, Nitrate → Nitrato). Most neutral molecules that are ligands carry their normal name. The few exceptions are the first four on the chart: ammine, aqua, carbonyl, and nitrosyl. The number of ligands present in the complex is indicated with the prefixes di, tri, etc. The exceptions are polydentates that have a prefix already in their name are the most common). When indicating how many of these are present in a coordination complex, put the ligand's name in parentheses and use (for two ligands), (for three ligands), and (for four ligands). Prefixes always go before the ligand name; they are taken into account when putting ligands in alphabetical order. Note that "mono" often is not used. For example, \(\ce{[FeCl(CO)2(NH3)3]^{2+}}\) would be called triamminedicarbonylchloroiron(III) ion. Remember that ligands are always , before the metal is. What is the name of this complex ion: \(\ce{[CrCl2(H2O)4]^{+}}\)? Let's start by identifying the ligands. The ligands here are Cl and H O. Therefore, we will use the monodentate ligand names of "chloro" and "aqua". Alphabetically, aqua comes before chloro, so this will be their order in the complex's name. There are 4 aqua's and 2 chloro's, so we will add the number prefixes before the names. Since both are monodentate ligands, we will say "tetra[aqua]di[chloro]". Now that the ligands are named, we will name the metal itself. The metal is Cr, which is chromium. Therefore, this coordination complex is called tetraaquadichlorochromium(III) ion. See the next section for an explanation of the (III). What is the name of this complex ion: \(\ce{[CoCl_2(en)_2]^{+}}\)? We take the same approach. There are two chloro and ethylenediamine ligands. The metal is Co, cobalt. We follow the same steps, except that \(en\) is a polydentate ligand with a prefix in its name (ethylene amine), so "bis" is used instead of "di", and parentheses are added. Therefore, this coordination complex is called dichlorobis(ethylenediamine)cobalt(III) ion. When naming the metal center, you must know the formal metal name and the oxidation state. To show the oxidation state, we use Roman numerals inside parenthesis. For example, in the problems above, chromium and cobalt have the oxidation state of +3, so that is why they have (III) after them. Copper, with an oxidation state of +2, is denoted as copper(II). If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. Copper +2 will change into cuprate(II). The following change to their Latin names when part of an anion complex: The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present). Finally, when a complex has an overall charge, "ion" is written after it. This is not necessary if it is neutral or part of a coordination compound (Example \(\Page {3}\)). Here are some examples with determining oxidation states, naming a metal in an anion complex, and naming coordination compounds. What is the name of [Cr(OH) ] ? Immediately we know that this complex is an anion. There is only one monodentate ligand, hydroxide. There are four of them, so we will use the name "tetrahydroxo". The metal is chromium, but since the complex is an anion, we will have to use the "-ate" ending, yielding "chromate". The oxidation state of the metal is 3 (x+(-1)4=-1). Write this with Roman numerals and parentheses (III) and place it after the metal to get tetrahydroxochromate(III) ion. What is the name of \(\ce{[CuCl4]^{2-}}\)? tetrachlorocuprate(II) ion A last little side note: when naming a coordination compound, it is important that you name the cation first, then the anion. You base this on the charge of the ligand. Think of NaCl. Na, the positive cation, comes first and Cl, the negative anion, follows. What is the name of \([\ce{Pt(NH3)4},\ce{Pt(Cl)4}]\)? NH is neutral, making the first complex positively charged overall. Cl has a -1 charge, making the second complex the anion. Therefore, you will write the complex with NH first, followed by the one with Cl (the same order as the formula). This coordination compound is called tetraammineplatinum(II) tetrachloroplatinate(II). What is the name of \(\ce{[CoCl(NO2)(NH3)4]^{+}}\) ? This coordination complex is called tetraamminechloronitrito-N-cobalt(III). N comes before the O in the symbol for the nitrite ligand, so it is called nitrito-N. If an O came first, as in [CoCl(ONO)(NH ) ] , the ligand would be called nitrito-O, yielding the name tetraamminechloronitrito-O-cobalt(III). Nitro (for NO ) and nitrito (for ONO) can also be used to describe the nitrite ligand, yielding the names tetraamminechloronitrocobalt(III) and tetraamminechloronitritocobalt(III). While chemistry typically follow the nomenclature rules for naming complexes and compounds, there is disagreement with the rules for constructing formulas of inorganic complex. The order of ligand names in their formula has been ambiguous with different conventions being used (charged vs neutral, number of each ligand, etc.). In 2005, adopted the recommendation that all ligand names in formulas be listed (in the same way as in the naming convention) irrespective of the charge or number of each ligand type. However, this rule is not adhered to in many chemistry laboratories. For practice, the order of the ligands in chemical formulas does not matter as long as you write the transition metal first, which is the stance taken here. Write the chemical formulas for: Write the chemical formulas for | 6,882 | 3,249 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomerism_in_Disubstituted_Cyclohexanes |
The distinction between configurational stereoisomers and the conformers they may assume is well-illustrated by the disubstituted cyclohexanes. The following discussion uses the various isomers of dichlorocyclohexane as examples. The 1,1-dichloro isomer is omitted because it is an unexceptional of the others, and has no centers of chirality (asymmetric carbon atoms). The 1,2- and 1,3-dichlorocyclohexanes each have two centers of chirality, bearing the same set of substituents. The cis & trans-1,4-dichlorocyclohexanes do not have any chiral centers, since the two ring groups on the substituted carbons are identical. There are three of 1,2-dichlorocyclohexane and three of 1,3-dichlorocyclohexane. These are shown in the following table. All the 1,2-dichloro isomers are isomers of the 1,3-dichloro isomers. In each category (1,2- & 1,3-), the ( )-trans isomer and the ( )-trans isomer are enantiomers. The cis isomer is a diastereomer of the trans isomers. Finally, all of these isomers may exist as a mixture of two (or more) conformational isomers, as shown in the table. The chair conformer of the cis 1,2-dichloro isomer is chiral. It exists as a 50:50 mixture of enantiomeric conformations, which interconvert so rapidly they cannot be resolved (ie. separated). Since the cis isomer has two centers of chirality (asymmetric carbons) and is optically inactive, it is a meso-compound. The corresponding trans isomers also exist as rapidly interconverting chiral conformations. The diequatorial conformer predominates in each case, the ( ) conformations being mirror images of the ( ) conformations. All these conformations are diastereomeric with the cis conformations. The diequatorial chair conformer of the cis 1,3-dichloro isomer is achiral. It is the major component of a fast equilibrium with the diaxial conformer, which is also achiral. This isomer is also a meso compound. The corresponding trans isomers also undergo a rapid conformational interconversion. For these isomers, however, this interconversion produces an identical conformer, so each enantiomer ( ) and ( ) has predominately a single chiral conformation. These enantiomeric conformations are diastereomeric with the cis conformations. The 1,4-dichlorocyclohexanes may exist as cis or trans stereoisomers. Both are achiral, since the disubstituted six-membered ring has a plane of symmetry. These isomers are diastereomers of each other, and are constitutional isomers of the 1,2- and 1,3- isomers. All the chair conformers of these isomers are achiral, and the diequatorial conformer of the trans isomer is the predominate species at equilibrium. | 2,649 | 3,250 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/08%3A_Gravimetric_Methods/8.02%3A_Precipitation_Gravimetry |
In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,251 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,254 |
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Using the rule "like dissolves like" with the formation of ionic solutions, we must assess first assess two things: 1) the strength of the ion-dipole forces of attraction between water and the ionic compound and 2) the strength of the interionic bond of the ionic compound. For an ionic compound to form a solution, the ion-dipole forces between water and ionic compound must be greater than the interionic bonds. Therefore, to form a compound: ion-dipole forces > interionic bonds When the ionic compound is surrounded by water, the water dipoles surround the crystal's clustered structure. The water's negative ends of the dipole will be attracted to the positive dipoles of the ion and the positive ends of the water's dipole will be attracted to the negative dipoles of the ion. If the force of this attraction is stronger than the interionic bonds, the crystal's interionic bonds will be broken, then surrounded by the water molecules or hydrated . There is a 3-step process that we can use to approach the energy involved in ionic solution formation. 1) Breaking apart the ionic compound is endothermic and requires energy. 2) Hydrating cation is exothermic and therefore releases energy. 3) Hydrating the anion is exothermic and also releases energy. The sum of these 3 steps will then give us the enthalpy of the solution. 1) CaCl (s) -> Ca (g) + Cl (g) energy > 0 2) Ca (g) > Ca (aq) energy < 0 3) Cl (g) > Cl (aq) energy < 0 CaCl (s) > Ca (aq) + Cl (aq) energy > 0 The dissolution is endothermic because in the formation of ionic solutions, you must take into account entropy in addition to the enthalpy of the solution to determine whether it will occur spontaneously. | 1,705 | 3,256 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidizing_and_Reducing_Agents |
Oxidizing and reducing agents are key terms used in describing the reactants in redox reactions that transfer electrons between reactants to form products. This page discusses what defines an oxidizing or reducing agent, how to determine an oxidizing and reducing agent in a chemical reaction, and the importance of this concept in real world applications. An , or , electrons and is reduced in a chemical reaction. Also known as the electron acceptor, the oxidizing agent is normally in one of its higher possible oxidation states because it will gain electrons and be reduced. Examples of oxidizing agents include halogens, potassium nitrate, and nitric acid. A or , electrons and is oxidized in a chemical reaction. A reducing agent is typically in one of its lower possible oxidation states, and is known as the electron donor. A reducing agent is oxidized, because it loses electrons in the redox reaction. Examples of reducing agents include the earth metals, formic acid, and sulfite compounds. To help eliminate confusion, there is a mnemonic device to help determine oxidizing and reducing agents. Identify the reducing and oxidizing agents in the balanced redox reaction: Identify the oxidizing agent and the reducing agent in the following redox reaction: \[\ce{MnO4^{-} + SO3^{2-} -> Mn^{+2} + SO4^{2-}}\nonumber\] \(\ce{SO3^{2-}}\) is the reducing agent and \(\ce{MnO4^{-}}\) is the oxidizing agent. Note that while a specific atom typically has an odization state changes, the agents are the actual species, not the atoms. Oxidizing and reducing agents are important in industrial applications. They are used in processes such as purifying water, bleaching fabrics, and storing energy (such as in batteries and gasoline). Oxidizing and reducing agents are especially crucial in biological processes such as metabolism and photosynthesis. For example, organisms use electron acceptors such as NAD to harvest energy from redox reactions as in the hydrolysis of glucose: \[C_6H_{12}O_6 + 2ADP + 2P + 2NAD^+ \rightarrow 2CH_3COCO_2H + 2ATP + 2NADH \nonumber\] All combustion reactions are also examples of redox reactions. A combustion reaction occurs when a substance reacts with oxygen to create heat. One example is the combustion of octane, the principle component of gasoline: \[2 C_8H_{18} (l) + 25 O_2 (g) \rightarrow 16 CO_2 (g) + 18 H_2O (g) \nonumber\] Combustion reactions are a major source of energy for modern industry. By looking at each element's oxidation state on the reactant side of a chemical equation compared with the same element's oxidation state on the product side, one can determine if the element is reduced or oxidized, and can therefore identify the oxidizing and reducing agents of a chemical reaction. \(NO_3^-\), \(NO\), \(N_2H_4\), \(NH_3\) | 2,812 | 3,259 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/05%3A_Alkenes_and_Alkynes |
When a carbon is bonded to one or more electronegative atoms, it takes on a partial positive charge and it is . Such electrophilic carbons can undergo or reactions, or both, depending upon the structures of the reacting molecules, the strength of the nucleophile, and the type of solvent in which the reaction occurs. Now, we turn to reactions that electron-rich carbon species can undergo. Both alkenes and alkynes are “unsaturated,” which means that they contain double or triple carbon-carbon bonds. The term unsaturated comes from the fact that more \(\mathrm{H}\) atoms can be added to these molecules across the double or triple bonds. A simple alkene contains a pair of carbons linked by a double bond; this double bond consists of a sigma bond and a pi bond. The sigma bond is formed by end-to-end overlap of \(\mathrm{sp}^{2}\) hybrid orbitals, and the pi bond by side-to-side overlap of the p orbitals. A pi bond has two lobes of electron density above and below the plane of the molecule. There are a number of consequences to this arrangement: Rotation around a double bond requires breaking the overlap of the pi bond and its subsequent reformation. As with all bond-breaking phenomena, the bond-breaking step requires energy; in fact, significantly more energy than is required to bring about rotation around a single bond where no bond-breaking occurs. As we will see, these three factors have a marked effect on the behavior of alkenes. Alkynes are compounds that contain triple bonds. The triple bond consists of one sigma bond formed from end-to-end overlap of sp-hybrid orbitals and two pi bonds formed from side to side overlap. The carbons are \(\mathrm{sp}\)-hybridized and the molecule is linear in the region of the triple bond; again rotation around a triple bond is constrained—two pi bonds must be broken for it to occur (which requires an input of energy). This bonding arrangement results in a very electron rich \(\mathrm{C-C}\) region with the sigma bond inside what looks like a cylinder of pi electron density. Since alkenes have restricted rotation around the \(\mathrm{C=C}\) group, they can exist as stereoisomers. For example, in 2-butene there is a methyl and an \(\mathrm{H}\) bonded to each of the double-bonded carbons (carbons 2 and 3 of the molecule). Because the \(\mathrm{C=C}\) group is planar, the \(\mathrm{CH}_{3}\) groups can be on either the same (“cis”) or opposite (“trans”) sides of the double bond (\(\rightarrow\)); this cis/trans nomenclature is similar to that we used with cyclohexane rings. As the groups attached to each carbon get more complex, such nomenclature quickly becomes confusing. To cope, we turn to another established naming scheme; in this case, the Cahn-Ingold-Prelog convention we previously used with chiral centers. This involves ranking the groups linked to each double-bond carbon. If the high groups are together (same side), the name is prefixed by Z (from the German word for together: zusammen). If they are on opposite sides, they are labeled E (entgegen; away). E and Z isomers are diastereoisomers: they have the same connectivity but neither can be superimposed on its mirror image. In E-3-bromo-2-pentene, the \(\mathrm{CH}_{3}\) and \(\mathrm{CH}_{2} \mathrm{CH}_{3}\) groups are closer to one another than they are in Z-3-bromo-2-pentene; the result is that they have different physical and chemical properties. These differences make it possible to separate E and Z isomers (and cis/trans since they are just a special case of E/Z) from one another. Elimination reactions that produce alkenes tend to favor the most substituted alkene as the major product. The relative stabilities of various alkenes can be determined by reacting the alkene with hydrogen and determining the enthalpy change (\(\Delta \mathrm{H}\)). For example, shown (\(\rightarrow\)), the three different alkenes produce the same product, and therefore the differences in the energy released must arise from the fact that the initial alkenes have different energies. The more alkyl groups attached to the double bond, the more stable (less reactive) the alkene is, and therefore a lower amount of energy is released. Molecular stability in alkenes is attributed to the same causes as the relative stabilities of carbocations; alkyl groups stabilize the pi bond by hyperconjugation and induction.
| 4,378 | 3,260 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./25.E%3A_Organic_and_Biological_Chemistry_(Exercises) |
. In addition to these individual basis; please contact Classify each compound as organic or inorganic. Which compound is likely organic and which is likely inorganic? Classify each compound as organic or inorganic. Classify each compound as organic or inorganic. Which member of each pair has a higher melting point? Which member of each pair has a higher melting point? What is the functional group of an alkene? An alkyne? Does CH CH CH CH CH CH CH CH CH CH CH have a functional group? Explain. carbon-to-carbon double bond; carbon-to-carbon triple bond No; it has nothing but carbon and hydrogen atoms and all single bonds. What is the functional group of 1-butanol (CH CH CH CH OH)? What is the functional group of butyl bromide, CH CH CH CH Br? OH | 795 | 3,261 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation-Reduction_Reactions |
An oxidation-reduction (redox) reaction is a type of chemical reaction that involves a transfer of electrons between two species. An oxidation-reduction reaction is any chemical reaction in which the oxidation number of a molecule, atom, or ion changes by gaining or losing an electron. Redox reactions are common and vital to some of the basic functions of life, including photosynthesis, respiration, combustion, and corrosion or rusting. The oxidation state (OS) of an element corresponds to the number of electrons, e , that an atom loses, gains, or appears to use when joining with other atoms in compounds. In determining the oxidation state of an atom, there are seven guidelines to follow: The of the oxidation states is equal to zero for neutral compounds and equal to the for polyatomic ion species. Determine the Oxidation States of each element in the following reactions: Determine the oxidation states of the phosphorus atom bold element in each of the following species: Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each): An atom is oxidized if its oxidation number increases, the reducing agent, and an atom is reduced if its oxidation number decreases, the oxidizing agent. The atom that is oxidized is the reducing agent, and the atom that is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound). Redox reactions are comprised of two parts, a reduced half and an oxidized half, that occur together. The reduced half gains electrons and the oxidation number decreases, while the oxidized half loses electrons and the oxidation number increases. Simple ways to remember this include the mnemonic devices meaning " " and " " There is no net change in the number of electrons in a redox reaction. Those given off in the oxidation half reaction are taken up by another species in the reduction half reaction. The two species that exchange electrons in a redox reaction are given special names: Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent. (Note: the oxidizing and reducing agents can be the same element or compound, as in disproportionation reactions discussed below). A good example of a redox reaction is the in which iron atoms in ferric oxide lose (or give up) \(\ce{O}\) atoms to \(\ce{Al}\) atoms, producing \(\ce{Al2O3}\). \[\ce{Fe2O3(s) + 2Al(s) \rightarrow Al2O3(s) + 2Fe(l)} \nonumber \] Determine what is the oxidizing and reducing agents in the following reaction. \[\ce{Zn + 2H^{+} -> Zn^{2+} + H2} \nonumber \] The oxidation state of \(\ce{H}\) changes from +1 to 0, and the oxidation state of \(\ce{Zn}\) changes from 0 to +2. Hence, \(\ce{Zn}\) is oxidized and acts as the . \(\ce{H^{+}}\) ion is reduced and acts as the . reactions are among the simplest redox reactions and, as the name suggests, involves "combining" elements to form a chemical compound. As usual, oxidation and reduction occur together. The general equation for a combination reaction is given below: \[\ce{ A + B -> AB} \nonumber \] Consider the combination reaction of hydrogen and oxygen \[\ce{H2 + O2 -> H2O } \nonumber \] 0 + 0 → In this reaction both H and O are free elements; following , their oxidation states are 0. The product is H O, which has a total oxidation state of 0. According to , the oxidation state of oxygen is usually -2. Therefore, the oxidation state of H in H O must be +1. A reaction is the reverse of a combination reaction, the breakdown of a chemical compound into individual elements: \[\ce{AB -> A + B} \nonumber \] Consider the following reaction: \[\ce{H2O -> H2 + O2}\nonumber \] This follows the definition of the decomposition reaction, where water is "decomposed" into hydrogen and oxygen. (2)(+1) + (-2) = 0 → 0 + 0 As in the previous example the \(\ce{H2O}\) has a total oxidation state of 0; thus, according to Rule #6 the oxidation state of oxygen is usually -2, so the oxidation state of hydrogen in \(\ce{H2O}\) must be +1. Note that the autoionization reaction of water is a redox nor decomposition reaction since the oxidation states do not change for any element: \[\ce{H2O -> H^{+} + OH^{-}}\nonumber \] A reaction involves the "replacing" of an element in the reactants with another element in the products: \[\ce{A + BC -> AB + C} \nonumber \] Equation: \[\ce{Cl_2 + Na\underline{Br} \rightarrow Na\underline{Cl} + Br_2 } \nonumber \] Calculation: (0) + ((+1) + (-1) = 0) -> ((+1) + (-1) = 0) + 0 In this equation, \(\ce{Br}\) is replaced with \(\ce{Cl}\), and the \(\ce{Cl}\) atoms in \(\ce{Cl2}\) are reduced, while the \(\ce{Br}\) ion in \(\ce{NaBr}\) is oxidized. A double replacement reaction is similar to a single replacement reaction, but involves "replacing" two elements in the reactants, with two in the products: \[\ce{AB + CD -> AD + CB} \nonumber \] An example of a double replacement reaction is the reaction of magnesium sulfate with sodium oxalate \[\ce{MgSO4(aq) + Na2C2O4(aq) -> MgC2O4(s) + Na2SO4(aq)} \nonumber \] Combustion is the formal terms for "burning" and typically involves a substance reacts with oxygen to transfer energy to the surroundings as light and heat. Hence, combustion reactions are almost always exothermic. For example, internal combustion engines rely on the combustion of organic hydrocarbons \(\ce{C_{x}H_{y}}\) to generate \(\ce{CO2}\) and \(\ce{H2O}\): \[\ce{C_{x}H_{y} + O2 -> CO2 + H2O}\nonumber \] Although combustion reactions typically involve redox reactions with a chemical being oxidized by oxygen, many chemicals can "burn" in other environments. For example, both titanium and magnesium metals can burn in nitrogen as well: \[\ce{ 2Ti(s) + N2(g) -> 2TiN(s)} \nonumber \] \[\ce{ 3 Mg(s) + N2(g) -> Mg3N2(s)} \nonumber \] Moreover, chemicals can be oxidized by other chemicals than oxygen, such as \(\ce{Cl2}\) or \(\ce{F2}\); these processes are also considered combustion reactions. Which of the following are combustion reactions? Both reaction b and reaction d are combustion reactions, although with different oxidizing agents. Reaction b is the conventional combustion reaction using \(\ce{O2}\) and reaction uses \(\ce{N2}\) instead. In disproportionation reactions, a single substance can be both oxidized and reduced. These are known as disproportionation reactions, with the following general equation: \[\ce{2A -> A^{+n} + A^{-n}} \nonumber \] where \(n\) is the number of electrons transferred. Disproportionation reactions do not need begin with neutral molecules, and can involve more than two species with differing oxidation states (but rarely). Disproportionation reactions have some practical significance in everyday life, including the reaction of hydrogen peroxide, \(\ce{H2O2}\) poured over a cut. This a decomposition reaction of hydrogen peroxide, which produces oxygen and water. Oxygen is present in all parts of the chemical equation and as a result it is both oxidized and reduced. The reaction is as follows: \[\ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \nonumber \] On the reactant side, \(\ce{H}\) has an oxidation state of +1 and \(\ce{O}\) has an oxidation state of -1, which changes to -2 for the product \(\ce{H2O}\) (oxygen is reduced), and 0 in the product \(\ce{O2}\) (oxygen is oxidized). Which element undergoes a bifurcation of oxidation states in this disproportionation reaction: \[\ce{HNO2 -> HNO3 + NO + H2O} \nonumber\] The \(\ce{N}\) atom undergoes disproportionation. You can confirm that by identifying the oxidation states of each atom in each species. | 7,637 | 3,262 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Nonstandard_Conditions%3A_The_Nernst_Equation |
The refer to cells in which all dissolved substances are at unit , which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know. Suppose, for example, that we reduce the concentration of \(Zn^{2+}\) in the \(Zn/Cu\) cell from its standard effective value of 1 M to an to a much smaller value: Zn | Zn || Cu | Cu This will reduce the value of (Q\) for the cell reaction Zn + Cu → Zn + Cu thus making it more spontaneous, or "driving it to the right" as the would predict, and making its free energy change \(\Delta G\) more negative than \(\Delta G°\), so that would be more than . The relation between the actual cell potential E and the standard potential E° is developed in the following way. We begin with the which relates the standard free energy change (for the complete conversion of products into reactants) to the standard potential \[\Delta G° = –nFE° \] By analogy we can write the more general equation \[\Delta G = –nFE\] which expresses the change in free energy for any extent of reaction— that is, for any value of the \(Q\). We now substitute these into the expression that relates \(\Delta G\) and \(\Delta G°\) which you will recall from the chapter on chemical equilibrium: \[\Delta G = \Delta G° + RT \ln Q\] which gives \[–nFE = –nFE° + RT \ln Q \] which can be rearranged to \[ E=E° -\dfrac{RT}{nF} \ln Q \label{1}\] This is the that relates the cell potential to the standard potential and to the activities of the electroactive species. Notice that the cell potential will be the same as \(E°\) only if \(Q\) is unity. The Nernst equation is more commonly written in base-10 log form and for 25 °C: \[ E=E° -\dfrac{0.059}{n} \log_{10} Q \label{2}\] The equation above indicates that the electrical potential of a cell depends upon the reaction quotient \(Q\) of the reaction. As the redox reaction proceeds, reactants are consumed, thus concentration of reactants decreases. Conversely, the products concentration increases due to the increased in products formation. As this happens, cell potential gradually decreases until the reaction is at , at which \(\Delta{G} = 0\). The Nernst equation tells us that a half-cell potential will change by 59 millivolts per 10-fold change in the concentration of a substance involved in a one-electron oxidation or reduction; for two-electron processes, the variation will be 28 millivolts per decade concentration change. Thus for the dissolution of metallic copper Cu → Cu + 2 the potential = (– 0.337) – .0295 log [Cu ] becomes more positive (the reaction has a greater tendency to take place) as the cupric ion concentration decreases. This, of course, is exactly what the predicts; the more dilute the product, the greater the extent of the reaction. Consider the Zn-Cu redox reaction: \[Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} E^{o}_{cell} = +1.10 \; V\] Initially, [Cu ] = [Zn ] = 1.0 M at standard T = 298K As the reaction proceeds, [Cu ] decreases as [Zn ] increases. Lets say after one minute, [Cu ] = 0.05 M while [Zn ] = 5.0 M. According to Nernst, cell potential after 1 minute is: \[E = E^o - \dfrac{0.0592 V}{n} \log Q\] \[E = 1.10V - \dfrac{0.0592 V}{2} \log\dfrac{5.0 \; M}{.05 \; M}\] \[E = 1.04 \; V\] As you can see, the initial cell potential is \(E = 1.10\, V\), after 1 minute, the potential drops to 1.04 V. As the reaction continues to progress, more Cu will be consumed and more Zn will be generated. As a result, the cell potential continues to decrease and when the cell potential drops down to 0, the concentration of reactants and products stops changing. This is when the reaction is at equilibrium. At equilibrium, the reaction quotient \(Q = K_{eq}\). Also, at equilibrium, \(\Delta{G} = 0\) and \(\Delta{G} = -nFE\), so \(E = 0\). Therefore, substituting \(Q = K_{eq}\) and \(E = 0\) into the Nernst equation, we have: \[0 = E^o - \dfrac{RT}{nF} \ln K_{eq}\] At standard conditions, the equation above simplifies into: \[0 = E^o - \dfrac{0.0592}{n} \log K_{eq}\] This equation can be rearranged into: \[log K_{eq} = \dfrac{nE^o}{0.0592}\] The equation above indicates that the equilibrium constant of the reaction. Specifically, when: This result fits , which states that when a system at equilibrium experiences a change, the system will minimize that change by shifting the equilibrium in the opposite direction. | 4,650 | 3,263 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Chromatography/Gas_Chromatography |
Gas chromatography is a term used to describe the group of analytical separation techniques used to analyze volatile substances in the gas phase. In gas chromatography, the components of a sample are dissolved in a solvent and vaporized in order to separate the analytes by distributing the sample between two phases: a stationary phase and a mobile phase. The mobile phase is a chemically inert gas that serves to carry the molecules of the analyte through the heated column. Gas chromatography is one of the sole forms of chromatography that does not utilize the mobile phase for interacting with the analyte. The stationary phase is either a solid adsorbant, termed gas-solid chromatography (GSC), or a liquid on an inert support, termed gas-liquid chromatography (GLC). In early 1900s, Gas chromatography (GC) was discovered by Mikhail Semenovich Tsvett as a separation technique to separate compounds. In organic chemistry, liquid-solid column chromatography is often used to separate organic compounds in solution. Among the various types of gas chromatography, gas-liquid chromatography is the method most commonly used to separate organic compounds. The combination of gas chromatography and mass spectrometry is an invaluable tool in the identification of molecules. A typical gas chromatograph consists of an injection port, a column, carrier gas flow control equipment, ovens and heaters for maintaining temperatures of the injection port and the column, an integrator chart recorder and a detector. To separate the compounds in gas-liquid chromatography, a solution sample that contains organic compounds of interest is injected into the sample port where it will be vaporized. The vaporized samples that are injected are then carried by an inert gas, which is often used by helium or nitrogen. This inert gas goes through a glass column packed with silica that is coated with a liquid. Materials that are less soluble in the liquid will increase the result faster than the material with greater solubility.The purpose of this module is to provide a better understanding on its separation and measurement techniques and its application. In GLC, the liquid stationary phase is adsorbed onto a solid inert packing or immobilized on the capillary tubing walls. The column is considered packed if the glass or metal column tubing is packed with small spherical inert supports. The liquid phase adsorbs onto the surface of these beads in a thin layer. In a capillary column, the tubing walls are coated with the stationary phase or an adsorbant layer, which is capable of supporting the liquid phase. However, the method of GSC, has limited application in the laboratory and is rarely used due to severe peak tailing and the semi-permanent retention of polar compounds within the column. Therefore, the method of gas-liquid chromatography is simply shortened to gas chromatography and will be referred to as such here.The purpose of this module is to provide a better understanding on its separation and measurement techniques and its application. A sample port is necessary for introducing the sample at the head of the column. Modern injection techniques often employ the use of heated sample ports through which the sample can be injected and vaporized in a near simultaneous fashion. A calibrated microsyringe is used to deliver a sample volume in the range of a few microliters through a rubber septum and into the vaporization chamber. Most separations require only a small fraction of the initial sample volume and a sample splitter is used to direct excess sample to waste. Commercial gas chromatographs often allow for both split and splitless injections when alternating between packed columns and capillary columns. The vaporization chamber is typically heated 50 °C above the lowest boiling point of the sample and subsequently mixed with the carrier gas to transport the sample into the column. The carrier gas plays an important role, and varies in the GC used. Carrier gas must be dry, free of oxygen and chemically inert mobile-phase employed in gas chromatography. Helium is most commonly used because it is safer than, but comprable to hydrogen in efficiency, has a larger range of flow rates and is compatible with many detectors. Nitrogen, argon, and hydrogen are also used depending upon the desired performance and the detector being used. Both hydrogen and helium, which are commonly used on most traditional detectors such as Flame Ionization(FID), thermal conductivity (TCD) and Electron capture (ECD), provide a shorter analysis time and lower elution temperatures of the sample due to higher flow rates and low molecular weight. For instance, hydrogen or helium as the carrier gas gives the highest sensitivity with TCD because the difference in thermal conductivity between the organic vapor and hydrogen/helium is greater than other carrier gas. Other detectors such as mass spectroscopy, uses nitrogen or argon which has a much better advantage than hydrogen or helium due to their higher molecular weights, in which improve vacuum pump efficiency. All carrier gasses are available in pressurized tanks and pressure regulators, gages and flow meters are used to meticulously control the flow rate of the gas. Most gas supplies used should fall between 99.995% - 99.9995% purity range and contain a low levels (< 0.5 ppm) of oxygen and total hydrocarbons in the tank. The carrier gas system contains a molecular sieve to remove water and other impurities. Traps are another option to keep the system pure and optimum sensitive and removal traces of water and other contaminants. A two stage pressure regulation is required to use to minimize the pressure surges and to monitor the flow rate of the gas. To monitor the flow rate of the gas a flow or pressure regulator was also require onto both tank and chromatograph gas inlet. This applies different gas type will use different type of regulator.The carrier gas is preheated and filtered with a molecular sieve to remove impurities and water prior to being introduced to the vaporization chamber. A carrier gas is typically required in GC system to flow through the injector and push the gaseous components of the sample onto the GC column, which leads to the detector ( see more detail in detector section). The thermostatted oven serves to control the temperature of the column within a few tenths of a degree to conduct precise work. The oven can be operated in two manners: isothermal programming or temperature programming. In isothermal programming, the temperature of the column is held constant throughout the entire separation. The optimum column temperature for isothermal operation is about the middle point of the boiling range of the sample. However, isothermal programming works best only if the boiling point range of the sample is narrow. If a low isothermal column temperature is used with a wide boiling point range, the low boiling fractions are well resolved but the high boiling fractions are slow to elute with extensive band broadening. If the temperature is increased closer to the boiling points of the higher boiling components, the higher boiling components elute as sharp peaks but the lower boiling components elute so quickly there is no separation. In the temperature programming method, the column temperature is either increased continuously or in steps as the separation progresses. This method is well suited to separating a mixture with a broad boiling point range. The analysis begins at a low temperature to resolve the low boiling components and increases during the separation to resolve the less volatile, high boiling components of the sample. Rates of 5-7 °C/minute are typical for temperature programming separations. Open tubular columns, which are also known as capillary columns, come in two basic forms. The first is a wall-coated open tubular (WCOT) column and the second type is a support-coated open tubular (SCOT) column. WCOT columns are capillary tubes that have a thin later of the stationary phase coated along the column walls. In SCOT columns, the column walls are first coated with a thin layer (about 30 micrometers thick) of adsorbant solid, such as diatomaceous earth, a material which consists of single-celled, sea-plant skeletons. The adsorbant solid is then treated with the liquid stationary phase. While SCOT columns are capable of holding a greater volume of stationary phase than a WCOT column due to its greater sample capacity, WCOT columns still have greater column efficiencies. Most modern WCOT columns are made of glass, but T316 stainless steel, aluminum, copper and plastics have also been used. Each material has its own relative merits depending upon the application. Glass WCOT columns have the distinct advantage of chemical etching, which is usually achieved by gaseous or concentrated hydrochloric acid treatment. The etching process gives the glass a rough surface and allows the bonded stationary phase to adhere more tightly to the column surface. One of the most popular types of capillary columns is a special WCOT column called the fused-silica wall-coated (FSWC) open tubular column. The walls of the fused-silica columns are drawn from purified silica containing minimal metal oxides. These columns are much thinner than glass columns, with diameters as small as 0.1 mm and lengths as long as 100 m. To protect the column, a polyimide coating is applied to the outside of the tubing and bent into coils to fit inside the thermostatted oven of the gas chromatography unit. The FSWC columns are commercially available and currently replacing older columns due to increased chemical inertness, greater column efficiency and smaller sampling size requirements. It is possible to achieve up to 400,000 theoretical plates with a 100 m WCOT column, yet the world record for the largest number of theoretical plates is over 2 million plates for 1.3 km section of column. Packed columns are made of a glass or a metal tubing which is densely packed with a solid support like diatomaceous earth. Due to the difficulty of packing the tubing uniformly, these types of columns have a larger diameter than open tubular columns and have a limited range of length. As a result, packed columns can only achieve about 50% of the efficiency of a comparable WCOT column. Furthermore, the diatomaceous earth packing is deactivated over time due to the semi-permanent adsorption of impurities within the column. In contrast, FSWC open tubular columns are manufactured to be virtually free of these adsorption problems. Different types of columns can be applied for different fields. Depending on the type of sample, some GC columns are better than the others. For example, the FSWC column shown in Figure 5 is designed specially for blood alcohol analysis. It produces fast run times with baseline resolution of key components in under 3 minutes. Moreover, it displays enhanced resolutions of ethanol and acetone peaks, which helps with determining the BAC levels. This particular column is known as Zebron-BAC and it made with polyimide coating on the outside and the inner layer is made of fused silica and the inner diameter ranges from .18 mm to .25 mm. There are also many other Zebron brand columns designed for other purposes. Another example of a Zebron GC column is known as the Zebron-inferno. Its outer layer is coated with a special type of polyimide that is designed to withstand high temperatures. As shown in figure 6, it contains an extra layer inside. It can withstand up to 430 °C to be exact and it is designed to provide true boiling point separation of hydrocarbons distillation methods. Moreover, it is also used for acidic and basic samples. The detector is the device located at the end of the column which provides a quantitative measurement of the components of the mixture as they elute in combination with the carrier gas. In theory, any property of the gaseous mixture that is different from the carrier gas can be used as a detection method. These detection properties fall into two categories: bulk properties and specific properties. Bulk properties, which are also known as general properties, are properties that both the carrier gas and analyte possess but to different degrees. Specific properties, such as detectors that measure nitrogen-phosphorous content, have limited applications but compensate for this by their increased sensitivity. Each detector has two main parts that when used together they serve as transducers to convert the detected property changes into an electrical signal that is recorded as a chromatogram. The first part of the detector is the sensor which is placed as close the the column exit as possible in order to optimize detection. The second is the electronic equipment used to digitize the analog signal so that a computer may analyze the acquired chromatogram. The sooner the analog signal is converted into a digital signal, the greater the signal-to-noise ratio becomes, as analog signal are easily susceptible to many types of interferences. An ideal GC detector is distinguished by several characteristics. The first requirement is adequate sensitivity to provide a high resolution signal for all components in the mixture. This is clearly an idealized statement as such a sample would approach zero volume and the detector would need infinite sensitivity to detect it. In modern instruments, the sensitivities of the detectors are in the range of 10 to 10 g of solute per second. Furthermore, the quantity of sample must be reproducible and many columns will distort peaks if enough sample is not injected. An ideal column will also be chemically inert and and should not alter the sample in any way. Optimized columns will be able to withstand temperatures in the range of -200 °C to at least 400 °C. In addition, such a column would have a short linear response time that is independent of flow rate and extends for several orders of magnitude. Moreover, the detector should be reliable, predictable and easy to operate. Understandably, it is not possible for a detector meet all of these requirements. The next subsections will discuss some of the more common types of gas chromatography detectors and the relative advantages and/or disadvantages of each. Type of Detector Applicable Samples Detection Limit Mass Spectrometer (MS) Tunable for any sample .25 to 100 pg Flame Ionization (FID) Hydrocarbons 1 pg/s Thermal Conductivity (TCD) Universal 500 pg/ml Electron-Capture (ECD) Halogenated hydrocarbons 5 fg/s Atomic Emission (AED) Element-selective 1 pg Chemiluminescence (CS) Oxidizing reagent Dark current of PMT Photoionization (PID) Vapor and gaseous Compounds .002 to .02 µg/L (MS) detectors are most powerful of all gas chromatography detectors. In a GC/MS system, the mass spectrometer scans the masses continuously throughout the separation. When the sample exits the chromatography column, it is passed through a transfer line into the inlet of the mass spectrometer . The sample is then ionized and fragmented, typically by an electron-impact ion source. During this process, the sample is bombarded by energetic electrons which ionize the molecule by causing them to lose an electron due to electrostatic repulsion. Further bombardment causes the ions to fragment. The ions are then passed into a mass analyzer where the ions are sorted according to their m/z value, or mass-to-charge ratio. Most ions are only singly charged. The Chromatogram will point out the retention times and the mass spectrometer will use the peaks to determine what kind of molecules are exist in the mixture. The figure below represents a typical mass spectrum of water with the absorption peaks at the appropriate m/z ratios. One of the most common types of mass analyzer in GC/MS is the quadrupole ion-trap analyzer, which allows gaseous anions or cations to be held for long periods of time by electric and magnetic fields. A simple quadrupole ion-trap consists of a hollow ring electrode with two grounded end-cap electrodes as seen in figure #. Ions are allowed into the cavity through a grid in the upper end cap. A variable radio-frequency is applied to the ring electrode and ions with an appropriate m/z value orbit around the cavity. As the radio-frequency is increased linearly, ions of a stable m/z value are ejected by mass-selective ejection in order of mass. Ions that are too heavy or too light are destabilized and their charge is neutralized upon collision with the ring electrode wall. Emitted ions then strike an electron multiplier which converts the detected ions into an electrical signal. This electrical signal is then picked up by the computer through various programs. As an end result, a chromatogram is produced representing the m/z ratio versus the abundance of the sample. GC/MS units are advantageous because they allow for the immediate determination of the mass of the analyte and can be used to identify the components of incomplete separations. They are rugged, easy to use and can analyze the sample almost as quickly as it is eluted. The disadvantages of mass spectrometry detectors are the tendency for samples to thermally degrade before detection and the end result of obliterating all the sample by fragmentation. Flame ionization detectors (FID) are the most generally applicable and most widely used detectors. In a FID, the sample is directed at an air-hydrogen flame after exiting the column. At the high temperature of the air-hydrogen flame, the sample undergoes pyrolysis, or chemical decomposition through intense heating. Pyrolized hydrocarbons release ions and electrons that carry current. A high-impedance picoammeter measures this current to monitor the sample's elution. It is advantageous to use FID because the detector is unaffected by flow rate, noncombustible gases and water. These properties allow FID high sensitivity and low noise. The unit is both reliable and relatively easy to use. However, this technique does require flammable gas and also destroys the sample. Thermal conductivity detectors (TCD) were one the earliest detectors developed for use with gas chromatography. The TCD works by measuring the change in carrier gas thermal conductivity caused by the presence of the sample, which has a different thermal conductivity from that of the carrier gas. Their design is relatively simple, and consists of an electrically heated source that is maintained at constant power. The temperature of the source depends upon the thermal conductivities of the surrounding gases. The source is usually a thin wire made of platinum, gold or . The resistance within the wire depends upon temperature, which is dependent upon the thermal conductivity of the gas. TCDs usually employ two detectors, one of which is used as the reference for the carrier gas and the other which monitors the thermal conductivity of the carrier gas and sample mixture. Carrier gases such as helium and hydrogen has very high thermal conductivities so the addition of even a small amount of sample is readily detected. The advantages of TCDs are the ease and simplicity of use, the devices' broad application to inorganic and organic compounds, and the ability of the analyte to be collected after separation and detection. The greatest drawback of the TCD is the low sensitivity of the instrument in relation to other detection methods, in addition to flow rate and concentration dependency. Figure 13 represents a standard chromatogram produced by a TCD detector. In a standard chromatogram regardless of the type detector, the x-axis is the time and the y-axis is the abundance or the absorbance. From these chromatograms, retention times and the peak heights are determined and used to further investigate the chemical properties or the abundance of the samples. Electron-capture detectors (ECD) are highly selective detectors commonly used for detecting environmental samples as the device selectively detects organic compounds with moieties such as halogens, peroxides, quinones and nitro groups and gives little to no response for all other compounds. Therefore, this method is best suited in applications where traces quantities of chemicals such as pesticides are to be detected and other chromatographic methods are unfeasible. The simplest form of ECD involves gaseous electrons from a radioactive ? emitter in an electric field. As the analyte leaves the GC column, it is passed over this ? emitter, which typically consists of nickle-63 or tritium. The electrons from the ? emitter ionize the nitrogen carrier gas and cause it to release a burst of electrons. In the absence of organic compounds, a constant standing current is maintained between two electrodes. With the addition of organic compounds with electronegative functional groups, the current decreases significantly as the functional groups capture the electrons. The advantages of ECDs are the high selectivity and sensitivity towards certain organic species with electronegative functional groups. However, the detector has a limited signal range and is potentially dangerous owing to its radioactivity. In addition, the signal-to-noise ratio is limited by radioactive decay and the presence of O2 within the detector. Atomic emission detectors (AED), one of the newest addition to the gas chromatographer's arsenal, are element-selective detectors that utilize plasma, which is a partially ionized gas, to atomize all of the elements of a sample and excite their characteristic atomic emission spectra. AED is an extremely powerful alternative that has a wider applicability due to its based on the detection of atomic emissions.There are three ways of generating plasma: microwave-induced plasma (MIP), inductively coupled plasma (ICP) or direct current plasma (DCP). MIP is the most commonly employed form and is used with a positionable diode array to simultaneously monitor the atomic emission spectra of several elements. The components of the Atomic emission detectors include 1) an interface for the incoming capillary GC column to induce plasma chamber,2) a microwave chamber, 3) a cooling system, 4) a diffration grating that associated optics, and 5) a position adjustable photodiode array interfaced to a computer. Chemiluminescence spectroscopy (CS) is a process in which both qualitative and quantitative properties can be be determined using the optical emission from excited chemical species. It is very similar to AES, but the difference is that it utilizes the light emitted from the energized molecules rather than just excited molecules. Moreover, chemiluminescence can occur in either the solution or gas phase whereas AES is designed for gaseous phases. The light source for chemiluminescence comes from the reactions of the chemicals such that it produces light energy as a product. This light band is used instead of a separate source of light such as a light beam. Like other methods, CS also has its limitations and the major limitation to the detection limits of CS concerns with the use of a photomultiplier tube (PMT). A PMT requires a dark current in it to detect the light emitted from the analyte. Another different kind of detector for GC is the photoionization detector which utilizes the properties of chemiluminescence spectroscopy. Photoionization detector (PID) is a portable vapor and gas detector that has selective determination of aromatic hydrocarbons, organo-heteroatom, inorganice species and other organic compounds. PID comprise of an ultrviolet lamp to emit photons that are absorbed by the compounds in an ionization chamber exiting from a GC column. Small fraction of the analyte molecules are actually ionized, nondestructive, allowing confirmation analytical results through other detectors. In addition, PIDs are available in portable hand-held models and in a number of lamp configurations. Results are almost immediate. PID is used commonly to detect VOCs in soil, sediment, air and water, which is often used to detect contaminants in ambient air and soil. The disavantage of PID is unable to detect certain hydrocarbon that has low molecular weight, such as methane and ethane. Gas chromatography is a physical separation method in where volatile mixtures are separated. It can be used in many different fields such as pharmaceuticals, cosmetics and even environmental toxins. Since the samples have to be volatile, human breathe, blood, saliva and other secretions containing large amounts of organic volatiles can be easily analyzed using GC. Knowing the amount of which compound is in a given sample gives a huge advantage in studying the effects of human health and of the environment as well. Air samples can be analyzed using GC. Most of the time, air quality control units use GC coupled with FID in order to determine the components of a given air sample. Although other detectors are useful as well, FID is the most appropriate because of its sensitivity and resolution and also because it can detect very small molecules as well. GC/MS is also another useful method which can determine the components of a given mixture using the retention times and the abundance of the samples. This method be applied to many pharmaceutical applications such as identifying the amount of chemicals in drugs. Moreover, cosmetic manufacturers also use this method to effectively measure how much of each chemical is used for their products. “Height equivalent to a theoretical plate” (HETP) use to calculate the flow rate by usingthe total number of theoretical plates (N) and column length (L). Some application, HETP concepts is used in industrial practice to convert number of theoretical plates to packing height. HETP can be calculate with the Van Deemter equation, which is given by \[ HETP= A + \dfrac{B}{υ} + Cv \tag{1}\] | 25,854 | 3,264 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Introduction_to_Chemical_Bonding |
Chemical bonding is one of the most basic fundamentals of chemistry that explains other concepts such as molecules and reactions. Without it, scientists wouldn't be able to explain why atoms are attracted to each other or how products are formed after a chemical reaction has taken place. To understand the concept of bonding, one must first know the basics behind atomic structure. A common contains a nucleus composed of protons and neutrons, with electrons in certain energy levels revolving around the nucleus. In this section, the main focus will be on these electrons. Elements are distinguishable from each other due to their "electron cloud," or the area where electrons move around the nucleus of an atom. Because each element has a distinct electron cloud, this determines their chemical properties as well as the extent of their reactivity (i.e. noble gases are inert/not reactive while alkaline metals are highly reactive). In chemical bonding, only valence electrons, electrons located in the of the outermost energy level (valence shell) of an element, are involved. Lewis diagrams are graphical representations of elements and their valence electrons. Valance electrons are the electrons that form the outermost shell of an atom. In a Lewis diagram of an element, the symbol of the element is written in the center and the valence electrons are drawn around it as dots. The position of the valence electrons drawn is unimportant. However, the general convention is to start from 12o'clock position and go clockwise direction to 3 o'clock, 6 o'clock, 9 o'clock, and back to 12 o'clock positions respectively. Generally the Roman numeral of the group corresponds with the number of valance electrons of the element. Below is the periodic table representation of the number of valance electrons. The alkali metals of Group IA have one valance electron, the alkaline-earth metals of Group IIA have 2 valance electrons, Group IIIA has 3 valance electrons, and so on. The nonindicated transition metals, lanthanoids, and actinoids are more difficult in terms of distinguishing the number of valance electrons they have; however, this section only introduces bonding, hence they will not be covered in this unit. To draw the lewis diagrams for molecular compounds or ions, follow these steps below (we will be using H O as an example to follow): Count the number of valance electrons of the molecular compound or ion. Remember, if there are two or more of the same element, then you have to double or multiply by however many atoms there are of the number of valance electrons. Follow the roman numeral group number to see the corresponding number of valance electrons there are for that element. Valance electrons: Oxygen (O)--Group VIA: therefore, there are 6 valance electrons Hydrogen (H)--Group IA: therefore, there is 1 valance electron Total: 6 + 2 = If the molecule in question is an ion, remember to add or subract the respective number of electrons to the total from step 1. For ions, if the ion has a negative charge (anion), add the corresponding number of electrons to the total number of electrons (i.e. if NO has a negative charge of 1-, then you 1 extra electron to the total; 5 + 3(6)= 23 +1 = ). A sign mean the molecule has an overall negative charge, so it must have this extra electron. This is because anions have a higher electron affinity (tendency to gain electrons). Most anions are composed of nonmetals, which have high electronegativity. If the ion has a positive charge (cation), the corresponding number of electrons to the total number of electrons (i.e. H O has a positive charge of 1+, so you 1 extra electron to the total; 6 + 1(3) = 9 - 1 = ). A sign means the molecule has an overall positive charge, so it must be missing one electron. Cations are positive and have weaker electron affinity. They are mostly composed of metals; their atomic radii are larger than the nonmetals. This consequently means that shielding is increased, and electrons have less tendency to be attracted to the "shielded" nucleus. From our example, water is a neutral molecule, therefore no electrons need to be added or subtracted from the total. Write out the symbols of the elements, making sure all atoms are accounted for (i.e. H O, write out O and 2 H's on either side of the oxygen). Start by adding single bonds (1 pair of electrons) to all possible atoms while making sure they follow the octet rule (with the exceptions of the duet rule and other elements mentioned above). If there are any leftover electrons, then add them to the central atom of the molecule (i.e. XeF has 4 extra electrons after being distributed, so the 4 extra electrons are given to Xe: like so. Finally, rearrange the electron pairs into double or triple bonds if possible. Most elements follow the octet rule in chemical bonding, which means that an element should have contact to eight valence electrons in a bond or exactly fill up its valence shell. Having eight electrons total ensures that the atom is stable. This is the reason why noble gases, a valence electron shell of 8 electrons, are chemically inert; they are already stable and tend to not need the transfer of electrons when bonding with another atom in order to be stable. On the other hand, alkali metals have a valance electron shell of one electron. Since they want to complete the octet rule they often simply lose one electron. This makes them quite reactive because they can easily donate this electron to other elements. This explains the highly reactive properties of the Group IA elements. Some elements that are exceptions to the octet rule include Aluminum(Al), Phosphorus(P), Sulfur(S), and Xenon(Xe). Hydrogen(H) and Helium(He) follow the duet rule since their valence shell only allows two electrons. There are no exceptions to the duet rule; hydrogen and helium will always hold a maximum of two electrons. Ionic bonding is the process of not sharing electrons between two atoms. It occurs between a nonmetal and a metal. Ionic bonding is also known as the process in which electrons are "transferred" to one another because the two atoms have different levels of electron affinity. In the picture below, a sodium (Na) ion and a chlorine (Cl) ion are being combined through ionic bonding. Na has less electronegativity due to a large atomic radius and essentially does not want the electron it has. This will easily allow the more electronegative chlorine atom to gain the electron to complete its 3rd energy level. Throughout this process, the transfer of the electron releases energy to the atmosphere. Another example of ionic bonding is the crystal lattice structure shown above. The ions are arranged in such a way that shows unifomity and stablity; a physical characteristic in crystals and solids. Moreover, in a concept called "the sea of electrons," it is seen that the molecular structure of metals is composed of stabilized positive ions (cations) and "free-flowing" electrons that weave in-between the cations. This attributes to the metal property of conductivity; the flowing electrons allow the electric current to pass through them. In addition, this explains why strong electrolytes are good conductors. Ionic bonds are easily broken by water because the polarity of the water molecules shield the anions from attracting the cations. Therefore, the ionic compounds dissociate easily in water, and the metallic properties of the compound allow conductivity of the solution. Covalent bonding is the process of sharing of electrons between two atoms. The bonds are typically between a nonmetal and a nonmetal. Since their electronegativities are all within the high range, the electrons are attracted and pulled by both atom's nuceli. In the case of two identical atoms that are bonded to each other (also known as a nonpolar bond, explained later below), they both emit the same force of pull on the electrons, thus there is equal attraction between the two atoms (i.e. oxygen gas, or O , have an equal distribution of electron affinity. This makes covalent bonds harder to break. There are three types of covalent bonds: single, double, and triple bonds. A single bond is composed of 2 bonded electrons. Naturally, a double bond has 4 electrons, and a triple bond has 6 bonded electrons. Because a triple bond will have more strength in electron affinity than a single bond, the attraction to the positively charged nucleus is increased, meaning that the distance from the nucleus to the electrons is less. Simply put, the more bonds or the greater the bond strength, the shorter the bond length will be. In other words: Bond length: triple bond < double bond < single bond Polar covalent bonding is the process of unequal sharing of electrons. It is considered the middle ground between ionic bonding and covalent bonding. It happens due to the differing electronegativity values of the two atoms. Because of this, the more electronegative atom will attract and have a stronger pulling force on the electrons. Thus, the electrons will spend more time around this atom. The symbols above indicate that on the flourine side it is slightly negitive and the hydrogen side is slightly positive. Polarity is the competing forces between two atoms for the electrons. It is also known as the polar covalent bond. A molecule is polar when the electrons are attracted to a more electronegative atom due to its greater electron affinity. A nonpolar molecule is a bond between two identical atoms. They are the ideal example of a covalent bond. Some examples are nitrogen gas (N ), oxygen gas (O ), and hydrogen gas (H ). One way to figure out what type of bond a molecule has is by determining the difference of the electronegativity values of the molecules. If the difference is between 0.0-0.3, then the molecule has a non-polar bond. If the difference is between 0.3-1.7, then the molecule has a polar bond. If the difference is 1.7 or more, then the molecule has an . | 10,008 | 3,265 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./04.E%3A_Reactions_in_Aqueous_Solutions_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact ♦ Acetaminophen (molar mass = 151 g/mol) is an analgesic used as a substitute for aspirin. If a child’s dose contains 80.0 mg of acetaminophen/5.00 mL of an ethanol-water solution, what is the molar concentration? Acetaminophen is frequently packaged as an ethanol-water solution rather than as an aqueous one. Why? ♦ Lead may have been the first metal ever recovered from its ore by humans. Its cation (Pb ) forms a precipitate with Cl according to the following equation: When PbCl is dissolved in hot water, its presence can be confirmed by its reaction with CrO , with which it forms a yellow precipitate: The precipitate is used as a rust inhibitor and in pigments. ♦ Reactions that affect buried marble artifacts present a problem for archaeological chemists. Groundwater dissolves atmospheric carbon dioxide to produce an aqueous solution of carbonic acid: This weakly acidic carbonic acid solution dissolves marble, converting it to soluble calcium bicarbonate: Evaporation of water causes carbon dioxide to be driven off, resulting in the precipitation of calcium carbonate: The reprecipitated calcium carbonate forms a hard scale, or incrustation, on the surface of the object. How many Maalox tablets are needed to neutralize 5.00 mL of 0.100 M HCl stomach acid if each tablet contains 200 mg Mg(OH) + 200 mg Al(OH) ? Each Rolaids tablet contains 412 mg CaCO + 80.0 mg Mg(OH) . How many Rolaids tablets are needed? Suggest another formula (and approximate composition) for an effective antacid tablet. Citric acid (C H O , molar mass = 192.12 g/mol) is a triprotic acid extracted from citrus fruits and pineapple waste that provides tartness in beverages and jellies. How many grams of citric acid are contained in a 25.00 mL sample that requires 38.43 mL of 1.075 M NaOH for neutralization to occur? What is the formula of the calcium salt of this compound? ♦ A method for determining the molarity of a strongly acidic solution has been developed based on the fact that a standard solution of potassium iodide and potassium iodate yields iodine when treated with acid: Starch is used as the indicator in this titration because starch reacts with iodine in the presence of iodide to form an intense blue complex. The amount of iodine produced from this reaction can be determined by subsequent titration with thiosulfate: The endpoint is reached when the solution becomes colorless. ♦ Sewage processing occurs in three stages. Primary treatment removes suspended solids, secondary treatment involves biological processes that decompose organic matter, and tertiary treatment removes specific pollutants that arise from secondary treatment (generally phosphates). Phosphate can be removed by treating the HPO solution produced in the second stage with lime (CaO) to precipitate hydroxyapatite [Ca (PO ) OH]. Calcium hydroxide and calcium carbonate are effective in neutralizing the effects of acid rain on lakes. Suggest other compounds that might be effective in treating lakes. Give a plausible reason to explain why Ca(OH) and CaCO are used. ♦ Approximately 95% of the chlorine produced industrially comes from the electrolysis of sodium chloride solutions (brine): Chlorine is a respiratory irritant whose presence is easily detected by its penetrating odor and greenish-yellow color. One use for the chlorine produced is in the manufacture of hydrochloric acid. ♦ The lead/acid battery used in automobiles consists of six cells that produce a 12 V electrical system. During discharge, lead(IV) oxide, lead, and aqueous sulfuric acid react to form lead(II) sulfate and water. ♦ The use of iron, which is abundant in Earth’s crust, goes back to prehistoric times. In fact, it is believed that the Egyptians used iron implements approximately 5000 years ago. One method for quantifying the iron concentration in a sample involves three steps. The first step is to dissolve a portion of the sample in concentrated hydrochloric acid to produce ferric chloride; the second is to reduce Fe to Fe using zinc metal; and the third is to titrate Fe with permanganate, producing Mn (aq) and ferric iron in the form of Fe O . Baking powder, which is a mixture of tartaric acid and sodium bicarbonate, is used in baking cakes and bread. Why does bread rise when you use baking powder? What type of reaction is involved? An activity series exists for the halogens, which is based on the ease of the diatomic halogen molecule (X ) to X . Experimentally, it is found that fluorine is the easiest halogen to reduce (i.e., F is the best oxidant), and iodine is the hardest halogen to reduce (i.e., I is the worst oxidant). Consequently, the addition of any diatomic halogen, Y , to solutions containing a halide ion (X ) that lies below Y in the periodic table will result in the reduction of Y to Y and the oxidation of X . Describe what you would expect to occur when Bromide is present in naturally occurring salt solutions called . Based on your answers, propose an effective method to remove bromide from brine. ♦ Marble is composed of mostly calcium carbonate. Assuming that acid rain contains 4.0 × 10 M H SO , approximately what volume of rain is necessary to dissolve a 250 lb marble statue? ♦ One of the “first-aid” measures used to neutralize lakes whose pH has dropped to critical levels is to spray them with slaked lime (Ca(OH) ) or limestone (CaCO ). (A slower but effective alternative is to add limestone boulders.) How much slaked lime would be needed to neutralize the acid in a lake that contains 4.0 × 10 M H SO and has a volume of 1.2 cubic miles (5.0 × 10 L)? Recall from Section 4.3 "Stoichiometry of Reactions in Solution" that the reaction of ethanol with dichromate ion in acidic solution yields acetic acid and Cr (aq): Balance the equation for this reaction using oxidation states. ( : the oxidation state of carbon in the CH group remains unchanged, as do the oxidation states of hydrogen and oxygen.) 0.106 M acetaminophen; acetaminophen is an organic compound that is much more soluble in ethanol than water, so using an ethanol/water mixture as the solvent allows a higher concentration of the drug to be used. 2.646 g citric acid, Ca (C H O ) | 6,345 | 3,267 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Solutions_Solvation_and_Dissociation |
means the process of dissolving or forming a . When dissolution happens, the separates into ions or molecules, and each ion or molecule is surrounded by molecules of . The interactions between the solute particles and the solvent molecules is called . A solvated ion or molecule is surrounded by solvent. Technically a solvent can mean anything that is the more abundant component of a homogeneous mixture, but usually it means a volatile liquid that things can easily dissolve in. (Volatile means that it can easily evaporate, like water or alcohol.) The most common solvent is water. When you scuba dive in the ocean, you will need to rinse your gear with water afterwards to remove the salt. The salt dissolves in the water, gets washed away, and then the water evaporates, leaving the gear clean. This is the typical action of a solvent. Solvents are either polar or non-polar. A has partial negative and positive charges. For instance, water has a partial negative charge on O and a partial positive charge on H. The symbol δ means a partial charge, less than the charge on one proton or electron, such as δ+ or δ–. This helps the solvent interact with (solvate) ions and polar molecules through Coulomb interactions. A is one that is electrically neutral all over, or almost so. Oil, or the gas in your car, are examples of non-polar liquids that could be used as solvents. Non-polar solvents are only good for dissolving non-polar solutes, which is why water, salt and sugar don't mix into oil. | 1,530 | 3,269 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Carbohydrates_Fundamentals/Carbohydrate_Overview |
The chemistry of carbohydrates most closely resembles that of alcohol, aldehyde, and ketone functional groups. As a result, the modern definition of a CARBOHYDRATE is that the compounds are polyhydroxy aldehydes or ketones. The chemistry of carbohydrates is complicated by the fact that there is a functional group (alcohol) on almost every carbon. In addition, the carbohydrate may exist in either a straight chain or a ring structure. Ring structures incorporate two additional functional groups: the hemiacetal and acetal. A major part of the carbon cycle occurs as carbon dioxide is converted to carbohydrates through photosynthesis. Carbohydrates are utilized by animals and humans in metabolism to produce energy and other compounds. Photosynthesis is a complex series of reactions carried out by algae, phytoplankton, and the leaves in plants, which utilize the energy from the sun. The simplified version of this chemical reaction is to utilize carbon dioxide molecules from the air and water molecules and the energy from the sun to produce a simple sugar such as glucose and oxygen molecules as a by product. The simple sugars are then converted into other molecules such as starch, fats, proteins, enzymes, and DNA/RNA i.e. all of the other molecules in living plants. All of the "matter/stuff" of a plant ultimately is produced as a result of this photosynthesis reaction. French word for "malt"; a disaccharide containing two units of glucose; found in germinating grains, used to make beer. occurs in animals and humans after the ingestion of organic plant or animal foods. In the cells a series of complex reactions occurs with oxygen to convert for example glucose sugar into the products of carbon dioxide and water and ENERGY. This reaction is also carried out by bacteria in the decomposition/decay of waste materials on land and in the water. occurs when any organic material is reacted (burned) in the presence of oxygen to give off the products of carbon dioxide and water and ENERGY. The organic material can be any fossil fuel such as natural gas (methane), oil, or coal. Other organic materials that combust are wood, paper, plastics, and cloth. The whole purpose of both processes is to convert chemical energy into other forms of energy such as heat. | 2,297 | 3,272 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Reaction_Equations |
Changes in a material or system are called , and they are divided into chemical and physical reactions. is the driving force of all changes, both physical and chemical reactions. Energy is always involved in these reactions. If a system is more stable by losing some energy, a reaction takes place, releasing energy. Such a reaction is said to be . Supplying energy to a system also causes a reaction. Energy absorbing reactions are called . Sometimes, the amount of energy involved in a reaction may be so small that the change in energy is not readily noticeable. An equation can be used to describe a , which involves a change of states. For example, , and can be represented as follows. In these equations, (s) stands for solid, (l) for liquid (l), and (g) for gas. In these changes, no chemical bonds are broken or formed, and the molecular identities of the substances have not changed. Is the between graphite and diamond a chemical or physical reaction? \(\mathrm{C(graphite) \rightarrow C(diamond)}\). The crystal structures of diamond and graphite are very different, and bonding between the carbon atoms is also different in the two solid states. Because chemical bonds are broken and new bonds are formed, the phase transition of diamond and graphite is a chemical reaction. Chemicals or substances change converting to one or more other substances, and these changes are called . At the molecular level, atoms or groups of atoms rearrange resulting in breaking and forming some chemical bonds in a chemical reaction. The substances undergoing changes are called , whereas substances newly formed are called . Physical appearances of products are often different from reactants. Chemical reactions are often accompanied by the appearance of gas, fire, precipitate, color, light, sound, or odor. These phenomena are related to energy and properties of the reactants and products. For example, the oxidation of propane releases heat and light, and a rapid reaction is an explosion, \(\mathrm{C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O}\) A balanced equation also shows a macroscopic quantitative relationship. This balanced reaction equation shows that five moles of oxygen reacts with one mole of propane generating three moles of carbon dioxide and four moles of water, a total of 7 moles of products in the combustion reaction. At the molecular level, this equation shows that for each propane molecule, 5 oxygen molecules are required. The three carbon atoms are converted to three molecules of carbon dioxide, whereas the 8 hydrogen atoms in propane are oxidized to 4 water molecules. The numbers of \(\ce{H}\), \(\ce{C}\), and \(\ce{O}\) atoms are the same on both sides of the equation. We study properties of substances so that we know how to make use of them. Tendencies of a substance to react, either by itself or with others, are important chemical properties. Via properties, we understand chemical reactions, which are best studied by experimentation and observation. After you have performed many experiments, you may generalize certain rules and facts. Knowing these rules and facts enables you to solve problems that you have not yet encountered. The most important aspect of a chemical reaction is to know what are the reactants and what are the products. For this, the best description of a reaction is to write an equation for the reaction. A gives the reactants and products, and a shows the mole relationships of reactants and products. Often, the amount of energy involved in the reaction is given. Dealing with the quantitative aspect of chemical reactions is called . For example, when clamshells, \(\ce{CaCO3}\), are heated, a gas \(\ce{CO2}\) will be released, leaving a white powder (solid \(\ce{CaO}\)) behind. The equation of the reaction is written as: \[\mathrm{CaCO_3 \rightarrow CaO + CO_2}\] The equation indicates that one mole of \(\ce{CaCO3}\) gives one mole each of \(\ce{CaO}\) and \(\ce{CO2}\). Amounts of substances represented by chemical formulas have been introduced on the two previous pages, and these concepts should help to figure out the stoichiometry of reactions when a reaction equation is given. When 10.0 g pure calcium carbonate is heated and converted to solid calcium oxide \(\ce{CaO}\), how much calcium oxide should be obtained? If only 5.0 grams \(\ce{CaO}\) is obtained, what is the actual yield? Under ideal conditions, amounts of substance in the reaction equation is as indicated below: \(\begin{alignat}{2}
\ce{&CaCO_3 \rightarrow &&CaO + &&CO_2}\\
&\:100.0 &&\:\:56 &&\:44\:\:\: \mathrm{g/mol\: (formula\: weights)}
\end{alignat}\) \[\mathrm{10.0\: g\: CaCO_3\times\dfrac{1\: mol\: CaCO_3}{100\: g\: CaCO_3}\times\dfrac{1\: mol\: CaO}{1\: mol\: CaCO_3}\times\dfrac{56\: g\: CaO}{1\: mol\: CaO}= 5.6\: g\: CaO}\] An inefficient conversion is given here, but the method shows the details of consideration. If the amount of \(\ce{CaO}\) obtained is not 5.6 g, one can conclude that the sample may not be pure. When 10.0 g pure calcium carbonate is heated and converted to solid calcium oxide \(\ce{CaO}\), how much \(\ce{CO2}\) at standard condition is released? \[\mathrm{CaCO_3 \rightarrow CaO + CO_2}\] \[\mathrm{10.0\: g\: CaCO_3\times\dfrac{1\: mol\: CO_2}{100\: g\: CaCO_3}\times\dfrac{22.4\: L\: CO_2}{1\: mol\: CO_2}= 2.24\: L\: CO_2}\] We have taken a short cut in this formulation compared to Example 1. Examples 1 and 2 illustrate the evaluation of quantities in g and in L. truly represent changes of materials. For many reactions, we may only be able to write equations for the overall reactions. For example, common sense tells us that when sugar is fully oxidized, carbon dioxide and water are the final products. The oxidation reaction is the same as the combustion reaction. Thus we write \[\ce{C12H22O11 + 12 O2 \rightarrow 12 CO2 + 11 H2O}\] This illustrates the methods used for writing balanced reaction equations: The compound \(\ce{N2O5}\) is unstable at room temperature. It decomposes yielding a brown gas \(\ce{NO2}\) and oxygen. Write a balanced chemical reaction equation for its decomposition. The first step is to write an unbalanced equation indicating only the reactant and products: \[\ce{N2O5 \rightarrow NO2 + O2}\] A \(\ce{N2O5}\) molecule decomposes into two \(\ce{NO2}\) molecule, and half of \(\ce{O2}\). \[\ce{N2O5 \rightarrow 2 NO2 + \dfrac{1}{2}O2}\] In order to give whole number to the equation, we multiply all the stoichiometric coefficients by 2. \[\ce{2 N2O5 \rightarrow 4 NO2 + O2}\] This example illustrates the steps used in writing a balanced equation for a chemical reaction. This balanced equation does not tell us how a \(\ce{N2O5}\) molecule decomposes, it only illustrates the overall reaction. When solutions of \(\ce{CaCl2}\) and \(\ce{AgNO3}\) are mixed, a white precipitate is formed. The same precipitate is also observed when \(\ce{NaCl}\) solution is mixed with \(\ce{AgCH3CO2}\) solution. Write a balanced equation for the reaction between \(\ce{CaCl2}\) and \(\ce{AgNO3}\). The common ions between \(\ce{NaCl}\) and \(\ce{CaCl2}\) are \(\ce{Cl-}\) ions, and \(\ce{Ag+}\) ions are common between the two silver containing compounds. The question illustrates a scientific deduction used in the determination of products. The product is \(\ce{AgCl}\), and the balanced reaction is \(\ce{CaCl2 + 2 AgNO3 \rightarrow 2 AgCl + Ca(NO3)2}\)
In reality, solutions of salts contain ions. In this case, the solutions contain \(\ce{Ca^2+}\), \(\ce{Cl-}\), \(\ce{Ag+}\), and \(\ce{NO3-}\) ions. The \(\ce{Cl-}\) and \(\ce{Ag+}\) ions form an insoluble solid, and a precipitate is formed, \(\ce{Cl- + Ag+ \rightarrow AgCl(s)}\) \(\ce{Ca^2+}\) and \(\ce{NO3-}\) are One of the most important topics in chemistry is . In this page, we only concentrate on the stoichiometry conveyed by reaction equations. Other topics related to chemical reactions are: Balancing oxidation and reduction reaction equations is a little more complicated than what we discussed here. You have to have the skills to assign oxidation states, explain oxidation and reduction in terms of oxidation-state change, and write half reaction equations. Then you will be able to balance redox reactions. All these are given in the next module on Chemical Reactions. Hint: \(\ce{CO2}\)
Combustion of \(\ce{C}\)-containing compounds converts all \(\ce{C}\) to \(\ce{CO2}\). Hint: \(\ce{H2S}\)
Sulfur and oxygen are group 6 elements, and they form \(\ce{H2O}\) and \(\ce{H2S}\). Hint: 56.7 g
How much (in g) oxygen is required? Hint: 24.9 g
For the reaction: \(\ce{2 KClO3 \rightarrow 2 KCl + 3 O2}\) the formulation suggestion is: \(\mathrm{0.50\: mol\: O_2 \times\dfrac{2\: mole\: KCl}{3\: mol\: O_2}\times\dfrac{74.6\: g\: KCl}{1\: mol\: KCl}=\: ??.?\: g\: KCl}\) \(\ce{H}\) \(\ce{O}\) \(\ce{S}\) \(\ce{Cl}\) \(\ce{Ba}\) Hint: 0.113 mol
How much (in g) \(\ce{BaCl2}\) is present in the solution?
How much silver nitrate is required to precipitate all the chloride ions?
The reaction is: \(\mathrm{BaCl_2 + H_2SO_4 \rightarrow BaSO_4 + 2 H^+ + 2 Cl^-}\). 0.0566 mole of \(\ce{Ba}\) correspond to 0.113 mol of \(\ce{Cl-}\) in \(\ce{BaCl2}\).
\(\mathrm{13.2\: g\times\dfrac{1\: mol\: BaSO_4}{233.39\: g}\times\dfrac{2\: mol\: Cl^-}{1\: mol\: BaSO_4}= 0.113\: mol}\) Hint: 1998 kg
The molecular weight of \(\ce{SO2}\) is about twice the atomic weight of \(\ce{S}\). Thus the weight of \(\ce{SO2}\) is twice that of \(\ce{S}\).
How much (in mole and L) \(\ce{SO2}\) is generated per day?
If all \(\ce{SO2}\) is converted to \(\ce{H2SO4}\), how much (in mol and kg) sulfuric acid is produced? (3055 kg) Hint: 4 moles; \(\ce{C3H8 + 5 O2 \rightarrow 3 CO2 + 4 H2O}\)
Work out a balanced reaction equation.
How many grams of water will be produced?
How many moles of \(\ce{CO2}\) will be produced? Hint: 66.0%
The problem illustrates a strategy for chemical analysis. Hint: 36.9% \(\ce{N}\) \(\ce{O}\) \(\ce{Cl}\) \(\ce{Ca}\) \(\ce{Ag}\)
This problem also illustrates a strategy for chemical analysis. | 10,070 | 3,273 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/03%3A_Hazards/3.03%3A_Case_Study_-_Badger_Army_Ammunition_Plant |
Environmental cleanup of the 7,400-acre Badger Army Ammunition Plant will be greater than $250 million. This is one of the 40 contaminated military sites in Wisconsin which the Defense Environmental Restoration Account cites as the most contaminated. In fact, 32 areas are polluted with dangerous levels of solvents, metals and explosive/incendiary waste. The water beneath the plant is contaminated with mutagenic chemicals that include carbon tetrachloride, trichloroethylene and dinitrotoluenes. One part of the site, known as the Propellant Burning Grounds, is the source of a three-mile plume of contaminated groundwater that has migrated offsite, completely contaminating private drinking water wells in addition to the Wisconsin River. In 1990, the Wisconsin Division of Health conducted a health survey. It concluded that communities near the Badger plant have a significantly higher incidence of cancer and deaths. In spite of these alarming findings, the State refused to take any action. In 1995, the Division of Health responded because of pressure from CSWAB (Citizens for Safe Water Around Badger) and reopened the community health study. On October 26, 1998, CSWAB concluded “there was entirely inadequate contact with our community – the population being studied.” Prior to September 1998, no press releases were published, no public meetings were held, and no interviews were conducted. Despite several requests, virtually no resources were devoted to interviewing residents about current health problems and concerns regarding their exposure to air, dust, emissions, and surface soils. Assessment of risk from cleanup activities was also absent. CSWAB also determined that the Wisconsin DOH focused on death studies when many health problems and community concerns have been nonlethal, such as respiratory illnesses or reproductive problems. CSWAB appealed to ATSDR (Agency for Toxic Substances and Disease Registry in D.C.) to discuss the lack of adequate community participation in this and similar health assessments across the nation. The Wisconsin Department of Health conducted the assessment under a cooperative agreement with ATSDR. The U.S. Army proposed to abandon and severely reduce cleanup of two priority areas within the plant. The Settling Ponds and Spoils Disposal area — a series of lagoons that run the length of the 7,000-acre facility — is contaminated with high levels of lead and dinitrotoluenes. Maintenance of the Badger plant costs in excess of $17 million per year. In 1991, only $3 million was allocated for environmental studies. Since 1975, there have been over 56 chemical spills and incidents. Moreover, there is no national strategic need for maintaining the Badger plant. A February 20, 1997 Government Accounting Office report concludes BAAP and three other military plants could be eliminated “because alternative sources exist…to provide the capabilities these plants provide.” | 2,944 | 3,274 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.06%3A_Entropy_and_Disorder |
A common interpretation of entropy is that it is somehow a measure of chaos or randomness. There is some utility in that concept. Given that entropy is a measure of the dispersal of energy in a system, the more chaotic a system is, the greater the dispersal of energy will be, and thus the greater the entropy will be. Ludwig Boltzmann (1844 – 1906) (O'Connor & Robertson, 1998) understood this concept well, and used it to derive a statistical approach to calculating entropy. Boltzmann proposed a method for calculating the entropy of a system based on the number of energetically equivalent ways a system can be constructed. Boltzmann proposed an expression, which in its modern form is: \[S = k_b \ln(W) \label{Boltz} \] This rather famous equation is etched on Boltzmann’s grave marker in commemoration of his profound contributions to the science of thermodynamics (Figure \(\Page {1}\)). Calculate the entropy of a carbon monoxide crystal, containing 1.00 mol of \(\ce{CO}\), and assuming that the molecules are randomly oriented in one of two equivalent orientations. Using the Boltzmann formula (Equation \ref{Boltz}): \[S = nK \ln (W) \nonumber \] And using \(W = 2\), the calculation is straightforward. \[ \begin{align*} S &= \left(1.00 \, mol \cot \dfrac{6.022\times 10^{23}}{1\,mol} \right) (1.38 \times 10^{-23} J/K) \ln 2 \\ &= 5.76\, J/K \end{align*} \] | 1,383 | 3,276 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/09_Chemical_Bonding_and_Molecular_Energy_Levels |
Our basis for understanding chemical bonding and the structures of molecules is the electron orbital description of the structure and valence of atoms, as provided by quantum mechanics. We assume an understanding of the periodicity of the elements based on the nuclear structure of the atom and our deductions concerning valence based on electron orbitals. Our model of valence describes a chemical bond as resulting from the sharing of a pair of electrons in the valence shell of the bonded atoms. This sharing allows each atom to complete an octet of electrons in its valence shell, at least in the sense that we count the shared electrons as belonging to both atoms. However, it is not clear that this electron counting picture has any basis in physical reality. What is meant, more precisely, by the sharing of the electron pair in a bond, and why does this result in the bonding of two atoms together? Indeed, what does it mean to say that two atoms are bound together? Furthermore, what is the significance of sharing a pair of electrons? Why aren't chemical bonds formed by sharing one or three electrons, for example? We seek to understand how the details of chemical bonding are related to the properties of the molecules formed, particularly in terms of the strengths of the bonds formed. We began our analysis of the energies and motions of the electrons in atoms by observing the properties of the simplest atom, hydrogen, with a single electron. Similarly, to understand the energies and motions of electrons which lead to chemical bonding, we begin our observations with the simplest particle with a chemical bond, which is the \(\ce{H_2^+}\) molecular ion. Each hydrogen nucleus has a charge of +1. An \(\ce{H_2^+}\) molecular ion therefore has a single electron. It seems inconsistent with our notions of valence that a single electron, rather than an electron pair, can generate a chemical bond. However, these concepts have been based on observations on molecules, not molecular ions like \(\ce{H_2^+}\). And it is indeed found that \(\ce{H_2^+}\) is a stable bound molecular ion. What forces and motions hold the two hydrogen nuclei close together in the \(\ce{H_2^+}\) ion? It is worth keeping in mind that the two nuclei must repel one another, since they are both positively charged. In the absence of the electron, the two nuclei would accelerate away from one another, rather than remaining in close proximity. What is the role of the electron? Clearly, the electron is attracted to both nuclei at the same time, and, in turn, each nucleus is attracted to the electron. The effect of this is illustrated in Figure 9.1. In Figure 9.1a, the electron is "outside" of the two nuclei. In this position, the electron is primarily attracted to the nucleus on the left, to which it is closer. More importantly, the nucleus on the right feels a greater repulsion from the other nucleus than attraction to the electron, which is farther away. As a result, the nucleus on the right experiences a strong force driving it away from the hydrogen atom on the left. This arrangement does not generate chemical bonding, therefore. By contrast, in Figure 9.1b, the electron is between the two nuclei. In this position, the electron is roughly equally attracted to the two nuclei, and very importantly, each nucleus feels an attractive force to the electron which is greater than the repulsive force generated by the other nucleus. Focusing on the electron's energy, the proximity of the two nuclei provides it a doubly attractive environment with a very low potential energy. If we tried to pull one of the nuclei away, this would raise the potential energy of the electron, since it would lose attraction to that nucleus. Hence, to pull one nucleus away requires us to add energy to the molecular ion. This is what is meant by a chemical bond: the energy of the electrons is lower when the atoms are in close proximity than when the atoms are far apart. This "holds" the nuclei close together, since we must do work (add energy) to take the nuclei apart. a. b. Note that the chemical bond in Figure 9.1b results from the electron's position between the nuclei. On first thought, this appears to answer our question of what we mean by "sharing an electron pair" to form a chemical bond. An electrons positioned between two nuclei is "shared" to the extent that its potential energy is lowered due to attraction to both nuclei simultaneously. On second thought, though, this description must be inaccurate. We have learned in our study of energy levels in atoms that an electron must obey the uncertainty principle and that, as a consequence, the electron does not have a definite position, between the nuclei or otherwise. We can only hope to specify a probability for observing an electron in a particular location. This probability is, from quantum mechanics, provided by the wave function. What does this probability distribution look like for the \(\ce{H_2^+}\) molecular ion? To answer this question, we begin by experimenting with a distribution that we know: the \(1s\) electron orbital in a hydrogen atom. This we recall has the symmetry of a sphere, with equal probability in all directions away from the nucleus. To create an \(\ce{H_2^+}\) molecular ion from a hydrogen atom, we must add a bare second hydrogen nucleus (an \(\ce{H^+}\) ion). Imagine bringing this nucleus closer to the hydrogen atom from a very great distance (see Figure 9.2a). As the \(\ce{H^+}\) ion approaches the neutral atom, both the hydrogen atom's nucleus and electron respond to the electric potential generated by the positive charge. The electron is attracted and the hydrogen atom nucleus is repelled. As a result, the distribution of probability for the electron about the nucleus must become distorted, so that the electron has a greater probability of being near the \(\ce{H^+}\) ion and the nucleus has a greater probability of being farther from the ion. This distortion, illustrated in Figure 9.2b, is called "polarization": the hydrogen atom has become like a "dipole", with greater negative charge to one side and greater positive charge to the other. a. b. c. This polarization must increase as the \(\ce{H^+}\) ion approaches the hydrogen atom until, eventually, the electron orbital must be sufficiently distorted that there is equal probability for observing the electron in proximity to either hydrogen nucleus (see Figure 9.2c). The electron probability distribution in Figure 9.2c now describes the motion of the electron, not in a hydrogen atom, but in an \(\ce{H_2^+}\) molecular ion. As such, we refer to this distribution as a "molecular orbital". We note that the molecular orbital in Figure 9.2c is more delocalized than the atomic orbital in Figure 9.2a, and this is also important in producing the chemical bond. We recall from the discussion of atomic energy levels that the energy of an electron in an orbital is determined, in part, by the compactness of the orbital. The more the orbital confines the motion of the electron, the higher is the kinetic energy of the electron, an effect we referred to as the "confinement energy". Applying this concept to the orbitals in Figure 9.2, we can conclude that the confinement energy is lowered when the electron is delocalized over two nuclei in a molecular orbital. This effect contributes significantly to the lowering of the energy of an electron resulting from sharing by two nuclei. Recall that the electron orbitals in the hydrogen atom are described by a set of quantum numbers. One of these quantum numbers is related to the symmetry or shape of the atomic orbital and is generally depicted by a letter. Recall that an \(s\) orbital is spheric in shape, and a \(p\) orbital has two lobes aligned along one axis. Similarly, the molecular orbitals for the \(\ce{H_2^+}\) molecular ion are described by a set of numbers which give the symmetry (or shape) of the orbital. For our purposes, we need only one of these descriptors, based on the symmetry of the orbital along the bond: if the molecular orbital has the symmetry of a cylinder, we refer to it as a "\(\sigma\) orbital". The orbital in Figure 9.2c satisfies this condition. We conclude that chemical bonding results from an electron in a molecular orbital which has substantial probability for the electron to be between two nuclei. However, this example illustrates chemical bonding with a single electron. Our rules of valence indicate that bonding typically occurs with a pair of electrons, rather than a single electron. Furthermore, this model of bonding does not tell us how to handle molecules with many electrons (say, \(\ce{F_2}\)) where most of the electrons do not participate in the bonding at all. We now consider molecules with more than one electron. These are illustrated most easily by diatomic molecules (molecules with only two atoms) formed by like atoms, beginning with the hydrogen molecule, \(\ce{H_2}\). The most direct experimental observation of a chemical bond is the amount of energy required to break it. This is called the bond energy, or somewhat less precisely, the bond strength. Experimentally, it is observed that the bond energy of the hydrogen molecule \(\ce{H_2}\) is \(458 \: \text{kJ/mol}\). By contrast, the bond energy of the \(\ce{H_2^+}\) molecular ion is \(269 \: \text{kJ/mol}\). Therefore, the bond in \(\ce{H_2}\) is stronger than the bond in \(\ce{H_2^+}\). Thus, the pair of shared electrons in \(\ce{H_2}\) generates a stronger attractive force than does the single electron in \(\ce{H_2^+}\). Before deducing an explanation of this in terms of electron orbitals, we first recall the valence shell electron pair description of the bonding in \(\ce{H_2}\). Each hydrogen atom has a single electron. By sharing these two electrons, each hydrogen atom can fill its valence shell, attaining the electron configuration of helium. How does this translate into the electron orbital picture of electron sharing that we have just described for the \(\ce{H_2^+}\) molecular ion? There are two ways to deduce the answer to this question, and, since they are both useful, we will work through them both. The first way is to imagine that we form an \(\ce{H_2}\) molecule by starting with an \(\ce{H_2^+}\) molecular ion and adding an electron to it. As a simple approximation, we might imagine that the first electron's probability distribution (its orbital) is not affected by the addition of the second electron. The second electron must have a probability distribution describing its location in the molecule as well. We recall that, in atoms, it is possible to put two electrons into a single electron orbital, provided that the two electrons have opposite values of the spin quantum number, \(m_s\). Therefore, we expect this to be true for molecules as well, and we place the added second electron in \(\ce{H_2}\) into the same \(\sigma\) orbital as the first. This results in two electrons in the region between the two nuclei, thus adding to the force of attraction of the two nuclei into the bond. This explains our observation that the bond energy of \(\ce{H_2}\) is almost (although not quite) twice the bond energy of \(\ce{H_2^+}\). The second way to understand the electron orbital picture of \(\ce{H_2}\) is to imagine that we form the molecule by starting with two separated hydrogen atoms. Each of these atoms has a single electron in a \(1s\) orbital. As the two atoms approach one another, each electron orbital is polarized in the direction of the other atom. Once the atoms are close enough together, these two orbitals become superimposed. Now we must recall that these orbitals describe the wave-like motion of the electron, so that, when these two wave functions overlap, they must interfere, either constructively or destructively. In Figure 9.3, we see the consequences of constructive and destructive interference. We can deduce that, in \(\ce{H_2}\) the electron orbitals from the atoms must constructively interfere, because that would increase the electron probability in the region between the nuclei, resulting in bonding as before. Therefore, the \(\sigma\) molecular orbital describing the two electrons in \(\ce{H_2}\) can be understood as resulting from the constructive overlap of two atomic \(1s\) electron orbitals. We now add to our observations of diatomic molecules by noting that, of the diatomic molecules formed from like atoms of the first ten elements, \(\ce{H_2}\), \(\ce{Li_2}\), \(\ce{B_2}\), \(\ce{C_2}\), \(\ce{N_2}\), \(\ce{O_2}\), and \(\ce{F_2}\) are stable molecules with chemical bonds, whereas \(\ce{He_2}\), \(\ce{Be_2}\), and \(\ce{Ne_2}\) are not bound. In examining the electron configurations of the atoms of these elements, we discover a correspondence with which diatomic molecules are bound and which ones are not. \(\ce{H}\), \(\ce{Li}\), \(\ce{B}\), \(\ce{N}\), and \(\ce{F}\) all have odd numbers of electrons, so that at least one electron in each atom is unpaired. By contrast, \(\ce{He}\), \(\ce{Be}\), and \(\ce{Ne}\) all have even numbers of electrons, none of which are unpaired. The other atoms, \(\ce{C}\) and \(\ce{O}\), both have an even number of electrons. However, as deduced in our understanding of the electron configurations in atoms, electrons will, when possible, distribute themselves into different orbitals of the same energy so as to reduce the effect of their mutual repulsion. Thus, in \(\ce{C}\) and \(\ce{O}\), there are three \(2p\) orbitals into which 2 and 4 electrons are placed, respectively. Therefore, in both atoms, there are two unpaired electrons. We conclude that bonds will form between atoms if and only if there are unpaired electrons in these atoms. In \(\ce{H_2}\), the unpaired electrons from the separated atoms become paired in a molecular orbital formed from the overlap of the \(1s\) atomic electron orbitals. In the case of a hydrogen atom, then, there are of course no paired electrons in the atom to worry about. In all other atoms, there certainly are paired electrons, regardless of whether there are or are not unpaired electrons. For example, in a lithium atom, there are two paired electrons in a \(1s\) orbital and an unpaired electron in the \(2s\) orbital. To form \(\ce{Li_2}\), the unpaired electron from each atom can be placed into a molecular orbital formed from the overlap of the \(2s\) atomic electron orbitals. However, what becomes of the two electrons paired in the \(1s\) orbital in a \(\ce{Li}\) atom during the bonding of \(\ce{Li_2}\)? To answer this question, we examine \(\ce{He_2}\), in which each atom begins with only the two \(1s\) electrons. As we bring the two \(\ce{He}\) atoms together from a large distance, these \(1s\) orbitals should become polarized, as in the hydrogen atom. When the polarized \(1s\) orbitals overlap, constructive interference will again result in a \(\sigma\) molecular orbital, just as in \(\ce{H_2}\). Yet, we observe that \(\ce{He_2}\) is not a stable bound molecule. The problem which prevents bonding for \(\ce{He_2}\) arises from the Pauli Exclusion Principle: only two of the four electrons in \(\ce{He_2}\) can be placed into this \(\sigma\) bonding molecular orbital. The other two must go into a different orbital with a different probability distribution. To deduce the form of this new orbital, we recall that the bonding orbital discussed so far arises from the constructive interference of the atomic orbitals, as shown in Figure 9.3. We could, instead, have assumed destructive interference of these orbitals. Destructive interference of two waves eliminates amplitude in the region of overlap of the waves, also shown in Figure 9.3. In the case of the atomic orbitals, this means that the molecular orbital formed from destructive interference decreases the probability for the electron to be in between the nuclei. Therefore, it increases probability for the electron to be outside the nuclei, as in Figure 9.1a. As discussed there, this arrangement for the electron does not result in bonding; instead, the nuclei repel each other and the atoms are forced apart. This orbital is thus called an anti-bonding orbital. This orbital also has the symmetry of a cylinder along the bond axis, so it is also a \(\sigma\) orbital; to indicate that it is an anti-bonding orbital, we designate it with an asterisk, \(\sigma^*\). a. b. In \(\ce{He_2}\), both the bonding and the anti-bonding orbitals must be used in order to accommodate four electrons. The two electrons in the bonding orbital lower the energy of the molecule, but the two electrons in the anti-bonding orbital raise it. Since two \(\ce{He}\) atoms will not bind together, then the net effect must be that the anti-bonding orbital more than offsets the bonding orbital. We have now deduced an explanation for why the paired electrons in an atom do not contribute to bonding. Both bonding and anti-bonding orbitals are always formed when two atomic orbitals overlap. When the electrons are already paired in the atomic orbitals, then there are too many electrons for the bonding molecular orbital. The extra electrons must go into the anti-bonding orbital, which raises the energy of the molecule, preventing the bond from forming. Returning to the \(\ce{Li_2}\) example discussed above, we can develop a simple picture of the bonding. The two \(1s\) electrons from each atom do not participate in the bonding, since the anti-bonding more than offsets the bonding. Thus, the paired "core" electrons remain in their atomic orbitals, unshared, and we can ignore them in describing the bond. The bond is formed due to overlap of the \(2s\) orbitals and sharing of these electrons only. This is also consistent with our earlier view that the core electrons are closer to the nucleus, and thus unlikely to be shared by two atoms. The model we have constructed seems to describe fairly well the bonding in the bound diatomic molecules listed above. For example, in a fluorine atom, the only unpaired electron is in a \(2p\) orbital. Recall that a \(2p\) orbital has two lobes, directed along one axis. If these lobes are assumed to lie along the axis between the two nuclei in \(\ce{F_2}\), then we can overlap them to form a bonding orbital. Placing the two unpaired electrons into this orbital then results in a single shared pair of electrons and a stable molecular bond. The energies of electrons in molecular orbitals can be observed directly by measuring the ionization energy. This is the energy required to remove an electron, in this case, from a molecule: \[\ce{H_2} \left( g \right) \rightarrow \ce{H_2^+} \left( g \right) + \ce{e^-} \left( g \right)\] The measured ionization energy of \(\ce{H_2}\) is \(1488 \: \text{kJ/mol}\). This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is \(1312 \: \text{kJ/mol}\). Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom, so we can conclude that the electron has a lower energy in the molecule. If we attempt to pull the atoms apart, we must raise the energy of the electron. Hence, energy is required to break the bond, so the molecule is bound. We conclude that a bond is formed when the energy of the electrons in the molecule is lower than the energy of the electrons in the separated atoms. This conclusion seems consistent with our previous view of shared electrons in bonding molecular orbitals. As a second example, we consider the nitrogen molecule, \(\ce{N_2}\). We find that the ionization energy of molecular nitrogen is \(1503 \: \text{kJ/mol}\), and that of atomic nitrogen is \(1402 \: \text{kJ/mol}\). Once again, we conclude that the energy of the electrons in molecular nitrogen is lower than that of electrons in the separated atoms, so the molecule is bound. As a third example, we consider fluorine, \(\ce{F_2}\). In this case, we find that the ionization energy of molecular fluorine is \(1515 \: \text{kJ/mol}\), which is smaller than the ionization energy of a fluorine atom, \(1681 \: \text{kJ/mol}\). This seems inconsistent with the bonding orbital concept we have developed above, which states that the electrons in the bond have a lower energy than in the separated atoms. If the electron being ionized has a higher energy in \(\ce{F_2}\) than in \(\ce{F}\), why is \(\ce{F_2}\) a stable molecule? Apparently, we need a more complete description of the molecular orbital concept of chemical bonding. To proceed further, we compare bond energies in several molecules. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. We observe that the bond energy of \(\ce{N_2}\) is \(956 \: \text{kJ/mol}\). This is very much larger than the bond energy of \(\ce{H_2}\), \(458 \: \text{kJ/mol}\), and of \(\ce{F_2}\), which is \(160 \: \text{kJ/mol}\). We can account for the unusually strong bond in nitrogen using both our valence shell electron pair sharing model and our electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell, because the three \(2p\) electrons distribute themselves over the three \(2p\) orbitals, each oriented along a different axis. Each of these unpaired electrons is available for sharing with a second nitrogen atom. The result, from valence shell electron pair sharing concepts, is that three pairs of electrons are shared between two nitrogen atoms, and we called the bond in \(\ce{N_2}\) a "triple bond". It is somewhat intuitive that the triple bond in \(\ce{N_2}\) should be much stronger than the single bond in \(\ce{H_2}\) or in \(\ce{F_2}\). Now consider the molecular orbital description of bonding in \(\ce{N_2}\). Each of the three \(2p\) atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, if we are to accommodate three electron pairs. Each \(2p\) orbital is oriented along a single axis. One \(2p\) orbital from each atom is oriented in the direction of the other atom, that is, along the bond axis. When these two atomic orbitals overlap, they form a molecular orbital which has the symmetry of a cylinder and which is therefore a \(\sigma\) orbital. Of course, they also form a \(\sigma^*\) orbital. The two electrons are then paired in the bonding orbital. The other two \(2p\) orbitals on each nitrogen atom are perpendicular to the bond axis. The constructive overlap between these orbitals from different atoms must therefore result in a molecular orbital somewhat different than what we have discussed before. As shown in Figure 9.4, the molecular orbital which results now does not have the symmetry of a cylinder, and in fact, looks something more like a cylinder cut into two pieces. This we call a \(\pi\) orbital. There are two such \(\pi\) orbitals since there are two sets of \(p\) orbitals perpendicular to the bond axis. Figure 9.4 also shows that an anti-bonding orbital is formed from the destructive overlap of \(2p\) orbitals, and this is called a \(\pi^*\) orbital. There are also two \(\pi^*\) orbitals formed from destructive overlap of \(2p\) orbitals. In \(\ce{N_2}\), the three shared electron pairs are thus in a single \(\sigma\) orbital and in two \(\pi\) orbitals. Each of these orbitals is a bonding orbital, therefore all six electrons have their energy lowered in comparison to the separated atoms. a. b. c. This is depicted in Figure 9.5 in what is called a "molecular orbital energy diagram". Each pair of atomic orbitals, one from each atom, is overlapped to form a bonding and an anti-bonding orbital. The three \(2p\) orbitals from each atom form one \(\sigma\) and \(\sigma^*\) pair and two \(\pi\) and \(\pi^*\) pairs. The lowering of the energies of the electrons in the \(\sigma\) and \(\pi\) orbitals is apparent. The ten \(n = 2\) electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the Pauli Exclusion Principle, each pair in a single orbital consists of one spin up and one spin down electron. Recall now that we began the discussion of bonding in \(\ce{N_2}\) because of the curious result that the ionization energy of an electron in \(\ce{F_2}\) is less than that of an electron in an \(\ce{F}\) atom. By comparing the molecular orbital diagrams for \(\ce{N_2}\) and \(\ce{F_2}\) we are now prepared to answer this puzzle. There are five \(p\) electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Figure 9.6. (Note that the ordering of the bonding \(2p\) orbitals differ between \(\ce{N_2}\) and \(\ce{F_2}\).) We place two electrons in the \(\sigma\) orbitals, four more in the two \(\pi\) orbitals, and four more in the two \(\pi^*\) orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since \(\ce{F_2}\) is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital. This also explains why the ionization energy of \(\ce{F_2}\) is less than that of an \(\ce{F}\) atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Figure 9.6 clearly shows that the highest energy electrons in \(\ce{F_2}\) are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic \(2p\) orbital, because the energy of an anti-bonding orbital is higher than that of the atomic orbitals. (Recall that this is why an anti-bonding orbital is, indeed, anti-bonding.) Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the anti-bonding orbitals. A particularly interesting case is the oxygen molecule, \(\ce{O_2}\). In completing the molecular orbital energy level diagram for oxygen, we discover that we must decide whether to pair the last two electrons in the same \(2p\) \(\pi^*\) orbital, or whether they should be separated into two different \(2p\) \(\pi^*\) orbitals. To determine which, we note that oxygen molecules are paramagnetic, meaning that they are strongly attracted to a magnetic field. To account for this paramagnetism, we recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each "spin up" electron there is a "spin down" electron and their magnetic fields cancel out. When all electrons are paired, the molecule is diamagnetic, meaning that it responds only weakly to a magnetic field. If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy diagram for an \(\ce{O_2}\) molecule is shown in Figure 9.7. In comparing these three diatomic molecules, we recall that \(\ce{N_2}\) has the strongest bond, followed by \(\ce{O_2}\) and \(\ce{F_2}\). We have previously accounted for this comparison with Lewis structures, showing that \(\ce{N_2}\) is a triple bond, \(\ce{O_2}\) is a double bond, and \(\ce{F_2}\) is a single bond. The molecular orbital energy level diagrams in Figures 9.5 to 9.7 cast a new light on this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for \(\ce{N_2}\), 4 for \(\ce{O_2}\), and 6 for \(\ce{F_2}\). Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, we now define the bond order as \[\text{Bond order} = \frac{1}{2} \left( \text{# bonding electrons} - \text{# antibonding electrons} \right)\] Note that, defined this way, the bond order for \(\ce{N_2}\) is 3, for \(\ce{O_2}\) is 2, and for \(\ce{F_2}\) is 1, which agrees with our conclusions from Lewis structures. We conclude that we can predict the relative strengths of bonds by comparing bond orders. Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this affect be measured experimentally? Explain why the bond in an \(\ce{H_2}\) molecule is almost twice as strong as the bond in the \(\ce{H_2^+}\) ion. Explain why the \(\ce{H_2}\) bond is less than twice as strong as the \(\ce{H_2^+}\) bond. \(\ce{Be_2}\) is not a stable molecule. What information can we determine from this observation about the energies of molecular orbitals? Less energy is required to remove an electron from an \(\ce{F_2}\) molecule than to remove an electron from an \(\ce{F}\) atom. Therefore, the energy of that electron is higher in the molecule than in the atom. Explain why, nevertheless, \(\ce{F_2}\) is a stable molecule, i.e., the energy of an \(\ce{F_2}\) molecule is less than the energy of two \(\ce{F}\) atoms. Why do the orbitals of an atom "hybridize" when forming a bond? Calculate the bond orders of the following molecules and predict which molecule in each pair has the stronger bond:
1. \(\ce{C_2}\) or \(\ce{C_2^+}\)
2. \(\ce{B_2}\) or \(\ce{B_2^+}\)
3. \(\ce{F_2}\) or \(\ce{F_2^-}\)
4. \(\ce{O_2}\) or \(\ce{O_2^+}\) Which of the following diatomic molecules are paramagnetic: \(\ce{CO}\), \(\ce{Cl_2}\), \(\ce{NO}\), \(\ce{N_2}\)? \(\ce{B_2}\) is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for \(\ce{B_2}\). ; Chemistry) | 30,070 | 3,277 |
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Complete the changes in concentrations for each of the following reactions: \(\begin{alignat}{3}
&\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\\
& &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\\
& && &&x\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\\
& &&\underline{\hspace{45px}} &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{AgI}(s)⟶&&\ce{Ag+}(aq)\,+\,&&\ce{I-}(aq)\\
& &&x &&\underline{x}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{CaCO3}(s)⟶&&\ce{Ca^2+}(aq)\,+\,&&\ce{CO3^2-}(aq)\\
& &&\underline{x} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg(OH)2}(s)⟶&&\ce{Mg^2+}(aq)\,+\,&&\ce{2OH-}(aq)\\
& &&x &&\underline{2x}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Mg3(PO4)2}(s)⟶&&\ce{3Mg^2+}(aq)\,+\,&&\ce{2PO4^3-}(aq)\\
& &&\underline{3x} &&2x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca5(PO4)3OH}(s)⟶&&\ce{5Ca^2+}(aq)\,+\,&&\ce{3PO4^3-}(aq)\,+\,&&\ce{OH-}(aq)\\
& &&\underline{5x} &&\underline{3x} &&x
\end{alignat}\) Complete the changes in concentrations for each of the following reactions: \(\begin{alignat}{3}
&\ce{BaSO4}(s)⟶&&\ce{Ba^2+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ag2SO4}(s)⟶&&\ce{2Ag+}(aq)\,+\,&&\ce{SO4^2-}(aq)\\
& &&\underline{\hspace{45px}} &&x
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Al(OH)3}(s)⟶&&\ce{Al^3+}(aq)\,+\,&&\ce{3OH-}(aq)\\
& &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Pb(OH)Cl}(s)⟶&&\ce{Pb^2+}(aq)\,+\,&&\ce{OH-}(aq)\,+\,&&\ce{Cl-}(aq)\\
& &&\underline{\hspace{45px}} &&x &&\underline{\hspace{45px}}
\end{alignat}\) \(\begin{alignat}{3}
&\ce{Ca3(AsO4)2}(s)⟶&&\ce{3Ca^2+}(aq)\,+\,&&\ce{2AsO4^3-}(aq)\\
& &&3x &&\underline{\hspace{45px}}
\end{alignat}\) How do the concentrations of Ag and \(\ce{CrO4^2-}\) in a saturated solution above 1.0 g of solid Ag CrO change when 100 g of solid Ag CrO is added to the system? Explain. There is no change. A solid has an activity of 1 whether there is a little or a lot. How do the concentrations of Pb and S change when K S is added to a saturated solution of PbS? What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised? The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve. Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO , CuI, PbCO , PbCl , Tl S, KClO ? Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO , CaF , Hg I , MnCO , ZnS, PbS? CaF , MnCO , and ZnS Write the ionic equation for dissolution and the solubility product ( ) expression for each of the following slightly soluble ionic compounds: Write the ionic equation for the dissolution and the expression for each of the following slightly soluble ionic compounds: The gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. The gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a)1.77 × 10 ; 1.6 × 10 ; 2.2 × 10 ; 7.91 × 10 Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF , Hg Cl , PbI , or Sn(OH) . Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: 2 × 10 ; 1.3 × 10 ; 2.27 × 10 ; 2.2 × 10 Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. 7.2 × 10 = [Ag ], [Cl ] = 0.025 Check: \(\dfrac{7.2×10^{−9}\:M}{0.025\:M}×100\%=2.9×10^{−5}\%\), an insignificant change; 2.2 × 10 = [Ca ], [F ] = 0.0013 Check: \(\dfrac{2.25×10^{−5}\:M}{0.00133\:M}×100\%=1.69\%\). This value is less than 5% and can be ignored. 0.2238 = \(\ce{[SO4^2- ]}\); [Ag ] = 2.30 × 10 Check: \(\dfrac{1.15×10^{−9}}{0.2238}×100\%=5.14×10^{−7}\); the condition is satisfied. [OH ] = 2.8 × 10 ; 5.7 × 10 = [Zn ] Check: \(\dfrac{5.7×10^{−12}}{2.8×10^{−3}}×100\%=2.0×10^{−7}\%\); is less than 5% of [OH ] and is, therefore, negligible. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that it is not appropriate to neglect the changes in the initial concentrations of the common ions. Check: \(\dfrac{7.6×10^{−3}}{0.025}×100\%=30\%\) This value is too large to drop . Therefore solve by using the quadratic equation: Check: \(\dfrac{1.7×10^{−3}}{0.0313}×100\%=5.5\%\) This value is too large to drop x, and the entire equation must be solved. \(\ce{[C2O4^2- ]}=3.5×10^{−3}\) Check: \(\dfrac{3.5×10^{−3}}{0.02444}×100\%=14\%\) This value is greater than 5%, so the quadratic equation must be used: Check: \(\dfrac{3.15×10^{−3}}{0.050}×100\%=6.28\%\) This value is greater than 5%, so a more exact method, such as successive approximations, must be used. Explain why the changes in concentrations of the common ions in can be neglected. Explain why the changes in concentrations of the common ions in cannot be neglected. The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change. Calculate the solubility of aluminum hydroxide, Al(OH) , in a solution buffered at pH 11.00. Refer to for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter. CaSO ∙2H O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract. This use of BaSO is possible because of its low solubility. Calculate the molar solubility of BaSO and the mass of barium present in 1.00 L of water saturated with BaSO . Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10 ) of \(\ce{SO4^2-}\) because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO (“gyp” water) as a result or passing through soil containing gypsum, CaSO •2H O, meet these standards? What is \(\ce{SO4^2-}\) in such water? 4.9 × 10 = \(\ce{[SO4^2- ]}\) = [Ca ]; Since this concentration is higher than 2.60 × 10 , “gyp” water does not meet the standards. Perform the following calculations: The solubility product of CaSO •2H O is 2.4 × 10 . What mass of this salt will dissolve in 1.0 L of 0.010 \(\ce{SO4^2-}\)? Mass (CaSO •2H O) = 0.34 g/L Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products). Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Table E3 for solubility products). [Ag ] = [I ] = 1.2 × 10 ; [Ag ] = 2.86 × 10 , \(\ce{[SO4^2- ]}\) = 1.43 × 10 ; [Mn ] = 2.2 × 10 , [OH ] = 4.5 × 10 ; [Sr ] = 4.3 × 10 , [OH ] = 8.6 × 10 ; [Mg ] = 1.6 × 10 , [OH ] = 3.1 × 10 . The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate for each of the slightly soluble solids indicated: The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate for each of the slightly soluble solids indicated: 2.0 × 10 ; 5.1 × 10 ; 1.35 × 10 ; 1.18 × 10 ; 1.08 × 10 Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for values.) Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for values.) Calculate the concentration of Tl when TlCl just begins to precipitate from a solution that is 0.0250 in Cl . Calculate the concentration of sulfate ion when BaSO just begins to precipitate from a solution that is 0.0758 in Ba . 1.42 × 10 Calculate the concentration of Sr when SrF starts to precipitate from a solution that is 0.0025 in F . Calculate the concentration of \(\ce{PO4^3-}\) when Ag PO starts to precipitate from a solution that is 0.0125 in Ag . 9.2 × 10 Calculate the concentration of F required to begin precipitation of CaF in a solution that is 0.010 in Ca . Calculate the concentration of Ag required to begin precipitation of Ag CO in a solution that is 2.50 × 10 in \(\ce{CO3^2-}\). [Ag ] = 1.8 × 10 What [Ag ] is required to reduce \(\ce{[CO3^2- ]}\) to 8.2 × 10 by precipitation of Ag CO ? What [F ] is required to reduce [Ca ] to 1.0 × 10 by precipitation of CaF ? 6.2 × 10 A volume of 0.800 L of a 2 × 10 - Ba(NO ) solution is added to 0.200 L of 5 × 10 Li SO . Does BaSO precipitate? Explain your answer. Perform these calculations for nickel(II) carbonate. 2.28 L; 7.3 × 10 g Iron concentrations greater than 5.4 × 10 in water used for laundry purposes can cause staining. What [OH ] is required to reduce [Fe ] to this level by precipitation of Fe(OH) ? A solution is 0.010 in both Cu and Cd . What percentage of Cd remains in the solution when 99.9% of the Cu has been precipitated as CuS by adding sulfide? 100% of it is dissolved A solution is 0.15 in both Pb and Ag . If Cl is added to this solution, what is [Ag ] when PbCl begins to precipitate? What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 with respect to each ion? In some cases it may be necessary to control the pH. (Hint: Consider the values given in .) A solution contains 1.0 × 10 mol of KBr and 0.10 mol of KCl per liter. AgNO is gradually added to this solution. Which forms first, solid AgBr or solid AgCl? A solution contains 1.0 × 10 mol of KI and 0.10 mol of KCl per liter. AgNO is gradually added to this solution. Which forms first, solid AgI or solid AgCl? AgI will precipitate first. The calcium ions in human blood serum are necessary for coagulation. Potassium oxalate, K C O , is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC O •H O. It is necessary to remove all but 1.0% of the Ca in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca per 100 mL of serum, what mass of K C O is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K value for CaC O in serum is the same as in water.) About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca (PO ) . The normal mid range calcium content excreted in the urine is 0.10 g of Ca per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form? 4 × 10 The pH of normal urine is 6.30, and the total phosphate concentration (\(\ce{[PO4^3- ]}\) + \(\ce{[HPO4^2- ]}\) + \(\ce{[H2PO4- ]}\) + [H PO ]) is 0.020 . What is the minimum concentration of Ca necessary to induce kidney stone formation? (See for additional information.) Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions: \(\ce{Mg^2+}(aq)+\ce{Ca(OH)2}(aq)⟶\ce{Mg(OH)2}(s)+\ce{Ca^2+}(aq)\) \(\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)⟶\ce{MgCl2}(s)+\ce{2H2O}(l)\) \(\ce{MgCl2}(l)\xrightarrow{\ce{electrolysis}}\ce{Mg}(s)+\ce{Cl2}(g)\) Sea water has a density of 1.026 g/cm and contains 1272 parts per million of magnesium as Mg ( ) by mass. What mass, in kilograms, of Ca(OH) is required to precipitate 99.9% of the magnesium in 1.00 × 10 L of sea water? 3.99 kg Hydrogen sulfide is bubbled into a solution that is 0.10 in both Pb and Fe and 0.30 in HCl. After the solution has come to equilibrium it is saturated with H S ([H S] = 0.10 ). What concentrations of Pb and Fe remain in the solution? For a saturated solution of H S we can use the equilibrium: \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) (Hint: The \(\ce{[H3O+]}\) changes as metal sulfides precipitate.) Perform the following calculations involving concentrations of iodate ions: 3.1 × 10 ; [Cu ] = 2.6 × 10 ; \(\ce{[IO3- ]}\) = 5.3 × 10 Calculate the molar solubility of AgBr in 0.035 NaBr ( = 5 × 10 ). How many grams of Pb(OH) will dissolve in 500 mL of a 0.050- PbCl solution ( = 1.2 × 10 )? 1.8 × 10 g Pb(OH) Use the from the earlier Link to Learning to complete the following exercise:. Using 0.01 g CaF , give the K values found in a 0.2- solution of each of the salts. Discuss why the values change as you change soluble salts. How many grams of Milk of Magnesia, Mg(OH) ( ) (58.3 g/mol), would be soluble in 200 mL of water. = 7.1 × 10 . Include the ionic reaction and the expression for in your answer. ( = 1 × 10 = [H O ,OH ]) \[\ce{Mg(OH)2}(s)⇌\ce{Mg^2+}+\ce{2OH-} \] \[K_\ce{sp}=\ce{[Mg^2+,OH- ]^2}\] \[1.14 × 10 g Mg(OH) Two hypothetical salts, LM and LQ, have the same molar solubility in H O. If for LM is 3.20 × 10 , what is the value for LQ? Which of the following carbonates will form first? Which of the following will form last? Explain. SrCO will form first, since it has the smallest value it is the least soluble. BaCO will be the last to precipitate, it has the largest value. How many grams of Zn(CN) ( ) (117.44 g/mol) would be soluble in 100 mL of H O? Include the balanced reaction and the expression for in your answer. The value for Zn(CN) ( ) is 3.0 × 10 . Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water? when the amount of solid is so small that a saturated solution is not produced Explain why the addition of NH or HNO to a saturated solution of Ag CO in contact with solid Ag CO increases the solubility of the solid. Calculate the cadmium ion concentration, [Cd ], in a solution prepared by mixing 0.100 L of 0.0100 Cd(NO ) with 1.150 L of 0.100 NH ( ). 2.35 × 10 Explain why addition of NH or HNO to a saturated solution of Cu(OH) in contact with solid Cu(OH) increases the solubility of the solid. Sometimes equilibria for complex ions are described in terms of dissociation constants, . For the complex ion \(\ce{AlF6^3-}\) the dissociation reaction is: \[\ce{AlF6^3- ⇌ Al^3+ + 6F-}\) and \(K_\ce{d}=\ce{\dfrac{[Al^3+,F- ]^6}{[AlF6^3- ]}}=2×10^{−24}\] Calculate the value of the formation constant, , for \(\ce{AlF6^3-}\). 5 × 10 Using the value of the formation constant for the complex ion \(\ce{Co(NH3)6^2+}\), calculate the dissociation constant. Using the dissociation constant, = 7.8 × 10 , calculate the equilibrium concentrations of Cd and CN in a 0.250- solution of \(\ce{Cd(CN)4^2-}\). [Cd ] = 9.5 × 10 ; [CN ] = 3.8 × 10 Using the dissociation constant, = 3.4 × 10 , calculate the equilibrium concentrations of Zn and OH in a 0.0465- solution of \(\ce{Zn(OH)4^2-}\). Using the dissociation constant, = 2.2 × 10 , calculate the equilibrium concentrations of Co and NH in a 0.500- solution of \(\ce{Co(NH3)6^3+}\). [Co ] = 3.0 × 10 ; [NH ] = 1.8 × 10 Using the dissociation constant, = 1 × 10 , calculate the equilibrium concentrations of Fe and CN in a 0.333 M solution of \(\ce{Fe(CN)6^3-}\). Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve 2.0 × 10 mol of silver cyanide, AgCN. 1.3 g Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve 3.0 × 10 mol of silver bromide. A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of Na S O •5H O (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as \(\ce{Ag(S2O3)2^3-}\) ( = 4.7 × 10 )? 0.80 g We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, H ) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions. Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: (a) ; (b) ; (c) ; (d) ; (e) Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each: Using Lewis structures, write balanced equations for the following reactions: (a) ; \(\ce{H3O+ + CH3- ⟶ CH4 + H2O}\) ; \(\ce{CaO + SO3 ⟶ CaSO4}\) ; \(\ce{NH4+ + C2H5O- ⟶ C2H5OH + NH3}\) Calculate \(\ce{[HgCl4^2- ]}\) in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100- HgCl solution. In a titration of cyanide ion, 28.72 mL of 0.0100 AgNO is added before precipitation begins. [The reaction of Ag with CN goes to completion, producing the \(\ce{Ag(CN)2-}\) complex.] Precipitation of solid AgCN takes place when excess Ag is added to the solution, above the amount needed to complete the formation of \(\ce{Ag(CN)2-}\). How many grams of NaCN were in the original sample? 0.0281 g What are the concentrations of Ag , CN , and \(\ce{Ag(CN)2-}\) in a saturated solution of AgCN? In dilute aqueous solution HF acts as a weak acid. However, pure liquid HF (boiling point = 19.5 °C) is a strong acid. In liquid HF, HNO acts like a base and accepts protons. The acidity of liquid HF can be increased by adding one of several inorganic fluorides that are Lewis acids and accept F ion (for example, BF or SbF ). Write balanced chemical equations for the reaction of pure HNO with pure HF and of pure HF with BF . \(\ce{HNO3}(l)+\ce{HF}(l)⟶\ce{H2NO3+}+\ce{F-}\); \(\ce{HF}(l)+\ce{BF3}(g)⟶\ce{H+}+\ce{BF4}\) The simplest amino acid is glycine, H NCH CO H. The common feature of amino acids is that they contain the functional groups: an amine group, –NH , and a carboxylic acid group, –CO H. An amino acid can function as either an acid or a base. For glycine, the acid strength of the carboxyl group is about the same as that of acetic acid, CH CO H, and the base strength of the amino group is slightly greater than that of ammonia, NH . Write the Lewis structures of the ions that form when glycine is dissolved in 1 HCl and in 1 KOH. Write the Lewis structure of glycine when this amino acid is dissolved in water. (Hint: Consider the relative base strengths of the –NH and \(\ce{−CO2-}\) groups.) Boric acid, H BO , is not a Brønsted-Lowry acid but a Lewis acid. \(\ce{H3BO3 + H2O ⟶ H4BO4- + H+}\); The electronic and molecular shapes are the same—both tetrahedral. The tetrahedral structure is consistent with hybridization. A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium? Calculate the equilibrium concentration of Ni in a 1.0- solution [Ni(NH ) ](NO ) . 0.014 Calculate the equilibrium concentration of Zn in a 0.30- solution of \(\ce{Zn(CN)4^2-}\). Calculate the equilibrium concentration of Cu in a solution initially with 0.050 Cu and 1.00 NH . 1.0 × 10 Calculate the equilibrium concentration of Zn in a solution initially with 0.150 Zn and 2.50 CN . Calculate the Fe equilibrium concentration when 0.0888 mole of K [Fe(CN) ] is added to a solution with 0.0.00010 CN . 9 × 10 Calculate the Co equilibrium concentration when 0.100 mole of [Co(NH ) ](NO ) is added to a solution with 0.025 NH . Assume the volume is 1.00 L. The equilibrium constant for the reaction \(\ce{Hg^2+}(aq)+\ce{2Cl-}(aq)⇌\ce{HgCl2}(aq)\) is 1.6 × 10 . Is HgCl a strong electrolyte or a weak electrolyte? What are the concentrations of Hg and Cl in a 0.015- solution of HgCl ? 6.2 × 10 = [Hg ]; 1.2 × 10 = [Cl ]; The substance is a weak electrolyte because very little of the initial 0.015 HgCl dissolved. Calculate the molar solubility of Sn(OH) in a buffer solution containing equal concentrations of NH and \(\ce{NH4+}\). Calculate the molar solubility of Al(OH) in a buffer solution with 0.100 NH and 0.400 \(\ce{NH4+}\). [OH ] = 4.5 × 10 ; [Al ] = 2.1 × 10 (molar solubility) What is the molar solubility of CaF in a 0.100- solution of HF? for HF = 7.2 × 10 . What is the molar solubility of BaSO in a 0.250- solution of NaHSO ? for \(\ce{HSO4-}\) = 1.2 × 10 . What is the molar solubility of Tl(OH) in a 0.10- solution of NH ? What is the molar solubility of Pb(OH) in a 0.138- solution of CH NH ? A solution of 0.075 CoBr is saturated with H S ([H S] = 0.10 ). What is the minimum pH at which CoS begins to precipitate? \(\ce{CoS}(s)⇌\ce{Co^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.5×10^{−27}\) \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) A 0.125- solution of Mn(NO ) is saturated with H S ([H S] = 0.10 ). At what pH does MnS begin to precipitate? \(\ce{MnS}(s)⇌\ce{Mn^2+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_\ce{sp}=4.3×10^{−22}\) \(\ce{H2S}(aq)+\ce{2H2O}(l)⇌\ce{2H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K=1.0×10^{−26}\) 3.27 Calculate the molar solubility of BaF in a buffer solution containing 0.20 HF and 0.20 NaF. Calculate the molar solubility of CdCO in a buffer solution containing 0.115 Na CO and 0.120 NaHCO To a 0.10- solution of Pb(NO ) is added enough HF( ) to make [HF] = 0.10 . Calculate the concentration of Cd resulting from the dissolution of CdCO in a solution that is 0.010 in H CO . 1 × 10 Both AgCl and AgI dissolve in NH . Calculate the volume of 1.50 CH CO H required to dissolve a precipitate composed of 350 mg each of CaCO , SrCO , and BaCO . 0.0102 L (10.2 mL) Even though Ca(OH) is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH) ? What mass of NaCN must be added to 1 L of 0.010 Mg(NO ) in order to produce the first trace of Mg(OH) ? 5 × 10 g Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.) The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: \(\ce{MgF2}(s)⇌\ce{Mg^2+}(aq)+\ce{2F-}(aq)\) In a saturated solution of MgF at 18 °C, the concentration of Mg is 1.21 × 10 . The equilibrium is represented by the preceding equation. = [Mg ,F ] = (1.21 × 10 )(2 × 1.21 × 10 ) = 7.09 × 10 ; 7.09 × 10 Determine the concentration of Mg and F that will be present in the final volume. Compare the value of the ion product [Mg ,F ] with . If this value is larger than , precipitation will occur. 0.1000 L × 3.00 × 10 Mg(NO ) = 0.3000 L × Mg(NO ) Mg(NO ) = 1.00 × 10 0.2000 L × 2.00 × 10 NaF = 0.3000 L × NaF NaF = 1.33 × 10 ion product = (1.00 × 10 )(1.33 × 10 ) = 1.77 × 10 This value is smaller than , so no precipitation will occur. MgF is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Chatelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic. Which of the following compounds, when dissolved in a 0.01- solution of HClO , has a solubility greater than in pure water: CuCl, CaCO , MnS, PbBr , CaF ? Explain your answer. Which of the following compounds, when dissolved in a 0.01- solution of HClO , has a solubility greater than in pure water: AgBr, BaF , Ca (PO ) , ZnS, PbI ? Explain your answer. BaF , Ca (PO ) , ZnS; each is a salt of a weak acid, and the \(\ce{[H3O+]}\) from perchloric acid reduces the equilibrium concentration of the anion, thereby increasing the concentration of the cations What is the effect on the amount of solid Mg(OH) that dissolves and the concentrations of Mg and OH when each of the following are added to a mixture of solid Mg(OH) and water at equilibrium? What is the effect on the amount of CaHPO that dissolves and the concentrations of Ca and \(\ce{HPO4-}\) when each of the following are added to a mixture of solid CaHPO and water at equilibrium? Effect on amount of solid CaHPO , [Ca ], [OH ]: increase, increase, decrease; decrease, increase, decrease; no effect, no effect, no effect; decrease, increase, decrease; increase, no effect, no effect Identify all chemical species present in an aqueous solution of Ca (PO ) and list these species in decreasing order of their concentrations. (Hint: Remember that the \(\ce{PO4^3-}\) ion is a weak base.) A volume of 50 mL of 1.8 NH is mixed with an equal volume of a solution containing 0.95 g of MgCl . What mass of NH Cl must be added to the resulting solution to prevent the precipitation of Mg(OH) ? 7.1 g | 27,632 | 3,278 |
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This page describes the structure of metals, and relates that structure to the physical properties of the metal. Metals are giant structures of atoms held together by metallic bonds. "Giant" implies that large but variable numbers of atoms are involved - depending on the size of the bit of metal. Most metals are close packed - that is, they fit as many atoms as possible into the available volume. Each atom in the structure has 12 touching neighbors. Such a metal is described as 12-coordinated. Each atom has 6 other atoms touching it in each layer. There are also 3 atoms touching any particular atom in the layer above and another 3 in the layer underneath. This second diagram shows the layer immediately above the first layer. There will be a corresponding layer underneath (there are actually two different ways of placing the third layer in a close packed structure, but that goes beyond the topic here). Some metals (notably those in Group 1 of the Periodic Table) are packed less efficiently, having only 8 touching neighbors. These are 8-coordinated. The left hand diagram shows that no atoms are touching each other within a particular layer . They are only touched by the atoms in the layers above and below. The right hand diagram shows the 8 atoms (4 above and 4 below) touching the darker colored one. It would be misleading to suppose that all the atoms in a piece of metal are arranged in a regular way. Any piece of metal is made up of a large number of "crystal grains", which are regions of regularity. At the grain boundaries atoms have become misaligned. Metals tend to have high melting and boiling points because of the strength of the metallic bond. The strength of the bond varies from metal to metal and depends on the number of electrons which each atom delocalizes into the sea of electrons, and on the packing. Group 1 metals like sodium and potassium have relatively low melting and boiling points mainly because each atom only has one electron to contribute to the bond - but there are other problems as well: Metals conduct electricity. The delocalized electrons are free to move throughout the structure in 3-dimensions. They can cross grain boundaries. Even though the pattern may be disrupted at the boundary, as long as atoms are touching each other, the metallic bond is still present. Liquid metals also conduct electricity, showing that although the metal atoms may be free to move, the delocalization remains in force until the metal boils. Metals are good conductors of heat. Heat energy is picked up by the electrons as additional kinetic energy (it makes them move faster). The energy is transferred throughout the rest of the metal by the moving electrons. Metals are described as malleable (can be beaten into sheets) and ductile (can be pulled out into wires). This is because of the ability of the atoms to roll over each other into new positions without breaking the metallic bond. If a small stress is put onto the metal, the layers of atoms will start to roll over each other. If the stress is released again, they will fall back to their original positions. Under these circumstances, the metal is said to be elastic. If a larger stress is put on, the atoms roll over each other into a new position, and the metal is permanently changed. This rolling of layers of atoms over each other is hindered by grain boundaries because the rows of atoms don't line up properly. It follows that the more grain boundaries there are (the smaller the individual crystal grains), the harder the metal becomes. Offsetting this, because the grain boundaries are areas where the atoms aren't in such good contact with each other, metals tend to fracture at grain boundaries. Increasing the number of grain boundaries not only makes the metal harder, but also makes it more brittle. If you have a pure piece of metal, you can control the size of the grains by heat treatment or by working the metal. Heating a metal tends to shake the atoms into a more regular arrangement - decreasing the number of grain boundaries, and so making the metal softer. Banging the metal around when it is cold tends to produce lots of small grains. Cold working therefore makes a metal harder. To restore its workability, you would need to reheat it. You can also break up the regular arrangement of the atoms by inserting atoms of a slightly different size into the structure. such as brass (a mixture of copper and zinc) are harder than the original metals because the irregularity in the structure helps to stop rows of atoms from slipping over each other. Jim Clark ( ) | 4,604 | 3,279 |
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The include fluorine, chlorine, bromine, and iodine. Astatine is also in the group, but is radioactive and will not be considered here. A summary of atomic properties of the halogens is given in the following table. The free elemental halogens all consist of diatomic molecules X , where X may be fluorine, chlorine, bromine, or iodine (recall the microscopic picture of bromine). They are strong oxidizing agents and are readily reduced to the X ions, and so the halogens form numerous ionic compounds. Fluorine, the most electronegative element, has no positive oxidation states, but the other halogens commonly exhibit +1, +3, +5, and +7 states. Most compounds containing halogens in positive oxidation states are good oxidizing agents, however, reflecting the strong tendency of these elements to gain electrons. Density/ g cm Electro- negativity Melting Point (in °C) -220 There is some variation among their physical properties and appearance. Fluorine and chlorine are both gases at room temperature, the former very pale yellow, and the latter yellow-green in color. Bromine is a red-brown liquid which vaporizes rather easily. Iodine forms shiny dark crystals and, when heated, sublimes (changes directly from solid to gas) to a beautiful violet vapor. All the gases produce a choking sensation when inhaled. Chlorine was used to poison soldiers on European battlefields in 1915 to 1918. Halogens are put to more humane uses such as to disinfect public water supplies by means of chlorination and to treat minor cuts by using an alcohol solution (tincture) of iodine. These applications depend on the ability of the halogens to destroy microorganisms which are harmful to humans. All halogens are quite reactive, and in the natural world they always occur combined with other elements. Fluorine reacts so readily with almost any substance it contacts that chemists were not successful in isolating pure fluorine until 1886, although its existence in compounds had been known for many years. Chlorine, bromine, and iodine are progressively less reactive but still form compounds with most other elements, especially metals. A good example is mercury, whose reaction with bromine was discussed in the section covering macroscopic and microscopic views of a chemical reaction. Mercury reacts with other halogens in the same way: \[\text{Hg}(l) + \text{X}_2(g, l, or s) \rightarrow \text{HgX}_2 (s) \nonumber \] X = F, Cl, Br, or I Already covered in the section on alkali metals, halogens react readily with alkali metals with the general form of: \[\text{2M} + \text{X}_2 \rightarrow \text{2MX} \nonumber \] M = Li, Na, K, Rb, or Cs and X = F, Cl, Br, I Iodine combines less vigorously with alkali metals than other halogens, but its reactions are analogous to the reactions of alkali metals with florine, chlorine and bromine. Compounds of an alkali metal and a halogen, such as sodium chloride, potassium fluoride, lithium bromide, or cesium iodide, have closely related properties. (All taste salty, for example.) They belong to a general category called , all of whose members are similar to ordinary table salt, sodium chloride. The term halogen is derived from Greek words meaning “salt former.” Halogens also react with alkaline-earth metals in the general reaction: \[\text{M} + \text{X}_2 \rightarrow \text{MCl}_2 \nonumber \] M = Be, Mg, Ca, Sr, Ba, or Ra and X = F, Cl, Br, I Another vigorous reaction occurs when certain compounds containing carbon and hydrogen contact the halogens. Turpentine, C H , reacts quite violently. In the case of fluorine and chlorine the equation is \[\text{C}_{10}\text{H}_{16}(l) + \text{8X}_2(g) \rightarrow \text{10C}(s) + \text{16HX}(g) \nonumber \] X = F, Cl but the products are different when bromine and iodine react. Before the advent of the automobile, veterinarians used solid iodine and turpentine to disinfect wounds in horses’ hooves. This may have been because of the superior antiseptic qualities of the mixture. However, a more likely reason is the profound impression made on the owner of the horse by the great clouds of violet iodine vapor which sublimed as a result of the increase in temperature when the reaction occurred! Below is a video of this impressive reaction: The violent reaction is due to α-pinene in turpentine. The relief of ring strain is highly exothermic. This temperature increase causes the sublimation leading to the impressive violet iodine vapor. The halogens also react directly with hydrogen, yielding the hydrogen halides: \[\text{H}_2 + \text{X}_2 \rightarrow \text{2HX} \nonumber \] X = F, Cl, Br, I These compounds are all gases, are water soluble, and, except for HF, are strong acids in aqueous solution. They are conveniently prepared in the laboratory by acidifying the appropriate sodium or other halide: \[\text{NaCl}(s) + \text{H}_3\text{O}^{+}(aq) \xrightarrow{\Delta} \text{Na}^{+}(aq) + \text{H}_2\text{O}(l) + \text{HCl}(g) \label{6} \] The acid must be nonvolatile so that heating will drive off only the gaseous hydrogen halide. In the case of fluorides and chlorides, H SO will do, but bromides and iodides are oxidized to Br or I by hot H SO and so H PO is used instead. A reaction similar to Eq. \(\ref{6}\) occurs when phosphate rock containing fluorapatite is treated with H SO to make fertilizer: \[\text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + \text{7H}_2\text{SO}_4 + \text{3H}_2\text{O} \rightarrow \text{3Ca(H}_2\text{PO}_4)_2•\text{H}_2\text{O} + \text{7CaSO}_4 + \text{2HF} \nonumber \] The HF produced in this reaction can cause significant air-pollution problems. Fluorides are also emitted to the atmosphere in steelmaking and aluminum production. There is some evidence that fluorides, rather than sulfur dioxide, may have been responsible for human deaths in air-pollution episodes at Donora, Pennsylvania, and the Meuse Valley in Belgium. The relative oxidizing strengths of the halogens can be illustrated nicely in the laboratory. If, for example, a solution of Cl in H O is combined with a solution of NaI, the dark color of I can be observed, showing that the Cl has oxidized the I : This very reaction is shown in the following video: The video starts out with four solutions. The experimental solution is on the far left, and contain Cl in water, which is covered by a layer of hexane, a nonpolar solvent which is immiscible with H O. The three other solutions, from left to right are a Cl solution, a Br solution, and an I solution. When a solution with iodide ions is added to the experimental solution, nonpolar I molecules are formed. They concentrate in the hexane layer, and a beautiful violet color can be observed, the same as I solution. From such experiments it can be shown that the strongest oxidizing agent is F (at the top of the group). F will react with Cl , Br , and I . The weakest oxidizing agent, I , does not react with any of the halide ions. The extremely high oxidizing power of F makes it the only element which can combine directly with a noble gas. The reactions \(\text{Xe}(g) + \text{F}_2(g) \rightarrow \text{XeF}_2(s)\) \(\text{XeF}_2(s) + \text{F}_2(g) \rightarrow \text{XeF}_4(s)\) \(\text{XeF}_4(s) + \text{F}_2(g) \rightarrow \text{XeF}_6(s)\) may be used to synthesize the three xenon fluorides, all of which are strong oxidizing agents. When an electrical discharge is passed through a mixture of Kr and F at a low temperature, KrF can be formed. This is the only compound of Kr, and it decomposes slowly at room temperature. Fluorine is also set apart from the other halogens because of its ability to oxidize water: \[\text{3F}_2 + \text{6H}_2\text{O} \rightarrow \text{4H}_3\text{O}^{+} + \text{4F}^{-} + \text{O}_2 \nonumber \] Chlorine is also capable of oxidizing water, but it does so very slowly. Instead the reaction \[\text{Cl}_2 + \text{2H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^{+} + \text{Cl}^{-} + \text{HOCl} \nonumber \] goes partway to completion. Hypochlorous acid, HOCl, is a weak acid. Small concentrations of hypobromous and hypoiodous acids can also be obtained in this way. In basic solution the halogen is completely consumed, producing the hypohalite anion: \[\text{Cl}_2 + \text{2OH}^{-} \rightarrow \text{Cl}^{-} + \text{H}_2\text{O} + \text{OCl}^{-} \nonumber \] Since hypochlorite, OCl , could also be supplied from an ionic compound such as NaOCl, the latter is often used to chlorinate swimming pools. Hypohalite ions disproportionate in aqueous solution: \[\text{3OCl}^{-} \rightarrow \text{2Cl}^{-} + \text{ClO}_3^{-} \nonumber \] This reaction is rather slow for hypochlorite unless the temperature is above 75°C, but OBr and OI are consumed immediately at room temperature. Chlorate, ClO , bromate, BrO , and iodate, IO , salts can be precipitated from such solutions. All are good oxidizing agents. Potassium chlorate, KClO , decomposes, giving O when heated in the presence of a catalyst: \[\text{2KClO}_{3} \xrightarrow[\text{MnO}_{\text{2}}\text{ catalyst}]{\Delta } \text{2KCl} + \text{3O}_{2} \nonumber \] This is a standard laboratory reaction for making O . If KClO is heated without a catalyst, potassium perchlorate, KClO , may be formed. Perchlorates oxidize organic matter rapidly and often uncontrollably. They are notorious for exploding unexpectedly and should be handled with great care. One other interesting group of compounds is the interhalogens, in which one halogen bonds to another. Some interhalogens, such as BrCl, are diatomic, but the larger halogen atoms have room for several smaller ones around them. Thus compounds such as ClF , BrF and BrF , and IF , ICl , IF , and IF can be synthesized. Note that the largest halogen atom I can accommodate three chlorines and up to seven fluorines around it. The following video showcases a reaction which involves some of these interhalogens: The video begins with a test tube containing a layer of KI aqueous solution on top of CCl below it. Chlorine is bubbled through the KI layer. As seen in the video on oxidizing strength of the halogens, Cl reacts with I to form iodine, according to the reaction: \[\text{2I}^{-}(aq ) + \text{Cl}_2(aq ) \rightarrow \text{I}_2(aq) + \text{2Cl}^{-}(aq) \nonumber \] A brown triiodide ion is also formed in the aqueous layer, according to the reaction: \[\text{I}^{-}(aq ) + \text{I}_2(aq ) \rightarrow \text{I}_3^{-}(aq) \nonumber \] A purple solution begins to form in the CCl layer, as iodine dissolves in it. The iodine in the aqueous layer also reacts with the excess Cl to form the red ICl, according to the following reaction: \[\text{I}_2(aq ) + \text{Cl}_2(aq ) \rightarrow \text{2ICl}(aq) \nonumber \] The final reaction takes place as more Cl is added, which reacts with ICl, to form the yellow ICl . This reaction causes the aqueous solution to decolorize. This goes according to the reaction: \[\text{ICl}(aq ) + \text{Cl}_2(aq ) \rightarrow \text{ICl}_3(aq) \nonumber \] At the end of the video, the layers have decolorized, with a red portion in the CCl which is, due to its color, most likely remaining ICl. | 11,120 | 3,281 |
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A characteristic of the diagram is that it has been assigned an identifying label, namely, 1 . This enables us to distinguish it from other wave patterns the electron could possibly adopt if it moved about the nucleus with a higher energy. Each of these three-dimensional wave patterns is different in shape, size, or orientation from all the others and is called an . The word is used in order to make a distinction between these wave patterns and the circular or elliptical of the Bohr picture shown in . In the case of a , the energy was determined by a positive whole number . Much the same situation prevails in the case of the hydrogen atom. An integer called the , also designated by the symbol , is used to label each orbital. The larger the value of , the greater the energy of the electron and the larger the average distance of the electron cloud from the nucleus. The energy increases with n, in part, because the total number of nodes is n-1 for each wavefunction in shell n. The next quantum number, represented by and called the "angular quantum number," can be any value in the range 0, 1, 2, ... - 1. As we have seen in the case of 2-dimensional drum vibrations, specifies the number of planar nodes in the wavefunction. This number represents the angular momentum of the orbital, and is important because it determines the shape of the orbital. This number is responsible for the s, p, d, f, etc., character of the orbital. = 0 corresponds to an s orbital, = 1 denotes a p orbital, and so forth. The "magnetic quantum number" corresponds to the projection of the orbital along an axis, i.e. when in three-dimensional space, along the x, y, or z axis. This value falls in the range of - , - + 1, ... -1, 0, 1, ... - 1, . The fourth quantum number, known as the "spin quantum number," refers to the intrinsic "spin" of the electron. This quantum number may hold only two values, either -1/2 or +1/2. states that each electron must have a set of four quantum numbers, so if two electrons are paired together in an orbital, they share three quantum numbers and must have opposite spin quantum numbers. This electron spin property is what causes a substance to be or , because a moving charge always creates a magnetic field. Substances whose atoms, molecules, or ions contain unpaired electrons (which must be in different orbitals) are weakly attracted into a magnetic field, a property known as . This is because the Spin Quantum Number for the substance will not be zero since each electron will not have a partner to cancel. Paramagnetism is typically 0.1% as strong as the familiar "ferromagnetism" of common magnets. Most substances have all their electrons paired. This means that each electron's spin number will be canceled by another electron (although they're usually in the same orbital, they need not be). The net spin will be zero for the substance, and it will not be attracted into a magnetic field, but actually repelled slightly. The repulsion is typically 0.1% as great as paramagnetic attraction. This property is known as . Hence measurement of magnetic properties can tell us whether all electrons are paired or not. \(\Page {2}\) There are several other videos on YouTube showing | 3,264 | 3,282 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,283 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,284 |
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You are expected to be able to define and explain the significance of terms identified in . Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly distributed throughout the water. What you will see is the reverse of this process, in which the tea would be sucked up and re-absorbed by the teabag. The making of tea, like all changes that take place in the world, possesses a “natural” direction. Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as or changes. In many cases our everyday life experiences teach us the direction in which spontaneous change can occur, and anything that runs counter to these expectations is immediately sensed as weird. In other cases, including that of most chemical change, we often have no obvious guidelines, and must learn how to apply the laws of thermodynamics which ultimately govern all spontaneous change. In order to answer the above question, let's begin by thinking about the outcomes of the following four experiments, each of which illustrates a A stack of one hundred coins is thrown into the air. After they have come to rest on the floor, the numbers that land “heads up” and “tails up” are noted. ordered coins → randomized coins (roughly equal numbers of heads and tails.) no relevant net change in energy Simple statistics shows that the probability that the coins will land in one particular arrangement of the huge number that are possible is vanishingly small. Two identical blocks of copper, one at 200°C and the other at 100°C, are brought into contact in a thermally-insulated environment. Eventually the temperatures of both blocks reach 150°C. block1 (200°) + block2 (100°) → combined blocks (150°) Heat (randomized molecular kinetic energy) flows from the warmer block to the cooler one until their temperatures are identical. Dispersal of kinetic energy amongst the copper atoms is a random process; the chances that such a process would lead to a non-uniform sharing of the energy are even smaller than in the case of the 100 coins because of the much greater number (around 10 ) of particles involved. A book or some other solid object is held above a table top, and is then allowed to fall. book in air book on table top; potential energy organized kinetic energy thermal energy. At the instant just before the end of its fall, the potential energy the object acquired when it was raised will exist entirely as kinetic energy /2 in which is the mass of the object and is its velocity. Each atom of which the object is composed will of course possess a proportionate fraction of this energy, again with its principal velocity component pointing down. Superimposed on this, however, will be minute thermal displacements that vary randomly in magnitude and direction from one instant to the next. The sum total of these constitutes the thermal energy contained in the object. When the object strikes the table top, its motion ceases and we say its kinetic energy is zero. Energy is supposed to be conserved, so where did it disappear to? The shock of impact has resulted in its dispersal into greatly augmented thermal motions of the atoms, both of the object itself and of the area of the table top where the impact occurred. In other words, the kinetic energy of organized motion the object had just before its motion stopped has been transformed into kinetic energy of random or disorganized motion (thermal energy) which spreads rapidly away from the point of impact. Once the kinetic energy of the book has been dispersed amongst the molecules of the book and the table top, the probability of these randomized motions reappearing at the surface where the two objects are in contact and then acting in concert to propel the object back into the air) is negligible. One mole of gas, initially at 300 K and 2 atm pressure, is allowed to expand to double its volume, keeping the temperature constant. : Increase in volume of gas. : No change in energy if the gas behaves ideally. Simple statistics: the probability that N randomly moving objects (flies in a bottle, for example,) will at any time all be located in one half of the container is (1/2) . For chemically-significant values of N (10 , say) this probability is indistinguishable from zero. , meaning that What determines the direction in which spontaneous change will occur? , since in most cases cited above the energy of the system did not change. Even in the case of the falling book, in which the potential energy of the system (the book) falls, energy is conserved overall; if there is no net loss of energy when these processes operate in the forward or natural direction, it would not require any expenditure of energy for them to operate in reverse. In other words, contrary to what all too many people appear to believe, the cannot predict the direction of a natural process. thus the cannot predict the direction of a natural process In our examination of the processes described above, we saw that although the total energy of the system and the surroundings (and thus, of the world) is unchanged, . How can we express disorder quantitatively? From the example of the coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins. Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its members are heads-up constitutes a of the system. Since we do not care which coins are heads-up, there are clearly numerous configurations of the individual coins which can result in this “macrostate”. Each of these configurations specifies a of the system. The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate. To see what this means, consider the possible outcomes of a toss of four coins (Table \(\Page {1}\) ): A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the table. The second column gives the number of “ways”— that is, the number of head/tail configurations of the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss resulting in a particular macrostate is proportional to the number of microstates corresponding to the macrostate, and is equal to this number, divided by the total number of possible microstates (in this example, 2 =16). An important assumption here is that all microstates are equally probable; that is, the toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an entirely random way. The greater the number of that correspond to a given , the greater the probability of that . Thus coins and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to become more scattered about a teenager’s room during the course of daily living. But there are some important differences between these large-scale mechanical, or macro systems, and the collections of sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically as . Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move around. Thus, , The importance of these last two points is far greater than you might at first think, but to fully appreciate this, you must recall the various ways in which thermal energy is stored in molecules— hence the following brief review. is the portion of a molecule's energy that is proportional to its , and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call . Since there are three directions in space, all molecules possess three modes of translational motion. For polyatomic molecules, two additional kinds of motions are possible. One of these is ; a linear molecule such as CO in which the atoms are all laid out along the -axis can rotate along the - and -axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal . For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated. The relative populations of the translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at a given temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. It is readily apparent that virtually all the thermal energy resides in translational states. Notice the greatly different spacing of the three kinds of energy levels. This is extremely important because it determines the number of energy quanta that a molecule can accept, and, as the following illustration shows, the number of different ways this energy can be distributed amongst the molecules. . The more closely spaced the quantized energy states of a molecule, the greater will be the number of ways in which a given quantity of thermal energy can be shared amongst a collection of these molecules. The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the of its energy states. This is a feature of the particle-in-a-box model, which predicts that the separation of the energy states of a gas confined within a box depends on the inverse square of the box length (and on the inverse of the particle mass as well.) At the atomic and molecular level, all energy is ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute (roughly 10 J) that translational kinetic energy can be regarded as (non-quantized) for most practical purposes. The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here: Each of these ten possibilities represents a distinct that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled will be observed 60% of the time, while will occur only 10% of the time. As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected. : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to Increasing temperature increases the number of microstates in a state and hence the entropy of the state. Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top. What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, The profundity of this conclusion was recognized around 1900, when it was first described as the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever change. Not a happy thought! Everybody knows that a gas, if left to itself, will tend to expand so as to fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of energy states (and thus microstates) its thermal energy can occupy. Since all such microstates within the thermally accessible range of energies are equally probable, the expansion of the gas can viewed as a consequence of the tendency of energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable. The same can in fact be said for even other highly probable distributions, such as having 49.999% of the molecules in the left half of the container and 50.001% in the right half. Even though the number of possible configurations that would yield this distribution of molecules is inconceivably great, it is essentially negligible compared to the number that would correspond to an exact 50-percent distribution. Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler always operates in the direction “warmer-to-cooler” because this allows thermal energy to occupy a larger number of energy states as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”. When two bodies at different temperatures are placed in thermal contact, thermal energy flow from the warmer into the cooler one until they reach the same temperature. As you might expect, the increase in the amount of energy spreading and sharing is proportional to amount of heat transferred , but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat passes into a system at temperature , the degree of dilution of the thermal energy is given by To understand why we have to divide by the temperature, consider the effect of very large and very small values of in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are occupied, so the amount of energy spreading can be very great. Conversely, if the temperature is initially large, the number of new thermal energy states that become occupied will be negligible compared to the number already active. When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved: The number of thermally accessible energy states (indicated by the shading in the diagram below) increases with temperature, but because 2 moles of H possess twice as many translational states as one mole of H , dissociation becomes increasingly favored at higher temperatures as more of these H states become thermally accessible. This argument can be generalized to all molecules, illustrating an important principle: All molecules spontaneously absorb heat dissociate at high temperatures. The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen. The result is exactly what the predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures. This is all very well for helping you understand the direct connection between energy spreading when a chemical reaction occurs, but it is of little help in achieving our goal of predicting the direction and extent of chemical change. For this, | 18,806 | 3,285 |
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Formation of an ionic bond by complete transfer of an electron from one atom to another is possible only for a fairly restricted set of elements. Covalent bonding, in which neither atom loses complete control over its valence electrons, is much more common. In a the electrons occupy a region of space the two nuclei and are said to be by them. The simplest example of a covalent bond is the bond between the two H atoms in a molecule of H . Suppose that two H atoms approach each other until their two 1 electron clouds interpenetrate. You can see what would happen to each 1s electron in the figure below by clicking on 1s left atom to show a 1s atomic orbital on the left H atom. In such a situation the electron does not continue to move about its own nucleus only. The electron initially associated with the left-hand H atom, for example, will feel the attractive pull of the right-hand nucleus as well as the left-hand nucleus. Since both nuclei have the same charge, the electron is unable to discriminate between them. Accordingly it adopts a new probability cloud. You can see this by clicking on the Sigma 1 button in of the figure. (Also click on the 1s button to remove the 1s atomic orbital.) The same argument applies to the electron initially associated with the right-hand H atom. (Use the figure to show what happens to that electron when it is attracted by two H nuclei.) Each electron is said to be over both nuclei and each electron has an equal probablility of being found in the vicinity of either nucleus. An orbital, like Sigma 1, that extends over a whole molecule rather than being restricted to a single atom is called a molecular orbital. We can consider it to be the result of a combination or overlap of the two 1s atomic orbitals. In fact, if you follow the instructions for part c of the figure, you can see how the overall molecular orbital overlaps with the densities of the two Sigma electrons, and how the molecular orbital has adopted a new shape from the densities shown by the two 1s orbitals. A molecular orbital formed in this way must conform to the Pauli exclusion principle. Only two electrons of opposite spin can occupy each orbital, denoted here as the two electrons in the Sigma orbital. Since the second electron (seen by clicking on the "Sigma 2" button) from the other H atom is available, it occupies the molecular orbital together with the original (Sigma 1) electron. The result is a shared pair of electrons moving around both nuclei and holding them together (the orange color). The overlap of two 1 electron clouds and their spreading over both nuclei when the H molecule forms has the effect of concentrating the electron density the protons. When the molecule forms, the negative charges move to positive charges than before. This is represented by the fact that the extent of dots in the molecular orbital is less than the spread due to the individual 1 electrons. There is thus a in potential energy. Since the guarantees that a reduction in the energy means that the total energy (kinetic + potential) is also reduced, we can conclude that the H molecule is and hence more stable under normal conditions than two separate H atoms. Alternatively we can say that energy is to break apart an H molecule and separate it into two H atoms. This energy (called the ) can be measured experimentally. It is found to have a value of 436 kJ mol for the H molecule. Below is a Jmol applet, a 3-D interactive view of H . There are a set of commands to the left where you can play around with appearances, viewing the molecule as ball and stick, a wire, or see the van der Waals radii. You can also label the atoms, radii, and bond length. Take a bit to play around with the applet if you are unfamiliar with Jmol. Now focus on the molecular orbital commands to the right. These allow you to visual the orbitals discussed earlier on this page. Set the MO cut off to 0.05. This will produce a surface that encloses 95% (0.95) of the electron density. (a) on HOMO. This stands for ighest ccupied olecular rbital. This is the highest-energy orbital with any electrons in it. The HOMO contains two electrons, one from each H atom. Try to see the relationship between the two-dimensional dot-density diagram above and the three-dimensional representation in the Jmol window. The Jmol shows a surface that encloses 95% of the electron density (95% of the dots in the figure above). Rotate the HOMO diagram until you are looking along a line that passes through both H atoms (end-on to the molecule). What can you observe about the shape of the MO? Now rotate the diagram so that you are looking along a line perpendicular to the H-H bond. What do you observe now? What does the shape of the HOMO tell you about H ? (b) Now click on LUMO. this stands for owest noccupied olecular rbital. This is the next higher energy molecular orbital in the molecule--the orbital that electrons would fill if more electrons were added. Based on the shape of the LUMO, what would be the consequence of placing electrons in this orbital? a) The orbital is symmetrical about the center of the H molecule. As was discussed on this page, the two electrons in this orbital are now equally shared by both atoms. The orbital is centered between the two atoms, so electron density is highest there. Again, as said above, this reduced the total energy of the system, and electrons in this orbital will thus lead to a covalent bond between the two atoms. This orbital is referred to as a bonding orbital. b)With the LUMO of H there is electron density half way between the two hydrogen atoms. The two protons repel each other, and without much electron density between them to attract them together the LUMO has a higher energy than the two separate hydrogen atoms. If electrons occupied this orbital, it would lead to the repulsion of the atoms and breaking of the covalent bond. The point exactly half way the two atoms where the electron denisty is zero is called a node. Because the atoms would fly apart if the LUMO were occupied, this molecular orbital with a node is referred to as an anti-bonding orbital. | 6,190 | 3,286 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,287 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,288 |
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The early application of iron to the manufacture of tools and weapons was possible because of the wide distribution of iron ores and the ease with which iron compounds in the ores could be reduced by carbon. For a long time, charcoal was the form of carbon used in the reduction process. The production and use of iron became much more widespread about 1620, when coke was introduced as the reducing agent. Coke is a form of carbon formed by heating coal in the absence of air to remove impurities. The overall reaction for the production of iron in a blast furnace is as follows: \[\mathrm{Fe_2O_3(s) +3C(s)\xrightarrow{\Delta}2Fe(l) +3CO(g)} \label{23.2.3}\] The actual reductant is CO, which reduces Fe O to give Fe(l) and CO (g) (Equation \(\ref{23.2.3}\)); the CO is then reduced back to CO by reaction with excess carbon. As the ore, lime, and coke drop into the furnace (Figure \(\Page {1}\)), any silicate minerals in the ore react with the lime to produce a low-melting mixture of calcium silicates called slag, which floats on top of the molten iron. Molten iron is then allowed to run out the bottom of the furnace, leaving the slag behind. Originally, the iron was collected in pools called pigs, which is the origin of the name pig iron. The first step in the metallurgy of iron is usually roasting the ore (heating the ore in air) to remove water, decomposing carbonates into oxides, and converting sulfides into oxides. The oxides are then reduced in a blast furnace that is 80–100 feet high and about 25 feet in diameter (Figure \(\Page {2}\)) in which the roasted ore, coke, and limestone (impure CaCO ) are introduced continuously into the top. Molten iron and slag are withdrawn at the bottom. The entire stock in a furnace may weigh several hundred tons. Near the bottom of a furnace are nozzles through which preheated air is blown into the furnace. As soon as the air enters, the coke in the region of the nozzles is oxidized to carbon dioxide with the liberation of a great deal of heat. The hot carbon dioxide passes upward through the overlying layer of white-hot coke, where it is reduced to carbon monoxide: \[\ce{CO2}(g)+\ce{C}(s)⟶\ce{2CO}(g)\] The carbon monoxide serves as the reducing agent in the upper regions of the furnace. The individual reactions are indicated in Figure \(\Page {2}\). The iron oxides are reduced in the upper region of the furnace. In the middle region, limestone (calcium carbonate) decomposes, and the resulting calcium oxide combines with silica and silicates in the ore to form slag. The slag is mostly calcium silicate and contains most of the commercially unimportant components of the ore: \[\ce{CaO}(s)+\ce{SiO2}(s)⟶\ce{CaSiO3}(l)\] Just below the middle of the furnace, the temperature is high enough to melt both the iron and the slag. They collect in layers at the bottom of the furnace; the less dense slag floats on the iron and protects it from oxidation. Several times a day, the slag and molten iron are withdrawn from the furnace. The iron is transferred to casting machines or to a steelmaking plant (Figure \(\Page {3}\)). Much of the iron produced is refined and converted into steel. is made from iron by removing impurities and adding substances such as manganese, chromium, nickel, tungsten, molybdenum, and vanadium to produce alloys with properties that make the material suitable for specific uses. Most steels also contain small but definite percentages of carbon (0.04%–2.5%). However, a large part of the carbon contained in iron must be removed in the manufacture of steel; otherwise, the excess carbon would make the iron brittle. ). Jim Clark ( ) | 3,662 | 3,289 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.09%3A_The_Amount_of_Substance |
Chemists use the mole so often to measure how much of a substance is present that it is convenient to have a name for the quantity which this unit measures. In the International System this quantity is called the and is given the symbol . Here again a common English word has been given a very specific scientific meaning. Although might refer to volume or mass in everyday speech, in chemistry it does not. When a chemist asks what amount of Br was added to a test tube, an answer like “0.0678 mol Br ” is expected. This would indicate that 0.0678 × 6.022 × 10 or 4.08 × 10 , Br molecules had been added to the test tube. | 642 | 3,290 |
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