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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.09%3A_Bonding_in_Coordination_Complexes |
Make sure you thoroughly understand the following essential ideas: Complexes such as Cu(NH ) have been known and studied since the mid-nineteenth century. and their structures had been mostly worked out by 1900. Although the hybrid orbital model was able to explain how neutral molecules such as water or ammonia could bond to a transition metal ion, it failed to explain many of the special properties of these complexes. Finally, in 1940-60, a model known as was developed that is able to organize and explain most of the observed properties of these compounds. Since that time, coordination complexes have played major roles in cellular biochemistry and inorganic catalysis. If you have taken a lab course in chemistry, you have very likely admired the deep blue color of crystals, CuSO ·5H O. The proper name of this substance is copper(II) sulfate pentahydrate, and it is typical of many salts that incorporate into their crystal structures. It is also a , a term used by chemists to describe a substance composed of two other substances (in this case, CuSO and H O) each of which is capable of an independent existence. The binding between the components of a complex is usually weaker than a regular chemical bond; thus most solid hydrates can be decomposed by heating, driving off the water and yielding the salt: \[\underbrace{\ce{CuSO4 \cdot 5 H2O}}_{\text{blue}} \rightarrow \underbrace{\ce{CuSO_{4 (s)}}}_{\text{white}} + 5 H_2O\] Driving off the water in this way also destroys the color, turning it from a beautiful deep blue to a nondescript white. If the anhydrous salt is now dissolved in water, the blue color now pervades the entire solution. It is apparent that the presence of water is somehow necessary for the copper(II) ion to take on a blue color, but why should this be? A very common lab experiment that most students carry out is to add some dilute ammonia to a copper sulfate solution. At first, the solution turns milky as the alkaline ammonia causes the precipitation of copper hydroxide: \[\ce{Cu^{2+} + 2 OH^{–} \rightarrow Cu(OH)2 (s)}\] However, if more ammonia is added, the cloudiness disappears and the solution assumes an intense deep blue color that makes the original solution seem pale by comparison. The equation for this reaction is usually given as \[\ce{Cu^{2+} + 6 NH3 \rightarrow Cu(NH3)6^{2+} } \label{ammine}\] The new product is commonly known as the , or more officially, hexamminecopper(II) complex ion. Equation \(\ref{ammine}\) is somewhat misleading, however, in that it implies the formation of a new complex where none existed before. In fact, since about 1895, it has been known that the ions of most transition metals dissolve in water to form complexes with water itself, so a better representation of the reaction of dissolved copper with ammonia would be \[\ce{Cu(H2O)6^{2+} + 6 NH3 \rightarrow Cu(NH3)6^{2+} + 6 H2O}\] In effect, the ammonia binds more tightly to the copper ion than does water, and it thus displaces the latter when it comes into contact with the hexaaquocopper(II) ion, as the dissolved form of Cu is properly known. Most transition metals dissolve in water to form complexes with water itself. Although our primary focus in this unit is on bonding, the topic of coordination complexes is so important in chemistry and biochemistry that some of their basic features are worth knowing about, even if their detailed chemistry is beyond the scope of this course. These complexes play an especially crucial role in physiology and biochemistry. Thus heme, the oxygen-carrying component of red blood cells (and the source of the red color) is basically a complex of iron, and the part of chlorophyll that converts sunlight into chemical energy within green plants is a magnesium complex. We have already defined a as a substance composed of two or more components capable of an independent existence. A is one in which a or ion is joined to one or more (Latin , to tie) through what is called a in which both of the bonding electrons are supplied by the ligand. In such a complex the central atom acts as an electron-pair acceptor ( — think of H which has no electrons at all, but can accept a pair from something like Cl ) and the ligand as an electron-pair donor ( ). The central atom and the ligands coordinated to it constitute the . Thus the salt [Co(NH ) Cl]Cl is composed of the complex ion [Co(NH ) Cl] and two Cl ions; components within the square brackets are inside the coordination sphere, whereas the two chloride ions are situated outside the coordination sphere. These latter two ions could be replaced by other ions such as NO without otherwise materially changing the nature of the salt. The central atoms of coordination complexes are most often (positive ions), but may in some cases be neutral atoms, as in nickel carbonyl Ni(CO) . composed of ions such as F or small molecules such as H O or CN possess more than one set of lone pair electrons, but only one of these pairs can coordinate with a central ion. Such ligands are said to be (“one tooth”.) Larger ligands may contain more than one atom capable of coordinating with a single central ion, and are described as . Thus ethylenediamine (shown below) is a ligand. Polydentate ligands whose geometry enables them to occupy more than one coordinating position of a central ion act as (Greek χελος, , claw) and tend to form extremely stable complexes known as . Complexes such as Cu(NH ) have been known and studied since the mid-nineteenth century. Why they should form, or what their structures might be, were complete mysteries. At that time all inorganic compounds were thought to be held together by ionic charges, but ligands such as water or ammonia are of course electrically neutral. A variety of theories such as the existence of “secondary valences” were concocted, and various chain-like structures such as CuNH -NH -NH -NH -NH -NH were proposed. Finally, in the mid-1890s, after a series of painstaking experiments, the chemist Alfred Werner (Swiss, 1866-1919) presented the first workable theory of complex ion structures. Werner claimed that his theory first came to him in a flash after a night of fitful sleep; by the end of the next day he had written his landmark paper that eventually won him the 1913 Nobel Prize in Chemistry. Werner was able to show, in spite of considerable opposition, that transition metal complexes consist of a central ion surrounded by ligands in a square-planar, tetrahedral, or octahedral arrangement. This was an especially impressive accomplishment at a time long before X-ray diffraction and other methods had become available to observe structures directly. His basic method was to make inferences of the structures from a careful examination of the chemistry of these complexes and particularly the existence of structural isomers. For example, the existence of two different compounds AX having the same composition shows that its structure must be square-planar rather than tetrahedral. An understanding of the nature of the bond between the central ion and its ligands would have to await the development of Lewis’ and Pauling’s picture. We have already shown how of the orbitals of the central ion creates vacancies able to accommodate one or more pairs of unshared electrons on the ligands. Although these models correctly predict the structures of many transition metal complexes, they are by themselves unable to account for several of their special properties: act as tiny magnets; if a substance that contains unpaired electrons is placed near an external magnet, it will undergo an attraction that tends to draw it into the field. Such substances are said to be , and the degree of paramagnetism is directly proportional to the number of unpaired electrons in the molecule. Magnetic studies have played an especially prominent role in determining how electrons are distributed among the various orbitals in transition metal complexes. Studies of this kind are carried out by placing a sample consisting of a solution of the complex between the poles of an electromagnet. The sample is suspended from the arm of a sensitive balance, and the change in apparent weight is measured with the magnet turned on and off. An increase in the weight when the magnet is turned on indicates that the sample is attracted to the magnet ( ) and must therefore possess one or more unpaired electrons. The precise number can be determined by calibrating the system with a substance whose electron configuration is known. The current model of bonding in coordination complexes developed gradually between 1930-1950. In its initial stages, the model was a purely electrostatic one known as which treats the ligand ions as simple point charges that interact with the five atomic orbitals of the central ion. It is this theory which we describe below. It is remarkable that this rather primitive model, quite innocent of quantum mechanics, has worked so well. However, an improved and more complete model that incorporates molecular orbital theory is known as . In an isolated transition metal atom the five outermost orbitals all have the same energy which depends solely on the spherically symmetric electric field due to the nuclear charge and the other electrons of the atom. Suppose now that this atom is made into a and is placed in solution, where it forms a hydrated species in which six H O molecules are coordinated to the central ion in an arrangement. An example of such an ion might be . The ligands (H O in this example) are bound to the central ion by electron pairs contributed by each ligand. Because the six ligands are located at the corners of an octahedron centered around the metal ion, these electron pairs are equivalent to clouds of negative charge that are directed from near the central ion out toward the corners of the octahedron. We will call this an octahedral electric field, or the . The differing shapes of the five kinds of d orbitals cause them to interact differently with the electric fields created by the coordinated ligands. This diagram (from a Purdue U. chemistry site) shows outlines of five kinds of d orbitals. Although the five orbitals of the central atom all have the same energy in a spherically symmetric field, their energies will not all be the same in the octahedral field imposed by the presence of the ligands. The reason for this is apparent when we consider the different geometrical properties of the five d orbitals. Two of the orbitals, designated and , have their electron clouds pointing directly toward ligand atoms. We would expect that any electrons that occupy these orbitals would be subject to repulsion by the electron pairs that bind the ligands that are situated at corresponding corners of the octahedron. As a consequence, the energies of these two orbitals will be raised in relation to the three other orbitals whose lobes are not directed toward the octahedral positions. The number of electrons in the orbital of the central atom is easily determined from the location of the element in the periodic table, taking in account, of course, of the number of electrons removed in order to form the positive ion. The effect of the octahedral ligand field due to the ligand electron pairs is to split the orbitals into two sets whose energies differ by a quantity denoted by ("delta") which is known as the . Note that both sets of central-ion orbitals are repelled by the ligands and are both raised in energy; the upper set is simply raised by a greater amount. Both the total energy shift and Δ are strongly dependent on the particular ligands. Returning to our example of Ti(H O) , we note that Ti has an outer configuration of 4s 3d , so that Ti will be a ion. This means that in its ground state, one electron will occupy the lower group of orbitals, and the upper group will be empty. The -orbital splitting in this case is 240 kJ per mole which corresponds to light of blue-green color; absorption of this light promotes the electron to the upper set of orbitals, which represents the of the complex. If we illuminate a solution of Ti(H O) with white light, the blue-green light is absorbed and the solution appears violet in color. The magnitude of the orbital splitting depends strongly on the nature of the ligand and in particular on how strong an electrostatic field is produced by its electron pair bond to the central ion. If Δ is not too large then the electrons that occupy the orbitals do so with their spins unpaired until a configuration is reached, just as occurs in the normal sequence for atomic electron configurations. Thus a weak-field ligand such as H O leads to a “high spin” complex with Fe(II). In contrast to this, the cyanide ion acts as a strong-field ligand; the orbital splitting is so great that it is energetically more favorable for the electrons to pair up in the lower group of orbitals rather than to enter the upper group with unpaired spins. Thus hexacyanoiron(II) is a “low spin” complex— actually zero spin, in this particular case. Different orbital splitting patterns occur in square planar and tetrahedral coordination geometries, so a very large number of arrangements are possible. In most complexes the value of Δ corresponds to the absorption of visible light, accounting for the colored nature of many such compounds in solution and in solids such as \(\ce{CuSO4·5H2O}\) ()Figure \(\Page {1}\). Approximately one-third of the chemical elements are present in living organisms. Many of these are metallic ions whose function within the cell depends on the formation of -orbital coordination complexes with small molecules such as (see below). These complexes are themselves bound within proteins ( ) which provide a local environment that is essential for their function, which is either to transport or store diatomic molecule (oxygen or nitric oxide), to transfer electrons in oxidation-reduction processes, or to catalyze a chemical reaction. The most common of these utilize complexes of Fe and Mg, but other micronutrient metals including Cu, Mn, Mo, Ni, Se, and Zn are also important. Hemoglobin is one of a group of heme proteins that includes myoglobin, cytochrome-c, and catalase. Hemoglobin performs the essential task of transporting dioxygen molecules from the lungs to the tissues in which it is used to oxidize glucose, this oxidation serving as the source of energy required for cellular metabolic processes. Hemoglobin consists of four protein subunits (depicted by different colors in this diagram) joined together by weak intermolecular forces. Each of these subunits contains, buried within it, a molecule of , which serves as the active site of oxygen transport. itself consists of an iron atom coordinated to a tetradentate . When in the ferrous (Fe state) the iron binds to oxygen and is converted into Fe . Because a bare heme molecule would become oxidized by the oxygen without binding to it, the adduct must be stabilized by the surrounding globin protein. In this environment, the iron becomes octahedrally-coordinated through binding to a component of the protein in a fifth position, and in the sixth position either by an oxygen molecule or by a water molecule, depending on whether the hemoglobin is in its oxygenated state (in arteries) or deoxygenated state (in veins). The heme molecule (purple) is enfolded within the polypeptide chain as shown here. The complete hemoglobin molecule contains four of these subunits, and all four must be present for it to function. The binding of O to heme in hemoglobin is not a simple chemical equilibrium; the binding efficiency is regulated by the concentrations of H , CO , and organic phosphates. It is remarkable that the binding sites for these substances are on the outer parts of the globin units, far removed from the heme. The mechanism of this exquisite molecular-remote-control arises from the fact that the Fe ion is too large to fit inside the porphyrin, so it sits slightly out of the porphyrin plane. This Fe radius diminishes when it is oxygenated, allowing it to move into the plane. In doing so, it pulls the protein component to which it is bound with it, triggering a sequence of structural changes that extend throughout the protein. Myoglobin is another important heme protein that is found in muscles. Unlike hemoglobin, which consists of four protein subunits, myoglobin is made up of only one unit. Its principal function is to act as an oxygen storage reservoir, enabling vigorous muscle activity at a rate that could not be sustained by delivery of oxygen through the bloodstream. Myoglobin is responsible for the red color of meat. Cooking of meat releases the O and oxidizes the iron to the +3 state, changing the color to brown. Other ligands, notably cyanide ion and , are able to bind to hemoglobin much more strongly than does iron, thereby displacing it and rendering hemoglobin unable to transport oxygen. Air containing as little as 1 percent CO will convert hemoglobin to carboxyhemoglobin in a few hours, leading to loss of consciousness and death. Even small amounts of carbon monoxide can lead to substantial reductions in the availability of oxygen. The 400-ppm concentration of CO in cigarette smoke will tie up about 6% of the hemoglobin in heavy smokers; the increased stress this places on the heart as it works harder to compensate for the oxygen deficit is believed to be one reason why smokers are at higher risk for heart attacks. CO binds to hemoglobin 200 times more tightly than does \(O_2\). Chlorophyll is the light-harvesting pigment present in green plants. Its name comes from the Greek word χλορος ( ), meaning “green”- the same root from which chlorine gets its name. Chlorophyll consists of a ring-shaped tetradentate ligand known as a coordinated to a central magnesium ion. A histidine residue from one of several types of associated proteins forms a fifth coordinate bond to the Mg atom. The light energy trapped by chlorophyll is utilized to drive a sequence of reactions whose net effect is to bring about the reduction of CO to glucose (C H O ) in a process known as photosynthesis which serves as the fuel for all life processes in both plants and animals. | 18,343 | 3,560 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltage_Amperage_and_Resistance_Basics |
To discuss electrochemistry meaningfully, the fundamental properties of electricity must be defined. The between two points is a short name for the electrical force that would drive an electric current between those points. In the case of static electric fields, the voltage between two points is equal to the between those points. In the more general case with electric and magnetic fields that vary with time, the terms are no longer synonymous. Electric potential is the energy required to move a unit electric charge to a particular place in a static electric field. The first is , usually abbreviated "V" and measured in (also abbreviated "V".) Voltage, also sometimes called or , refers to the amount of potential energy the electrons have in an object or circuit. In some ways, you can think of this as the amount of "push" the electrons are making to try to get towards a positive charge. The more energy the electrons have, the stronger the voltage. The means the rate of flow of electric charge. This flowing electric charge is typically carried by moving electrons, in a conductor such as wire; in an electrolyte, it is instead carried by ions. The unit for measuring the rate of flow of electric charge is the ampere. Electric current is measured using an ammeter. is usually abbreviated "I" ("C" is reserved for the principle of , the most fundamental building block of electricity.) Current is measured in or , abbreviation "A". Current refers to how much electricity is flowing--how many electrons are moving through a circuit in a unit of time. The of an object is a measure of its opposition to the passage of a steady electric current. An object of uniform cross section will have a resistance proportional to its length and inversely proportional to its cross-sectional area, and proportional to the resistivity of the material. Discovered by Georg Ohm in 1827, electrical resistance shares some conceptual parallels with the mechanical notion of friction. The unit of electrical resistance is the ohm (Ω). Resistance refers to how much the material that is conducting electricity opposes the flow of electrons. The higher the resistance, the harder it is for the electrons to push through. If we draw an analogy to a waterfall, the voltage would represent the height of the waterfall: the higher it is, the more potential energy the water has by virtue of its distance from the bottom of the falls, and the more energy it will possess as it hits the bottom. Then current represents how much water was going over the edge of the falls each second . Resistance refers to any obstacles that slows down the flow of water over the edge of the falls (e.g. rocks in the river before the edge). These voltage, current and resistance are related via a principle known as : \[ V = I * R \] which states that the voltage of a circuit is equal to the current through the circuit times its resistance. Another way of stating Ohm's Law, that is often easier to understand, is: \[ I = V / R \] which means that the current through a circuit is equal to the voltage divided by the resistance. This makes sense, if you think about our waterfall example: the higher the waterfall, the more water will want to rush through, but it can only do so to the extent that it is able to as a result of any opposing forces. If you tried to fit Niagara Falls through a garden hose, you'd only get so much water every second, no matter how high the falls, and no matter how much water was waiting to get through! And if you replace that hose with one that is of a larger diameter, you will get more water in the same amount of time. | 3,663 | 3,561 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.07%3A_Disaccharides |
Combinations of two or more of the simple sugars through glycoside linkages give substances known as . They also are called if made from two to ten sugar units. The simplest oligosaccharides are disaccharides made of two molecules of simple sugars that can be joined with \(\ce{O}\)-glycoside links, and it probably is easiest to visualize these as shown in the "stripped-down" formulas, \(22\) and \(23\): You should look at \(22\) and \(23\) carefully to be sure that you recognize the difference between them.\(^5\) In \(22\), sugar A is acting as a simple hydroxy compound, the aglycone of the sugar G to which it is linked by an \(\ce{O}\)-glycoside linkage.\(^6\) Hydrolysis of \(22\) at the glycoside link then will proceed as follows: Disaccharides such as \(22\) are like glucose in being reducing sugars ( ), because Component A has the hemiacetal grouping that is opened easily to the aldehyde form in the mildly alkaline conditions used for the Tollen's and Fehling's solution oxidations. Because there is a free hemiacetal group, reducing sugars also form osazones and they mutarotate ( and ). Disaccharides of type \(23\) are different in that each sugar, G and G', is acting as both a glycoside sugar as an aglycone. The linkage between them is that of a double-barreled acetal, \(\ce{-O-C-O-C-O}-\), and there is no hemiacetal grouping in the molecule. Therefore these are sugars as far as the standard tests go. However, hydrolysis of the \(\ce{O}\)-glycoside linkages of \(23\) does generate reducing sugars with hemiacetal carbons: In general, we find that the nonreducing disaccharides give none of the carbonyl reactions observed for glucose, such as mutarotation and osazone formation, except when the conditions are sufficiently acidic to hydrolyze the acetal linkage. Among the more important disaccharides are sucrose, \(24\), maltose, \(25\), cellobiose, \(26\), and lactose, \(27\): Sucrose and lactose occur widely as the free sugars, lactose in the milk of mammals, and sucrose in fruit and plants (especially in sugar cane and sugar beet). Maltose is the product of enzymatic hydrolysis of starch, and cellobiose is a product of hydrolysis of cellulose. To fully establish the structure of a disaccharide, we must determine (1) the identity of the component monosaccharides; (2) the type of ring junction, furanose or pyranose, in each monosaccharide, as it exists in the disaccharide; (3) the positions that link one monosaccharide with the other; and (4) the anomeric configuration (\(\alpha\) or \(\beta\)) of this linkage. Hydrolysis of disaccharides with enzymes is very helpful in establishing anomeric configurations, because enzymes are highly specific catalysts for hydrolysis of the different types of glycosidic linkages. For instance, \(\alpha\)-\(D\)-glucosidase (maltase) catalyzes hydrolysis of \(\alpha\)-\(D\)-glycosides more rapidly than of \(\beta\)-\(D\)-glycosides. The enzyme emulsin (found in bitter almonds) in contrast shows a strong preference for \(\beta\)-\(D\)-glycosides over \(\alpha\)-\(D\)-glycosides. Yeast invertase catalyzes hydrolysis of \(\beta\)-\(D\)-fructosides. We know that sucrose consists of the two monosaccharides glucose and fructose because hydrolysis with acids or enzymes gives equal amounts of each hexose. Further, sucrose is not a reducing sugar, it forms no phenylosazone derivative, and it does not mutarotate. Therefore the anomeric carbons of both glucose and fructose must be linked through an in sucrose. Thus sucrose is a or, equally, a . Because sucrose is hydrolyzed by enzymes that specifically assist hydrolysis of both \(\alpha\) glycosides (such as yeast \(\alpha\)-glucosidase) and \(\beta\)-fructosides (such as invertase), it is inferred that the glucose residue is present as an \(\alpha\) and the fructose residue as a \(\beta\) . If so, the remaining uncertainty in the structure of sucrose is the size of the rings in the glucose and fructose residues. The size of the sugar rings in sucrose has been determined by the reactions shown in Figure 20-5. Methylation of sucrose with dimethyl sulfate in basic solution followed by hydrolysis of the octamethyl derivative gives 2,3,4,6-tetra-\(\ce{O}\)-methyl-\(D\)-glucopyranose ( ) and a tetra-\(\ce{O}\)-methyl-\(D\)-fructose. This establishes the glucose reside in sucrose as a . The fructose residue must be a because periodate oxidation of sucrose consumes three moles of periodate, whereby one mole of methanoic acid and one mole of a tetraaldehyde are formed. On bromine oxidation followed by acid hydrolysis, the tetraaldehyde gives 3-hydroxy-2-oxopropanoic acid (hydroxypyruvic acid, \(\ce{HOCH_2COCO_2H}\)), oxoethanoic acid (glyoxylic acid, \(\ce{OCHCO_2H}\)), and \(D\)-glyceric acid \(\left( \ce{HOCH_2CHOHCO_2H} \right)\). Sucrose therefore has structure \(24\), and this structure was confirmed by synthesis (R. Lemieux in 1953). \(^5\)For now, we will ignore the possibility of different anomers of the disaccharide or their component sugars. \(^6\)The manner in which sugars are linked together to form oligosaccharides was elucidated by W. N. Haworth, who received the Nobel Prize in chemistry in 1937 for this and other contributions to research on the structures and reactions of carbohydrates. and (1977) | 5,313 | 3,562 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.01%3A_The_Empirical_Gas_Laws |
A number of important relationships describing the nature of gas samples have been derived completely empirically (meaning based solely on observation rather making an attempt to define the theoretical reason these relationships may exist. These are the . One of the important relationships governing gas samples that can be modeled mathematically is the relationship between pressure and volume. Robert Boyle (1627 – 1691) (Hunter, 2004) did experiments to confirm the observations of Richard Towneley and Henry Powers to show that for a fixed sample of gas at a constant temperature, pressure and volume are inversely proportional. \[ pV = \text{constant} \nonumber \] or \[p_1V_2=p_2V_2 \nonumber \] Boyle used a glass u-tube that was closed at one end and with the lower portion filled with mercury (trapping a sample of air in the closed end.) By adding mercury to the open end, he was able to observe and quantify the compression of the trapped air. Charles’ Law states that the volume of a fixed sample of gas at constant pressure is proportional to the temperature. For this law to work, there must be an absolute minimum to the temperature scale since there is certainly an absolute minimum to the volume scale! \[\dfrac{V}{T} = \text{constant} \nonumber \] or \[\dfrac{V_1}{T_2} = \dfrac{V_1}{T_2} \nonumber \] The second law of thermodynamics also predicts an absolute minimum temperature, but that will be developed in a later chapter. Gay-Lussac’s Law states that the pressure of a fixed sample of gas is proportional to the temperature. As with Charles’ Law, this suggests the existence of an absolute minimum to the temperature scale since the pressure can never be negative. \[\dfrac{p}{T} = \text{constant} \nonumber \] or \[\dfrac{p_1}{T_2} = \dfrac{p_1}{T_2} \nonumber \] Boyle’s, Charles’, and Gay-Lussac’s Laws can be combined into a single empirical formula that can be useful. For a given amount of gas, the following relationship must hold: \[\dfrac{pV}{T} = \text{constant} \nonumber \] or \[\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2} \nonumber \] Amedeo Avogadro (1776-1856) (Encycolopedia, 2016) did extensive work with gases in his studies of matter. In the course of his work, he noted an important relationship between the number of moles in a gas sample. Avogadro’s Law (Avogadro, 1811) states that at the same temperature and pressure, any sample of gas has the same number of molecules per unit volume. \[\dfrac{n}{V} = \text{constant} \nonumber \] or \[\dfrac{n_1}{V_1} = \dfrac{n_2}{V_2} \nonumber \] | 2,552 | 3,563 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Countercurrent_Separations |
In 1949, Lyman Craig introduced an improved method for separating analytes with similar distribution ratios. The technique, which is known as a countercurrent liquid–liquid extraction, is outlined in Figure A16.1 and discussed in detail below. In contrast to a sequential liquid–liquid extraction, in which we repeatedly extract the sample containing the analyte, a countercurrent extraction uses a serial extraction of both the sample and the extracting phases. Although countercurrent separations are no longer common—chromatographic separations are far more efficient in terms of resolution, time, and ease of use—the theory behind a countercurrent extraction remains useful as an introduction to the theory of chromatographic separations. To track the progress of a countercurrent liquid-liquid extraction we need to adopt a labeling convention. As shown in Figure A16.1, in each step of a countercurrent extraction we first complete the extraction and then transfer the upper phase to a new tube containing a portion of the fresh lower phase. Steps are labeled sequentially beginning with zero. Extractions take place in a series of tubes that also are labeled sequentially, starting with zero. The upper and lower phases in each tube are identified by a letter and number, with the letters U and L representing, respectively, the upper phase and the lower phase, and the number indicating the step in the countercurrent extraction in which the phase was first introduced. For example, U is the upper phase introduced at step 0 (during the first extraction), and L is the lower phase introduced at step 2 (during the third extraction). Finally, the partitioning of analyte in any extraction tube results in a fraction remaining in the upper phase, and a fraction remaining in the lower phase. Values of are calculated using equation A16.1, which is identical to equation 7.26 in Chapter 7. \[(q_{aq})_1 = \dfrac{(\text{moles aq})_1 }{(\text{moles aq})_0 }= \dfrac{V_{aq}}{(DV_{org} + V_{aq})} \tag{A16.1}\] The fraction , of course is equal to 1 – . Typically and are equal in a countercurrent extraction, although this is not a requirement. Let’s assume that the analyte we wish to isolate is present in an aqueous phase of 1 M HCl, and that the organic phase is benzene. Because benzene has the smaller density, it is the upper phase, and 1 M HCl is the lower phase. To begin the countercurrent extraction we place the aqueous sample containing the analyte in tube 0 along with an equal volume of benzene. As shown in Figure A16.1a, before the extraction all the analyte is present in phase L . When the extraction is complete, as shown in Figure A16.1b, a fraction of the analyte is present in phase U , and a fraction is in phase L . This completes step 0 of the countercurrent extraction. If we stop here, there is no difference between a simple liquid–liquid extraction and a countercurrent extraction. After completing step 0, we remove phase U and add a fresh portion of benzene, U , to tube 0 (see Figure A16.1c). This, too, is identical to a simple liquid-liquid extraction. Here is where the power of the countercurrent extraction begins—instead of setting aside the phase U , we place it in tube 1 along with a portion of analyte-free aqueous 1 M HCl as phase L (see Figure A16.1c). Tube 0 now contains a fraction of the analyte, and tube 1 contains a fraction of the analyte. Completing the extraction in tube 0 results in a fraction of its contents remaining in the upper phase, and a fraction remaining in the lower phase. Thus, phases U and L now contain, respectively, fractions and of the original amount of analyte. Following the same logic, it is easy to show that the phases U and L in tube 1 contain, respectively, fractions and of analyte. This completes step 1 of the extraction (see Figure A16.1d). As shown in the remainder of Figure A16.1, the countercurrent extraction continues with this cycle of phase transfers and extractions. In a countercurrent liquid–liquid extraction, the lower phase in each tube remains in place, and the upper phase moves from tube 0 to successively higher numbered tubes. We recognize this difference in the movement of the two phases by referring to the lower phase as a stationary phase and the upper phase as a mobile phase. With each transfer some of the analyte in tube moves to tube + 1, while a portion of the analyte in tube – 1 moves to tube . Analyte introduced at tube 0 moves with the mobile phase, but at a rate that is slower than the mobile phase because, at each step, a portion of the analyte transfers into the stationary phase. An analyte that preferentially extracts into the stationary phase spends proportionally less time in the mobile phase and moves at a slower rate. As the number of steps increases, analytes with different values of eventually separate into completely different sets of extraction tubes. We can judge the effectiveness of a countercurrent extraction using a histogram showing the fraction of analyte present in each tube. To determine the total amount of analyte in an extraction tube we add together the fraction of analyte present in the tube’s upper and lower phases following each transfer. For example, at the beginning of step 3 (see Figure A16.1g) the upper and lower phases of tube 1 contain fractions and 2 of the analyte, respectively; thus, the total fraction of analyte in the tube is 3 . Table A16.1 summarizes this for the steps outlined in Figure A16.1. A typical histogram, calculated assuming distribution ratios of 5.0 for analyte A and 0.5 for analyte B, is shown in Figure A16.2. Although four steps is not enough to separate the analytes in this instance, it is clear that if we extend the countercurrent extraction to additional tubes, we will eventually separate the analytes. Figure A16.1 and Table A16.1 show how an analyte’s distribution changes during the first four steps of a countercurrent extraction. Now we consider how we can generalize these results to calculate the amount of analyte in any tube, at any step during the extraction. You may recognize the pattern of entries in Table A16.1 as following the binomial distribution \[f(r,n) = \dfrac{n!}{(n−r)!r!} p^rq^{n−r} \tag{A16.2}\] where ( , ) is the fraction of analyte present in tube at step of the countercurrent extraction, with the upper phase containing a fraction × ( , ) of analyte and the lower phase containing a fraction × ( , ) of the analyte. The countercurrent extraction shown in Figure A16.2 is carried out through step 30. Calculate the fraction of analytes A and B in tubes 5, 10, 15, 20, 25, and 30. To calculate the fraction, , for each analyte in the lower phase we use equation A6.1. Because the volumes of the lower and upper phases are equal, we get = 1 / ( + 1) = 1 / (5 + 1) = 0.167 = 1 / ( + 1) = 1 / (4 + 1) = 0.200 Because we know that + = 1, we also know that is 0.833 and that is 0.333. For analyte A, the fraction in tubes 5, 10, 15, 20, 25, and 30 after the 30 step are (5,30) = (30! / ((30−5)!5!))(0.833) (0.167) = 2.1×10 ≈ 0 (10,30) = (30! / ((30−10)!10!))(0.833) (0.167) = 1.4×10 ≈ 0 (15,30) = (30! / ((30−15)!15!))(0.833) (0.167) = 2.2×10 ≈ 0 (20,30) = (30! / ((30−20)!20!))(0.833) (0.167) = 0.013 (25,30) = (30! / ((30−25)!25!))(0.833) (0.167) = 0.192 (30,30) = (30! / ((30−30)!30!))(0.833) (0.167) = 0.004 The fraction of analyte B in tubes 5, 10, 15, 20, 25, and 30 is calculated in the same way, yielding respective values of 0.023, 0.153, 0.025, 0, 0, and 0. Figure A16.3, which provides the complete histogram for the distribution of analytes A and B, shows that 30 steps is sufficient to separate the two analytes. Constructing a histogram using equation A16.2 is tedious, particularly when the number of steps is large. Because the fraction of analyte in most tubes is approximately zero, we can simplify the histogram’s construction by solving equation A16.2 only for those tubes containing an amount of analyte exceeding a threshold value. For a binomial distribution, we can use the mean and standard deviation to determine which tubes contain a significant fraction of analyte. The properties of a binomial distribution were covered in , with the mean, μ, and the standard deviation, s, given as μ = σ = – Furthermore, if both and are greater than 5, the binomial distribution is closely approximated by the normal distribution and we can use the properties of a normal distribution to determine the location of the analyte and its recovery. The fraction, , of each analyte remaining in the lower phase is calculated using equation A16.1. Because the volumes of the lower and upper phases are equal, we find that = 1 / ( + 1) = 1 / (9 + 1) = 0.10 = 1 / ( + 1) = 1 / (4 + 1) = 0.20 Because we know that + = 1, we also know that is 0.90 and is 0.80. After 100 steps, the mean and the standard deviation for the distribution of analytes A and B are µ = = (100)(0.90) = 90 and σ = ) = (100)(0.90)(0.10)) = 3 µ = = (100)(0.80) = 80 and σ = ) = (100)(0.80)(0.20)) = 4 Given that , , , and are all greater than 5, we can assume that the distribution of analytes follows a normal distribution and that the confidence interval for the tubes containing each analyte is = µ ± σ where is the tube’s number and the value of is determined by the desired significance level. For a 99% confidence interval the value of is 2.58 ( ); thus, = 90 ± (2.58)(3) = 90 ± 8 = 80 ± (2.58)(4) = 80 ± 10 Because the two confidence intervals overlap, a complete separation of the two analytes is not possible using a 100 step countercurrent extraction. The complete distribution of the analytes is shown in Figure A16.4. From Example A16.2 we know that after 100 steps of the countercurrent extraction, analyte A is normally distributed about tube 90 with a standard deviation of 3. To determine the fraction of analyte A in tubes 85–99, we use the single-sided normal distribution in to determine the fraction of analyte in tubes 0–84, and in tube 100. The fraction of analyte A in tube 100 is determined by calculating the deviation = ( − µ) / σ = (99 − 90) / 3 = 3 and using the table in to determine the corresponding fraction. For = 3 this corresponds to 0.135% of analyte A. To determine the fraction of analyte A in tubes 0–84 we again calculate the deviation = ( − µ) / σ = (85 − 90) / 3 = –1.67 From we find that 4.75% of analyte A is present in tubes 0–84. Analyte A’s recovery, therefore, is 100% – 4.75% – 0.135% ≈ 95% To calculate the separation factor we determine the recovery of analyte B in tubes 85.99 using the same general approach as for analyte A, finding that approximately 89.4% of analyte B remains in tubes 0.84 and that essentially no analyte B is in tube 100. The recover for B, therefore, is 100% – 89.4% – 0% ≈ 10.6% and the separation factor is = / = 10.6 / 95 = 0.112 | 10,977 | 3,564 |
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On a chemical level, an important clue to the unraveling of is the existence of noble gases, which are almost completely unable to form chemical compounds. This lack of reactivity suggests that the atoms of these elements have structures which do not permit interaction with the structures of other atoms. A second clue is the close correspondence between the valence of an element and the extent to which its atomic number differs from that of the nearest noble gas. Elements which have a valence of , for instance, have atomic numbers or than that of a noble gas. Thus the atoms of Li, Na, K, Rb, and Cs all contain one more than the corresponding noble gases He, Ne, Ar, Kr, and Xe, while atoms of hydrogen H and F, Cl, Br, and I all contain one electron less. Similar remarks apply to a valence of . Be, Mg, Ca, Sr, and Ba all contain two electrons more than a noble-gas atom, while the elements O, S, Se, and Te all contain two electrons less. Exactly the same pattern of behavior also extends to elements with a valence of 3 or 4. As early as 1902, Lewis began to suggest (in his lectures to general chemistry students, no less) that the behavior just described could be explained by assuming that the electrons in atoms were arranged in , all electrons in the same shell being approximately the same distance from the nucleus. The pictures which he eventually developed for helium, chlorine, and potassium atoms are illustrated below. In the helium atom the two electrons occupy only one shell, in the chlorine atom the 17 electrons are arranged in three shells, and in the potassium atom the 19 electrons occupy four shells. Lewis suggested that each shell can only accommodate so many electrons. Once this number has been reached, the shell must be regarded as , and any extra electrons are accommodated in the next shell, somewhat farther from the nucleus. Once a shell is filled, moreover, it is assumed to have a particularly stable structure which prevents the electrons in the shell from any involvement with other atoms. Thus it is only the electrons in the (called ) that have any chemical importance. Furthermore, if the outermost shell is filled, then the resulting atom will have little or no tendency to react with other atoms and form compounds with them. Since this is exactly the behavior exhibited by the noble gases, Lewis concluded that the characteristic feature of an atom of a noble gas is a of electrons. Assuming that the noble gases all contain an outermost filled shell, it is now quite simple to work out how many electrons can be accommodated in each shell. Since the first noble gas helium has two electrons, we know that only two electrons are needed to fill the first shell. A further eight electrons brings us to the next noble gas neon ( = 10). Accordingly we deduce that the second shell can accommodate a maximum of eight electrons. A similar argument leads to the conclusion that the third shell also requires eight electrons to fill it and that an atom of argon has two electrons in the first shell, eight in the second, and eight in the third, a total of 18 electrons. | 3,148 | 3,565 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Inductive_Effects_of_Alkyl_Groups |
A substituent on a benzene ring can effect the placement of additional substituents on that ring during . How do we know where an additonal substituent will most likely be placed? The answer to this is through inductive and resonance effects. Inductive effects are directly correlated with electronegativity. Substituents can either be meta directing or ortho-para directing. The three general positions of a disubstituted benzene ring are , and . The first scenario for adding an electrophile to a monosubstituted benzene ring is when the substituent is an . Electron donating groups are alkyl groups, phenyl groups or substituents that have a lone pair of electrons on the atom directly bonded to the ring. Electron donating groups are donating by induction (Activating and Deactivating Benzene Rings) and resonance. Examples of electron donating groups: -CH -OCH , -OH, -NH Electron donating groups cause the second subtituent to add on to the or position on the benzene ring. The reason for this can be explained by the different carbocation resonance structures of the ortho, meta and para positions. Some electron donating groups have an extra resonance form in which there is a double bond between the atom and the carbon on the benzene. This is a very stable resonance form. This is due to directing effects of substituents in conjugation with the benzene ring. When the electrophile is added to the ortho position, three different resonance forms are possible. Carbocation forms 1 and 2 are secondary carbocations, but position 3 forms a tertiary carbocation and the positive charge is on the carbon directly attached to the electron donating group, which is the most stable. This carbocation is also stablized by the electrons from the electron donating group. This is the reason that the ortho position is one of the major products. If the electrophile is added to the monosubstituted benzene ring in the para position one of the three resonance forms of the carbocations will be a tertiary carbocation which is highly stable because of the +I effect if the three -CH . This carbocation intermediate is the same as the one formed from ortho substitution. For the meta substituted carbocation resonance structures, there are three possible resonance froms that are secondary carbocations. These forms are not as stable as the tertiary carbocation form in the ortho and para substituted carbocations. Therefore, the two major products of the reaction of a monosubstituted benzene ring with an electron donating group and additional electrophile are the and positions. H-NMR spectroscopy can be used to determine whether or not a compound has a second substituent at the ortho or para position. At the ortho position there are four distinct signals, but for the para position there are only two signals because the molecule is symmetrical. Electron donating groups on a benzene ring are said to be , because they increase the rate of the second substitution so that it is than that of standard benzene. Electron donating groups are said to be ortho/para directing and they are activators. The other circumstance is when you have add an additional electrophile to a monosubstituted benzene ring with an on it. Electron withdrawing groups have an atom with a slight positive or full positive charge directly attached to a benzene ring. Examples of electron withdrawing groups: -CF COOH, -CN. Electron withdrawing groups only have one major product, the second substituent adds in the position. Again, this can be explained by the resonance forms of the carbocation intermediates. When the second electrophile is added on to the benzene ring in the ortho position, the same three resonance forms of the carbocation are produced. Again, one form is a tertiary carbocation with the positive charge on the carbon directly attached to the electron withdrawing group. Unlike in the case with an electron donating group, this resonance form is . This is due to the electron withdrawing group pulling away electrons from the carbon, creating an even stronger positive charge. This situation holds true for the para substituted tertiary carbocation resonace form as well. For the meta position, all the carbocations formed are secondary. Although these are not entirely stable, they are more favored than the resonance forms of the ortho and para positions. The major product of a monosubstituted benzene ring with an electron withdrawing group and an additional electrophile is a product with substitution. In contrast to electron donating groups, electron withdrawing groups are . This means that the rate of the second substitution is lower than that of standard benzene. Halogens are very electronegative. This means that inductively they are electron withdrawing. However, because of their ability to donate a lone pair of electrons in resonance forms, they are activators and ortho/para directing. Resonance forms win out in directing. Because they are electron withdrawing, halogens are very weak activators. Electron withdrawing groups are meta directors and they are deactivators. | 5,132 | 3,566 |
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A number of hydrocarbons, called or , have triple bonds between carbon atoms.\(^3\) They conform to the general formula \(C_nH_{2n-2}\) for one triple bond. The IUPAC system for naming alkynes employs the ending instead of the used for naming of the corresponding saturated hydrocarbon: The numbering system for locating the triple bond and substituent groups is analogous to that used for the corresponding alkenes: Hydrocarbons with more than one triple bond are called , , and so on, according to the number of triple bonds Hydrocarbons with both double and triple bonds are called ( alkynenes). The chain always should be numbered to give the multiple bonds the lowest possible numbers, and when there is a choice, double bonds are given lower numbers than triple bonds. For example, The hydrocarbon substituents derived from alkynes are called groups: \(^3\)Alkyne rhymes with "mine" and "thine." and (1977) | 945 | 3,568 |
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After completing this section, you should be able to Make certain that you can define, and use in context, the key terms below. The two alkenes, -CH CH=CHCH and (CH ) C=CH have similar heats of hydrogenation (−120 kJ/mol and −119 kJ/mol, respectively), and are therefore of similar stability. However, they are both less stable than -CH CH=CHCH (−116 kJ/mol). You may wonder why an - bond is stronger than an - bond. Bond strength depends on the efficiency with which orbitals can overlap. In general, orbitals overlap more efficiently than do orbitals; therefore, the - bond in the hydrogen molecule is stronger than the - bond in fluorine. In hybrid orbitals, the greater the character of the orbital, the more efficiently it can overlap: an orbital, which has a 33% character, can overlap more effectively than an orbital, with only 25% character. Alkene hydrogenation is the of hydrogen gas (H ) to an the bond and forms an alkane. Alkene hydrogenation reactions require a transition metal catalyst, such as Pt or Pd, to speed up the reaction. The hydrogenation reaction is used in this section to investigate the stability of alkenes, however, it will be discussed in greater detail in . Hydrogenation reactions are exothermic and the enthalpy change in this reaction is called the heat of ° . Since the double bond is breaking in this reaction, the energy released during hydrogenation is proportional to the energy in the double bond of the molecule. By comparing the heat of hydrogenations from a series of alkenes that produce the same alkane, a quantitative measure of relative alkene stabilities can be produced. These experiments will lead to an general understanding of structural features which tend to stabilize or destabilize alkenes. A increases the reaction rate by lowering the activation energy of the reaction. Although the catalyst is not consumed in the reaction, it is required to accelerate the reaction sufficiently to be observed in a reasonable amount of time. Catalysts commonly used in alkene hydrogenation are: platinum, palladium, and nickel. The metal catalyst acts as a surface on which the reaction takes place. This increases the rate by putting the reactants in close proximity to each other, facilitating interactions between them. With this catalyst present, the sigma bond of H breaks, and the two hydrogen atoms instead bind to the metal (see #2 in the figure below). The \(\pi\) bond of the alkene weakens as it also interacts with the metal (see #3 below). Since both the reactants are bound to the metal catalyst, the hydrogen atoms can easily add, one at a time, to the previously double-bonded carbons (see #4 and #5 below). The position of both of the reactants bound to the catalyst makes it so the hydrogen atoms are only exposed to one side of the alkene. This explains why the hydrogen atoms add to same side of the molecule, called syn-addition. The catalyst remains intact and unchanged throughout the reaction. The stability of alkene can be determined by measuring the amount of energy associated with the hydrogenation of the molecule. Since the double bond is breaking in this reaction, the energy released in hydrogenation is proportional to the energy in the double bond of the molecule. This is a useful tool because heats of hydrogenation can be measured very accurately. The \(\Delta H^o\) is usually around -30 kcal/mol for alkenes. Stability is simply a measure of energy. Lower energy molecules are more stable than higher energy molecules. More substituted alkenes are more stable than less substituted ones due to . They have a lower heat of hydrogenation. The following illustrates stability of alkenes with various substituents: Between cis and trans isomers of an alkene, the cis isomer tends to be less stable due to the molecular crowding created nonbonding interaction between two alky groups on the same side of the double bond. The crowding creates steric strain which distorts bond angles creating less effective bond orbital overlap and desabilizing the molecule. Steric strain has previously been seen in gauche interactions in Newman projections ( ) and 1,3-diaxial interactions in substituted cyclohexanes See the following isomers of butene: In general, the stability of an alkene increases with the number of alkyl substituents. This effect is due by the combination of two factors: In classical valence-bond theory, electron delocalization can only occur by the parallel overlap of adjacent p orbitals. According to hyperconjugation theory, electron delocalization could also occur by the parallel overlap of p orbitals with adjacent hybridized orbitals participating in sigma bonds. This electron delocalization serves to stabilize the alkene. As the number of alkyl substituents increases, the number of sigma bonds available for hyperconjugation increases, and the alkene tends to become more stabilized. In the example of propene shown below, a p orbital from a sp hybridized carbon involved in the double bond interacts with a sp hybridized orbital participating in an adjacent C-H sigma bond. In a molecular orbital description of hyperconjugation, the electrons in sigma molecular orbitals (C-H or C-C) of alkyl substituens, interact with adjacent unpopulated non-bonding or antibonding molecular orbitals from the double bond. The interaction creates a bonding molecular orbital which extends over the four atom chain (C=C-C-H) involved in hyperconjugation. The expanded molecular orbital helps to stabilize the double bond. Bond strengths play an important part in determining the overall stability of a molecule. A C-C bond between a sp carbon and a sp carbon is slightly stronger than a C-C bond between two sp carbons. Increasing the number alkyl substituents of a double bond also increases the number of sp -sp C-C bonds making the alkene more stable. This is idea can be clearly seen when comparing the isomers 1-butene and 2-butene. The molecule 1-butene is monosubstituted and contains a sp -sp C-C and a sp -sp C-C bond. The disubstituted, 2-butene, contains 2 sp -sp C-C bonds which contributes to its greater stability. In cycloalkenes smaller than cyclooctene, the cis isomers are more stable than the trans as a result of ring strain. 1) Of the three following isomers which would be expected to be the most stable? a) b) c) 1) a) b) c) 3-Bromobut-1-ene reacts with hydrogen gas in the presence of a platinum catalyst. What is the name of the product? 2-Bromobutane ( ) Cyclohexene reacts with hydrogen gas in the presence of a palladium catalyst. What is the name of the product? Cyclohexane What is the stereochemistry (syn or anti addition) of an alkene hydrogenation reaction? Syn-addition When looking at their heats of hydrogenation, is the cis or the trans isomer generally more stable? Trans Show the product for the following | 6,904 | 3,569 |
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An important class of crystal structures is found in many metals and also in the solidified noble gases where the atoms (which are all the same) are packed together as closely as possible. Most of us are familiar with the process of packing spheres together, either from playing with marbles or BB’s as children or from trying to stack oranges or other round fruit into a pyramid. On a level surface we can easily arrange a collection of spheres of the same size into a very compact hexagonal layer in which each sphere is touching six of its fellows, as seen Figure \(\Page {1}\). Then we can add a second layer so that each added sphere snuggles into a depression between three spheres in the layer below. Within this second layer each sphere also contacts six neighbors, and the layer is identical to the first one. It appears that we can add layer after layer indefinitely, or until we run out of spheres. Each sphere will be touching of its fellows since it is surrounded by six in the same plane and nestles among three in the plane above and three in the plane below. We say that each sphere has a of 12. It is impossible to make any other structure with a larger coordination number, that is, to pack more spheres within a given volume. Accordingly the structure just described is often referred to as a . Nickel, shown below, is an example of a metal with a closest-packed structure. In part of Figure \(\Page {2}\) the first layer of spheres has been labeled and the to indicate that spheres in the second layer are not directly above those in the fist. The third layer is directly above the first, and so it is labeled . If we continue in the fashion shown, adding alternately , then , then layers, we obtain a structure whose unit cell (shown in part ) has two equal sides with an angle of 120° between them. Other angles are 90°, and so the cell belongs to the hexagonal crystal system. Hence this structure is called (hpc). An example of a metal with an HPC structure is Magnesium, which is shown below. Also shown in Figure \(\Page {3}\) is the unit cell of a structure called (bcc). This is similar to the fcc structure except that, instead of spheres in the faces, there is a single sphere in the center of the cube. This central sphere is surrounded by eight neighbors at the corners of the unit cell, giving a coordination number of 8. Hence the bcc structure is not as compact as the closest-packed structures which had a coordination number 12. Nevertheless, some metals are found to have bcc structures. Count the number of spheres in the unit cell of (a) a face-centered cubic structure, and (b) a body-centered cubic structure. Referring to the last figure and using the equation: \[N=N_{\text{body}}\text{ + }\frac{N_{\text{face}}}{\text{2}}\text{ + }\frac{N_{\text{edge}}}{\text{4}}\text{ + }\frac{N_{\text{corner}}}{\text{8}} \nonumber \] we find \[N=\text{1 + 0 + 0 + }\frac{\text{8}}{\text{8}}=\text{2} \nonumber \] Silicon has the same crystal structure as . Techniques are now available for growing crystals of this element which are virtually flawless. Analysis on some of these perfect crystals found the side of the unit cell to be 543.102064 pm long. The unit cell is a cube containing eight Si atoms, but is ont one of the simple cubic cells discussed already. From the isotope make up, molar mass and density of the crystals, it was determined that one mole of Si in this crystal form has a volume of 12.0588349×10 m . Determine from this data. This problem uses knowledge of silicon crystal structure to determine . From the edge length, we can obtain the volume of the cubic unit cell. We know that the unit contains eight atoms, and since we know the volume of one mole, we can calculate , with the Avogadro constant defined as the number of particles per unit amount of substance. \[N_{A}= \frac{N*V_{\text{m}}}{V_{\text{unit cell}}}=\frac{8\times{12.0588349}\times{10}^{-6}\text{m}^{3}}{({ 543.102064}\times{10}^{-12}\text{m})^{3}}={6.02214179}\times{10}^{23} \nonumber \] The values used to determine this value were taken from crystals using X Ray Crystal Density(XRCD), to determine side length. These values were used in the most recent analysis published by the Committee on Data for Science and Technology(CODATA) , which standardizes definitions of important scientific constants and units. The value you just calculated is therefore the most accurate determination of Avogadro's constant at this time. | 4,490 | 3,570 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Compounds/Coordinative_Unsaturation |
is effectively the maximum coordination number that a metal ion can adopt. Coordinative unsaturation is based on ionic radii (ultimately atomic radii from 0.1), where the atomic size increases from right to left (lower nuclear charge) and top to bottom (increasing n level being filled). For any metal, ionization readily occurs to generate a cation, which will yield ionic radii smaller than atomic radii. In this case, the metal cation is considered a Lewis acid with Lewis basic ligands surrounding it through coordinate bonds. The number of Lewis bases surrounding the Lewis acidic metal cation is the coordination number. The metal cation charge does complicate comparison of say K+ to Ca2+, where Ca2+ is smaller despite being to the right of K+. In this comparison, the coordinative unsaturation is roughly equivalent between K+ and Ca2+ because the additional charge on Ca2+ compensates for the decreased radius to yield both ions tending up to a coordination number of 8. So the coordinative unsaturation increases as you go down a group: Be2+, 4; Mg2+, 6; Ca2+, 8; Sr2+, 8, Ba2+, 8. | 1,111 | 3,572 |
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The independent behavior of each type of ion in solution was illustrated in by means of precipitation reactions. is a process in which a solute separates from a supersaturated solution. In a chemical laboratory it usually refers to a solid crystallizing from a liquid solution, but in weather reports it applies to liquid or solid water separating from supersaturated air. A typical precipitation reaction occurs when an aqueous solution of barium chloride is mixed with one containing sodium sulfate. The equation \[\ce{Ba^{2+}(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}\label{1} \] can be written to describe what happens, and such an equation is useful in making chemical calculations. However, Equation \(\ref{1}\) does not really represent the microscopic particles (that is, the ions) present in the solution. Thus we might write \[\ce{Ba^{2+}(aq) + 2Cl^{-}(aq) + 2Na^{+}(aq) + SO4^{2-}(aq) -> BaSO4(s) + 2Na^{+} + 2Cl^{-}(aq)}\label{2} \] Equation \(\ref{2}\) is rather cumbersome and includes so many different ions that it may be confusing. In any case, we are often interested in the independent behavior of ions, not the specific compound from which they came. A precipitate of BaSO ( ) will form when solution containing Ba ( ) is mixed with solution containing SO ( ) (provided concentrations are not extremely small). This happens independently of the Cl ( ) and Na ( ) ions in Eq. \(\ref{2}\). These ions are called because they do not participate in the reaction (see the figure above). When we want to emphasize the independent behavior of ions, a is written, omitting the spectator ions. For precipitation of BaSO the net ionic equation is \[\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO{4}(s)}\label{3} \] When a solution of AgNO is added to a solution of CaCl , insoluble AgCl precipitates. Write three equations to describe this process. Both AgNO and CaCl are soluble ionic compounds, and so they are strong electrolytes. The three equations are \(\ce{2AgNO3(aq) + CaCl2(aq) -> 2AgCl(s) + Ca(NO3)2(aq)}\) \(\ce{2Ag^+(aq) + 2NO3^{-}(aq) + Ca^{2+}(aq) + Cl^{-}(aq) -> 2AgCl(s) + Ca^{2+}(aq) + 2NO3^{-}(aq)}\) \(\ce{Ag^+(aq) + Cl^{-} (aq) -> AgCl(s)}\) The occurrence or nonoccurrence of precipitates can be used to detect the presence or absence of various species in solution. BaCl solution, for instance, is often used as a test for SO ( ) ion. There are several insoluble salts of Ba, but they all dissolve in dilute acid except for BaSO . Thus, if BaCl solution is added to an unknown solution which has previously been acidified, the occurrence of a white precipitate is proof of the presence of the SO ion. AgNO solution is often used in a similar way to test for halide ion. If AgNO solution is added to an acidified unknown solution, a white precipitate indicates the presence of Cl ions, a cream-colored precipitate indicates the presence of Br ions, and a yellow precipitate indicates the presence of I ions. Further tests can then be made to see whether perhaps a mixture of these ions is present. When AgNO is added to tap water, a white precipitate is almost always formed. The Cl ions in tap water usually come from the Cl which is added to municipal water supplies to kill microorganisms. Precipitates are also used for quantitative analysis of solutions, that is, to determine the amount of solute or the mass of solute in a given solution. For this purpose it is often convenient to use the first of the three types of equations described above. Then the rules of stoichiometry may be applied. When a solution of 0.1 AgNO is added to 50.0 cm of a CaCl solution of unknown concentration, 2.073 g AgCl precipitates. Calculate the concentration of the unknown solution. We know the volume of the unknown solution, and so only the amount of solute is needed to calculate the concentration. This can be found using Eq. (2 ) in Example 1. From the equation the stoichiometric ratio (CaCl /AgCl) may be obtained. A road map to the solution of the problem is \(m_{AgCl} \underset{\text{M_{AgCl}}}{\mathop{\rightarrow}}\, n_{AgCl} \underset{\text{M_{CaCL_2}}}{\mathop{\rightarrow}}\, n_{CaCl_2}\) \(n_{CaCl-2} = 2.073 g AgCl * \frac{1mol AgCl}{143.32 g AgCl} * \frac{1 mol CaCl_2}{2 mol AgCl} = 7.23 * 10^{-3} \text{mol CaCl}_2\) \(c_{CaCl_2} = \frac{n_{CaCl_2}}{V_{soln}} = \frac{7.23 * 10^{-3} mol CaCl_2}{50.0 cm^3} * \frac{10^3 cm^3}{1 dm^3} = 0.145 \frac{mol}{dm^3}\) Thus the concentration of the unknown solution is 0.145 . Because of the general utility of precipitates in chemistry, it is worth having at least a rough idea of which common classes of compounds can be precipitated from solution and which cannot. Table \(\Page {1}\) gives a list of rules which enable us to predict the solubility of the most commonly encountered substances. Use of this table is illustrated in the following example. Write balanced net ionic equations to describe any reactions which occur when the following solutions are mixed: If any precipitate forms, it will be either a combination of Na ions and I ions, namely, NaI, or a combination of ammonium ions, NH , and sulfate ions, SO , namely, (NH ) SO . From Table 11.2 we find that NaI and (NH ) SO are both soluble. Thus no precipitation reaction will occur, and there is no equation to write. Possible precipitates are KCl and SrCO . From Table 11.2 we find that SrCO is insoluble. Accordingly we write the net ionic equation as \(\ce{Sr^{2+}(aq) + CO3^{2-}(aq) -> SrCO3(s)}\) omitting the spectator ions K and Cl . Possible precipitates are Fe(OH) and BaSO . Both are insoluble. The net ionic equation is thus \(\ce{Fe^{2+}(aq) + SO4^{2-}(aq) + Ba^{2+}(aq) + 2OH^{-}(aq) -> Fe(OH)2(s) + BaSO4(s)}\) | 5,742 | 3,573 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.4%3A_Strong_Acids_and_Strong_Bases |
Acids and bases that are completely ionized when dissolved in water are called and There are only a few strong acids and bases, and everyone should know their names and properties. These acids are often used in industry and everyday life. The concentrations of acids and bases are often expressed in terms of pH, and as an educated person, you should have the skill to convert concentrations into pH and pOH. The pH is an indication of the hydrogen ion concentration, \(\ce{[H+]}\). are acids that are completely or nearly 100% ionized in their solutions; Table \(\Page {1}\) includes some common strong acids. Hence, the ionization in Equation \(\ref{gen ion}\) for a strong acid HA can be represented with a single arrow: \[\ce{HA(aq) + H2O(l) \rightarrow H3O^{+}(aq) + A^{-}(aq)} \label{gen ion}\] Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid HA, and the hydronium ion is the conjugate acid of water. By definition, a strong acid yields 100% of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water. Table \(\Page {1}\) lists several strong acids. For a strong acid, \(\ce{[H+]}\) = \(\ce{[A^{-}]}\) = concentration of acid if the concentration is much higher than \(1 \times 10^{-7}\, M\). However, for a very dilute strong acid solution with concentration less than \(1 \times 10^{-7}\, M\), the pH is dominated by the \[\ce{H2O \rightleftharpoons H+ + OH-}\] Calculate the pH of a solution with \(1.2345 \times 10^{-4}\; M \ce{HCl}\), a strong acid. The solution of a strong acid is completely ionized. That is, this equation goes to completion \[\ce{HCl(aq) -> H(aq) + Cl^{-}(aq)} \nonumber\] Thus, \(\ce{[H+]} = 1.2345 \times 10^{-4}\). \[\ce{pH} = -\log(1.2345 \times 10^{-4}) = 3.90851 \nonumber\] What is the pH for a solution containing 1.234 M \(\ce{[HCl]}\)? pH = -0.0913 Calculate the pH of a stock \(\ce{HCl}\) solution that is 32% by mass \(\ce{HCl}\).
The density of such a solution is needed before we can calculate the pH. Since the density is not on the label, we need to find it from the Material Safety Data Sheet, which gives the specific gravity of 1.150. Thus, the amount of acid in 1.0 L is 1150 g. \[\begin{align*}
\textrm{The amount of HCl} &= 1000\times1.150\times0.32\\
&= \mathrm{368\: g\: \left(\dfrac{1\: mol}{36.5\: g} \leftarrow molar\: mass\: of\: HCl\right)}\\
&= \mathrm{10.08\: M}\\
&= \ce{[H+]}
\end{align*}\] \[\ce{pH} = -\log(10.08) = -1.003 \nonumber\] Yes, pH have negative values if \(\ce{[H+]} > 1.0\) Check out the information on nitric acid, a strong acid, and calculate the pH of a stock nitric acid solution. Calculate the pH of a solution containing \(1.00 \times 10^{-7}\; M\) of the strong acid \(\ce{HCl}\). \(\ce{[H+]} = 1.0 \times 10^{-7}\; M\) from the strong acid, and if is the amount from the ionization of water, then we have the equilibrium due to the of water. We can model this with an . \[H_2O \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)} \nonumber\] Recall that \(K_{w} = \ce{[H+] [OH- ]} = 1 \times 10^{-14}\), due to the ionization equilibrium of water in the solution: \[\begin{align*} (1.00 \times 10^{-7} +x) x &= 1 \times 10^{-14} \\[4pt] x^2 + 1.00 \times 10^{-7}x - 1.00 \times 10^{-14} &= 0 \end{align*}\] Solving this equation for \(x\) from the quadratic equation results in \[\begin{align*} x &= \dfrac{-1.00 \times 10^{-7} \pm \sqrt{1.00 \times 10^{-14} + (4)(1) (1.00 \times 10^{-14})}}{2}\\
&= 0.61 \times 10^{-7} \end{align*}\] only the additive root is physical (positive concentration); therefore \[ \begin{align*} \ce{[H+]} &= (1.00 + 0.61) \times 10^{-7}\; M \\[4pt] \ce{pH} &= -\log(1.61 \times 10^{-7}) \\[4pt] &= 6.79 \end{align*}\] If you require only 1 significant figure, the pH is about 7. are completely ionized in solution. Table \(\Page {1}\) includes some common strong bases. For example, \(\ce{KOH}\) dissolves in water in the reaction \[\ce{KOH \rightarrow K+ + OH-} \nonumber\] Relative to the number of strong acids, there are fewer number of strong bases and most are alkali hydroxides. Calcium hydroxide is considered a strong base, because it is completely, almost completely, ionized. However, the solubility of calcium hydroxide is very low. When \(\ce{Ca(OH)2}\) dissolves in water, the ionization reaction is as follows: \[\ce{Ca(OH)2 \rightarrow Ca^2+ + 2 OH-} \nonumber\] Because of the stoichiometry of calcium hydroxide, upon dissociation, the concentration of \(\ce{OH-}\) will be twice the concentration of \(\ce{Ca^2+}\): \[\mathrm{[OH^-] = 2 [Ca^{2+}]} \nonumber\] Calculate the pOH of a solution containing \(1.2345 \times 10^{-4}\; M \; \ce{Ca(OH)2}\). Based on complete ionization of \[\ce{Ca(OH)2 \rightarrow Ca^{+2} + 2OH-} \nonumber\] \[\begin{align*} \ce{[OH^{-}]} &= 2 \times 1.2345 \times 10^{-4} \\[4pt] &= 2.4690 \times 10^{-4}\; M \\[4pt] \ce{pOH} &= -\log( 2.4690 \times 10^{-4})\\[4pt] &= 3.6074 \end{align*}\] The molar solubility of calcium hydroxide is 0.013 M \(\ce{Ca(OH)2}\). Calculate the pOH. pOH = 1.58 Calculating pH in Strong Acid or Strong Base Solutions: \(\ce{HF}\), \(\ce{HNO2}\), \(\ce{H2CO3}\), \(\ce{H2S}\), \(\ce{HSO4-}\), \(\ce{Cl-}\), \(\ce{HNO3}\), \(\ce{HCN}\)
\(\mathrm{[OH^-] = \dfrac{0.80}{40} = 0.020\: M}\); \([H^+] = \dfrac{1.0 \times 10^{-14}}{0.020} = 5\times 10^{-13} M\). The pH is 12.30. | 5,367 | 3,574 |
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We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of and atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both N and O have fairly similar Lewis structures that contain lone pairs of electrons. Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations. | 1,144 | 3,575 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.13%3A_The_Molecular_View_of_Equilibrium |
In other sections, we show that exist, how they can be determined, and how they can be used, but we have not explained why they are sometimes large and sometimes small. In order to answer this and related questions we must switch our point of view from the macroscopic to the microscopic level. The cis-trans isomerism of difluoroethene provides a simple example for examination on the molecular level. As we saw earlier, if 1 mol cis isomer of this compound is heated to 623 K, it is gradually transformed into the equilibrium mixture of ⅔ mol cis isomer and ⅓ mol trans isomer. From a macroscopic point of view, we might conclude that the reaction stops once equilibrium is attained, but on the microscopic level, such a statement is obviously untrue. As cis molecules move around a container at 623 K, they occasionally suffer a collision with sufficient energy to flip them over to the trans conformation ( for a microscopic view of this process, see Figure \(\Page {1}\) ). This conversion process does not suddenly stop when equilibrium has been attained. There are plenty of cis molecules present in the equilibrium mixture, and they continue to collide with their fellow molecules and hence to flip over from the cis to the trans conformation. Thus the forward reaction \[cis-\text{C}_2\text{H}_2\text{F}_2 \rightarrow trans-\text{C}_2\text{H}_2\text{F}_2 \nonumber \] must in the equilibrium mixture. If this were the only reaction taking place in the flask, the eventual result would be that all the cis isomer would be converted to the trans isomer and no equilibrium mixture would result. This does not happen because the \[trans-\ce{C_{2}H_{2}F_{2} \rightarrow} cis-\ce{C_{2}H_{2}F_{2}} \nonumber \] is also occurring at the same time. The trans molecules in the equilibrium mixture are also continually being bombarded by their fellow molecules. Many of these collisions have sufficient energy to flip the trans molecule back into the cis form. Thus cis molecules are not only being consumed but also being produced in the flask. The concentration of the cis molecules remains constant because the rate at which they are being consumed is exactly balanced by the rate at which they are being produced. In other words, for every cis molecule which flips to the trans conformation in one part of the container, there will be on the average a molecule of trans isomer flipping in the reverse direction in another part of the container. This constant reshuffle of molecules between the cis and the trans forms can then continue indefinitely without any net change in the concentration of either species. This molecular interpretation of equilibrium is not confined to the particular example of cis-trans isomerism or even just to chemical reactions. In we describe vapor-liquid equilibria in which the condensation of vapor just balanced the evaporation of liquid. Any Since these rates are equal, there is no net change in the concentration of any of the reactants or products with time. Because an equilibrium corresponds to a balance between the forward and the reverse reaction in this way, we use double arrows (\(\rightleftharpoons\)) in the equation. As another example of chemical equilibrium, let us take a 0.001 solution of acetic acid at 25°C. As seen in , slightly more than 10 percent of the acetic acid molecules in this solution have ionized according to the equation \[\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^{-} + \text{H}_3\text{O}^{+} \nonumber \] Although the concentrations of all species in this solution remain constant with time, this does not mean that acetic acid molecules have stopped transferring protons to water molecules. The concentrations remain constant because the reverse reaction (transfer of a proton from a hydronium ion to an acetate ion) is occurring at the same rate as the forward reaction. For every hydronium ion produced by a proton transfer somewhere in the solution, another hydronium ion is losing its proton somewhere else. The net result is that the concentration of the hydronium ion remains constant, and with it the concentrations of the other species involved. The microscopic view of equilibrium as a constant reshuffle of chemical species is often given a special name and referred to as a . It is important to realize that once a dynamic equilibrium has been set up, a particular atom will sometimes turn up as part of one of the product molecules and sometimes as part of one of the reactant molecules. In the ionization equilibrium of acetic acid just considered, suppose for the sake of argument that we could identify a particular oxygen atom in some way. If we could now follow the history of this atom, we would find that it would sometimes form part of an acetic acid molecule, CH CO*OH and sometimes part of an acetate ion, CH CO*O . Since the acid is about 10 percent ionized, we would find on the average that our labeled oxygen would spend only 10 percent of its time as part of an acetate ion and the remaining 90 percent as part of an acetic acid molecule. Figure \(\Page {1}\) gives a computer simulation of this dynamic view of equilibrium for the cis-trans isomerization equilibrium of difluoroethene at 623 K. Suppose that it were possible to color one of the molecules in the equilibrium mixture and to photograph it at regular intervals thereafter under a high-power microscope. The figure shows the kind of results we could expect. In each “photograph” the molecule appears at a different orientation― between photos it has been bounced around quite a bit by collisions with other molecules. A careful inspection reveals a more important fact. Three of the nine photographs show the screened molecule in the trans conformation, while in the other six the molecule has the cis conformation. If we went on taking a large number of photographs of this molecule, we would find on the average that this ratio would be preserved. On the average one-third of the photographs would show the trans conformation, while the other two-thirds would show the cis form. The situation is much as though we had a conventional six-sided cubic die with the word printed on two sides and the word cis printed on the other four. If we roll this die a sufficient number of times, it will come up cis twice as often as trans. We can now begin to see why it is that the equilibrium constant for this reaction is 0.5 at 623 K. Not only the screened molecule, but every other molecule in the equilibrium mixture as well, spends one-third of its time in the trans conformation and two-thirds in the cis conformation. At any given time, therefore, we will find a mixture in which one-third of the molecules are trans while the other two-thirds are cis. The ratio of trans to cis will be 1:2, or 0.5, as found experimentally. Such an explanation is still not the whole story. We still need to explain why the dice are loaded against the trans conformation of a molecule of difluoroethene. Why is it that this isomer is less likely to occur in the constant reshuffle between the two conformations caused by the continual random collisions of molecules with each other in the equilibrium mixture? The reason is that the trans isomer is slightly than the cis form. The enthalpy change Δ for the reaction \[cis-\text{C}_2\text{H}_2\text{F}_2 \rightarrow trans-\text{C}_2\text{H}_2\text{F}_2 \nonumber \] has the small, but nevertheless positive, value of +3.88 kJ mol . In any collection of molecules, energy is constantly being transferred from one molecule to another as they collide. In this continual interchange of energy, inevitably some molecules acquire more kinetic or potential energy than their fellow molecules. However, , the probability of a molecule’s acquiring a given energy depends on the magnitude of that energy. The higher the energy, the less likely it is to occur. In the constant reshuffle of energies, molecules which are even slightly higher in energy than their fellows occur slightly less often. In the case under consideration, since a molecule of -difluoroethene is 3.88 kJ mol in energy than the cis form, it will occur often than the cis form in an equilibrium mixture in which each molecule is constantly flipping back and forth between the two forms. In other words, the die is loaded against the trans configuration because it is higher in energy. We can now also understand why the position of equilibrium for this reaction varies with temperature. At a very low temperature it will be a rare occurrence for a molecule to acquire the extra 3.88 kJ mol needed if it is to have the trans configuration. Accordingly, we expect the equilibrium mixture to contain many fewer trans molecules than cis molecules and the equilibrium constant \[K_{c}=\frac{\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }} \nonumber \] to be quite small. Indeed, if it were possible to take the system down to absolute zero, no trans molecules would be possible and the value of would be zero. As the temperature is increased, the probability of a molecule acquiring enough energy to assume the trans configuration increases, and increases along with it. cannot increase indefinitely, though. If the temperature is high enough, the energy difference between the cis and trans configurations becomes insignificant compared with the average kinetic energy of a molecule. Both forms will then be almost equally probable, and the value of will be very slightly less than 1. Very few chemical equilibria are as simple as the cis-trans isomerism of difluoroethene. In almost all other cases the position of equilibrium is governed not only by the energy difference between reactants and products but also by a probability factor which is independent of the energy. A simple example of such a probability factor is provided by the equilibrium between the ring and chain forms of 1,4-butanediol. The projection formula for this alcohol is
\[K_{c}=\frac{\text{ }\!\![\!\!\text{ chain }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ ring }\!\!]\!\!\text{ }}=\text{1}\text{.22} \nonumber \] In order to see what this other factor is, let us suppose that we can somehow investigate our system at a temperature which is so high that the energy difference between the ring and the chain form is insignificant. (In practice, the carbon tetrachloride solvent would boil off long before this temperature was obtained.) At such a temperature, if the energy difference between the two forms were the only factor determining their relative probability, we would expect the ring and chain forms (seen below) to be equally probable. It is easy to see, though, why this would not actually be the case. The chain form of 1,4-butanediol is a very flexible molecule because of the free rotation which is possible around the three carbon-carbon bonds and the two carbon-oxygen bonds. As it collides with its fellow molecules, it can adopt a very large number of different conformations. The figure below illustrates some of these conformations, selected at random by a computer. Again we can take these as representing a series of photographs of the same molecule taken at regular intervals, say every 5 min. For the sake of clarity, though, we have kept the position of two of the carbon atoms the same for each photograph. In actual fact the molecule would be rotating as well as flexing, and the orientation of the bond joining these two carbon atoms would be continually changing. We should also remember that the molecule has undergone very many collisions between successive photographs. \(\Page {2}\) The chain form of butanediol. Because various segments of the molecule are free to rotate about carbon-carbon and carbon-oxygen bonds, the molecule can adopt a very large number of conformations. A random selection of nine of these is shown. Note that in none of these are the two oxygen atoms close enough to hydrogen bond together and form a ring. The ring form of butanediol. is at the bottom right. The two oxygen atoms at the top of the ring are hydrogen bonded together. From a strictly geometrical point of view this is a rather improbable conformation for the molecule. It is only because of the lowering in energy caused by the formation of a hydrogen bond that this ring form turns out to be slightly more probable in the equilibrium mixture than the chain form at low temperatures. The important thing to notice about this random selection of nine conformations shown in Figure \(\Page {2}\) is that in none of them are the two oxygen atoms close enough to allow the formation of a hydrogen bond. At this high temperature, because of the large number of ways in which a butanediol molecule can arrange itself in space, it will only occasionally adopt a ring conformation like that shown in Figure \(\Page {3}\) in which the two oxygen atoms are close enough to hydrogen bond to each other. In the equilibrium mixture at high temperature, therefore, there will be many more chain molecules than ring molecules and will be much larger than 1. This situation begins to change as the temperature is reduced. With decreasing temperature, a smaller and smaller fraction of the molecules will have enough energy to break open a hydrogen bond. In consequence the ring form will become more probable. Once having formed, a ring will be able to survive quite a few collisions with other molecules before it is broken open again. Indeed, if we could reduce the temperature of our system close to absolute zero without the solvent freezing, then we would find that virtually all the butanediol molecules were in the ring form. Once formed, a ring would almost never experience a collision which was sufficiently energetic to break it open. At such a low temperature the value of would be very close to zero. The two equilibria we have just described have both been very simple examples, especially since they each involved only one reactant and one product. Nevertheless the principles we have discovered in discussing them apply to all chemical equilibria. We can always interpret the equilibrium constant of a chemical reaction as the product of two factors, each of which has a meaning on the microscopic level: The takes account of the fact that either the products or the reactants are higher in energy and hence less probable in the constant reshuffling of energy that goes on between molecules. The takes account of the fact that even if there were no energy difference between products and reactants, one or the other would still be more probable. This greater probability arises because of the larger number of ways in which the products (or reactants) can be realized on the molecular level. In general the energy factor has the greatest effect on the value of at low temperatures. The lower the temperature, the very much lower the probability of a high-energy species occurring by chance in the constant energy reshuffle. At higher temperatures the effect of the energy factor becomes less pronounced and the probability factor becomes more important. If the temperature can be raised high enough, the probability factor eventually predominates. | 15,391 | 3,576 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.18%3A_Common_Oxidizing_Agents |
Oxidizing agents must be able to accept electrons readily. Elements with high electronegativity readily accept electrons, as can molecules or ions which contain relatively electronegative elements and even some metals which have high oxidation numbers. Bear these general rules in mind as we examine examples of common oxidizing agents in the following paragraphs. All four elemental halogens, F , Cl , Br , and I , are able to accept electrons according to the half-equation \[\ce{X2 + 2e^{–} -> 2X^{–}} \nonumber \] with \(X = F, Cl, Br, I\) As we might expect from the periodic variation of electronegativity, the oxidizing power of the halogens decreases in the order F > Cl > Br > I . Fluorine is such a strong oxidizing agent that it can react with water (water is very difficult to oxidize): \[\ce{2F2 + 6H2O -> 4H3O^+ + 4F^{–} + O2} \nonumber \] Chlorine also reacts with water, but only in the presence of sunlight. Bromine is weaker, and iodine has only mild oxidizing power. Oxygen gas, which constitutes about 20 percent of the earth’s atmosphere, is another electronegative element which is a good oxidizing agent. It is very slightly weaker than chlorine, but considerably stronger than bromine. Because the atmosphere contains such a strong oxidant, few substances occur in reduced form at the earth’s surface. An oxidized form of silicon, SiO , is one of the most plentiful constituents of the crust of the earth. Most metals, too, occur as oxides and must be reduced before they can be obtained in elemental form. When iron rusts, it forms the red-brown oxide Fe O • H O, seen below, which always contains an indeterminate amount of water. In aqueous solution NO , IO , MnO , Cr O , and a number of other oxyanions serve as convenient, strong oxidizing agents. The structure of the last oxyanion mentioned above is shown in Figure 1. The most strongly oxidizing oxyanions often contain an element in its highest possible oxidation state, that is, with an oxidation number equal to the periodic group number. For example, NO contains nitrogen in a +5 oxidation state, Cr O (seen below) contains chromium +6, and has manganese +7. The oxidizing power of the dichromate ion is employed in laboratory cleaning solution, a solution of Na Cr O in concentrated H SO . This readily oxidizes the organic compounds in grease to carbon dioxide. It is also highly corrosive, eats holes in clothing, and must be handled with care. Dark purple permanganate ion is another very common oxidizing agent (seen below). it is reduced to solid dark brown MnO . , however, it forms almost colorless Mn ( ). | 2,637 | 3,577 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.02%3A_AcidBase_Titrations |
Before 1800, most used H SO , HCl, or HNO as acidic titrants, and K CO or Na CO as basic titrants. A titration’s end point was determined using litmus as an indicator, which is red in acidic solutions and blue in basic solutions, or by the cessation of CO effervescence when neutralizing \(\text{CO}_3^{2-}\). Early examples of acid–base titrimetry include determining the acidity or alkalinity of solutions, and determining the purity of carbonates and alkaline earth oxides. The determination of acidity and alkalinity continue to be important applications of acid–base titrimetry. We will take a closer look at these applications later in this section. Three limitations slowed the development of acid–base titrimetry: the lack of a strong base titrant for the analysis of weak acids, the lack of suitable indicators, and the absence of a theory of acid–base reactivity. The introduction, in 1846, of NaOH as a strong base titrant extended acid–base titrimetry to the determination of weak acids. The synthesis of organic dyes provided many new indicators. Phenolphthalein, for example, was first synthesized by Bayer in 1871 and used as an indicator for acid–base titrations in 1877. Despite the increased availability of indicators, the absence of a theory of acid–base reactivity made it difficult to select an indicator. The development of equilibrium theory in the late 19th century led to significant improvements in the theoretical understanding of acid–base chemistry, and, in turn, of acid–base titrimetry. Sørenson’s establishment of the pH scale in 1909 provided a rigorous means to compare indicators. The determination of acid–base dissociation constants made it possible to calculate a theoretical titration curve, as outlined by Bjerrum in 1914. For the first time analytical chemists had a rational method for selecting an indicator, making acid–base titrimetry a useful alternative to gravimetry. In the overview to this chapter we noted that a titration’s end point should coincide with its equivalence point. To understand the relationship between an acid–base titration’s end point and its equivalence point we must know how the titrand’s pH changes during a titration. In this section we will learn how to calculate a titration curve using the equilibrium calculations from . We also will learn how to sketch a good approximation of any acid–base titration curve using a limited number of simple calculations. For our first titration curve, let’s consider the titration of 50.0 mL of 0.100 M HCl using a titrant of 0.200 M NaOH. When a strong base and a strong acid react the only reaction of importance is \[\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \label{9.1}\] Although we have not written reaction \ref{9.1} as an equilibrium reaction, it is at equilibrium; however, because its equilibrium constant is large—it is ( ) or \(1.00 \times 10^{14}\)—we can treat reaction \ref{9.1} as though it goes to completion. The first task is to calculate the volume of NaOH needed to reach the equivalence point, . At the equivalence point we know from reaction \ref{9.1} that \[\begin{aligned} \text { moles } \mathrm{HCl}=& \text { moles } \mathrm{NaOH} \\ M_{a} \times V_{a} &=M_{b} \times V_{b} \end{aligned} \nonumber\] where the subscript ‘ ’ indicates the acid, HCl, and the subscript ‘ ’ indicates the base, NaOH. The volume of NaOH needed to reach the equivalence point is \[V_{e q}=V_{b}=\frac{M_{a} V_{a}}{M_{b}}=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{(0.200 \ \mathrm{M})}=25.0 \ \mathrm{mL} \nonumber\] Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of unreacted HCl. At the start of the titration the solution is 0.100 M in HCl, which, because HCl is a strong acid, means the pH is \[\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left[\text{HCl} \right] = -\log (0.100)=1.00 \nonumber\] After adding 10.0 mL of NaOH the concentration of excess HCl is \[[\text{HCl}] = \frac {(\text{mol HCl})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] \[[\mathrm{HCl}]=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})-(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0500 \ \mathrm{M} \nonumber\] and the pH increases to 1.30. At the equivalence point the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water. \[\begin{array}{c}{K_{w}=1.00 \times 10^{-14}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.00 \times 10^{-7}}\end{array} \nonumber\] Thus, the pH at the equivalence point is 7.00. For volumes of NaOH greater than the equivalence point, the pH is determined by the concentration of excess OH . For example, after adding 30.0 mL of titrant the concentration of OH is \[[\text{OH}^-] = \frac {(\text{mol NaOH})_\text{added} - (\text{mol HCl})_\text{initial}} {\text{total volume}} = \frac {M_b V_b - M_a V_a} {V_a + V_b} \nonumber\] \[\left[\mathrm{OH}^{-}\right]=\frac{(0.200 \ \mathrm{M})(30.0 \ \mathrm{mL})-(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{30.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0125 \ \mathrm{M} \nonumber\] To find the concentration of H O we use the expression \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0125}=8.00 \times 10^{-13} \ \mathrm{M} \nonumber\] to find that the pH is 12.10. Table 9.2.1
and Figure 9.2.1
show additional results for this titration curve. You can use this same approach to calculate the titration curve for the titration of a strong base with a strong acid, except the strong base is in excess before the equivalence point and the strong acid is in excess after the equivalence point. Construct a titration curve for the titration of 25.0 mL of 0.125 M NaOH with 0.0625 M HCl. The volume of HCl needed to reach the equivalence point is \[V_{e q}=V_{a}=\frac{M_{b} V_{b}}{M_{a}}=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{(0.0625 \ \mathrm{M})}=50.0 \ \mathrm{mL} \nonumber\] Before the equivalence point, NaOH is present in excess and the pH is determined by the concentration of unreacted OH . For example, after adding 10.0 mL of HCl \[\begin{array}{c}{\left[\mathrm{OH}^{-}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})-(0.0625 \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0714 \ \mathrm{M}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0714 \ \mathrm{M}}=1.40 \times 10^{-13} \ \mathrm{M}}\end{array} \nonumber\] the pH is 12.85. For the titration of a strong base with a strong acid the pH at the equivalence point is 7.00. For volumes of HCl greater than the equivalence point, the pH is determined by the concentration of excess HCl. For example, after adding 70.0 mL of titrant the concentration of HCl is \[[\mathrm{HCl}]=\frac{(0.0625 \ \mathrm{M})(70.0 \ \mathrm{mL})-(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{70.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0132 \ \mathrm{M} \nonumber\] giving a pH of 1.88. Some additional results are shown here. For this example, let’s consider the titration of 50.0 mL of 0.100 M acetic acid, CH COOH, with 0.200 M NaOH. Again, we start by calculating the volume of NaOH needed to reach the equivalence point; thus \[\operatorname{mol} \ \mathrm{CH}_{3} \mathrm{COOH}=\mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[M_{a} \times V_{a}=M_{b} \times V_{b} \nonumber\] \[V_{e q}=V_{b}=\frac{M_{a} V_{a}}{M_{b}}=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{(0.200 \ \mathrm{M})}=25.0 \ \mathrm{mL} \nonumber\] Before we begin the titration the pH is that for a solution of 0.100 M acetic acid. Because acetic acid is a weak acid, we calculate the pH using the method outlined in \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] \[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^-\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=\frac{(x)(x)}{0.100-x}=1.75 \times 10^{-5} \nonumber\] finding that the pH is 2.88. Adding NaOH converts a portion of the acetic acid to its conjugate base, CH COO . \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \label{9.2}\] Because the equilibrium constant for reaction \ref{9.2} is quite large \[K=K_{\mathrm{a}} / K_{\mathrm{w}}=1.75 \times 10^{9} \nonumber\] we can treat the reaction as if it goes to completion. Any solution that contains comparable amounts of a weak acid, HA, and its conjugate weak base, A , is a buffer. As we learned in , we can calculate the pH of a buffer using the Henderson–Hasselbalch equation. \[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \nonumber\] Before the equivalence point the concentration of unreacted acetic acid is \[\left[\text{CH}_3\text{COOH}\right] = \frac {(\text{mol CH}_3\text{COOH})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] and the concentration of acetate is \[[\text{CH}_3\text{COO}^-] = \frac {(\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_b V_b} {V_a + V_b} \nonumber\] For example, after adding 10.0 mL of NaOH the concentrations of CH COOH and CH COO are \[\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})-(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}} = 0.0500 \text{ M} \nonumber\] \[\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=\frac{(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0333 \ \mathrm{M} \nonumber\] which gives us a pH of \[\mathrm{pH}=4.76+\log \frac{0.0333 \ \mathrm{M}}{0.0500 \ \mathrm{M}}=4.58 \nonumber\] At the equivalence point the moles of acetic acid initially present and the moles of NaOH added are identical. Because their reaction effectively proceeds to completion, the predominate ion in solution is CH COO , which is a weak base. To calculate the pH we first determine the concentration of CH COO \[\left[\mathrm{CH}_{3} \mathrm{COO}^-\right]=\frac{(\mathrm{mol} \ \mathrm{NaOH})_{\mathrm{added}}}{\text { total volume }}= \frac{(0.200 \ \mathrm{M})(25.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0667 \ \mathrm{M} \nonumber\] Alternatively, we can calculate acetate’s concentration using the initial moles of acetic acid; thus \[\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=\frac{\left(\mathrm{mol} \ \mathrm{CH}_{3} \mathrm{COOH}\right)_{\mathrm{initial}}}{\text { total volume }} = \frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}} = 0.0667 \text{ M} \nonumber\] Next, we calculate the pH of the weak base as shown earlier in \[\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q) \nonumber\] \[K_{\mathrm{b}}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}=\frac{(x)(x)}{0.0667-x}=5.71 \times 10^{-10} \nonumber\] \[x=\left[\mathrm{OH}^{-}\right]=6.17 \times 10^{-6} \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{6.17 \times 10^{-6}}=1.62 \times 10^{-9} \ \mathrm{M} \nonumber\] finding that the pH at the equivalence point is 8.79. After the equivalence point, the titrant is in excess and the titration mixture is a dilute solution of NaOH. We can calculate the pH using the same strategy as in the titration of a strong acid with a strong base. For example, after adding 30.0 mL of NaOH the concentration of OH is \[\left[\mathrm{OH}^{-}\right]=\frac{(0.200 \ \mathrm{M})(30.0 \ \mathrm{mL})-(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{30.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0125 \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0125}=8.00 \times 10^{-13} \ \mathrm{M} \nonumber\] giving a pH of 12.10. Table 9.2.2
and Figure 9.2.2
show additional results for this titration. You can use this same approach to calculate the titration curve for the titration of a weak base with a strong acid, except the initial pH is determined by the weak base, the pH at the equivalence point by its conjugate weak acid, and the pH after the equivalence point by excess strong acid. Construct a titration curve for the titration of 25.0 mL of 0.125 M NH with 0.0625 M HCl. The volume of HCl needed to reach the equivalence point is \[V_{a q}=V_{a}=\frac{M_{b} V_{b}}{M_{a}}=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{(0.0625 \ \mathrm{M})}=50.0 \ \mathrm{mL} \nonumber\] Before adding HCl the pH is that for a solution of 0.100 M NH . \[K_{\mathrm{b}}=\frac{[\mathrm{OH}^-]\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]}=\frac{(x)(x)}{0.125-x}=1.75 \times 10^{-5} \nonumber\] \[x=\left[\mathrm{OH}^{-}\right]=1.48 \times 10^{-3} \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{[\mathrm{OH}^-]}=\frac{1.00 \times 10^{-14}}{1.48 \times 10^{-3} \ \mathrm{M}}=6.76 \times 10^{-12} \ \mathrm{M} \nonumber\] The pH at the beginning of the titration, therefore, is 11.17. Before the equivalence point the pH is determined by an \(\text{NH}_3/\text{NH}_4^+\) buffer. For example, after adding 10.0 mL of HCl \[\left[\mathrm{NH}_{3}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})-(0.0625 \ \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0714 \ \mathrm{M} \nonumber\] \[\left[\mathrm{NH}_{4}^{+}\right]=\frac{(0.0625 \ \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0179 \ \mathrm{M} \nonumber\] \[\mathrm{pH}=9.244+\log \frac{0.0714 \ \mathrm{M}}{0.0179 \ \mathrm{M}}=9.84 \nonumber\] At the equivalence point the predominate ion in solution is \(\text{NH}_4^+\). To calculate the pH we first determine the concentration of \(\text{NH}_4^+\) \[\left[\mathrm{NH}_{4}^{+}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0417 \ \mathrm{M} \nonumber\] and then calculate the pH \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^{+}\right]}=\frac{(x)(x)}{0.0417-x}=5.70 \times 10^{-10} \nonumber\] obtaining a value of 5.31. After the equivalence point, the pH is determined by the excess HCl. For example, after adding 70.0 mL of HCl \[[\mathrm{HCl}]=\frac{(0.0625 \ \mathrm{M})(70.0 \ \mathrm{mL})-(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{70.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0132 \ \mathrm{M} \nonumber\] and the pH is 1.88. Some additional results are shown here. We can extend this approach for calculating a weak acid–strong base titration curve to reactions that involve multiprotic acids or bases, and mixtures of acids or bases. As the complexity of the titration increases, however, the necessary calculations become more time consuming. Not surprisingly, a variety of algebraic and spreadsheet approaches are available to aid in constructing titration curves. The following papers provide information on algebraic approaches to calculating titration curves: (a) Willis, C. J. , , 659–663; (b) Nakagawa, K. , , 673–676; (c) Gordus, A. A. , , 759–761; (d) de Levie, R. , , 209–217; (e) Chaston, S. , , 878–880; (f) de Levie, R. , , 585–590. The following papers provide information on the use of spreadsheets to generate titration curves: (a) Currie, J. O.; Whiteley, R. V. , , 923–926; (b) Breneman, G. L.; Parker, O. J. , , 46–47; (c) Carter, D. R.; Frye, M. S.; Mattson, W. A. , , 67–71; (d) Freiser, H. , CRC Press: Boca Raton, 1992. To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching an acid–base titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M CH COOH with 0.200 M NaOH to illustrate our approach. This is the same example that we used to develop the calculations for a weak acid–strong base titration curve. You can review the results of that calculation in and in . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 25.0 mL. Next we draw our axes, placing pH on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point volume, we draw a vertical line that intersects the -axis at 25.0 mL of NaOH. Figure 9.2.3
a shows the first step in our sketch. Before the equivalence point the titrand’s pH is determined by a buffer of acetic acid, CH COOH, and acetate, CH COO . Although we can calculate a buffer’s pH using the Henderson–Hasselbalch equation, we can avoid this calculation by making a simple assumption. You may recall from that a buffer operates over a pH range that extends approximately ±1 pH unit on either side of the weak acid’s p value. The pH is at the lower end of this range, pH = p – 1, when the weak acid’s concentration is \(10 \times\) greater than that of its conjugate weak base. The buffer reaches its upper pH limit, pH = p + 1, when the weak acid’s concentration is \(10 \times\) smaller than that of its conjugate weak base. When we titrate a weak acid or a weak base, the buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume. The actual values are 9.09% and 90.9%, but for our purpose, using 10% and 90% is more convenient; that is, after all, one advantage of an approximation! Figure 9.2.3
b shows the second step in our sketch. First, we superimpose acetic acid’s ladder diagram on the -axis, including its buffer range, using its p value of 4.76. Next, we add two points, one for the pH at 10% of the equivalence point volume (a pH of 3.76 at 2.5 mL) and one for the pH at 90% of the equivalence point volume (a pH of 5.76 at 22.5 mL). The third step is to add two points after the equivalence point. The pH after the equivalence point is fixed by the concentration of excess titrant, NaOH. Calculating the pH of a strong base is straightforward, as we saw earlier. Figure 9.2.3
c includes points (see ) for the pH after adding 30.0 mL and after adding 40.0 mL of NaOH. Next, we draw a straight line through each pair of points, extending each line through the vertical line that represents the equivalence point’s volume (Figure 9.2.3
d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.2.3
e). A comparison of our sketch to the exact titration curve (Figure 9.2.3
f) shows that they are in close agreement. Sketch a titration curve for the titration of 25.0 mL of 0.125 M NH with 0.0625 M HCl and compare to the result from . The figure below shows a sketch of the titration curve. The black dots and curve are the approximate sketch of the titration curve. The points in are the calculations from . The two black points before the equivalence point ( = 5 mL, pH = 10.24 and = 45 mL, pH= 8.24) are plotted using the p of 9.244 for \(\text{NH}_4^+\). The two black points after the equivalence point ( = 60 mL, pH = 2.13 and = 80 mL, pH= 1.75 ) are from the answer to . As shown in the following example, we can adapt this approach to any acid–base titration, including those where exact calculations are more challenging, including the titration of polyprotic weak acids and bases, and the titration of mixtures of weak acids or weak bases. Sketch titration curves for the following two systems: (a) the titration of 50.0 mL of 0.050 M H A, a diprotic weak acid with a p of 3 and a p of 7; and (b) the titration of a 50.0 mL mixture that contains 0.075 M HA, a weak acid with a p of 3, and 0.025 M HB, a weak acid with a p of 7. For both titrations, assume that the titrant is 0.10 M NaOH. Figure 9.2.4
a shows the titration curve for H A, including the ladder diagram for H A on the -axis, the two equivalence points at 25.0 mL and at 50.0 mL, two points before each equivalence point, two points after the last equivalence point, and the straight-lines used to sketch the final titration curve. Before the first equivalence point the pH is controlled by a buffer of H A and HA . An HA /A buffer controls the pH between the two equivalence points. After the second equivalence point the pH reflects the concentration of excess NaOH. Figure 9.2.4
b shows the titration curve for the mixture of HA and HB. Again, there are two equivalence points; however, in this case the equivalence points are not equally spaced because the concentration of HA is greater than that for HB. Because HA is the stronger of the two weak acids it reacts first; thus, the pH before the first equivalence point is controlled by a buffer of HA and A . Between the two equivalence points the pH reflects the titration of HB and is determined by a buffer of HB and B . After the second equivalence point excess NaOH determines the pH. Sketch the titration curve for 50.0 mL of 0.050 M H A, a diprotic weak acid with a p of 3 and a p of 4, using 0.100 M NaOH as the titrant. The fact that p falls within the buffer range of p presents a challenge that you will need to consider. The figure below shows a sketch of the titration curve. The titration curve has two equivalence points, one at 25.0 mL \((\text{H}_2\text{A} \rightarrow \text{HA}^-)\) and one at 50.0 mL (\(\text{HA}^- \rightarrow \text{A}^{2-}\)). In sketching the curve, we plot two points before the first equivalence point using the p of 3 for H A \[V_{\mathrm{HCl}}=2.5 \ \mathrm{mL}, \mathrm{pH}=2 \text { and } V_{\mathrm{HCl}}=22.5 \ \mathrm{mL}, \mathrm{pH}=4 \nonumber\] two points between the equivalence points using the p of 5 for HA \[V_{\mathrm{HCl}}=27.5 \ \mathrm{mL}, \mathrm{pH}=3, \text { and } V_{\mathrm{HCl}}=47.5 \ \mathrm{mL}, \mathrm{pH}=5 \nonumber\] and two points after the second equivalence point \[V_{\mathrm{HCl}}=70 \ \mathrm{mL}, \mathrm{pH}=12.22 \text { and } V_{\mathrm{HCl}}=90 \ \mathrm{mL}, \mathrm{pH}=12.46 \nonumber\] Drawing a smooth curve through these points presents us with the following dilemma—the pH appears to increase as the titrant’s volume approaches the first equivalence point and then appears to decrease as it passes through the first equivalence point. This is, of course, absurd; as we add NaOH the pH cannot decrease. Instead, we model the titration curve before the second equivalence point by drawing a straight line from the first point ( = 2.5 mL, pH = 2) to the fourth point ( = 47.5 mL, pH= 5), ignoring the second and third points. The results is a reasonable approximation of the exact titration curve. Earlier we made an important distinction between a titration’s end point and its equivalence point. The difference between these two terms is important and deserves repeating. An equivalence point, which occurs when we react stoichiometrically equal amounts of the analyte and the titrant, is a theoretical not an experimental value. A titration’s end point is an experimental result that represents our best estimate of the equivalence point. Any difference between a titration’s equivalence point and its corresponding end point is a source of determinate error. Earlier we learned how to calculate the pH at the equivalence point for the titration of a strong acid with a strong base, and for the titration of a weak acid with a strong base. We also learned how to sketch a titration curve with only a minimum of calculations. Can we also locate the equivalence point without performing any calculations. The answer, as you might guess, often is yes! For most acid–base titrations the inflection point—the point on a titration curve that has the greatest slope—very nearly coincides with the titration’s equivalence point. The red arrows in , for example, identify the equivalence points for the titration curves in . An inflection point actually precedes its corresponding equivalence point by a small amount, with the error approaching 0.1% for weak acids and weak bases with dissociation constants smaller than 10 , or for very dilute solutions [Meites, L.; Goldman, J. A. , , 472–479]. The principal limitation of an inflection point is that it must be present and easy to identify. For some titrations the inflection point is missing or difficult to find. Figure 9.2.5
, for example, demonstrates the affect of a weak acid’s dissociation constant, , on the shape of its titration curve. An inflection point is visible, even if barely so, for acid dissociation constants larger than 10 , but is missing when is 10 . An inflection point also may be missing or difficult to see if the analyte is a multiprotic weak acid or weak base with successive dissociation constants that are similar in magnitude. To appreciate why this is true let’s consider the titration of a diprotic weak acid, H A, with NaOH. During the titration the following two reactions occur. \[\mathrm{H}_{2} \mathrm{A}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HA}^{-}(a q) \label{9.3}\] \[\mathrm{HA}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{A}^{2-}(a q) \label{9.4}\] To see two distinct inflection points, reaction \ref{9.3} must essentially be complete before reaction \ref{9.4} begins. Figure 9.2.6
shows titration curves for three diprotic weak acids. The titration curve for maleic acid, for which is approximately \(20000 \times\) larger than , has two distinct inflection points. Malonic acid, on the other hand, has acid dissociation constants that differ by a factor of approximately 690. Although malonic acid’s titration curve shows two inflection points, the first is not as distinct as the second. Finally, the titration curve for succinic acid, for which the two values differ by a factor of only \(27 \times\), has only a single inflection point that corresponds to the neutralization of \(\text{HC}_2\text{H}_4\text{O}_4^-\) to \(\text{C}_2\text{H}_4\text{O}_4^{2-}\). In general, we can detect separate inflection points when successive acid dissociation constants differ by a factor of at least 500 (a \(\Delta\) of at least 2.7). The same holds true for mixtures of weak acids or mixtures of weak bases. To detect separate inflection points when titrating a mixture of weak acids, their p values must differ by at least a factor of 500. One interesting group of weak acids and weak bases are organic dyes. Because an organic dye has at least one highly colored conjugate acid–base species, its titration results in a change in both its pH and its color. We can use this change in color to indicate the end point of a titration provided that it occurs at or near the titration’s equivalence point. As an example, let’s consider an indicator for which the acid form, HIn, is yellow and the base form, In , is red. The color of the indicator’s solution depends on the relative concentrations of HIn and In . To understand the relationship between pH and color we use the indicator’s acid dissociation reaction \[\mathrm{HIn}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{In}^{-}(a q) \nonumber\] and its equilibrium constant expression. \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \label{9.5}\] Taking the negative log of each side of Equation \ref{9.5}, and rearranging to solve for pH leaves us with a equation that relates the solution’s pH to the relative concentrations of HIn and In . \[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{In}^-]}{[\mathrm{HIn}]} \label{9.6}\] If we can detect HIn and In with equal ease, then the transition from yellow-to-red (or from red-to-yellow) reaches its midpoint, which is orange, when the concentrations of HIn and In are equal, or when the pH is equal to the indicator’s p . If the indicator’s p and the pH at the equivalence point are identical, then titrating until the indicator turns orange is a suitable end point. Unfortunately, we rarely know the exact pH at the equivalence point. In addition, determining when the concentrations of HIn and In are equal is difficult if the indicator’s change in color is subtle. We can establish the range of pHs over which the average analyst observes a change in the indicator’s color by making two assumptions: that the indicator’s color is yellow if the concentration of HIn is \(10 \times\) greater than that of In and that its color is red if the concentration of HIn is \(10 \times\) smaller than that of In . Substituting these inequalities into Equation \ref{9.6} \[\begin{array}{l}{\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{1}{10}=\mathrm{p} K_{\mathrm{a}}-1} \\ {\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{10}{1}=\mathrm{p} K_{\mathrm{a}}+1}\end{array} \nonumber\] shows that the indicator changes color over a pH range that extends ±1 unit on either side of its p . As shown in Figure 9.2.7
, the indicator is yel-ow when the pH is less than p – 1 and it is red when the pH is greater than p + 1. For pH values between p – 1 and p + 1 the indicator’s color passes through various shades of orange. The properties of several common acid–base indicators are listed in Table 9.2.3
. You may wonder why an indicator’s pH range, such as that for phenolphthalein, is not equally distributed around its p value. The explanation is simple. Figure 9.2.7
presents an idealized view in which our sensitivity to the indicator’s two colors is equal. For some indicators only the weak acid or the weak base is colored. For other indicators both the weak acid and the weak base are colored, but one form is easier to see. In either case, the indicator’s pH range is skewed in the direction of the indicator’s less colored form. Thus, phenolphthalein’s pH range is skewed in the direction of its colorless form, shifting the pH range to values lower than those suggested by Figure 9.2.7
. The relatively broad range of pHs over which an indicator changes color places additional limitations on its ability to signal a titration’s end point. To minimize a determinate titration error, the indicator’s entire pH range must fall within the rapid change in pH near the equivalence point. For example, in Figure 9.2.8
we see that phenolphthalein is an appropriate indicator for the titration of 50.0 mL of 0.050 M acetic acid with 0.10 M NaOH. Bromothymol blue, on the other hand, is an inappropriate indicator because its change in color begins well before the initial sharp rise in pH, and, as a result, spans a relatively large range of volumes. The early change in color increases the probability of obtaining an inaccurate result, and the range of possible end point volumes increases the probability of obtaining imprecise results. Suggest a suitable indicator for the titration of 25.0 mL of 0.125 M NH with 0.0625 M NaOH. You constructed a titration curve for this titration in and . The pH at the equivalence point is 5.31 (see ) and the sharp part of the titration curve extends from a pH of approximately 7 to a pH of approximately 4. Of the indicators in , methyl red is the best choice because its p value of 5.0 is closest to the equivalence point’s pH and because the pH range of 4.2–6.3 for its change in color will not produce a significant titration error. An alternative approach for locating a titration’s end point is to monitor the titration’s progress using a sensor whose signal is a function of the analyte’s concentration. The result is a plot of the entire titration curve, which we can use to locate the end point with a minimal error. A pH electrode is the obvious sensor for monitoring an acid–base titration and the result is a . For example, Figure 9.2.9
a shows a small portion of the potentiometric titration curve for the titration of 50.0 mL of 0.050 M CH COOH with 0.10 M NaOH, which focuses on the region that contains the equivalence point. The simplest method for finding the end point is to locate the titration curve’s inflection point, which is shown by the arrow. This is also the least accurate method, particularly if the titration curve has a shallow slope at the equivalence point. See for more details about pH electrodes. . Titration curves for the titration of 50.0 mL of 0.050 M CH COOH with 0.10 M NaOH: (a) normal titration curve; (b) first derivative titration curve; (c) second derivative titration curve; (d) Gran plot. The arrows show the location of each titration’s end point. Another method for locating the end point is to plot the first derivative of the titration curve, which gives its slope at each point along the -axis. Examine Figure 9.2.9
a and consider how the titration curve’s slope changes as we approach, reach, and pass the equivalence point. Because the slope reaches its maximum value at the inflection point, the first derivative shows a spike at the equivalence point (Figure 9.2.9
b). The second derivative of a titration curve can be more useful than the first derivative because the equivalence point intersects the volume axis. Figure 9.2.9
c shows the resulting titration curve. Suppose we have the following three points on our titration curve: Mathematically, we can approximate the first derivative as \(\Delta \text{pH} / \Delta V\), where \(\Delta \text{pH}\) is the change in pH between successive additions of titrant. Using the first two points, the first derivative is \[\frac{\Delta \mathrm{pH}}{\Delta V}=\frac{6.10-6.00}{23.91-23.65}=0.385 \nonumber\] which we assign to the average of the two volumes, or 23.78 mL. For the second and third points, the first derivative is 0.455 and the average volume is 24.02 mL. We can approximate the second derivative as \(\Delta (\Delta \text{pH} / \Delta V) / \Delta V\), or \(\Delta^2 \text{pH} / \Delta V^2\). Using the two points from our calculation of the first derivative, the second derivative is \[\frac{\Delta^{2} \mathrm{p} \mathrm{H}}{\Delta V^{2}}=\frac{0.455-0.385}{24.02-23.78}=0.292 \nonumber\] which we assign to the average of the two volumes, or 23.90 mL. Note that calculating the first derivative comes at the expense of losing one piece of information (three points become two points), and calculating the second derivative comes at the expense of losing two pieces of information. Derivative methods are particularly useful when titrating a sample that contains more than one analyte. If we rely on indicators to locate the end points, then we usually must complete separate titrations for each analyte so that we can see the change in color for each end point. If we record the titration curve, however, then a single titration is sufficient. The precision with which we can locate the end point also makes derivative methods attractive for an analyte that has a poorly defined normal titration curve. Derivative methods work well only if we record sufficient data during the rapid increase in pH near the equivalence point. This usually is not a problem if we use an automatic titrator, such as the one seen earlier in . Because the pH changes so rapidly near the equivalence point—a change of several pH units over a span of several drops of titrant is not unusual—a manual titration does not provide enough data for a useful derivative titration curve. A manual titration does contain an abundance of data during the more gently rising portions of the titration curve before and after the equivalence point. This data also contains information about the titration curve’s equivalence point. Consider again the titration of acetic acid, CH COOH, with NaOH. At any point during the titration acetic acid is in equilibrium with H O and CH COO \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] for which the equilibrium constant is \[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\] Before the equivalence point the concentrations of CH COOH and CH COO are \[[\text{CH}_3\text{COOH}] = \frac {(\text{mol CH}_3\text{COOH})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] \[[\text{CH}_3\text{COO}^-] = \frac {(\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_b V_b} {V_a + V_b} \nonumber\] Substituting these equations into the expression and rearranging leaves us with \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(M_{b} V_{b}\right) /\left(V_{a}+V_{b}\right)}{\left\{M_{a} V_{a}-M_{b} V_{b}\right\} /\left(V_{a}+V_{b}\right)} \nonumber\] \[K_{a} M_{a} V_{a}-K_{a} M_{b} V_{b}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(M_{b} V_{b}\right) \nonumber\] \[\frac{K_{a} M_{a} V_{a}}{M_{b}}-K_{a} V_{b}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] V_{b} \nonumber\] Finally, recognizing that the equivalence point volume is \[V_{eq}=\frac{M_{a} V_{a}}{M_{b}} \nonumber\] leaves us with the following equation. \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times V_{b}=K_{\mathrm{a}} V_{eq}-K_{\mathrm{a}} V_{b} \nonumber\] For volumes of titrant before the equivalence point, a plot of \(V_b \times [\text{H}_3\text{O}^+]\) versus is a straight-line with an -intercept of and a slope of – . shows a typical result. This method of data analysis, which converts a portion of a titration curve into a straight-line, is a . Values of determined by this method may have a substantial error if the effect of activity is ignored. See for a discussion of activity. The reaction between an acid and a base is exothermic. Heat generated by the reaction is absorbed by the titrand, which increases its temperature. Monitoring the titrand’s temperature as we add the titrant provides us with another method for recording a titration curve and identifying the titration’s end point (Figure 9.2.10
). Before we add the titrant, any change in the titrand’s temperature is the result of warming or cooling as it equilibrates with the surroundings. Adding titrant initiates the exothermic acid–base reaction and increases the titrand’s temperature. This part of a thermometric titration curve is called the titration branch. The temperature continues to rise with each addition of titrant until we reach the equivalence point. After the equivalence point, any change in temperature is due to the titrant’s enthalpy of dilution and the difference between the temperatures of the titrant and titrand. Ideally, the equivalence point is a distinct intersection of the titration branch and the excess titrant branch. As shown in Figure 9.2.10
, however, a thermometric titration curve usually shows curvature near the equivalence point due to an incomplete neutralization reaction or to the excessive dilution of the titrand and the titrant during the titration. The latter problem is minimized by using a titrant that is 10–100 times more concentrated than the analyte, although this results in a very small end point volume and a larger relative error. If necessary, the end point is found by extrapolation. Although not a common method for monitoring an acid–base titration, a thermometric titration has one distinct advantage over the direct or indirect monitoring of pH. As discussed earlier, the use of an indicator or the monitoring of pH is limited by the magnitude of the relevant equilibrium constants. For example, titrating boric acid, H BO , with NaOH does not provide a sharp end point when monitoring pH because boric acid’s of \(5.8 \times 10^{-10}\) is too small (Figure 9.2.11
a). Because boric acid’s enthalpy of neutralization is fairly large, –42.7 kJ/mole, its thermometric titration curve provides a useful endpoint (Figure 9.2.11
b). Thus far we have assumed that the titrant and the titrand are aqueous solutions. Although water is the most common solvent for acid–base titrimetry, switching to a nonaqueous solvent can improve a titration’s feasibility. For an amphoteric solvent, SH, the autoprotolysis constant, , relates the concentration of its protonated form, \(\text{SH}_2^+\), to its deprotonated form, S \[\begin{aligned} 2 \mathrm{SH} &\rightleftharpoons\mathrm{SH}_{2}^{+}+\mathrm{S}^{-} \\ K_{\mathrm{s}} &=\left[\mathrm{SH}_{2}^{+}\right,\mathrm{S}^-] \end{aligned} \nonumber\] and the solvent’s pH and pOH are \[\begin{array}{l}{\mathrm{pH}=-\log \left[\mathrm{SH}_{2}^{+}\right]} \\ {\mathrm{pOH}=-\log \left[\mathrm{S}^{-}\right]}\end{array} \nonumber\] You should recognize that is just specific form of when the solvent is water. The most important limitation imposed by is the change in pH during a titration. To understand why this is true, let’s consider the titration of 50.0 mL of \(1.0 \times 10^{-4}\) M HCl using \(1.0 \times 10^{-4}\) M NaOH as the titrant. Before the equivalence point, the pH is determined by the untitrated strong acid. For example, when the volume of NaOH is 90% of , the concentration of H O is \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{M_{a} V_{a}-M_{b} V_{b}}{V_{a}+V_{b}} = \frac{\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})-\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(45.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+45.0 \ \mathrm{mL}} = 5.3 \times 10^{-6} \ \mathrm{M} \nonumber\] and the pH is 5.3. When the volume of NaOH is 110% of , the concentration of OH is \[\left[\mathrm{OH}^{-}\right]=\frac{M_{b} V_{b}-M_{a} V_{a}}{V_{a}+V_{b}} = \frac{\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(55.0 \ \mathrm{mL})-\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{55.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}} = 4.8 \times 10^{-6} \ \mathrm{M} \nonumber\] and the pOH is 5.3. The titrand’s pH is \[\mathrm{pH}=\mathrm{p} K_{w}-\mathrm{pOH}=14.0-5.3=8.7 \nonumber\] and the change in the titrand’s pH as the titration goes from 90% to 110% of is \[\Delta \mathrm{pH}=8.7-5.3=3.4 \nonumber\] If we carry out the same titration in a nonaqueous amphiprotic solvent that has a of \(1.0 \times 10^{-20}\), the pH after adding 45.0 mL of NaOH is still 5.3. However, the pH after adding 55.0 mL of NaOH is \[\mathrm{pH}=\mathrm{p} K_{s}-\mathrm{pOH}=20.0-5.3=14.7 \nonumber\] In this case the change in pH \[\Delta \mathrm{pH}=14.7-5.3=9.4 \nonumber\] is significantly greater than that obtained when the titration is carried out in water. Figure 9.2.12
shows the titration curves in both the aqueous and the nonaqueous solvents. Another parameter that affects the feasibility of an acid–base titration is the titrand’s dissociation constant. Here, too, the solvent plays an important role. The strength of an acid or a base is a relative measure of how easy it is to transfer a proton from the acid to the solvent or from the solvent to the base. For example, HF, with a of \(6.8 \times 10^{-4}\), is a better proton donor than CH COOH, for which is \(1.75 \times 10^{-5}\). The strongest acid that can exist in water is the hydronium ion, H O . HCl and HNO are strong acids because they are better proton donors than H O and essentially donate all their protons to H O, leveling their acid strength to that of H O . In a different solvent HCl and HNO may not behave as strong acids. If we place acetic acid in water the dissociation reaction \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] does not proceed to a significant extent because CH COO is a stronger base than H O and H O is a stronger acid than CH COOH. If we place acetic acid in a solvent that is a stronger base than water, such as ammonia, then the reaction \[\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3}\rightleftharpoons\mathrm{NH}_{4}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-} \nonumber\] proceeds to a greater extent. In fact, both HCl and CH COOH are strong acids in ammonia. All other things being equal, the strength of a weak acid increases if we place it in a solvent that is more basic than water, and the strength of a weak base increases if we place it in a solvent that is more acidic than water. In some cases, however, the opposite effect is observed. For example, the p for NH is 4.75 in water and it is 6.40 in the more acidic glacial acetic acid. In contradiction to our expectations, NH is a weaker base in the more acidic solvent. A full description of the solvent’s effect on the p of weak acid or the p of a weak base is beyond the scope of this text. You should be aware, however, that a titration that is not feasible in water may be feasible in a different solvent. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical acid–base titrimetric method. Although each method is unique, the following description of the determination of protein in bread provides an instructive example of a typical procedure. The description here is based on Method 13.86 as published in , 8th Ed., Association of Official Agricultural Chemists: Washington, D. C., 1955. This method is based on a determination of %w/w nitrogen using the Kjeldahl method. The protein in a sample of bread is oxidized to \(\text{NH}_4^+\) using hot concentrated H SO . After making the solution alkaline, which converts \(\text{NH}_4^+\) to NH , the ammonia is distilled into a flask that contains a known amount of HCl. The amount of unreacted HCl is determined by a back titration using a standard strong base titrant. Because different cereal proteins contain similar amounts of nitrogen—on average there are 5.7 g protein for every gram of nitrogen—we multiply the experimentally determined %w/w N by a factor of 5.7 gives the %w/w protein in the sample. Transfer a 2.0-g sample of bread, which previously has been air-dried and ground into a powder, to a suitable digestion flask along with 0.7 g of a HgO catalyst, 10 g of K SO , and 25 mL of concentrated H SO . Bring the solution to a boil. Continue boiling until the solution turns clear and then boil for at least an additional 30 minutes. After cooling the solution below room temperature, remove the Hg catalyst by adding 200 mL of H O and 25 mL of 4% w/v K S. Add a few Zn granules to serve as boiling stones and 25 g of NaOH. Quickly connect the flask to a distillation apparatus and distill the NH into a collecting flask that contains a known amount of standardized HCl. The tip of the condenser must be placed below the surface of the strong acid. After the distillation is complete, titrate the excess strong acid with a standard solution of NaOH using methyl red as an indicator (Figure 9.2.13
). 1. Oxidizing the protein converts all of its nitrogen to \(\text{NH}_4^+\). Why is the amount of nitrogen not determined by directly titrating the \(\text{NH}_4^+\) with a strong base? There are two reasons for not directly titrating the ammonium ion. First, because \(\text{NH}_4^+\) is a very weak acid (its is \(5.6 \times 10^{-10}\)), its titration with NaOH has a poorly-defined end point. Second, even if we can determine the end point with acceptable accuracy and precision, the solution also contains a substantial concentration of unreacted H SO . The presence of two acids that differ greatly in concentration makes for a difficult analysis. If the titrant’s concentration is similar to that of H SO , then the equivalence point volume for the titration of \(\text{NH}_4^+\) is too small to measure reliably. On the other hand, if the titrant’s concentration is similar to that of \(\text{NH}_4^+\), the volume needed to neutralize the H SO is unreasonably large. 2. Ammonia is a volatile compound as evidenced by the strong smell of even dilute solutions. This volatility is a potential source of determinate error. Is this determinate error negative or positive? Any loss of NH is loss of nitrogen and, therefore, a loss of protein. The result is a negative determinate error. 3. Identify the steps in this procedure that minimize the determinate error from the possible loss of NH . Three specific steps minimize the loss of ammonia: (1) the solution is cooled below room temperature before we add NaOH; (2) after we add NaOH, the digestion flask is quickly connected to the distillation apparatus; and (3) we place the condenser’s tip below the surface of the HCl to ensure that the NH reacts with the HCl before it is lost through volatilization. 4. How does K S remove Hg , and why is its removal important? Adding sulfide precipitates Hg as HgS. This is important because NH forms stable complexes with many metal ions, including Hg . Any NH that reacts with Hg is not collected during distillation, providing another source of determinate error. Although many quantitative applications of acid–base titrimetry have been replaced by other analytical methods, a few important applications continue to find use. In this section we review the general application of acid–base titrimetry to the analysis of inorganic and organic compounds, with an emphasis on applications in environmental and clinical analysis. First, however, we discuss the selection and standardization of acidic and basic titrants. The most common strong acid titrants are HCl, HClO , and H SO . Solutions of these titrants usually are prepared by diluting a commercially available concentrated stock solution. Because the concentration of a concentrated acid is known only approximately, the titrant’s concentration is determined by standardizing against one of the primary standard weak bases listed in Table 9.2.4
. The nominal concentrations of the concentrated stock solutions are 12.1 M HCl, 11.7 M HClO , and 18.0 M H SO . The actual concentrations of these acids are given as %w/v and vary slightly from lot-to-lot. (a) The end point for this titration is improved by titrating to the second equivalence point, boiling the solution to expel CO2, and retitrating to the second equivalence point. The reaction in this case is \[\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow \mathrm{CO}_{2}+2 \mathrm{Na}^{+}+3 \mathrm{H}_{2} \mathrm{O} \nonumber\] (b) -(hydroxymethyl)aminomethane often goes by the shorter name of TRIS or THAM. (c) Potassium hydrogen phthalate often goes by the shorter name of KHP. (d) Because it is not very soluble in water, dissolve benzoic acid in a small amount of ethanol before diluting with water. The most common strong base titrant is NaOH, which is available both as an impure solid and as an approximately 50% w/v solution. Solutions of NaOH are standardized against any of the primary weak acid standards listed in Table \(\Page [4|\). Using NaOH as a titrant is complicated by potential contamination from the following reaction between dissolved CO and OH . \[\mathrm{CO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}( l) \label{9.7}\] Any solution in contact with the atmosphere contains a small amount of CO from the equilibrium \[\mathrm{CO}_{2}(g)\rightleftharpoons\mathrm{CO}_{2}(a q) \nonumber\] During the titration, NaOH reacts both with the titrand and with CO , which increases the volume of NaOH needed to reach the titration’s end point. This is not a problem if the end point pH is less than 6. Below this pH the \(\text{CO}_3^{2-}\) from reaction \ref{9.7} reacts with H O to form carbonic acid. \[\mathrm{CO}_{3}^{2-}(a q)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q). \label{9.8}\] Combining reaction \ref{9.7} and reaction \ref{9.8} gives an overall reaction that does not include OH . \[\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l ) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber\] Under these conditions the presence of CO does not affect the quantity of OH used in the titration and is not a source of determinate error. If the end point pH is between 6 and 10, however, the neutralization of \(\text{CO}_3^{2-}\) requires one proton \[\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q) \nonumber\] and the net reaction between CO and OH is \[\mathrm{CO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{HCO}_{3}^{-}(a q) \nonumber\] Under these conditions some OH is consumed in neutralizing CO , which results in a determinate error. We can avoid the determinate error if we use the same end point pH for both the standardization of NaOH and the analysis of our analyte, although this is not always practical. Solid NaOH is always contaminated with carbonate due to its contact with the atmosphere, and we cannot use it to prepare a carbonate-free solution of NaOH. Solutions of carbonate-free NaOH are prepared from 50% w/v NaOH because Na CO is insoluble in concentrated NaOH. When CO is absorbed, Na CO precipitates and settles to the bottom of the container, which allow access to the carbonate-free NaOH. When pre- paring a solution of NaOH, be sure to use water that is free from dissolved CO . Briefly boiling the water expels CO ; after it cools, the water is used to prepare carbonate-free solutions of NaOH. A solution of carbonate-free NaOH is relatively stable if we limit its contact with the atmosphere. Standard solutions of sodium hydroxide are not stored in glass bottles as NaOH reacts with glass to form silicate; instead, store such solutions in polyethylene bottles. Acid–base titrimetry is a standard method for the quantitative analysis of many inorganic acids and bases. A standard solution of NaOH is used to determine the concentration of inorganic acids, such as H PO or H AsO , and inorganic bases, such as Na CO are analyzed using a standard solution of HCl. If an inorganic acid or base that is too weak to be analyzed by an aqueous acid–base titration, it may be possible to complete the analysis by adjusting the solvent or by an indirect analysis. For example, when analyzing boric acid, H BO , by titrating with NaOH, accuracy is limited by boric acid’s small acid dissociation constant of \(5.8 \times 10^{-10}\). Boric acid’s value increases to \(1.5 \times 10^{-4}\) in the presence of mannitol, because it forms a stable complex with the borate ion, which results is a sharper end point and a more accurate titration. Similarly, the analysis of ammonium salts is limited by the ammonium ion’ small acid dissociation constant of \(5.7 \times 10^{-10}\). We can determine \(\text{NH}_4^+\) indirectly by using a strong base to convert it to NH , which is removed by distillation and titrated with HCl. Because NH is a stronger weak base than \(\text{NH}_4^+\) is a weak acid (its is \(1.58 \times 10^{-5}\)), the titration has a sharper end point. We can analyze a neutral inorganic analyte if we can first convert it into an acid or a base. For example, we can determine the concentration of \(\text{NO}_3^-\) by reducing it to NH in a strongly alkaline solution using Devarda’s alloy, a mixture of 50% w/w Cu, 45% w/w Al, and 5% w/w Zn. \[3 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{Al}(s)+5 \mathrm{OH}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 8 \mathrm{AlO}_{2}^{-}(a q)+3 \mathrm{NH}_{3}(a q) \nonumber\] The NH is removed by distillation and titrated with HCl. Alternatively, we can titrate \(\text{NO}_3^-\) as a weak base by placing it in an acidic nonaqueous solvent, such as anhydrous acetic acid, and using HClO as a titrant. Acid–base titrimetry continues to be listed as a standard method for the determination of alkalinity, acidity, and free CO in waters and wastewaters. is a measure of a sample’s capacity to neutralize acids. The most important sources of alkalinity are OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), although other weak bases, such as phosphate, may contribute to the overall alkalinity. Total alkalinity is determined by titrating to a fixed end point pH of 4.5 (or to the bromocresol green end point) using a standard solution of HCl or H SO . Results are reported as mg CaCO /L. Although a variety of strong bases and weak bases may contribute to a sample’s alkalinity, a single titration cannot distinguish between the possible sources. Reporting the total alkalinity as if CaCO is the only source provides a means for comparing the acid-neutralizing capacities of different samples. When the sources of alkalinity are limited to OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), separate titrations to a pH of 4.5 (or the bromocresol green end point) and a pH of 8.3 (or the phenolphthalein end point) allow us to determine which species are present and their respective concentrations. Titration curves for OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\)are shown in Figure 9.2.14
. For a solution that contains OH alkalinity only, the volume of strong acid needed to reach each of the two end points is identical (Figure 9.2.14
a). When the only source of alkalinity is \(\text{CO}_3^{2-}\), the volume of strong acid needed to reach the end point at a pH of 4.5 is exactly twice that needed to reach the end point at a pH of 8.3 (Figure 9.2.14
b). If a solution contains \(\text{HCO}_3^-\) alkalinity only, the volume of strong acid needed to reach the end point at a pH of 8.3 is zero, but that for the pH 4.5 end point is greater than zero (Figure 9.2.14
c). A mixture of OH and \(\text{CO}_3^{2-}\) or a mixture of \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\) also is possible. Consider, for example, a mixture of OH and \(\text{CO}_3^{2-}\). The volume of strong acid to titrate OH is the same whether we titrate to a pH of 8.3 or a pH of 4.5. Titrating \(\text{CO}_3^{2-}\) to a pH of 4.5, however, requires twice as much strong acid as titrating to a pH of 8.3. Consequently, when we titrate a mixture of these two ions, the volume of strong acid needed to reach a pH of 4.5 is less than twice that needed to reach a pH of 8.3. For a mixture of \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\) the volume of strong acid needed to reach a pH of 4.5 is more than twice that needed to reach a pH of 8.3. Table 9.2.5
summarizes the relationship between the sources of alkalinity and the volumes of titrant needed to reach the two end points. A mixture of OH and \(\text{HCO}_3^-\) is unstable with respect to the formation of \(\text{CO}_3^{2-}\). Problem 15 in the end-of-chapter problems asks you to explain why this is true. is a measure of a water sample’s capacity to neutralize base and is divided into strong acid and weak acid acidity. Strong acid acidity from inorganic acids such as HCl, HNO , and H SO is common in industrial effluents and in acid mine drainage. Weak acid acidity usually is dominated by the formation of H CO from dissolved CO , but also includes contributions from hydrolyzable metal ions such as Fe , Al , and Mn . In addition, weak acid acidity may include a contribution from organic acids. Acidity is determined by titrating with a standard solution of NaOH to a fixed pH of 3.7 (or the bromothymol blue end point) and to a fixed pH of 8.3 (or the phenolphthalein end point). Titrating to a pH of 3.7 provides a measure of strong acid acidity, and titrating to a pH of 8.3 provides a measure of total acidity. Weak acid acidity is the difference between the total acidity and the strong acid acidity. Results are expressed as the amount of CaCO that can be neutralized by the sample’s acidity. An alternative approach for determining strong acid and weak acid acidity is to obtain a potentiometric titration curve and use a Gran plot to determine the two equivalence points. This approach has been used, for example, to determine the forms of acidity in atmospheric aerosols [Ferek, R. J.; Lazrus, A. L.; Haagenson, P. L.; Winchester, J. W. , , 315–324]. As is the case with alkalinity, acidity is reported as mg CaCO /L. Water in contact with either the atmosphere or with carbonate-bearing sediments contains free CO in equilibrium with CO ( ) and with aqueous H CO , \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\). The concentration of free CO is determined by titrating with a standard solution of NaOH to the phenolphthalein end point, or to a pH of 8.3, with results reported as mg CO /L. This analysis essentially is the same as that for the determination of total acidity and is used only for water samples that do not contain strong acid acidity. Free CO is the same thing as CO . Acid–base titrimetry continues to have a small, but important role for the analysis of organic compounds in pharmaceutical, biochemical, agricultur- al, and environmental laboratories. Perhaps the most widely employed acid–base titration is the for organic nitrogen. Examples of analytes determined by a Kjeldahl analysis include caffeine and saccharin in pharmaceutical products, proteins in foods, and the analysis of nitrogen in fertilizers, sludges, and sediments. Any nitrogen present in a –3 oxidation state is oxidized quantitatively to \(\text{NH}_4^+\). Because some aromatic heterocyclic compounds, such as pyridine, are difficult to oxidize, a catalyst is used to ensure a quantitative oxidation. Nitrogen in other oxidation states, such as nitro and azo nitrogens, are oxidized to N , which results in a negative determinate error. Including a reducing agent, such as salicylic acid, converts this nitrogen to a –3 oxidation state, eliminating this source of error. Table 9.2.6
provides additional examples in which an element is converted quantitatively into a titratable acid or base. Several organic functional groups are weak acids or weak bases. Carboxylic (–COOH), sulfonic (–SO H) and phenolic (–C H OH) functional groups are weak acids that are titrated successfully in either aqueous or non-aqueous solvents. Sodium hydroxide is the titrant of choice for aqueous solutions. Nonaqueous titrations often are carried out in a basic solvent, such as ethylenediamine, using tetrabutylammonium hydroxide, (C H ) NOH, as the titrant. Aliphatic and aromatic amines are weak bases that are titrated using HCl in aqueous solutions, or HClO in glacial acetic acid. Other functional groups are analyzed indirectly following a reaction that produces or consumes an acid or base. Typical examples are shown in Table 9.2.7
. [1]: (CH CO) O + ROH \(\rightarrow\) CH COOR + [2]: (CH CO) ) + H O \(\rightarrow\) 2 the species that is titrated is shown in for alcohols, reaction [1] is carried out in pyridine to prevent the hydrolysis of acetic anhydride by water. After reaction [1] is complete, water is added to covert any unreacted acetic anhydride to acetic acid (reaction [2]) Many pharmaceutical compounds are weak acids or weak bases that are analyzed by an aqueous or a nonaqueous acid–base titration; examples include salicylic acid, phenobarbital, caffeine, and sulfanilamide. Amino acids and proteins are analyzed in glacial acetic acid using HClO as the titrant. For example, a procedure for determining the amount of nutritionally available protein uses an acid–base titration of lysine residues [(a) Molnár-Perl, I.; Pintée-Szakács, M. , , 159–166; (b) Barbosa, J.; Bosch, E.; Cortina, J. L.; Rosés, M. , , 177–181]. The quantitative relationship between the titrand and the titrant is determined by the titration reaction’s stoichiometry. If the titrand is polyprotic, then we must know to which equivalence point we are titrating. The following example illustrates how we can use a ladder diagram to determine a titration reaction’s stoichiometry. A 50.00-mL sample of a citrus drink requires 17.62 mL of 0.04166 M NaOH to reach the phenolphthalein end point. Express the sample’s acidity as grams of citric acid, C H O , per 100 mL. Because citric acid is a triprotic weak acid, we first must determine if the phenolphthalein end point corresponds to the first, second, or third equivalence point. Citric acid’s ladder diagram is shown in Figure 9.2.15
a. Based on this ladder diagram, the first equivalence point is between a pH of 3.13 and a pH of 4.76, the second equivalence point is between a pH of 4.76 and a pH of 6.40, and the third equivalence point is greater than a pH of 6.40. Because phenolphthalein’s end point pH is 8.3–10.0 (see ), the titration must proceed to the third equivalence point and the titration reaction is \[ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q)+3 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \nonumber\] To reach the equivalence point, each mole of citric acid consumes three moles of NaOH; thus \[(0.04166 \ \mathrm{M} \ \mathrm{NaOH})(0.01762 \ \mathrm{L} \ \mathrm{NaOH})=7.3405 \times 10^{-4} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[7.3405 \times 10^{-4} \ \mathrm{mol} \ \mathrm{NaOH} \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{3 \ \mathrm{mol} \ \mathrm{NaOH}}= 2.4468 \times 10^{-4} \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \nonumber\] \[2.4468 \times 10^{-4} \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \times \frac{192.1 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}=0.04700 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \nonumber\] Because this is the amount of citric acid in a 50.00 mL sample, the concentration of citric acid in the citrus drink is 0.09400 g/100 mL. The complete titration curve is shown in Figure 9.2.15
b. Your company recently received a shipment of salicylic acid, C H O , for use in the production of acetylsalicylic acid (aspirin). You can accept the shipment only if the salicylic acid is more than 99% pure. To evaluate the shipment’s purity, you dissolve a 0.4208-g sample in water and titrate to the phenolphthalein end point, using 21.92 mL of 0.1354 M NaOH. Report the shipment’s purity as %w/w C H O . Salicylic acid is a diprotic weak acid with p values of 2.97 and 13.74. Because salicylic acid is a diprotic weak acid, we must first determine to which equivalence point it is being titrated. Using salicylic acid’s p values as a guide, the pH at the first equivalence point is between 2.97 and 13.74, and the second equivalence points is at a pH greater than 13.74. From , phenolphthalein’s end point is in the pH range 8.3–10.0. The titration, therefore, is to the first equivalence point for which the moles of NaOH equal the moles of salicylic acid; thus \[(0.1354 \ \mathrm{M})(0.02192 \ \mathrm{L})=2.968 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[2.968 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{138.12 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{\mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}=0.4099 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \nonumber\] \[\frac{0.4099 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{0.4208 \ \mathrm{g} \text { sample }} \times 100=97.41 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \nonumber\] Because the purity of the sample is less than 99%, we reject the shipment. In an indirect analysis the analyte participates in one or more preliminary reactions, one of which produces or consumes acid or base. Despite the additional complexity, the calculations are straightforward. The purity of a pharmaceutical preparation of sulfanilamide, C H N O S, is determined by oxidizing the sulfur to SO and bubbling it through H O to produce H SO . The acid is titrated to the bromothymol blue end point using a standard solution of NaOH. Calculate the purity of the preparation given that a 0.5136-g sample requires 48.13 mL of 0.1251 M NaOH. The bromothymol blue end point has a pH range of 6.0–7.6. Sulfuric acid is a diprotic acid, with a p of 1.99 (the first value is very large and the acid dissociation reaction goes to completion, which is why H SO4 is a strong acid). The titration, therefore, proceeds to the second equivalence point and the titration reaction is \[\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{4}^{2-}(a q) \nonumber\] Using the titration results, there are \[(0.1251 \ \mathrm{M} \ \mathrm{NaOH})(0.04813 \ \mathrm{L} \ \mathrm{NaOH})=6.021 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[6.012 \times 10^{-3} \text{ mol NaOH} \times \frac{1 \text{ mol} \mathrm{H}_{2} \mathrm{SO}_{4}} {2 \text{ mol NaOH}} = 3.010 \times 10^{-3} \text{ mol} \mathrm{H}_{2} \mathrm{SO}_{4} \nonumber\] \[3.010 \times 10^{-3} \ \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{4} \times \frac{1 \ \mathrm{mol} \text{ S}}{\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{4}} \times \ \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{\mathrm{mol} \text{ S}} \times \frac{168.17 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}= 0.5062 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S} \nonumber\] produced when the SO is bubbled through H O . Because all the sulfur in H SO comes from the sulfanilamide, we can use a conservation of mass to determine the amount of sulfanilamide in the sample. \[\frac{0.5062 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{0.5136 \ \mathrm{g} \text { sample }} \times 100=98.56 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S} \nonumber\] The concentration of NO in air is determined by passing the sample through a solution of H O , which oxidizes NO to HNO , and titrating the HNO with NaOH. What is the concentration of NO , in mg/L, if a 5.0 L sample of air requires 9.14 mL of 0.01012 M NaOH to reach the methyl red end point The moles of HNO produced by pulling the sample through H O is \[(0.01012 \ \mathrm{M})(0.00914 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HNO}_{3}}{\mathrm{mol} \ \mathrm{NaOH}}=9.25 \times 10^{-5} \ \mathrm{mol} \ \mathrm{HNO}_{3} \nonumber\] A conservation of mass on nitrogen requires that each mole of NO produces one mole of HNO ; thus, the mass of NO in the sample is \[9.25 \times 10^{-5} \ \mathrm{mol} \ \mathrm{HNO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{NO}_{2}}{\mathrm{mol} \ \mathrm{HNO}_{3}} \times \frac{46.01 \ \mathrm{g} \ \mathrm{NO}_{2}}{\mathrm{mol} \ \mathrm{NO}_{2}}=4.26 \times 10^{-3} \ \mathrm{g} \ \mathrm{NO}_{2} \nonumber\] and the concentration of NO is \[\frac{4.26 \times 10^{-3} \ \mathrm{g} \ \mathrm{NO}_{2}}{5 \ \mathrm{L} \text { air }} \times \frac{1000 \ \mathrm{mg}}{\mathrm{g}}=0.852 \ \mathrm{mg} \ \mathrm{NO}_{2} \ \mathrm{L} \text { air } \nonumber\] For a back titration we must consider two acid–base reactions. Again, the calculations are straightforward. The amount of protein in a sample of cheese is determined by a Kjeldahl analysis for nitrogen. After digesting a 0.9814-g sample of cheese, the nitrogen is oxidized to \(\text{NH}_4^+\), converted to NH with NaOH, and the NH distilled into a collection flask that contains 50.00 mL of 0.1047 M HCl. The excess HCl is back titrated with 0.1183 M NaOH, requiring 22.84 mL to reach the bromothymol blue end point. Report the %w/w protein in the cheese assuming there are 6.38 grams of protein for every gram of nitrogen in most dairy products. The HCl in the collection flask reacts with two bases \[\mathrm{HCl}(a q)+\mathrm{NH}_{3}(a q) \rightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{Cl}^{-}(a q) \nonumber\] \[\mathrm{HCl}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}^{-}(a q) \nonumber\] The collection flask originally contains \[(0.1047 \ \mathrm{M \ HCl})(0.05000 \ \mathrm{L \ HCl})=5.235 \times 10^{-3} \mathrm{mol} \ \mathrm{HCl} \nonumber\] of which \[(0.1183 \ \mathrm{M} \ \mathrm{NaOH})(0.02284 \ \mathrm{L} \ \mathrm{NaOH}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{NaOH}}=2.702 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] react with NaOH. The difference between the total moles of HCl and the moles of HCl that react with NaOH is the moles of HCl that react with NH . \[5.235 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl}-2.702 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} =2.533 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] Because all the nitrogen in NH comes from the sample of cheese, we use a conservation of mass to determine the grams of nitrogen in the sample. \[2.533 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \times \frac{1 \ \mathrm{mol} \ \mathrm{NH}_{3}}{\mathrm{mol} \ \mathrm{HCl}} \times \frac{14.01 \ \mathrm{g} \ \mathrm{N}}{\mathrm{mol} \ \mathrm{NH}_{3}}=0.03549 \ \mathrm{g} \ \mathrm{N} \nonumber\] The mass of protein, therefore, is \[0.03549 \ \mathrm{g} \ \mathrm{N} \times \frac{6.38 \ \mathrm{g} \text { protein }}{\mathrm{g} \ \mathrm{N}}=0.2264 \ \mathrm{g} \text { protein } \nonumber\] and the % w/w protein is \[\frac{0.2264 \ \mathrm{g} \text { protein }}{0.9814 \ \mathrm{g} \text { sample }} \times 100=23.1 \ \% \mathrm{w} / \mathrm{w} \text { protein } \nonumber\] Limestone consists mainly of CaCO , with traces of iron oxides and other metal oxides. To determine the purity of a limestone, a 0.5413-g sample is dissolved using 10.00 mL of 1.396 M HCl. After heating to expel CO , the excess HCl was titrated to the phenolphthalein end point, requiring 39.96 mL of 0.1004 M NaOH. Report the sample’s purity as %w/w CaCO . The total moles of HCl used in this analysis is \[(1.396 \ \mathrm{M})(0.01000 \ \mathrm{L})=1.396 \times 10^{-2} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] Of the total moles of HCl \[(0.1004 \ \mathrm{M} \ \mathrm{NaOH})(0.03996 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{NaOH}} =4.012 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] are consumed in the back titration with NaOH, which means that \[ 1.396 \times 10^{-2} \ \mathrm{mol} \ \mathrm{HCl}-4.012 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \\ =9.95 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] react with the CaCO . Because \(\text{CO}_3^{2-}\) is dibasic, each mole of CaCO consumes two moles of HCl; thus \[9.95 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \times \frac{1 \ \mathrm{mol} \ \mathrm{CaCO}_{3}}{2 \ \mathrm{mol} \ \mathrm{HCl}} \times \\ \frac{100.09 \ \mathrm{g} \ \mathrm{CaCO}_{3}}{\mathrm{mol} \ \mathrm{CaCO}_{3}}=0.498 \ \mathrm{g} \ \mathrm{CaCO}_{3} \nonumber\] \[\frac{0.498 \ \mathrm{g} \ \mathrm{CaCO}_{3}}{0.5143 \ \mathrm{g} \text { sample }} \times 100=96.8 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{CaCO}_{3} \nonumber\] Earlier we noted that we can use an acid–base titration to analyze a mixture of acids or bases by titrating to more than one equivalence point. The concentration of each analyte is determined by accounting for its contribution to each equivalence point. The alkalinity of natural waters usually is controlled by OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), present singularly or in combination. Titrating a 100.0-mL sample to a pH of 8.3 requires 18.67 mL of 0.02812 M HCl. A second 100.0-mL aliquot requires 48.12 mL of the same titrant to reach a pH of 4.5. Identify the sources of alkalinity and their concentrations in milligrams per liter. Because the volume of titrant to reach a pH of 4.5 is more than twice that needed to reach a pH of 8.3, we know from , that the sample’s alkalinity is controlled by \(\text{CO}_3^{2-}\) and \(\text{HCO}_3^-\). Titrating to a pH of 8.3 neutralizes \(\text{CO}_3^{2-}\) to \(\text{HCO}_3^-\) \[\mathrm{CO}_{3}^{2-}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \nonumber\] but there is no reaction between the titrant and \(\text{HCO}_3^-\) (see ). The concentration of \(\text{CO}_3^{2-}\) in the sample, therefore, is \[{(0.02812 \ \mathrm{M \ HCl})(0.01867 \ \mathrm{L \ HCl}) \times} {\frac{1 \ \mathrm{mol} \ \mathrm{CO}_3^{2-}}{\mathrm{mol} \ \mathrm{HCl}}=5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \nonumber\] \[\frac{5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-}}{0.1000 \ \mathrm{L}} \times \frac{60.01 \ \mathrm{g} \ \mathrm{CO}_{3}^{2-}}{\mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \times \frac{1000 \ \mathrm{mg}}{\mathrm{g}}=315.1 \ \mathrm{mg} / \mathrm{L} \nonumber\] Titrating to a pH of 4.5 neutralizes \(\text{CO}_3^{2-}\) to H CO and neutralizes \(\text{HCO}_3^-\) to H CO (see ). \[\begin{array}{l}{\mathrm{CO}_{3}^{2-}(a q)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+2 \mathrm{Cl}^{-}(a q)} \\ {\mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Cl}^{-}(a q)}\end{array} \nonumber\] Because we know how many moles of \(\text{CO}_3^{2-}\) are in the sample, we can calculate the volume of HCl it consumes. \[{5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-} \times \frac{2 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \times} {\frac{1 \ \mathrm{L} \ \mathrm{HCl}}{0.02812 \ \mathrm{mol} \ \mathrm{HCl}} \times \frac{1000 \ \mathrm{mL}}{\mathrm{L}}=37.34 \ \mathrm{mL} \ \mathrm{HCl}} \nonumber\] This leaves 48.12 mL–37.34 mL, or 10.78 mL of HCl to react with \(\text{HCO}_3^-\). The amount of \(\text{HCO}_3^-\) in the sample is \[{(0.02812 \ \mathrm{M \ HCl})(0.01078 \ \mathrm{L} \ \mathrm{HCl}) \times} {\frac{1 \ \mathrm{mol} \ \mathrm{H} \mathrm{CO}_{3}^{-}}{\mathrm{mol} \ \mathrm{HCl}}=3.031 \times 10^{-4} \ \mathrm{mol} \ \mathrm{HCO}_{3}^{-}} \nonumber\] The sample contains 315.1 mg \(\text{CO}_3^{2-}\)/L and 185.0 mg \(\text{HCO}_3^-\)/L Samples that contain a mixture of the monoprotic weak acids 2–methylanilinium chloride (C H NCl, p = 4.447) and 3–nitrophenol (C H NO , p = 8.39) can be analyzed by titrating with NaOH. A 2.006-g sample requires 19.65 mL of 0.200 M NaOH to reach the bromocresol purple end point and 48.41 mL of 0.200 M NaOH to reach the phenolphthalein end point. Report the %w/w of each compound in the sample. Of the two analytes, 2-methylanilinium is the stronger acid and is the first to react with the titrant. Titrating to the bromocresol purple end point, therefore, provides information about the amount of 2-methylanilinium in the sample. \[(0.200\ \mathrm{M} \ \mathrm{NaOH} )(0.01965 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{143.61 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{\mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}=0.564 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl} \nonumber\] \[\frac{0.564 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{2.006 \ \mathrm{g} \text { sample }} \times 100=28.1 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl} \nonumber\] Titrating from the bromocresol purple end point to the phenolphthalein end point, a total of 48.41 mL – 19.65 mL = 28.76 mL, gives the amount of NaOH that reacts with 3-nitrophenol. The amount of 3-nitrophenol in the sample, therefore, is \[(0.200 \ \mathrm{M} \ \mathrm{NaOH}) (0.02876 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{139.11 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}=0.800 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3} \nonumber\] \[\frac{0.800 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{2.006 \ \mathrm{g} \text { sample }} \times 100=39.8 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3} \nonumber\] shows how we can use an acid–base titration to determine the forms of alkalinity in waters and their concentrations. We can extend this approach to other systems. For example, if we titrate a sample to the methyl orange end point and the phenolphthalein end point using either a strong acid or a strong base, we can determine which of the following species are present and their concentrations: H PO , \(\text{H}_2\text{PO}_4^-\), \(\text{HPO}_4^{2-}\), \(\text{PO}_4^{3-}\), HCl, and NaOH. As outlined in Table 9.2.8
, each species or mixture of species has a unique relationship between the volumes of titrant needed to reach these two end points. Note that mixtures containing three or more these species are not possible. Use a ladder diagram to convince yourself that mixtures containing three or more of these species are unstable. and are, respectively, the volume of titrant at the phenolphthalein and methyl orange end points when no information is provided, the volume at each end point is zero In addition to a quantitative analysis and a qualitative analysis, we also can use an acid–base titration to characterize the chemical and physical properties of matter. Two useful characterization applications are the determination of a compound’s equivalent weight and the determination of its acid dissociation constant or its base dissociation constant. Suppose we titrate a sample of an impure weak acid to a well-defined end point using a monoprotic strong base as the titrant. If we assume the titration involves the transfer of protons, then the moles of titrant needed to reach the end point is \[\text { moles titrant }=\frac{n \text { moles titrant }}{\text { moles analyte }} \times \text { moles analyte } \nonumber\] If we know the analyte’s identity, we can use this equation to determine the amount of analyte in the sample \[\text { grams analyte }=\text { moles titrant } \times \frac{1 \text { mole analyte }}{n \text { moles analyte }} \times F W \text { analyte } \nonumber\] where is the analyte’s formula weight. But what if we do not know the analyte’s identify? If we titrate a pure sample of the analyte, we can obtain some useful information that may help us establish its identity. Because we do not know the number of protons that are titrated, we let = 1 and replace the analyte’s formula weight with its equivalent weight ( ) \[\text { grams analyte }=\text { moles titrant } \times \frac{1 \text { equivalent analyte }}{1 \text { mole analyte }}=E W \text { analyte } \nonumber\] where \[F W=n \times E W \nonumber\] A 0.2521-g sample of an unknown weak acid is titrated with 0.1005 M NaOH, requiring 42.68 mL to reach the phenolphthalein end point. Determine the compound’s equivalent weight. Which of the following compounds is most likely to be the unknown weak acid? The moles of NaOH needed to reach the end point is \[(0.1005 \ \mathrm{M} \ \mathrm{NaOH})(0.04268 \ \mathrm{L} \ \mathrm{NaOH})=4.289 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] The equivalents of weak acid are the same as the moles of NaOH used in the titration; thus, he analyte’s equivalent weight is \[E W=\frac{0.2521 \ \mathrm{g}}{4.289 \times 10^{-3} \text { equivalents }}=58.78 \ \mathrm{g} / \mathrm{equivalent} \nonumber\] The possible formula weights for the weak acid are 58.78 g/mol ( = 1), 117.6 g/mol ( = 2), and 176.3 g/mol ( = 3). If the analyte is a monoprotic weak acid, then its formula weight is 58.78 g/mol, eliminating ascorbic acid as a possibility. If it is a diprotic weak acid, then the analyte’s formula weight is either 58.78 g/mol or 117.6 g/mol, depending on whether the weak acid was titrated to its first or its second equivalence point. Succinic acid, with a formula weight of 118.1 g/mole is a possibility, but malonic acid is not. If the analyte is a triprotic weak acid, then its formula weight is 58.78 g/mol, 117.6 g/mol, or 176.3 g/mol. None of these values is close to the formula weight for citric acid, eliminating it as a possibility. Only succinic acid provides a possible match. Figure 9.2.16
shows the potentiometric titration curve for the titration of a 0.500-g sample an unknown weak acid. The titrant is 0.1032 M NaOH. What is the weak acid’s equivalent weight? The first of the two visible end points is approximately 37 mL of NaOH. The analyte’s equivalent weight, therefore, is \[(0.1032 \ \mathrm{M} \ \mathrm{NaOH})(0.037 \ \mathrm{L}) \times \frac{1 \text { equivalent }}{\mathrm{mol} \ \mathrm{NaOH}}=3.8 \times 10^{-3} \text { equivalents } \nonumber\] \[E W=\frac{0.5000 \ \mathrm{g}}{3.8 \times 10^{-3} \text { equivalents }}=1.3 \times 10^{2} \ \mathrm{g} / \mathrm{equivalent} \nonumber\] Another application of acid–base titrimetry is the determination of a weak acid’s or a weak base’s dissociation constant. Consider, for example, a solution of acetic acid, CH COOH, for which the dissociation constant is \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\] When the concentrations of CH COOH and CH COO are equal, the expression reduces to = [H O ], or pH = p . If we titrate a solution of acetic acid with NaOH, the pH equals the p when the volume of NaOH is approximately 1⁄2 . As shown in Figure 9.2.17
, a potentiometric titration curve provides a reasonable estimate of acetic acid’s p . Recall that pH = p is a step on a ladder diagram, which divides the pH axis into two regions, one where the weak acid is the predominate species, and one where its conjugate weak base is the predominate species. This method provides a reasonable estimate for a weak acid’s p if the acid is neither too strong nor too weak. These limitations are easy to appreciate if we consider two limiting cases. For the first limiting case, let’s assume the weak acid, HA, is more than 50% dissociated before the titration begins (a relatively large value); in this case the concentration of HA before the equivalence point is always less than the concentration of A and there is no point on the titration curve where [HA] = [A ]. At the other extreme, if the acid is too weak, then less than 50% of the weak acid reacts with the titrant at the equivalence point. In this case the concentration of HA before the equivalence point is always greater than that of A . Determining the p by the half-equivalence point method overestimates its value if the acid is too strong and underestimates its value if the acid is too weak. Use the potentiometric titration curve in to estimate the p values for the weak acid in . At 1⁄2 , or approximately 18.5 mL, the pH is approximately 2.2; thus, we estimate that the analyte’s p is 2.2. A second approach for determining a weak acid’s p is to use a Gran plot. For example, earlier in this chapter we derived the following equation for the titration of a weak acid with a strong base. \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times V_{b}=K_{a} V_{e q}-K_{a} V_{b} \nonumber\] A plot of [H O ] \(\times\) versus for volumes less than the equivalence point yields a straight line with a slope of – . Other linearizations have been developed that use the entire titration curve or that require no assumptions [(a) Gonzalez, A. G.; Asuero, A. G. , , 29–33; (b) Papanastasiou, G.; Ziogas, I.; Kokkindis, G. , , 119–135]. This approach to determining an acidity constant has been used to study the acid–base properties of humic acids, which are naturally occurring, large molecular weight organic acids with multiple acidic sites. In one study a humic acid was found to have six titratable sites, three which were identified as carboxylic acids, two which were believed to be secondary or tertiary amines, and one which was identified as a phenolic group [Alexio, L. M.; Godinho, O. E. S.; da Costa, W. F. , , 35–39]. Values of determined by this method may have a substantial error if the effect of activity is ignored. See for a discussion of activity. In an acid–base titration, the volume of titrant needed to reach the equivalence point is proportional to the moles of titrand. Because the pH of the titrand or the titrant is a function of its concentration, the change in pH at the equivalence point—and thus the feasibility of an acid–base titration—depends on their respective concentrations. Figure 9.2.18
, for example, shows a series of titration curves for the titration of several concentrations of HCl with equimolar solutions NaOH. For titrand and titrant concentrations smaller than 10 M, the change in pH at the end point is too small to provide an accurate and a precise result. Acid–base titrimetry is an example of a total analysis technique in which the signal is proportional to the absolute amount of analyte. See for a discussion of the difference between total analysis techniques and concentration techniques. A minimum concentration of 10 M places limits on the smallest amount of analyte we can analyze successfully. For example, suppose our analyte has a formula weight of 120 g/mol. To successfully monitor the titration’s end point using an indicator or a pH probe, the titrand needs an initial volume of approximately 25 mL. If we assume the analyte’s formula weight is 120 g/mol, then each sample must contain at least 3 mg of analyte. For this reason, acid–base titrations generally are limited to major and minor analytes. We can extend the analysis of gases to trace analytes by pulling a large volume of the gas through a suitable collection solution. We need a volume of titrand sufficient to cover the tip of the pH probe or to allow for an easy observation of the indicator’s color. A volume of 25 mL is not an unreasonable estimate of the minimum volume. One goal of analytical chemistry is to extend analyses to smaller samples. Here we describe two interesting approaches to titrating μL and pL samples. In one experimental design (Figure 9.2.19
), samples of 20–100 μL are held by capillary action between a flat-surface pH electrode and a stainless steel sample stage [Steele, A.; Hieftje, G. M. , , 2884–2888]. The titrant is added using the oscillations of a piezoelectric ceramic device to move an angled glass rod in and out of a tube connected to a reservoir that contains the titrant. Each time the glass tube is withdrawn an approximately 2 nL microdroplet of titrant is released. The microdroplets are allowed to fall onto the sample, with mixing accomplished by spinning the sample stage at 120 rpm. A total of 450 microdroplets, with a combined volume of 0.81–0.84 μL, is dispensed between each pH measurement. In this fashion a titration curve is constructed. This method has been used to titrate solutions of 0.1 M HCl and 0.1 M CH COOH with 0.1 M NaOH. Absolute errors ranged from a minimum of +0.1% to a maximum of –4.1%, with relative standard deviations from 0.15% to 4.7%. Samples as small as 20 μL were titrated successfully. Another approach carries out the acid–base titration in a single drop of solution [(a) Gratzl, M.; Yi, C. , , 2085–2088; (b) Yi, C.; Gratzl, M. , , 1976–1982; (c) Hui, K. Y.; Gratzl, M. , , 695–698; (d) Yi, C.; Huang, D.; Gratzl, M. , , 1580–1584; (e) Xie, H.; Gratzl, M. , , 3665–3669]. The titrant is delivered using a microburet fashioned from a glass capillary micropipet (Figure 9.2.20
). The microburet has a 1-2 μm tip filled with an agar gel membrane. The tip of the microburet is placed within a drop of the sample solution, which is suspended in heptane, and the titrant is allowed to diffuse into the sample. The titration’s progress is monitored using an acid–base indicator and the time needed to reach the end point is measured. The rate of the titrant’s diffusion from the microburet is determined by a prior calibration. Once calibrated the end point time is converted to an end point volume. Samples usually consist of picoliter volumes (10 liters), with the smallest sample being 0.7 pL. The precision of the titrations is about 2%. Titrations conducted with microliter or picoliter sample volumes require a smaller absolute amount of analyte. For example, diffusional titrations have been conducted on as little as 29 femtomoles (10 moles) of nitric acid. Nevertheless, the analyte must be present in the sample at a major or minor level for the titration to give accurate and precise results. When working with a macro–major or a macro–minor sample, an acid–base titration can achieve a relative error of 0.1–0.2%. The principal limitation to accuracy is the difference between the end point and the equivalence point. An acid–base titration’s relative precision depends primarily on the precision with which we can measure the end point volume and the precision in detecting the end point. Under optimum conditions, an acid–base titration has a relative precision of 0.1–0.2%. We can improve the relative precision by using the largest possible buret and by ensuring we use most of its capacity in reaching the end point. A smaller volume buret is a better choice when using costly reagents, when waste disposal is a concern, or when we must complete the titration quickly to avoid competing chemical reactions. An automatic titrator is particularly useful for titrations that require small volumes of titrant because it provides significantly better precision (typically about ±0.05% of the buret’s volume). The precision of detecting the end point depends on how it is measured and the slope of the titration curve at the end point. With an indicator the precision of the end point signal usually is ±0.03–0.10 mL. Potentiometric end points usually are more precise. For an acid–base titration we can write the following general analytical equation to express the titrant’s volume in terms of the amount of titrand \[\text { volume of titrant }=k \times \text { moles of titrand } \nonumber\] where , the sensitivity, is determined by the stoichiometry between the titrand and the titrant. Consider, for example, the determination of sulfurous acid, H SO , by titrating with NaOH to the first equivalence point \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l )+\mathrm{HSO}_{3}^{-}(a q) \nonumber\] At the equivalence point the relationship between the moles of NaOH and the moles of H SO is \[\mathrm{mol} \ \mathrm{NaOH}=\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] Substituting the titrant’s molarity and volume for the moles of NaOH and rearranging \[M_{\mathrm{NaOH}} \times V_{\mathrm{NNOH}}=\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] \[V_{\mathrm{NaOH}}=\frac{1}{M_{\mathrm{NaOH}}} \times \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] we find that is \[k=\frac{1}{M_{\mathrm{NaOH}}} \nonumber\] There are two ways in which we can improve a titration’s sensitivity. The first, and most obvious, is to decrease the titrant’s concentration because it is inversely proportional to the sensitivity, . The second approach, which applies only if the titrand is multiprotic, is to titrate to a later equivalence point. If we titrate H SO to its second equivalence point \[ \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+2 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{3}^{2-}(a q)\nonumber\] then each mole of H SO consumes two moles of NaOH \[\mathrm{mol} \ \mathrm{NaOH}=2 \times \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] and the sensitivity becomes \[k=\frac{2}{M_{\mathrm{NaOH}}} \nonumber\] In practice, however, any improvement in sensitivity is offset by a decrease in the end point’s precision if a larger volume of titrant requires us to refill the buret. For this reason, standard acid–base titrimetric procedures are written to ensure that a titration uses 60–100% of the buret’s volume. Acid–base titrants are not selective. A strong base titrant, for example, reacts with all acids in a sample, regardless of their individual strengths. If the titrand contains an analyte and an interferent, then selectivity depends on their relative acid strengths. Let’s consider two limiting situations. If the analyte is a stronger acid than the interferent, then the titrant will react with the analyte before it begins reacting with the interferent. The feasibility of the analysis depends on whether the titrant’s reaction with the interferent affects the accurate location of the analyte’s equivalence point. If the acid dissociation constants are substantially different, the end point for the analyte can be determined accurately. Conversely, if the acid dissociation constants for the analyte and interferent are similar, then there may not be an accurate end point for the analyte. In the latter case a quantitative analysis for the analyte is not possible. In the second limiting situation the analyte is a weaker acid than the interferent. In this case the volume of titrant needed to reach the analyte’s equivalence point is determined by the concentration of both the analyte and the interferent. To account for the interferent’s contribution to the end point, an end point for the interferent must be available. Again, if the acid dissociation constants for the analyte and interferent are significantly different, then the analyte’s determination is possible. If the acid dissociation constants are similar, however, there is only a single equivalence point and we cannot separate the analyte’s and the interferent’s contributions to the equivalence point volume. Acid–base titrations require less time than most gravimetric procedures, but more time than many instrumental methods of analysis, particularly when analyzing many samples. With an automatic titrator, however, concerns about analysis time are less significant. When performing a titration manually our equipment needs—a buret and, perhaps, a pH meter—are few in number, inexpensive, routinely available, and easy to maintain. Automatic titrators are available for between $3000 and $10 000. | 98,786 | 3,578 |
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Before we consider specific gravimetric methods, let’s take a moment to develop a broad survey of . Later, as you read through the descriptions of specific gravimetric methods, this survey will help you focus on their similarities instead of their differences. It is easier to understand a new analytical method when you can see its relationship to other similar methods. Suppose we are to determine the total suspended solids in the water released by a sewage-treatment facility. Suspended solids are just that: solid matter that has yet to settle out of its solution matrix. The analysis is easy. After collecting a sample, we pass it through a preweighed filter that retains the suspended solids, and then dry the filter and solids to remove any residual moisture. The mass of suspended solids is the difference between the filter’s final mass and its original mass. We call this a because the analyte—the suspended solids in this example—is the species that is weighed. Method 2540D in , 20th Edition (American Public Health Association, 1998) provides an approved method for determining total suspended solids. The method uses a glass-fiber filter to retain the suspended solids. After filtering the sample, the filter is dried to a constant weight at 103–105 C. What if our analyte is an aqueous ion, such as Pb ? Because the analyte is not a solid, we cannot isolate it by filtration. We can still measure the analyte’s mass directly if we first convert it into a solid form. If we suspend a pair of Pt electrodes in the sample and apply a sufficiently positive potential between them for a long enough time, we can convert the Pb to PbO , which deposits on the Pt anode. \[\mathrm{Pb}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{PbO}_{2}(s)+\mathrm{H}_{2}(g)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \nonumber\] If we weigh the anode before and after we apply the potential, its change in mass gives the mass of PbO and, from the reaction’s stoichiometry, the amount of Pb in the sample. This is a direct analysis because PbO contains the analyte. Sometimes it is easier to remove the analyte and let a change in mass serve as the analytical signal. Suppose we need to determine a food’s moisture content. One approach is to heat a sample of the food to a temperature that will vaporize water and capture the water vapor using a preweighed absorbent trap. The change in the absorbent’s mass provides a direct determination of the amount of water in the sample. An easier approach is to weigh the sample of food before and after we heat it and use the change in its mass to determine the amount of water originally present. We call this an because we determine the analyte, H O in this case, using a signal that is proportional its disappearance. Method 925.10 in , 18th Edition (AOAC International, 2007) provides an approved method for determining the moisture content of flour. A preweighed sample is heated for one hour in a 130 C oven and transferred to a desiccator while it cools to room temperature. The loss in mass gives the amount of water in the sample. The indirect determination of a sample’s moisture content is made by measuring a change in mass. The sample’s initial mass includes the water, but its final mass does not. We can also determine an analyte indirectly without its being weighed. For example, phosphite, \(\text{PO}_3^{3-}\), reduces Hg to \(\text{Hg}_2^{2+}\), which in the presence of Cl precipitates as Hg Cl . \[2 \mathrm{HgCl}_{2}(a q)+\mathrm{PO}_{3}^{3-}(a q) +3 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 \mathrm{Cl}^{-}(a q)+\mathrm{PO}_{4}^{3-}(a q) \nonumber\] If we add HgCl in excess to a sample that contains phosphite, each mole of \(\text{PO}_3^{3-}\) will produce one mole of Hg Cl . The precipitate’s mass, therefore, provides an indirect measurement of the amount of \(\text{PO}_3^{3-}\) in the original sample. The examples in the previous section illustrate four different ways in which a measurement of mass may serve as an analytical signal. When the signal is the mass of a precipitate, we call the method . The indirect determination of \(\text{PO}_3^{3-}\) by precipitating Hg Cl is an example, as is the direct determination of Cl by precipitating AgCl. In , we deposit the analyte as a solid film on an electrode in an electrochemical cell. The deposition as PbO at a Pt anode is one example of electrogravimetry. The reduction of Cu to Cu at a Pt cathode is another example of electrogravimetry. We will not consider electrogravimetry in this chapter. See on electrochemical methods of analysis for a further discussion of electrogravimetry. When we use thermal or chemical energy to remove a volatile species, we call the method . In determining the moisture content of bread, for example, we use thermal energy to vaporize the water in the sample. To determine the amount of carbon in an organic compound, we use the chemical energy of combustion to convert it to CO . Finally, in we determine the analyte by separating it from the sample’s matrix using a filtration or an extraction. The determination of total suspended solids is one example of particulate gravimetry. An accurate gravimetric analysis requires that the analytical signal—whether it is a mass or a change in mass—is proportional to the amount of analyte in our sample. For all gravimetric methods this proportionality involves a . If the method relies on one or more chemical reactions, then we must know the stoichiometry of the reactions. In the analysis of \(\text{PO}_3^{3-}\) described earlier, for example, we know that each mole of Hg Cl corresponds to a mole of \(\text{PO}_3^{3-}\). If we remove the analyte from its matrix, then the separation must be selective for the analyte. When determining the moisture content in bread, for example, we know that the mass of H O in the bread is the difference between the sample’s final mass and its initial mass. We will return to this concept of applying a conservation of mass later in the chapter when we consider specific examples of gravimetric methods. Except for particulate gravimetry, which is the most trivial form of gravimetry, you probably will not use gravimetry after you complete this course. Why, then, is familiarity with gravimetry still important? The answer is that gravimetry is one of only a small number of definitive techniques whose measurements require only base SI units, such as mass or the mole, and defined constants, such as Avogadro’s number and the mass of C. Ultimately, we must be able to trace the result of any analysis to a , such as gravimetry, that we can relate to fundamental physical properties [Valacárcel, M.; Ríos, A. , , 2291–2297]. Although most analysts never use gravimetry to validate their results, they often verifying an analytical method by analyzing a standard reference material whose composition is traceable to a definitive technique [(a) Moody, J. R.; Epstein, M. S. , , 1571–1575; (b) Epstein, M. S. , , 1583–1591]. Other examples of definitive techniques are coulometry and isotope-dilution mass spectrometry. Coulometry is discussed in . Isotope-dilution mass spectrometry is beyond the scope of this textbook; however, you will find some suggested readings in this chapter’s Additional Resources. | 7,393 | 3,579 |
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Analytical chemistry is more than a collection of techniques; it is the application of chemistry to the analysis of samples. As we will see in later chapters, almost all analytical methods use chemical reactivity to accomplish one or more of the following: dissolve a sample, separate analytes from interferents, transform an analyte into a more useful form, or provide a signal. Equilibrium chemistry and thermodynamics provide us with a means for predicting which reactions are likely to be favorable. The most important types of reactions are precipitation reactions, acid–base reactions, metal‐ligand complexation reactions, and oxidation–reduction reactions. In a precipitation reaction two or more soluble species combine to produce an insoluble precipitate, which we characterize using a solubility product. An acid–base reaction occurs when an acid donates a proton to a base. The reaction’s equilibrium position is described using either an acid dissociation constant, , or a base dissociation constant, . The product of and for an acid and its conjugate base is the dissociation constant for water, . When a ligand donates one or more pairs of electron to a metal ion, the result is a metal–ligand complex. Two types of equilibrium constants are used to describe metal–ligand complexation: stepwise formation constants and overall formation constants. There are two stepwise formation constants for the metal–ligand complex ML , each of which describes the addition of one ligand; thus, represents the addition of the first ligand to M, and represents the addition of the second ligand to ML. Alternatively, we can use a cumulative, or overall formation constant, \(\beta_2\), for the metal–ligand complex ML , in which both ligands are added to M. In an oxidation–reduction reaction, one of the reactants is oxidized and another reactant is reduced. Instead of using an equilibrium constants to characterize an oxidation–reduction reactions, we use the potential, positive values of which indicate a favorable reaction. The Nernst equation relates this potential to the concentrations of reactants and products. Le Châtelier’s principle provides a means for predicting how a system at equilibrium responds to a change in conditions. If we apply a stress to a system at equilibrium—by adding a reactant or product, by adding a reagent that reacts with a reactant or product, or by changing the volume—the system will respond by moving in the direction that relieves the stress. You should be able to describe a system at equilibrium both qualitatively and quantitatively. You can develop a rigorous solution to an equilibrium problem by combining equilibrium constant expressions with appropriate mass balance and charge balance equations. Using this systematic approach, you can solve some quite complicated equilibrium problems. If a less rigorous answer is acceptable, then a ladder diagram may help you estimate the equilibrium system’s composition. Solutions that contain relatively similar amounts of a weak acid and its conjugate base experience only a small change in pH upon the addition of a small amount of strong acid or of strong base. We call these solutions buffers. A buffer can also be formed using a metal and its metal–ligand complex, or an oxidizing agent and its conjugate reducing agent. Both the systematic approach to solving equilibrium problems and ladder diagrams are useful tools for characterizing buffers. A quantitative solution to an equilibrium problem may give an answer that does not agree with experimental results if we do not consider the effect of ionic strength. The true, thermodynamic equilibrium constant is a function of activities, , not concentrations. A species’ activity is related to its molar concentration by an activity coefficient, \(\gamma\). Activity coefficients are estimated using the extended Debye‐Hückel equation, making possible a more rigorous treatment of equilibria. acid activity coefficient base dissociation constant charge balance equation dissociation constant equilibrium formation constant Henderson–Hasselbalch equation Le Châtelier’s principle metal–ligand complex Nernst equation pH scale precipitate reduction steady state acid dissociation constant amphiprotic buffer
common ion effect enthalpy equilibrium constant Gibb’s free energy ionic strength ligand method of successive approximations oxidation polyprotic redox reaction standard‐state stepwise formation constant activity base buffer capacity cumulative formation constant entropy extended Debye‐Hückel equation half‐reaction ladder diagram mass balance equation monoprotic oxidizing agent potential reducing agent standard potential solubility product | 4,739 | 3,580 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.02%3A_Pressure-_The_Result_of_Particle_Collisions |
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure \(\Page {1}\)). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. Pressure is defined as the force exerted on a given area: \[P=\dfrac{F}{A} \label{9.2.1} \] Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either the amount of force or by the area over which it is applied. Correspondingly, pressure can be decreased by either the force or the area. Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure \(\Page {2}\).—the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in ), so the pressure exerted by each foot is about 14 lb/in : \[\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2} \] The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in , so the pressure exerted by each blade is about 30 lb/in : \[\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3} \] Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted: \[\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4} \] The SI unit of pressure is the , with 1 Pa = 1 N/m , where N is the newton, a unit of force defined as 1 kg m/s . One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch— —for example, in car tires. Pressure can also be measured using the unit , which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table \(\Page {1}\) provides some information on these and a few other common units for pressure measurements The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into: This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar? 0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a (Figure \(\Page {3}\)). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere. If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be \(\dfrac{1}{13.6}\) as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as , : \[p=hρg \label{9.2.5} \] where Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = \(13.6 \,g/cm^3\). The hydrostatic pressure is given by Equation \ref{9.2.5}, with \(h = 760 \,mm\), \(ρ = 13.6\, g/cm^3\), and \(g = 9.81 \,m/s^2\). Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:) \[\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber \] \[\begin {align*}
p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\\[4pt]
&\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \\[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber \] Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm . 10.3 m A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure \(\Page {3}\)) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere. The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in: The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.) The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in: Blood pressure is measured using a device called a sphygmomanometer (Greek = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure \(\Page {5}\)). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg). Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure \(\Page {5}\)) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide. In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events. The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure \(\Page {7}\): the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease. Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere. Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. | 12,164 | 3,581 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.03%3A_Gas_Mixtures |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The gas mixtures described in this chapter are assumed to be mixtures of nonreacting gaseous substances. The \(p_i\) of substance \(i\) in a gas mixture is defined as the product of its mole fraction in the gas phase and the pressure of the phase: \begin{gather} \s{ p_i \defn y_i p } \tag{9.3.1} \cond{(gas mixture)} \end{gather} The sum of the partial pressures of all substances in a gas mixture is \(\sum_i p_i = \sum_i y_i p = p\sum_i y_i\). Since the sum of the mole fractions of all substances in a mixture is \(1\), this sum becomes \begin{gather} \s{ \sum_i p_i = p } \tag{9.3.2} \cond{(gas mixture)} \end{gather} Thus, the sum of the partial pressures equals the pressure of the gas phase. This statement is known as . It is valid for any gas mixture, regardless of whether or not the gas obeys the ideal gas equation. As discussed in Sec. 3.5.1, an ideal gas (whether pure or a mixture) is a gas with negligible intermolecular interactions. It obeys the ideal gas equation \(p = nRT/V\) (where \(n\) in a mixture is the sum \(\sum_i n_i\)) and its internal energy in a closed system is a function only of temperature. The partial pressure of substance \(i\) in an ideal gas mixture is \(p_i = y_i p = y_i nRT/V\); but \(y_i n\) equals \(n_i\), giving \begin{gather} \s{ p_i = \frac{n_i RT}{V} } \tag{9.3.3} \cond{(ideal gas mixture)} \end{gather} Equation 9.3.3 is the ideal gas equation with the partial pressure of a constituent substance replacing the total pressure, and the amount of the substance replacing the total amount. The equation shows that the partial pressure of a substance in an ideal gas mixture is the pressure the substance by itself, with all others removed from the system, would have at the same \(T\) and \(V\) as the mixture. Note that this statement is only true for an gas mixture. The partial pressure of a substance in a real gas mixture is in general different from the pressure of the pure substance at the same \(T\) and \(V\), because the intermolecular interactions are different. We need to relate the chemical potential of a constituent of a gas mixture to its partial pressure. We cannot measure the absolute value of a chemical potential, but we can evaluate its value relative to the chemical potential in a particular reference state called the standard state. The \(i\) is the same as the standard state of the pure gas described in Sec. 7.7: It is the hypothetical state in which pure gaseous \(i\) has the same temperature as the mixture, is at the standard pressure \(p\st\), and behaves as an ideal gas. The standard chemical potential \(\mu_i\st\gas\) of gaseous \(i\) is the chemical potential of \(i\) in this gas standard state, and is a function of temperature. By combining Eqs. 9.3.12 and 9.3.16, we obtain \begin{gather} \s{ \mu_i(p')=\mu_i\st\gas +RT\ln\frac{p'_i}{p\st} +\int_0^{p'}\!\!\left( V_i-\frac{RT}{p}\right)\difp } \tag{9.3.19} \cond{(gas mixture,} \nextcond{constant \(T\))} \end{gather} which is the analogue for a gas mixture of Eq. 7.9.2 for a pure gas. Section 7.9 describes the procedure needed to obtain formulas for various molar quantities of a pure gas from Eq. 7.9.2. By following a similar procedure with Eq. 9.3.19, we obtain the formulas for differences between partial molar and standard molar quantities of a constituent of a gas mixture shown in the second column of Table 9.1. These formulas are obtained with the help of Eqs. 9.2.46, 9.2.48, 9.2.50, and 9.2.52. The equation of state of a real gas mixture can be written as the virial equation \begin{equation} pV/n=RT\left[ 1+\frac{B}{(V/n)}+\frac{C}{(V/n)^2}+\cdots \right] \tag{9.3.20} \end{equation} This equation is the same as Eq. 2.2.2 for a pure gas, except that the molar volume \(V\m\) is replaced by the mean molar volume \(V/n\), and the virial coefficients \(B, C, \ldots\) depend on composition as well as temperature. At low to moderate pressures, the simple equation of state \begin{equation} V/n=\frac{RT}{p}+B \tag{9.3.21} \end{equation} describes a gas mixture to a sufficiently high degree of accuracy (see Eq. 2.2.8). This is equivalent to a compression factor given by \begin{equation} Z \defn \frac{pV}{nRT} = 1 + \frac{Bp}{RT} \tag{9.3.22} \end{equation} From statistical mechanical theory, the dependence of the second virial coefficient \(B\) of a binary gas mixture on the mole fraction composition is given by \begin{gather} \s{ B = y\A^2 B\subs{AA} + 2y\A y\B B\subs{AB} + y\B^2 B\subs{BB} } \tag{9.3.23} \cond{(binary gas mixture)} \end{gather} where \(B\subs{AA}\) and \(B\subs{BB}\) are the second virial coefficients of pure A and B, and \(B\subs{AB}\) is a mixed second virial coefficient. \(B\subs{AA}\), \(B\subs{BB}\), and \(B\subs{AB}\) are functions of \(T\) only. For a gas mixture with any number of constituents, the composition dependence of \(B\) is given by \begin{gather} \s{ B = \sum_i \sum_j y_i y_j B_{ij} } \tag{9.3.24} \cond{(gas mixture, \(B_{ij}{=}B_{ji}\))} \end{gather} Here \(B_{ij}\) is the second virial of \(i\) if \(i\) and \(j\) are the same, or a mixed second virial coefficient if \(i\) and \(j\) are different. If a gas mixture obeys the equation of state of Eq. 9.3.21, the partial molar volume of constituent \(i\) is given by \begin{equation} V_i = \frac{RT}{p} + B'_i \tag{9.3.25} \end{equation} where the quantity \(B'_i\), in order to be consistent with \(V_i=\pd{V}{n_i}{T,p,n_{j\ne i}}\), is found to be given by \begin{equation} B'_i = 2\sum_j y_j B_{ij} - B \tag{9.3.26} \end{equation} For the constituents of a binary mixture of A and B, Eq. 9.3.26 becomes \begin{gather} \s{ B\A' = B\subs{AA}+(-B\subs{AA}+2B\subs{AB}-B\subs{BB})y\B^2 } \tag{9.3.27} \cond{(binary gas mixture)} \end{gather} \begin{gather} \s{ B\B' = B\subs{BB}+(-B\subs{AA}+2B\subs{AB}-B\subs{BB})y\A^2 } \tag{9.3.28} \cond{(binary gas mixture)} \end{gather} When we substitute the expression of Eq. 9.3.25 for \(V_i\) in Eq. 9.3.18, we obtain a relation between the fugacity coefficient of constituent \(i\) and the function \(B'_i\): \begin{equation} \ln\phi_i = \frac{B'_i p}{RT} \tag{9.3.29} \end{equation} The third column of Table 9.1 gives formulas for various partial molar quantities of constituent \(i\) in terms of \(B'_i\) and its temperature derivative. The formulas are the same as the approximate formulas in the third column of Table 7.5 for molar quantities of a gas, with \(B'_i\) replacing the second virial coefficient \(B\). | 13,823 | 3,585 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/10%3A_Spectroscopic_Methods/10.07%3A_Atomic_Emission_Spectroscopy |
The focus of this section is on the emission of ultraviolet and visible radiation following the thermal excitation of atoms. Atomic emission spectroscopy has a long history. Qualitative applications based on the color of flames were used in the smelting of ores as early as 1550 and were more fully developed around 1830 with the observation of atomic spectra generated by flame emission and spark emission [Dawson, J. B. , , 93–98]. Quantitative applications based on the atomic emission from electric sparks were developed by Lockyer in the early 1870 and quantitative applications based on flame emission were pioneered by Lundegardh in 1930. Atomic emission based on emission from a plasma was introduced in 1964. For an on-line introduction to much of the material in this section, see by Tomas Spudich and Alexander Scheeline, a resource that is part of the Analytical Sciences Digital Library. Atomic emission occurs when a valence electron in a higher energy atomic orbital returns to a lower energy atomic orbital. Figure 10.7.1
shows a portion of the energy level diagram for sodium, which consists of a series of discrete lines at wavelengths that correspond to the difference in energy between two atomic orbitals. The intensity of an atomic emission line, , is proportional to the number of atoms, \(N^*\), that populate the excited state, \[I_{e}=k N^* \label{10.1}\] where is a constant that accounts for the efficiency of the transition. If a system of atoms is in thermal equilibrium, the population of excited state is related to the total concentration of atoms, , by the Boltzmann distribution. For many elements at temperatures of less than 5000 K the Boltzmann distribution is approximated as \[N^* = N\left(\frac{g_{i}}{g_{0}}\right) e^{-E_i / k T} \label{10.2}\] where and are statistical factors that account for the number of equivalent energy levels for the excited state and the ground state, is the energy of the excited state relative to a ground state energy, , is Boltzmann’s constant (\(1.3807 \times 10^{-23}\) J/K), and is the temperature in Kelvin. From Equation \ref{10.2} we expect that excited states with lower energies have larger populations and more intense emission lines. We also expect emission intensity to increase with temperature. An atomic emission spectrometer is similar in design to the instrumentation for atomic absorption. In fact, it is easy to adapt most flame atomic absorption spectrometers for atomic emission by turning off the hollow cathode lamp and monitoring the difference between the emission intensity when aspirating the sample and when aspirating a blank. Many atomic emission spectrometers, however, are dedicated instruments designed to take advantage of features unique to atomic emission, including the use of plasmas, arcs, sparks, and lasers as atomization and excitation sources, and an enhanced capability for multielemental analysis. Atomic emission requires a means for converting into a free gaseous atom an analyte that is present in a solid, liquid, or solution sample. The same source of thermal energy used for atomization usually serves as the excitation source. The most common methods are flames and plasmas, both of which are useful for liquid or solution samples. Solid samples are analyzed by dissolving in a solvent and using a flame or plasma atomizer. Atomization and excitation in flame atomic emission is accomplished with the same nebulization and spray chamber assembly used in atomic absorption ( ). The burner head consists of a single or multiple slots, or a Meker-style burner. Older atomic emission instruments often used a total consumption burner in which the sample is drawn through a capillary tube and injected directly into the flame. A Meker burner is similar to the more common Bunsen burner found in most laboratories; it is designed to allow for higher temperatures and for a larger diameter flame. A is a hot, partially ionized gas that contains an abundant concentration of cations and electrons. The plasma used in atomic emission is formed by ionizing a flowing stream of argon gas, producing argon ions and electrons. A plasma’s high temperature results from resistive heating as the electrons and argon ions move through the gas. Because a plasma operates at a much higher temperature than a flame, it provides for a better atomization efficiency and a higher population of excited states. A schematic diagram of the inductively coupled plasma source (ICP) is shown in Figure 10.7.2
. The ICP torch consists of three concentric quartz tubes, surrounded at the top by a radio-frequency induction coil. The sample is mixed with a stream of Ar using a nebulizer, and is carried to the plasma through the torch’s central capillary tube. Plasma formation is initiated by a spark from a Tesla coil. An alternating radio-frequency current in the induction coil creates a fluctuating magnetic field that induces the argon ions and the electrons to move in a circular path. The resulting collisions with the abundant unionized gas give rise to resistive heating, providing temperatures as high as 10000 K at the base of the plasma, and between 6000 and 8000 K at a height of 15–20 mm above the coil, where emission usually is measured. At these high temperatures the outer quartz tube must be thermally isolated from the plasma. This is accomplished by the tangential flow of argon shown in the schematic diagram. Atomic emission spectroscopy is ideally suited for a multielemental analysis because all analytes in a sample are excited simultaneously. If the instrument includes a scanning monochromator, we can program it to move rapidly to an analyte’s desired wavelength, pause to record its emission intensity, and then move to the next analyte’s wavelength. This sequential analysis allows for a sampling rate of 3–4 analytes per minute. Another approach to a multielemental analysis is to use a multichannel instrument that allows us to monitor simultaneously many analytes. A simple design for a multichannel spectrometer, shown in Figure 10.7.3
, couples a monochromator with multiple detectors that are positioned in a semicircular array around the monochromator at positions that correspond to the wavelengths for the analytes. Atomic emission is used widely for the analysis of trace metals in a variety of sample matrices. The development of a quantitative atomic emission method requires several considerations, including choosing a source for atomization and excitation, selecting a wavelength and slit width, preparing the sample for analysis, minimizing spectral and chemical interferences, and selecting a method of standardization. Except for the alkali metals, detection limits when using an ICP are significantly better than those obtained with flame emission (Table 10.7.1
). Plasmas also are subject to fewer spectral and chemical interferences. For these reasons a plasma emission source is usually the better choice. Parsons, M. L.; Major, S.; Forster, A. R.; App. Spectrosc. 1983, 37, 411–418. The choice of wavelength is dictated by the need for sensitivity and the need to avoid interferences from the emission lines of other constituents in the sample. Because an analyte’s atomic emission spectrum has an abundance of emission lines—particularly when using a high temperature plasma source—it is inevitable that there will be some overlap between emission lines. For example, an analysis for Ni using the atomic emission line at 349.30 nm is complicated by the atomic emission line for Fe at 349.06 nm. A narrower slit width provides better resolution, but at the cost of less radiation reaching the detector. The easiest approach to selecting a wavelength is to record the sample’s emission spectrum and look for an emission line that provides an intense signal and is resolved from other emission lines. Flame and plasma sources are best suited for samples in solution and in liquid form. Although a solid sample can be analyzed by directly inserting it into the flame or plasma, they usually are first brought into solution by digestion or extraction. The most important spectral interference is broad, background emission from the flame or plasma and emission bands from molecular species. This background emission is particularly severe for flames because the temperature is insufficient to break down refractory compounds, such as oxides and hydroxides. Background corrections for flame emission are made by scanning over the emission line and drawing a baseline (Figure 10.7.4
). Because a plasma’s temperature is much higher, a background interference due to molecular emission is less of a problem. Although emission from the plasma’s core is strong, it is insignificant at a height of 10–30 mm above the core where measurements normally are made. Flame emission is subject to the same types of chemical interferences as atomic absorption; they are minimized using the same methods: by adjusting the flame’s composition and by adding protecting agents, releasing agents, or ionization suppressors. An additional chemical interference results from . Because the flame’s temperature is greatest at its center, the concentration of analyte atoms in an excited state is greater at the flame’s center than at its outer edges. If an excited state atom in the flame’s center emits a photon, then a ground state atom in the cooler, outer regions of the flame may absorb the photon, which decreases the emission intensity. For higher concentrations of analyte self-absorption may invert the center of the emission band (Figure 10.7.5
). Chemical interferences when using a plasma source generally are not significant because the plasma’s higher temperature limits the formation of nonvolatile species. For example, \(\text{PO}_4^{3-}\) is a significant interferent when analyzing samples for Ca by flame emission, but has a negligible effect when using a plasma source. In addition, the high concentration of electrons from the ionization of argon minimizes ionization interferences. From Equation \ref{10.1} we know that emission intensity is proportional to the population of the analyte’s excited state, \(N^*\). If the flame or plasma is in thermal equilibrium, then the excited state population is proportional to the analyte’s total population, , through the Boltzmann distribution (Equation \ref{10.2}). A calibration curve for flame emission usually is linear over two to three orders of magnitude, with ionization limiting linearity when the analyte’s concentrations is small and self-absorption limiting linearity at higher concentrations of analyte. When using a plasma, which suffers from fewer chemical interferences, the calibration curve often is linear over four to five orders of magnitude and is not affected significantly by changes in the matrix of the standards. Emission intensity is affected significantly by many parameters, including the temperature of the excitation source and the efficiency of atomization. An increase in temperature of 10 K, for example, produces a 4% increase in the fraction of Na atoms in the 3 excited state, an uncertainty in the signal that may limit the use of external standards. The method of internal standards is used when the variations in source parameters are difficult to control. To compensate for changes in the temperature of the excitation source, the internal standard is selected so that its emission line is close to the analyte’s emission line. In addition, the internal standard should be subject to the same chemical interferences to compensate for changes in atomization efficiency. To accurately correct for these errors the analyte and internal standard emission lines are monitored simultaneously. 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 sodium in salt substitutes provides an instructive example of a typical procedure. The description here is based on Goodney, D. E. , , 875–876. Salt substitutes, which are used in place of table salt for individuals on low-sodium diets, replaces NaCl with KCl. Depending on the brand, fumaric acid, calcium hydrogen phosphate, or potassium tartrate also are present. Although intended to be sodium-free, salt substitutes contain small amounts of NaCl as an impurity. Typically, the concentration of sodium in a salt substitute is about 100 μg/g The exact concentration of sodium is determined by flame atomic emission. Because it is difficult to match the matrix of the standards to that of the sample, the analysis is accomplished by the method of standard additions. A sample is prepared by placing an approximately 10-g portion of the salt substitute in 10 mL of 3 M HCl and 100 mL of distilled water. After the sample has dissolved, it is transferred to a 250-mL volumetric flask and diluted to volume with distilled water. A series of standard additions is prepared by placing 25-mL portions of the diluted sample into separate 50-mL volumetric flasks, spiking each with a known amount of an approximately 10 mg/L standard solution of Na , and diluting to volume. After zeroing the instrument with an appropriate blank, the instrument is optimized at a wavelength of 589.0 nm while aspirating a standard solution of Na . The emission intensity is measured for each of the standard addition samples and the concentration of sodium in the salt substitute is reported in μg/g. 1. Potassium ionizes more easily than sodium. What problem might this present if you use external standards prepared from a stock solution of 10 mg Na/L instead of using a set of standard additions? Because potassium is present at a much higher concentration than is sodium, its ionization suppresses the ionization of sodium. Normally suppressing ionization is a good thing because it increases emission intensity. In this case, however, the difference between the standard's matrix and the sample’s matrix means that the sodium in a standard experiences more ionization than an equivalent amount of sodium in a sample. The result is a determinate error. 2. One way to avoid a determinate error when using external standards is to match the matrix of the standards to that of the sample. We could, for example, prepare external standards using reagent grade KCl to match the matrix to that of the sample. Why is this not a good idea for this analysis? Sodium is a common contaminant in many chemicals. Reagent grade KCl, for example, may contain 40–50 μg Na/g. This is a significant source of sodium, given that the salt substitute contains approximately 100 μg Na/g. 3. Suppose you decide to use an external standardization. Given the previous questions, is the result of your analysis likely to underestimate or to overestimate the amount of sodium in the salt substitute? The solid black line in Figure 10.7.6
shows the ideal calibration curve, assuming we match the standard’s matrix to the sample’s matrix, and that we do so without adding any additional sodium. If we prepare the external standards without adding KCl, the emission for each standard decreases due to increased ionization. This is shown by the lower of the two dashed red lines. Preparing the standards by adding reagent grade KCl increases the concentration of sodium due to its contamination. Because we underestimate the actual concentration of sodium in the standards, the resulting calibration curve is shown by the other dashed red line. In both cases, the sample’s emission results in our overestimating the concentration of sodium in the sample. 4. One problem with analyzing salt samples is their tendency to clog the aspirator and burner assembly. What effect does this have on the analysis? Clogging the aspirator and burner assembly decreases the rate of aspiration, which decreases the analyte’s concentration in the flame. The result is a decrease in the emission intensity and a negative determinate error. To evaluate the method described in Representative Method 10.7.1, a series of standard additions is prepared using a 10.0077-g sample of a salt substitute. The results of a flame atomic emission analysis of the standards is shown here [Goodney, D. E. , , 875–876]. What is the concentration of sodium, in μg/g, in the salt substitute. Linear regression of emission intensity versus the concentration of added Na gives the standard additions calibration curve shown below, which has the following calibration equation. \[I_{e}=1.97+1.37 \times \frac{\mu \mathrm{g} \ \mathrm{Na}}{\mathrm{mL}} \nonumber\] The concentration of sodium in the sample is the absolute value of the calibration curve’s -intercept. Substituting zero for the emission intensity and solving for sodium’s concentration gives a result of 1.44 μgNa/mL. The concentration of sodium in the salt substitute is \[\frac{\frac{1.44 \ \mu \mathrm{g} \ \mathrm{Na}}{\mathrm{mL}} \times \frac{50.00 \ \mathrm{mL}}{25.00 \ \mathrm{mL}} \times 250.0 \ \mathrm{mL}}{10.0077 \ \mathrm{g} \text { sample }}=71.9 \ \mu \mathrm{g} \ \mathrm{Na} / \mathrm{g}\nonumber\] The scale of operations for atomic emission is ideal for the direct analysis of trace and ultratrace analytes in macro and meso samples. With appropriate dilutions, atomic emission can be applied to major and minor analytes. When spectral and chemical interferences are insignificant, atomic emission can achieve quantitative results with accuracies of 1–5%. For flame emission, accuracy frequently is limited by chemical interferences. Because the higher temperature of a plasma source gives rise to more emission lines, accuracy when using plasma emission often is limited by stray radiation from overlapping emission lines. For samples and standards in which the analyte’s concentration exceeds the detection limit by at least a factor of 50, the relative standard deviation for both flame and plasma emission is about 1–5%. Perhaps the most important factor that affect precision is the stability of the flame’s or the plasma’s temperature. For example, in a 2500 K flame a temperature fluctuation of \(\pm 2.5\) K gives a relative standard deviation of 1% in emission intensity. Significant improvements in precision are realized when using internal standards. Sensitivity is influenced by the temperature of the excitation source and the composition of the sample matrix. Sensitivity is optimized by aspirating a standard solution of analyte and maximizing the emission by adjusting the flame’s composition and the height from which we monitor the emission. Chemical interferences, when present, decrease the sensitivity of the analysis. Because the sensitivity of plasma emission is less affected by the sample matrix, a calibration curve prepared using standards in a matrix of distilled water is possible even for samples that have more complex matrices. The selectivity of atomic emission is similar to that of atomic absorption. Atomic emission has the further advantage of rapid sequential or simultaneous analysis of multiple analytes. Sample throughput with atomic emission is rapid when using an automated system that can analyze multiple analytes. For example, sampling rates of 3000 determinations per hour are possible using a multichannel ICP, and sampling rates of 300 determinations per hour when using a sequential ICP. Flame emission often is accomplished using an atomic absorption spectrometer, which typically costs between $10,000–$50,000. Sequential ICP’s range in price from $55,000–$150,000, while an ICP capable of simultaneous multielemental analysis costs between $80,000–$200,000. Combination ICP’s that are capable of both sequential and simultaneous analysis range in price from $150,000–$300,000. The cost of Ar, which is consumed in significant quantities, can not be overlooked when considering the expense of operating an ICP. | 19,948 | 3,586 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/08%3A_Nucleic_acids/8.08%3A_Genetic_engineering |
DNA can be isolated from the cell and cut at a specific place using enzymes called restriction enzymes. Fragments of DNA from different sources, e.g., an insulin gene from a human and a DNA from Escherichia coli (E. coli) cut using the same restriction enzyme, can be combined into a new DNA called . The recombinant DNA can be introduced into the cell, e.g., back into E. coli, where it propagates using the cellular machinery of the host cell. It is then used to produce proteins of interest, e.g., human insulin that is later used to treat diseases like diabetes. The DNA of host cells, e.g, small circular plasmids of DNA from E. coli, are isolated first. The host cells are soaked in a detergent solution that disrupts the plasma membrane and releases the plasmids. A restriction enzyme is used to cut the DNA of the host cell at a specific location. The same enzyme is used to cut a piece of donor DNA, e.g., an insulin gene from a human. The cut ends of the DNA are called . When the cut donor and host DNA are mixed, they join at the sticky ends as illustrated in Figure \(\Page {1}\). The resultant recombinant DNA is introduced to a fresh culture of E. Coli. The host cells that have taken up the recombinant DNA are selected and grown. When these bacteria grow and divide, they produce the protein encoded in the inserted gene, i.e., insulin in this example. The desired protein, insulin in this example, is extracted, purified, and used wherever needed, e.g., to treat diabetes. The recombinant DNA technology is used in food production, medicine, agriculture, and bioengineering. Some examples are listed below. olymerase chain reaction (PCR) is a technique used to produce millions of copies of a DNA segment of interest in a short time. The selected DNA is heated to separate the two strands and mixed with DNA polymerase, a short synthetic DNA called primer to select the segment to be amplified, and nucleotides to produce a complementary strand. The process is repeated several times to produce millions of copies of the desired DNA fragment, as illustrated in Figure \(\Page {2}\). Genetic testing is done in a laboratory to study an individual's DNA to diagnose a defective gene that may cause a genetic disease, ancestry studies, or forensic studies. For example, breast cancer may be related to defects in breast cancer genes BRCA1 and BRCA2. Patients are screened for these defects by using blood or saliva samples to extract the DNA and the PCR technique is applied to amplify and study the defective genes. Genetic testing is also used to study the DNA of tumors or cancer cases. Fingerprinting is a technique based on PCR that allows determining the nucleotide sequence of certain regions of human DNA that are unique to individuals. A small sample from flood, skin, saliva, or semen is used to extract the DNA for amplification by PCR. Fluorescent or radioactive isotopes are incorporated in the amplified DNA for easy monitoring. The DNA is cut into pieces of restriction enzymes, placed on a gel, separated by electrophoresis, and studied to identify the individual, as illustrated in Figure \(\Page {3}\). This technique is used for paternity tests, criminal investigations, and other forensic purposes. The human genome project that was meant to comprehensively study all of the human DNA (human genome) was launched in October 1990 and completed in April 2003. It shows that there are about 20,000 protein-coding genes in humans, which represent only about 1.5% of the human genome. The role of the rest of the genome is still been explored. It is mostly related to regulating genes and serving as a recognition site for proteins. These studies provide fundamental information to study human biology, defects in DNA that lead to genetic diseases, and find cures for diseases. | 3,825 | 3,593 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Solubility_and_Precipitation |
is the process of a compound coming out of solution. It is the opposite of dissolution or solvation. In dissolution, the solute particles separate from each other and are surrounded by solvent molecules. In precipitation, the solute particles find each other and form a solid together. This solid is called the precipitate or sometimes abbreviated "ppt". Precipitation and dissolution are a great example of a dynamic equilibrium (also described ). Any time there is a solution with a little bit of solid solute in it, both processes will be happening at once. Some molecules or ions will leave the solid and become solvated, and some solvated solute particles will bump into the solid and get stuck there. The rates of the 2 processes determine the overall effect: if precipitation happens faster, then a lot of solid can come out of the solution very quickly. If dissolution happens faster, than the solid will dissolve. As the solution becomes more concentrated, the rate of precipitation will increase and the rate of dissolution will decrease, so that eventually the concentration will stop changing, and this is equilibrium. When equilibrium is reached, the solution is , and that concentration defines the of the solute. Solubility is the maximum possible concentration, and it is given in M, g/L, or other units. Solubility changes with temperature, so if you look up solubility data it will specify the temperature. Precipitation can happen for various reasons, such as that you cooled a solution, or removed some solvent by evaporation, or both. (This is often used as a way to purify a compound.) You can also have a precipitation reaction, when you mix two solutions together and a new combination of ions is in the combined solution. For example, maybe you mixed a solution of silver(I) nitrate and sodium chloride. Silver(I) chloride is very insoluble, so it will precipitate, leaving soluble sodium nitrate in solution. Precipitation reactions can be a good way to prepare a salt you want from some other salts with the right anion and cation. Precipitation reactions can also be used to detect the presence of particular ions in solution. For instance, you might test for chloride, iodide and bromide in an unknown solution by adding silver(I) ions and looking for precipitation. Beginning chemistry students usually memorize a list of solubility rules. Here it is (these rules will be a little bit different in different textbooks, because people might not have exactly the same definition of soluble or insoluble): You can use this list to predict when precipitation reactions will occur. For this purpose, you don't usually have to worry about whether the compounds are strong or weak electrolytes, you can think of the ions as being separate. The reason is that usually some of the ions will be separate, and once those precipitate with a new partner, more of the original compound ions will separate from each other, and the process will continue. Chemists may write equations in different ways to emphasize the important parts. For instance, we might write an equation like this, which describes mixing 2 solutions of different soluble salts and getting a precipitate:
$$AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)$$
Alternately, we might write the same reaction just focusing on the part that forms the precipitate, and leaving out the that don't really do anything, just stay in solution:
$$Ag^{+}(aq) + Cl^{–}(aq) \rightarrow AgCl(s)$$
Chemistry students are sometimes asked to prove their understanding of dissociation by writing out all the ions separately, like this: \[Ag^{+}(aq) + NO_{3}^{–}(aq) + Cl^{–}(aq) + Na^{+} \rightarrow AgCl(s) + Na^{+} + NO_{3}^{–}(aq)\] No real chemist would be likely to do this because it is a nuisance. (It's also a little funny because many salts aren't strong electrolytes, so teachers might be telling their students to write an equation that doesn't show what's really happening.) However, it does help show what it means to be a spectator ion, since they are the same on both sides when you write it like this. Solubility depends on the relative stability of the solid and solvated states for a particular compound. For instance, if it has very strong interactions between molecules or ions in the solid state, then it won't be very soluble unless the solvation interations are also very strong. (Ionic salts are a good example: usually they have strong interactions in the solid and solvated states.) If the interactions in the solid are weak, the compound can still be insoluble in polar solvents if the interactions with the solvent are weaker than the Coulomb interactions of the solvent molecules with other solvent molecules. (This is why wax is insoluble in water: it is non-polar, so the wax-wax interactions are weak, but the wax-water interactions are weaker than the water-water interactions.) We can't explain what makes these interactions strong or weak well until after we study chemical bonding, but in general ionic compounds with larger charges on the ions and smaller ions are less soluble, because they can have stronger Coulomb interactions in the solid. | 5,188 | 3,595 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/08%3A_Conformational_Analysis_of_Alkanes/8.02%3A_Conformational_Analysis |
Conformational analysis is the study of the different energy levels associated with the different conformations of a molecule. are the different 3-dimensional arrangements that the molecule can acquire by freely rotating around σ-bonds. One must keep in mind that . Isomers are different molecules. Conformations are simply different structural arrangements of the same molecule. The example below illustrates four out of an infinite number of conformations that the n-butane molecule can acquire by freely rotating around the C -C bond, indicated in yellow. The red arrow indicates the direction of the rotation. The methyl carbons (C and C ) are indicated in red and all the hydrogen atoms in white. In this example C , C and C remain stationary while C gradually moves down from top right to bottom right. The type of drawing used at the bottom of the figure in the last page is called a , or representation. It is widely used to represent the conformations of open-chain alkanes that result as rotation around a sigma bond occurs. Another way to visualize Newman projections is to imagine them as the shadow a molecule would cast against a wall if a light beam was directed along the carbon-carbon axis under consideration. A concept strongly associated with Newman projections is . It is defined as the angle between any two atoms attached to neighboring carbons in the Newman projection. For example, the two methyl groups above (in red) are attached to neighboring carbons. The angle between the two as seen is zero (that is, they overlap, or” eclipse” each other). This is then the dihedral angle between those methyl groups. The relationship between molecular structure and potential energy is a major area of study in organic chemistry. It has applications in conformational analysis because we are interested in relating the structure of conformers to their energy, and therefore to their relative stabilities. Remember that the relationship between potential energy and stability is inverse. The higher the potential energy of a system the lower its stability. Also remember that these are always relative concepts. There is no absolute energy or stability. They are always measured relative to a previously agreed upon standard. There are and therefore decrease their stability. They are: - Crowding of alkyl groups or other substituents as they come too close together. - Tendency of s-bonds to rotate in order to acquire a more stable conformation. Increase in potential energy due to bond angles being forced to depart from ideal values in cycloalkanes and other rings. The term steric interactions refers to interference that occurs between atoms or groups of atoms by virtue of their size, or volume. When bulky groups or large molecules get too close to each other, steric interactions can become severe, raising the potential energy of the system. When these groups are allowed to drift far apart, steric interactions are relieved and the system gains stability. Three representations of steric interactions, or “crowding,” between the two methyl groups in the eclipsed conformation of -butane. This term is closely related to dihedral angle (also called torsional angle). Free rotation around sigma bonds changes the dihedral angle and therefore can bring bulky groups together or apart. When the dihedral angle is small and bulky groups interact (such as in the illustration above), there is a driving force for the molecule to rotate around the C -C axis to relieve the steric interactions between the methyl groups. If this drive is impeded and the molecule is forced to acquire the eclipsed conformation, or a small dihedral angles between bulky groups, then torsional strain results. This might be the case for example in cyclic systems, where free rotation around sigma bonds is limited or impeded. Obviously, torsional strain increases the energy of a system and therefore decreases its stability. As the molecule rotates around the C -C bond, steric crowding between the methyl groups gets relieved. | 4,054 | 3,596 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Stille_Coupling |
The Stille reaction, named after the late John Kenneth Stille, is a palladium-catalyzed cross coupling reaction. Heavily used in organic synthesis, it involves the coupling of an organic halide with an organotin compound. The reaction proceeds according to the following equation : The reaction proceeds using a palladium phosphine catalyst. In this reaction, R and R represent the organic chains that are to be coupled. These are usually chains containing an sp hybridized carbon (aryl, alkenyl, allyl.) X represents a leaving group such as a halide or triflate (Cl, Br, I, OTf, etc.) In some instances, it is possible to use sp hybridized alkyls and aromatics as R . It’s worth noting that, although air stable, tin reagents tend to be highly toxic. Prior to Stille’s work, palladium-catalyzed cross coupling reactions had been observed. The foundations were laid by Eaborn and Migita in the late 70’s. Eaborn’s process involved formation of diary product using an organotin reagent. This was expanded by Migita who coupled acyl chlorides with organotin reagents to form ketones with yields ranging from 53-87%. Later on, Migita was able to couple aryl and acyl halides with allyl tin reagents. He was able to do so at lower temperatures due to allyl’s ability to interact more efficiently with the palladium catalyst. Yields for both aryl and acyl halides were inconsistent ranging from as little as 4% to as much as 100%. Stille was able to build upon these foundations using a variety of alkyl tin reagents. In 1978, he published a report on the successful coupling of many alkyl tin reagents with myriad acyl and aryl halides. He was able to carry out these reactions under much more mild conditions and obtaining much higher yields (76-99%). Subsequent research focused on synthesizing a variety of compounds in this fashion. Stille coupling is a commonly used procedure because of its wide scope. There are ample choices for both nucleophile and electrophile. Furthermore, organotin reagents are air stable, commercially available or readily synthesized, making Stille coupling an easily accessible method for synthesis. A wide variety of aryl, vinyl and acyl halides or pseudo halides can be used as electrophiles. Furthermore, heterocyclic compounds are also viable choices. The example below shows that the Stille reaction can also work with purines to build more complex molecules.
The major exceptions to this are organic chlorides which are not reactive enough to undergo oxidative addition with the palladium catalyst. As a result, organic bromides or iodides are preferred. Similarly, the organotin reagent shows great versatility. As with the electrophile, many classes of organotin reagents work well in Stille coupling. Aryl, alkyl, vinyl and even heterocyclic stannanes are known to work. The following example shows N, S containing heterocyclic ring participating in Stille coupling. Limitations include very bulky or heavily substituted reagents. These tend to react very slowly and may require optimization, typically in the form of co-catalytic copper iodide. Nonetheless, the Stille reaction is a versatile one of great importance to organic synthesis. Stereochemistry is typically retained in Stille coupling. Regioselectivity, as with most other coupling reactions is hard to control though optimizations have been found to control regioselectivity to some extent. Despite advances by researchers in controlling stereoselectivity, mechanisms have not been fully elucidated and require further study. The synthetic applications of Stille coupling are immense. A reaction of this versatility and scope readily found its way into total syntheses of natural products. Stille coupling has been of great use towards the synthesis of many such products, many of these antibiotics or anticancer drugs. Among these are manzamine A, ircinal A, oxazolomycin and many others. The Stille reaction uses a palladium catalyst. It can use an 18- or 16-electron Pd (0) complex as a source of the catalyst, such as Pd(PPh ) , Pd(dba) . Then through ligand dissociation, it can be formed into a 14-electron Pd (0), the active catalyst. Upon oxidative addition, the electrophile will bind to palladium, turning it into a 16-electron square planer intermediate. The organostannane then comes in as a transmetalation agent to introduce the R’ group and also taking away the halide. Finally, the two R groups will be coupled by reductive elimination through various possible pathways, and turning the catalyst back to the 14-electron Pd (0). The alkyl halide would first perform oxidative addition to the palladium in a concerted fashion, resulting in a 16-electron Pd(II) intermediate. The cis square planer product is in a fast equilibrium with the trans product.The bulky ligands used on the catalyst makes the trans product more thermodynamically stable, therefore most of the intermediate will isomerize into the trans product . A trans- intermediate can also be formed when the catalyst reacts with an sp3 organohalide in a S 2 mechanism. The Stille coupling uses organostannane as a trans coupling reagent. The tin is usually bound to allyl, alkenyl, or aryl groups. The tin and the R’ group will form a four-member ring with the palladium center and the halide, forming an 18-electron transition state. Then the tin halide will leave and the R’ group stays bonded to palladium . The two trans- R groups must first isomerize back into a cis- conformation, then it can undergo a concerted reductive elimination . There are other two proposed mechanisms which involves the association or dissociation of the ligands. An extra ligand can be bond to palladium, forming an 18-electron trigonal bipyramidal structure and forces the two R groups to the equatorial position, which is a suitable conformation for them to form C-C bonds. Another proposed mechanism is that one of the ligands will dissociate first to form a T-shape 14-electron intermediate, which is able to speed up the reductive elimination process. If the ligands are bulky enough, like phosphines with large cone angles, it is possible to push the two R groups closer to each other into an appropriate coordination angle, thus speeding up the reductive elimination process . Various ligands are used as the 16- or 18- electron Pd catalyst, such as Pd(PPh ) and Pd(dba) . Some Pd(II) complexes can also be used as a source and can be reduced to the active Pd(0) catalyst, such as Pd(OAc) , PdCl (MeCN) , PdCl (PPh ) . Methyl and butyl group are often used as the other three R groups that binds to stannane, although these alkyl groups are not as reactive in the transmetalation process, there is still possibility that the alkyl coupling side product would form. This is called a” Non-transferable” ligand product. Though there are cases that shows benzylstannane are able to couple through Stille catalysis at higher temperatures. Both electron donating and electron withdrawing properties of the R’ group is favorable for transmetalation. The reactivity of the R group on organostannane have the following order : Alkynyl > Alkenyl > Aryl > Allyl = Benzyl > -alkoxyl > Alkyl Vinyl halides, aryl halides and heterocyclic halides can all be used as the electrophile for the reaction. Iodine and bromide are used for the halogen because the chlorides are too inert for oxidative addition, and iodides would react even faster than bromide, as can be seen from the reaction below : The stereochemistry is often retained during the catalytic reaction, unless it’s carried out under a harsh condition, so the Stille coupling reaction is quite good towards steroselectivity. LiCl is often used to enhance the reaction rate by stabilizing the transition state during oxidative addition. It can also improve the rate of transmetalation by increasing the solvent’s polarity . In addition, Cu(I) and Mn(II) salts are also used to increase reaction rate as well as improving selectivity. | 7,978 | 3,599 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/10%3A_Intro_to_Theory_of_Chemical_Reactions/10.09%3A_Activation_Energies |
The majority of chemical transformations do not amount to only breaking or forming bonds. The two processes go together, either simultaneously or sequentially. Since breaking bonds requires energy and forming bonds returns energy, most chemical reactions proceed with either a net output or a net input of energy. A few reactions can occur with no net energy imbalance either way. Lilkewise, bond breaking typically leads (runs ahead of) bond formation. An old bond must be at least partially broken before a new one can begin to form. The energy cost of breaking the old bond must be supplied before we can get a return from bond formation. This is the reason for the of many chemical reactions. The more the bond formation process lags behind bond breaking, the higher the activation energy requirement. However, one must be careful not to try to relate bond dissociation energies with activation energies. Unless an extremely simple process involving small atoms or molecules is under consideration, the two parameters are very different. One such process might be the breaking of the H-H bond in the hydrogen molecule to produce hydrogen atoms, or the breaking of the Cl-Cl bond in the chlorine molecule to produce chlorine atoms. these processes are taking place in the gas phase and the bonds are being broken homolytically, the activation energy is the same as the bond dissociation energy. But a large number of chemical reactions take place in solution, or with bond formation taking place as bond breaking is also occurring. A transition state is reached as two (or more) reactants approach each other and their rises to a maximum. At the potential energy maximum they have become so closely associated in a configuration that is so highly disorted that any further change will cause them to either revert to reactants, or to form products. Any factors (other than the bond dissociation energies) that stabilize or destabilize the transition state, such as involvement of the solvent (solvation) or the development (or relief) of steric strain can affect the energy level associated with this transition state. In a single step reaction the energy associated with the transition state is also the activation energy. In a multistep reaction, where several transition states are possible for each step, the energy associated with the highest transition state is the activation energy of the overall reaction. The activation energy can also be viewed as a kinetic parameter if it is defined as the minimum kinetic energy with which the reactants should approach each other for reaction to occur. We will come back to further elaborate on some of these points as we expand on the study of different types of reactions and examine the meaning of the Hammond postulate. | 2,788 | 3,600 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Lasers/Overview_of_Lasers |
LASER is an acronym for Light Amplification by Stimulated Emission of Radiation. Laser is a type of light source which has the unique characteristics of directionality, brightness, and monochromaticity. The goal of this module is to explain how a laser operates (stimulated or spontaneous emission), describe important components, and give some examples of types of lasers and their applications. The word LASER is an acronym for Light Amplification by Stimulated Emission of Radiation. In 1916, Albert Einstein discovered the physical principle responsible for this amplification, and the foundation principle is called stimulated emission. It was widely accepted at that time that laser would represent a big leap in science and technology, even before Theodore H. Maiman built the first one in 1960. The 1951 Nobel Prize in physics was shared by Charles H. Townes, Nokolay Basov, and Aleksandr Prokhorov, in citation, “For Fundamental work in the field of quantum electronics, which has led to the construction of oscillator and amplifiers based on the maser-laser principle”. The early lasers developed in the 1950s by Charles H. Townes and Arthur Shawlow were gas and solid-state lasers for use in spectroscopy. The principles of lasers were adapted from masers. MASER is an acronym that stands for Microwave Amplification by Stimulated Emission of Radiation. It uses the idea of stimulated emission and population inversion to produce a coherent amplified radiation of light in the microwave region. Stimulated emission is when an electron in an excited state falls back to ground state after absorbing energy from an incident photon. Amplified radiation or light is produced with the same direction and energy as the incident light. Population inversion is when you have a greater population of electrons in the excited state than in ground state. Population inversion is achieved through various pumping mechanisms. The laser uses these same ideas except that the electromagnetic wave created is in the visible light region. When emission begins the light oscillates within the resonant cavity and gains magnitude. Once enough light has been acquired, the laser beam is produced. This allows lasers to be used as a powerful light source. Three unique characteristics of a laser are its properties of monochromaticity, directionality, and brightness. The monochromaticity of lasers is due to the fact that lasers are highly selective in the wavelength of light produced, which in itself is due to the resonant frequency inside the active material. Resonant frequency means that the light is oscillating in a single mode creating a monochromatic beam of light. The property of directionality depends on the angle of which the light propagates out of the source. Since lasers have large spatial and temporal coherence directionality is maximized. Temporal coherence is when there are small fluctuations in the phase. Spatial coherence has small changes in the amplitude of the emitted light. Like monochromaticity, directionality is dependent on the resonant cavity of the active material. The property of brightness is a result of the directionality and the coherence of the light. Due to these properties, lasers today are used in simple laser pointers, cutting devices, the development of military technologies, spectroscopy, and medical treatments. Their direct application to spectroscopy has allowed scientists to measure lifetimes of excited state molecules, structural analysis, probing far regions of the atmosphere, photochemistry and their use as ionization sources. One of the most important characteristics of light is that it has wave-like properties and that it is an electromagnetic wave. Experiments on the blackbody radiation demonstrated a comprehensive idea of emission and absorption of electromagnetic waves. In 1900, Max Plank developed the theory that electromagnetic waves can only exist in distinct quantities of energy, which are directly proportional to a given frequency (\(\nu\)). In 1905, Albert Einstein proposed the dual nature of light, having both wave-like and particle-like properties. He used the photoelectric effect to show that light acts as a particle, with energy inversely proportional to the wavelength of light. This is important, because the number of particles is directly related to how intense a light beam will be. In 1915, Einstein introduced the idea of stimulated emission- a key concept to lasers. In 1957, Townes and Shawlow proposed the concept of lasers in the infrared and optical region by adapting the concept of masers to produce monochromatic and coherent radiation. In 1953, Townes was the first to build a maser with an ammonia gas source. Masers use stimulated emission to generate microwaves. Townes and other scientists wanted to develop the optical maser to generate light. Optical masers would soon adopt the name LASER: Light Amplification by Stimulated Emission of Radiation. An optical maser would need more energy than what can be provided by microwave frequencies and a resonant cavity of the order of 1μm or less. Townes and Shawlow proposed the use of a Fabry-Pérot interferometer equipped with parallel mirrors, where Interference of radiation which is traveling back and forth between parallel mirrors in the cavity allowed for selection of certain wavelengths. Townes built an optical maser with potassium gas. That failed because the mirrors degraded over time. In 1957, Gordon Gould improved upon Townes' and Shawlow's laser concept. It was Gould who renamed the optical maser to laser. In April 1959, Gould filed a patent for the laser and later in March 1960, Townes and Shawlow had also made a request for a patent. Since Gould’s notebook was officially dated the idea was his first, but he did not receive the patent until 1977. A laser consists of three main components: a lasing medium, a resonant or optical cavity, and an output coupler. The lasing medium consists of a group of atoms, molecules , or ions in solid, liquid or gaseous form, which acts as an amplifier for light waves. For amplification, the medium has to achieve population inversion, which means in a state in which the number of atoms in the upper energy level is greater than the number of atoms in the lower energy level. The output coupler serves as energy source which provides for obtaining such a state of population inversion between a pair of energy levels of the atomic system. When the active medium is placed inside an optical resonator, the system acts as an oscillator. The lasing medium is the component used to achieve lasing, such as chromium in the aluminum oxide crystal-found in a ruby laser. Helium and neon gas are two materials most commonly used in gas lasers. These are only a few examples of lasing mediums or materials that have been used in the past and present states of the laser. For further information about different types of lasing mediums please refer to the section where is discussed. Rays of light moving along an optical path tend to diverge over time, because the energy of radiation has very high frequency. Therefore an optical cavity is needed to refocus the light. , represents the basics of an optical cavity where the light inside moves back and forth between two mirrors. These redirect and focus the light each time it hits the surface of the mirrors. There are two types of cavities: stable cavities and unstable cavities. A stable cavity is when the ray of light does not diverge far from the optical axis. An unstable cavity is when the ray of light bounces off and away from a mirrors surface. The importance of the cavities is that it allows for the laser to have properties of directionality, monochromaticity and brightness. Light oscillating between the first mirror (M ) and the second mirror (M ) separated by distance, d, will have a round-trip phase shift (RTPS) of 2θ=2kd=q2π- ϕ. In , a round-trip can be described as the beam traveling from M to M back to M . Resonance occurs in the cavity because the light propagating between the two mirrors is uniform. The ABCD law describes that an optical cavity has a field distribution, because it reproduces itself as it is making these round-trips between the two parallel mirrors. The ABCD law was first applied to a Gaussian beam with a beam parameter, q, which is described as \[q_2=\dfrac {(Aq_1+B)}{(Cq_1+D)}\] This law states that the beam, which is oscillating through an optical system, will experience some changing as it moves in the cavity. Fields that are created in an optical cavity have analogous shape and phase as they make each trip back and forth. However, the one thing that changes is the size of the field because the electromagnetic wave is unrestricted, unlike a wave in a short-circuited coaxial cable used to build a resonator or a microwave cavity mode. Since there is a field distribution of E at the surface of M it can be said that there is a field distribution at the surface of M . Since there will be a change in size of the field this means that the electromagnetic wave will have change in amplitude by ρ *ρ and a phase factor of \(e^-jk2d\), creating additional fields. This is an example of a phasorial addition of all fields between M and M creating a total field E ( ). Phasorial addition is described by RTPS, where each additional E will have a delay of angle ϕ which is related to kd. E will always be greater than E only if ρ and ρ are not greater than 1 and ϕ=0. In this case, when ϕ=0 resonance is enhanced because all factors such as E travelling between M and M , the intensity of the electromagnetic waves, number of photons traveling between M and M and the amount of energy that is stored are maximized. The resonant wavelength can also be determined by using the relationship between RTPS, and because \(k=\dfrac {\omega n}{c}=\dfrac {2\pi}{\lambda}\) Using \(2\theta=2kd=q2\pi\) \(\dfrac{2\pi2d}{\lambda}=q2π\) \(d=\dfrac {q\lambda}{2}\) Where the wavelength of interest is given by \(\lambda=\lambda_0/n\), where n is the index of refraction and \(\lambda_0\) is the free-space wavelength . Since we are dealing with light as a wave, the light in the resonant cavity can be described in terms of frequency, ν. Where \(k2d=\omega\dfrac {2nd}{c}=2\pi \nu \dfrac {2nd}{c}=q(2\pi)\) \(\nu=q\dfrac {c}{2nd}\) A Fabry-Pérot interferometer is a prime example of an optical cavity used in a laser. The Fabry-Pérot is equipped with two parallel mirrors, one that is completely reflective and the other that is partially reflective. As light is accumulating in the cavity after taking several round trips between the two mirrors, some light is transmitted through the partially reflective mirror and a laser beam is produced. The beam can be in pulsed mode or continuous-wave (CW) mode. To increase the performance of the resonant cavity, the length of the cavity (d) must be considered as a way to avoid a decrease in the laser beam intensity due to any diffraction losses. The size of the aperture of the cavity is also important because it determines the strength or the intensity of the laser beam. In fact, determining the best length of a resonant cavity will enhance the coupling conditions of the output coupler by producing a frequency that is stable, which ultimately generates a laser beam that is coherent and has high power. There are essentially six stages in the lasing process. First is the ground state where there is no excitation of the lasing medium. Second is pumping, which is applied to the medium where spontaneous emission occurs. Then the third stage is when emitted photons collide with an excited molecule where stimulated emission occurs. In the fourth stage the photons are produced in multiples, however those moving parallel in the cavity will hit a mirror and then hit the second mirror. During the fifth stage this process continues until there is an accumulation of light that is coherent and of a specific frequency. Finally, the sixth stage is when the light or laser beam exits the partially reflective mirror which is also known as the output coupler. An output coupler is the last important component of a laser because it must be efficient to produce an output of light with maximum intensity. If the output coupler is too transparent than there is much more loss of electromagnetic waves and this will decrease lasing significantly because population inversion will no longer be maintained. If the output coupler or partially reflective mirror is too reflective, then all the accumulated light that is built up in the resonant cavity will be trapped in the cavity. The beam will not pass through the output coupler, producing little to no light making the laser ineffective. Lasers create a high energy beam of light by stimulated emission or spontaneous emission. Within in a molecule there are discrete energy levels. A simple molecular description has a low energy ground state (E ) and a high energy excited state (E ). When an electromagnetic wave, referred to as the incident light, irradiates a molecule there are two processes that can occur: absorption and stimulated emission. Absorption occurs when the energy of the incident light matches the energy difference between the ground and excited state, causing the population in the ground state to be promoted to the excited state. The rate of absorption is given by the equation: \(\dfrac {dN_1}{dt}=-W_{12} N_1\) Where N is the population in E and W is the probability of this transition. The probability of the transition can also be related to the photon flux (intensity of incident light): \(W_{12}=\sigma_{12} F\) Where F is the photon flux and σ is the cross section of the transition with units of area. When absorption occurs photons are removed from the incident light and the intensity of the light is decreased. Stimulated emission is the reverse of absorption. Stimulated emission has two main requirements: there must be population in the excited state and the energy of the incident light must match the difference between the excited and ground state. When these two requirements are met, population from the excited state will move to the ground energy level. During this process a photon is emitted with the same energy and direction as the incident light. Unlike absorption, stimulated emission adds to the intensity of the incident light. The rate for stimulated emission is similar to the rate of absorption, except that it uses the population of the higher energy level: \(W_{21}=\sigma_{21} F\) Like absorption the probability of the transition is related to the photon flux of the incident light through the equation: \(\dfrac {dN_2}{dt}=-W_{21} N_2\) When absorption and stimulated emission occur simultaneously in a system the photon flux of the incident light can increase or decrease. The change in the photon flux is a combination of the rate equations for absorption and stimulated emission. This is given by the equation: \[dF=\sigma F(N_2-N_1 )d\tau\] Spontaneous emission has the same characteristics as stimulated emission except that no incident light is required to cause the transition from the excited to ground state. Population in the excited state is unstable and will decay to the ground state through several processes. Most decays involve non-radiative vibrational relaxation, but some molecules will decay while emitting a photon matching the energy of the energy difference between the two states. The rate of spontaneous emission is given by: \[\dfrac {dN_2}{dt}=-AN_2\] Where A is the spontaneous emission probability which depends on the transition involved. The coefficient A is an Einstein coefficient obtained from the spontaneous emission lifetime. Since spontaneous emission is not competing with absorption, the photon flux is based solely on the rate of spontaneous emission. The population ratio of a molecule or atom is found using the Boltzmann distribution and the energy of the ground state (E ) and the excited state (E ): \[ \dfrac{N_2}{N_1} = e^{\dfrac{-(E_2-E_1)}{kT}}\] Under normal conditions, the majority, if not all, of the population is in the lower energy level (E ). This is because the energy of the excited is greater than the ground state. Normal thermal energy available (kT) is not enough to overcome the difference, and the ratio of population favors the ground state. For example, if the difference in energy between two states absorbes light at 500nm, the ratio of N to N is 5.1x10 :1. The photon flux of the incident light is directly proportional to the difference in populations. Since the ground state has more populations, the photon flux decreases: there is more absorption occurring than stimulated emission. In order to increase the photon flux there must be more population in the excited state than in the ground state, generally known as a population inversion. In a two level energy system it is impossible to create the population inversion needed for a laser. Instead three or four level energy systems are generally used ( ). Three level processes involve pumping of population from the lowest energy level to the highest, third energy state. The population can then decay down to the second energy level or back down to the first energy level. The population that makes it to the second energy level is available for stimulated emission. Light matching the energy difference between the second and first energy level will cause a stimulated emission. Four level systems follow roughly the same process except that population is moved from the lowest state to the highest fourth level. Then it decays to the third level and lasing happens when the incident light matches the energy between the third and second level. After lasing there is decay to the first level. Pumping is the movement of population from the ground state to a higher excited state. The general rate at which this is done is given by: \((\dfrac {dN_g}{dt})_p=W_p N_g\) Where N is the population in the ground level and W is the pump rate. Pumping can be done optically, electronically, chemically (see chemical laser), using gases at high flow rates, and nuclear fission. Only optical and electrical pumping will be discussed in detail. Optical pumping uses light to create the necessary population inversion for a laser. Usually high pressure xenon or krypton lamps are used to excite solid or liquid laser systems. The active material in the laser absorbs the light from the pump lamp, promoting the population from the ground state to the higher energy state. The material used in the laser can be continuously exposed to the pumping light which creates a continuous wave laser (CW). A pulsed laser can be created by using flashes of pumping light. In optical pumping there are three types of efficiency: transfer, lamp radiative, and pump quantum efficiency. Transfer efficiency is the ratio of the energy created by the lamp and the power of the light emitted by the laser. The lamp radiative efficiency is the measure of how much electrical power is converted into light in the optical lamp. Pump quantum efficiency accounts for the ratio of population that decays to the correct energy level and population that decays either back to the ground state or another incorrect energy level. For example, the overall pumping rate of the first ruby laser was around 1.1%. The average pump rate for optical pumping depends on the total efficiency of the pump (\(η_p\)), volume of the laser material (V), ground state population (N ), power input (P), and frequency of the lasing transition (ν ): \(〈W_p 〉=\eta_p (P/(VN_g ℏυ_0 ))\) Electrical pumping is a much more complicated process than optical pumping. Usually used for gas and semiconducting lasers, electrical pumping uses electrical current to excite and promote the ground state population. In a simple gas laser that contains only one species (A), current passes through the gas medium and creates electrons that collide with the gas molecules to produce excited state molecules (A ): \(A+e \longrightarrow A^*+e\) During electron impact either an ion or an excited state can be created. The ability to make the excited state depends mostly on the material used in the laser and not the electrical pumping source making it difficult to describe the efficiency of the pumping. Total efficiencies have been calculated and tabulated for most active materials used in electrical pumping. Where eficiencies range from a < 0.1% N gas laser to 70% for some CO gas lasers. Like the pumping rate of optical pumps, the rate of electrical pumping is found using the overall efficiency of the pump, power applied, and population of the ground state. However instead of using the frequency of the ground to upper state transition, electrical pumping uses the energy of the upper state (ħω ) and the volume of the electron discharge (V): \(〈W_p 〉=\eta_p (P/(VN_g ℏω_p ))\) The technique of Q switching allows the generation of laser pulses of short duration from a few nanoseconds to a few tens of nanoseconds and high peak power from a few megawatts to a few tens of megawatts. Suggest we put a shutter into the laser cavity. If the shutter is closed, laser action cannot occur and the population inversion can be very high. If the shutter is opened suddenly, the stored energy will be released in a short and intense light pulse. This technique is known as Q-switching. Q here denotes the ratio of the energy stored to the energy dissipated in the cavity. This technique is used in many types of solid-stat lasers and CO lasers to get a high-power pulsed output. To produce high inversion required for Q-switching, four requirements must be satisfied. The technique of mode locking allows the generation of laser pulses of ultrashort duration from less than a picosecond to femtoseconds and very high peak, a few gigawatts. Mode-locking is achieved by inducing the different longitudinal modes of a laser to a locked mode. When combining the electromagnetic waves modes with different frequencies and random phases, they produce a random and average output. When the modes are added in phase, they combine to produce a total amplitude and intensity output with a repeated pulse. There are many different types of lasers with a wide range applications, and below is a brief description of some of the main types. A solid-state laser is one that uses a solid active medium generally in a rod shape. Inside the active material is a dopant that acts as the light emitting source. Optical pumping is used to create population inversion of the active material. Solid-state lasers generally use stimulated emission as the mechanism for creating the high energy beam. The ruby laser was the first operating laser and was built in 1960. It has a three-level ( ) energy system that uses aluminum oxide with some of the aluminum atom replaced with chromium as its active material. The chromium in the aluminum oxide crystal is the active part of the laser. Electrons in the ground state of chromium absorb the incident light and become promoted to higher energy states. The short lived excited state relaxes down to a metastable state with a longer lifetime. Laser emission happens when there is relaxation from the metastable state back to the ground state. A xenon flash lamp emitting light at wavelengths of 6600Å and 4000Å (matching the energy needed to excite the chromium atoms). In order to create resonance of the incident light in the active material silver platting was put at both ends of the ruby rod. One end was completely covered while the other end was partially covered so lasing light could exit the system. Nd: YAG laser are the most popular type of solid state laser. The laser medium is a crystal of Y Al O which are commonly called YAG, an acronym for yttrium aluminum garnet. A simplified energy-level scheme for Nd:YAG is shown in Fig. 9. The λ=1.06 μm laser transition is the strongest of the 4F →4I transitions. The major application of the Nd laser is in various form of material processing: drilling, spot welding, and laser marking. Because they can be focused to a very small spot, the laser are also used in resistor trimming and in circuit mask , memory repair and also in cutting out specialized circuits. Medical applications include many types of surgeries. Many medical applications take advantage of low-loss optical fiber delivery systems that can be inserted into the body to wherever is needed. Nd lasers are also used in military applications such as range finding and target designation. High power pilsed versions are also used for X-ray spectral regions. In addition, Nd lasers are used in scientific lab as good sources for pumping dye laser and other types of lasers. The semiconductor laser is another type of solid state laser that uses a semiconducting material like germanium or silicon. When the temperature of the semiconducting material is increased, electrons move from the valence band to the conducting band creating holes in the valence band ( ). In between the conducting band and valence band is a region where there are no energy levels, called the band gap. Applying a voltage to the semiconductor causes electrons to move to the conduction band creating a population inversion. Irradiating a semiconductor with incident light matching the energy of the forbidden area causes a large transition from the conduction band to the valence band, increasing and amplifying the incident light. A gas laser contains active material composed of a mixture of gases with similar energy states inside a small gas chamber. Electrical pumping is used to create the population inversion where one gas is excited through collisions with electrons and in turn excites the other gas through collisions. The helium-neon laser was the first gas laser. It consists of a long narrow tube that contains He and Ne gas. Mirrors are placed at both ends of the gas tube to form the resonant cavity with one of the mirrors partially reflecting the incident light. Stimulated emission of the gas mixture is carried out by first exciting the He gas to a higher energy state through electron collision with electrons from the electronic pumping source (electrical pumping). Then the excited He atoms collide with the Ne atoms transferring their energy and exciting them to a higher energy level. The Ne atoms in the higher energy level will then relax to a lower metastable energy state. Lasing occurs when there is relaxation from the metastable state to a lower energy state causing spontaneous emission. The Ne gas then returns to the ground state when it collides with the outer walls of the gas tube ( ). The carbon dioxide laser is a gas laser that uses the energy difference between rotational-vibrational energy levels. Within the vibrational levels of CO there are rotational sub-energy levels. A mixture of N and CO gas are placed inside a chamber. The N atoms are excited through an electrical pumping mechanism. The excited atoms then collide with the CO atoms transfer energy. This transfer of energy causes the CO to go into a higher vibrational level. The excited CO molecules then go through spontaneous emission when they are relaxed to lower rotational-vibrational levels increasing the signal of the incident light ( ). Carbon dioxide lasers are extremely efficient, around 70%, and powerful compared to other gas lasers making them useful for welding and cutting. Liquid lasers consist of a liquid active material usually composed of an organic dye compound. The most common type of liquid laser uses rhodamine 6G ( ) dye mixed with alcohol and is excited by different types of lasers, such as an argon-ion laser or a nitrogen laser. Organic dyes are large compounds that have absorption bands in the UV or visible region with a strong intense fluorescence spectrum. The free π electrons of the dye are excited using an optical pumping source and the transition from the S to the S state creates the lasing light (see . Liquids are generally used because they can easily be tuned to emit a certain wavelength by changing the resonant frequency within the cavity. Wavelengths from the visible to the infrared can be covered. There are many benefits of liquid lasers, some include that they can be cooled in a relative amount of time, they cannot be damaged unlike a solid-laser, and their production is cost-effective. The efficiency of liquid lasers is low because the lifetime of the excited state is relatively short; there are many non-radiative decay processes, and the material degrades over time. Liquid lasers tend to be used only as a pulse laser when tunability is required. Liquid lasers can be used for high-resolution spectroscopy since they are easily tuned over a wide range of wavelengths. They can also be used because they have concentrations which are manageable when dissolved in solids or other liquids. Chemical lasers are different from other lasers because the population inversion is the direct product of a chemical reaction when energy is released as a result of an exothermic reaction. Usually reactions involve gases where the energy created is used to make vibrationally excited molecules. Light used for lasing is then created from vibrational-rotational relaxation like in the CO gas laser. An example of a chemical laser is the HF gas laser. Inside the gas chamber fluorine and hydrogen react to form an excited HF molecule: F + H → HF + H The excess energy from the reaction allows HF to stay in its excited state. As it relaxes, light is emitted through spontaneous emission. Deuterium can also be used in place of hydrogen. Deterium fluoride is useds for applications that require high-power. For example, MIRACL was built for military research and was known to produce 2.2 megawatts of power. The uniqueness of a chemical laser is that the power required for lasing is produced in the reaction itself. The applications of lasers are numerous and cover scientific and technological fields. In general, these applications are a direct consequence of the special characteristics of lasers. Below are a few examples of the laser applications, for a complete list please go to en. .org/wiki/List_of...ons_for_lasers Lidar is short for light detection and ranging which is an optical remote sensing technology can be used for monitoring the environment. A typical lidar system involves a transmitter of laser radiation and a receiver for the detection and analysis of backscattered light. A beam expander is usually used at transmitter to reduce divergence of the laser beam before it propagates into the atmosphere. The receiver includes a wavelength filter, a photo detector, and computers and electronics for data acquisition and analysis. Lidar system dates back to the 1930, because of the laser, it has become one of the primary tools in atmospheric and environmental research. Other than that, Lidar has been put into various uses. In agriculture, lidar can be used to create topographic map to help farmer to decide appropriate amount of fertilizing to achieve a better crop yield. In Archaeology, lidar can be used to create a geographic information system to help archaeologists to find sites. In transportation, lidar has been used in autonomous cruise control system to prevent road accident and policemen are also using lidar speed gun to enforce the speed limit regulation. The beam of a laser is usually a few millimeters in diagram. For most material processing applications, lenses are used to increase the intensity of the beam. The beam from a laser is either plane or spherical. After passing through a lens, the beam should get focused to a point. But in actual practice, diffraction effects have to be taken into consideration, the incoming will focus into a region of radius. If λ is the wavelength of the laser light, a is the radius of the beam, and f is the focal length of the lens, then the radius of the region is \(I =\dfrac { P }{ \pi b^2 }=\dfrac { P a^2 }{ \pi πλ^2f^2 }\) The high-power (P>100w) laser are widely used in material processing such as welding, drilling, cutting, surface treatment, and alloying. The main advantage of the laser beam can be summarized as follow: (1) The heating produced by the laser is less than that in conventional process. Material distortion is considerably reduced. (2) Possibility of working in inaccessible region. Any region which can be seen can be processed by a laser. (3) The process can be better controlled and easily automatized. However, against all these advantages, the disadvantages are: (1) high cost of the laser system. (2) Reliability and reproducibility problems of the laser system. (3) Safety problems. In field of medicine, the major use of lasers is for surgery such as laser eye surgery commonly known as LASIK. Besides that, there are also a few diagnosetic applications such as clinical use of flow microfluormeters, Doppler velocimetry to measure the blood velocity, laser fluorescence bronchoscope to detect tumors in their early phase. For surgery, the laser beams are used instead of a conventional scalpel. The infrared beam from the CO laser is strongly absorbed by water molecules in the tissue. It produces a rapid evaporation of these molecules, consequently cutting the tissue. The main advantage of laser beam surgery can be summarized as follows: (1) High precision. The incision can be made with a high precision particularly when the beam is directed by means of a microscope. (2) Possibility of operating in inaccessible region. Laser surgery can be operated in any region of the body which can be observed by means of an optical system. (4) Limited damage to blood vessel and adjacent tissue. However, the disadvantages are: (1) considerable cost. (2) Smaller velocity of the laser scalpel. (3) Reliability and safety problems associated with the laser procedure. | 33,929 | 3,601 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./02.E%3A_Atoms_Molecules_and_Ions_(Exercises) |
. In addition to these individual basis; please contact Which of the following elements exist as diatomic molecules? Which of the following elements exist as diatomic molecules? Why is it proper to represent the elemental form of helium as He but improper to represent the elemental form of hydrogen as H? Why is it proper to represent the elemental form of chlorine as Cl but improper to represent the elemental form of calcium as Ca ? To solve, determine the percent of sodium in each sample of sodium chloride. There is 4.36 g sodium for every 11.08 g of sodium chloride in the first experiment. The amount of sodium in the sodium chloride for the second experiment must be found. This is found by subtracted the known amount of reacted chlorine (4.20 g) from the amount of sodium chloride (6.92 g). 6.92 g sodium chloride - 4.20 g chlorine = 2.72 g sodium. Thus, the percent of sodium in each sample is represented below:
% Na = (4.36 g Na)/(11.08 g NaCl) x 100% = 39.4% Na % Na = (2.72 g Na)/(6.92 g NaCl) x 100% = 39.3%
The slight difference in compositions is due to significant figures: each percent has an uncertainty of .01% in either direction. The two samples of sodium chloride have the same composition. 36.0 g wood x (1lb)/(453.59237 g) = 0.0793664144 lb wood
38.4 g oxygen x (1 lb)/(453.59237 g) = .0846575087 lb oxygen Now two ratios equal to each other can be set up to determine the unknown mass of oxygen. (0.0793664144 lb wood)/(.0846575087 lb oxygen) = (8.00 lb wood)/(unknown mass oxygen)
Solving reveals that it requires 8.53 lb of oxygen to burn 8.00 lb of wood. Please be sure you are familiar with the topics discussed in before proceeding to the Numerical Problems. 1. Complete the following table for the missing elements, symbols, and numbers of electrons. 2. Complete the following table for the missing elements, symbols, and numbers of electrons. 3. Is the mass of an ion the same as the mass of its parent atom? Explain your answer. 4. What isotopic standard is used for determining the mass of an atom? 5. Give the symbol \(^A_Z X\) for these elements, all of which exist as a single isotope. 6. Give the symbol \(_Z^AX\) for these elements, all of which exist as a single isotope. 7. Identify each element, represented by X, that has the given symbols. Give the symbol \(_Z^AX\) for these elements. fluorine \(^{19}_{9} F\) helium \(^{4}_{2} He\) Identify each element, represented by X, that has the given symbols. Please be sure you are familiar with the topics discussed in before proceeding to the Numerical Problems. 1. The isotopes I and Co are commonly used in medicine. Determine the number of neutrons, protons, and electrons in a neutral atom of each. 2. Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope: 3. Both technetium-97 and americium-240 are produced in nuclear reactors. Determine the number of protons, neutrons, and electrons in the neutral atoms of each. 4. The following isotopes are important in archeological research. How many protons, neutrons, and electrons does a neutral atom of each contain? 5. Copper, an excellent conductor of heat, has two isotopes: Cu and Cu. Use the following information to calculate the average atomic mass of copper: 6. Silicon consists of three isotopes with the following percent abundance: Calculate the average atomic mass of silicon. 7. Complete the following table for neon. The average atomic mass of neon is 20.1797 amu. 8. Are \(_{28}^{63} X \) and \( _{29}^{62} X \) isotopes of the same element? Explain your answer. 9.Complete the following table: 10.Complete the following table: 11. Using a mass spectrometer, a scientist determined the percent abundances of the isotopes of sulfur to be 95.27% for S, 0.51% for S, and 4.22% for S. Use the atomic mass of sulfur from the periodic table (see ) and the following atomic masses to determine whether these data are accurate, assuming that these are the only isotopes of sulfur: 31.972071 amu for S, 32.971459 amu for S, and 33.967867 amu for S. 12. The percent abundances of two of the three isotopes of oxygen are 99.76% for O, and 0.204% for O. Use the atomic mass of oxygen given in the periodic table and the following data to determine the mass of O: 15.994915 amu for O and 17.999160 amu for O. 13. Which element has the higher proportion by mass in NaI? 14. Which element has the higher proportion by mass in KBr? Determine the number of protons, neutrons, and electrons in a neutral atom of each isotope: How many protons, neutrons, and electrons does a neutral atom of each contain? 1. Classify each element in Conceptual Problem 1 of section 2.4 as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. 2. Classify each element in Conceptual Problem 2 of section 2.4 as a metal, a nonmetal, or a semimetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. 3. Classify each element as a metal, a semimetal, or a nonmetal. If a metal, state whether it is an alkali metal, an alkaline earth metal, or a transition metal. 4. Which of these sets of elements are all in the same period? 5. Which of these sets of elements are all in the same period? 6. Which of these sets of elements are all in the same group? 7. Which of these sets of elements are all in the same group? 8. Indicate whether each element is a transition metal, a halogen, or a noble gas. 9. Which of the elements indicated in color in the periodic table shown below is most likely to exist as a monoatomic gas? As a diatomic gas? Which is most likely to be a semimetal? A reactive metal? 10. Based on their locations in the periodic table, would you expect these elements to be malleable? Why or why not? 11. Based on their locations in the periodic table, would you expect these elements to be lustrous? Why or why not? 7. Covalent compounds generally melt at lower temperatures than ionic compounds because the intermolecular interactions that hold the molecules together in a molecular solid are weaker than the electrostatic attractions that hold oppositely charged ions together in an ionic solid. 1. The structural formula for chloroform (CHCl ) was shown in Example 2.6.2. Based on this information, draw the structural formula of dichloromethane (CH Cl ). 2. What is the total number of electrons present in each ion? 3. What is the total number of electrons present in each ion? 4. Predict how many electrons are in each ion. 5. Predict how many electrons are in each ion. 6. Predict the charge on the most common monatomic ion formed by each element. 7. Predict the charge on the most common monatomic ion formed by each element. 8. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. 9. For each representation of a monatomic ion, identify the parent atom, write the formula of the ion using an appropriate superscript, and indicate the period and group of the periodic table in which the element is found. How many electrons are in each ion: The charge on the most common monatomic ion formed by each element: The charge on the most common monatomic ion formed by each element: 1. What are the differences and similarities between a polyatomic ion and a molecule? 2. Classify each compound as ionic or covalent. 3. Classify each compound as ionic or covalent. Which are organic compounds and which are inorganic compounds? 4. Generally, one cannot determine the molecular formula directly from an empirical formula. What other information is needed? 5. Give two pieces of information that we obtain from a structural formula that we cannot obtain from an empirical formula. 6. The formulas of alcohols are often written as ROH rather than as empirical formulas. For example, methanol is generally written as CH OH rather than CH O. Explain why the ROH notation is preferred. 7. The compound dimethyl sulfide has the empirical formula C H S and the structural formula CH SCH . What information do we obtain from the structural formula that we do not get from the empirical formula? Write the condensed structural formula for the compound. 8. What is the correct formula for magnesium hydroxide—MgOH or Mg(OH) ? Why? 9. Magnesium cyanide is written as Mg(CN) , not MgCN . Why? 10. Does a given hydrate always contain the same number of waters of hydration? Classify each compound as ionic or covalent. 1. Write the formula for each compound. 2. Write the formula for each compound. 3. Complete the following table by filling in the formula for the ionic compound formed by each cation-anion pair. 4. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. 5. Write the empirical formula for the binary compound formed by the most common monatomic ions formed by each pair of elements. 6. Write the empirical formula for each compound. 7. Write the empirical formula for each compound. The empirical formula for each compound: 1. Benzene (C H ) is an organic compound, and KCl is an ionic compound. The sum of the masses of the atoms in each empirical formula is approximately the same. How would you expect the two to compare with regard to each of the following? What species are present in benzene vapor? 2. Can an inorganic compound be classified as a hydrocarbon? Why or why not? 3. Is the compound NaHCO a hydrocarbon? Why or why not? 4. Name each compound. 5. Name each compound. 6. For each structural formula, write the condensed formula and the name of the compound. a. b. c. d. e. 7. For each structural formula, write the condensed formula and the name of the compound. a. b. c. d. 8. Would you expect PCl to be an ionic compound or a covalent compound? Explain your reasoning. 9. What distinguishes an aromatic hydrocarbon from an aliphatic hydrocarbon? 10. The following general formulas represent specific classes of hydrocarbons. Refer to and and and identify the classes. 11. Using R to represent an alkyl or aryl group, show the general structure of an 1. Write the formula for each compound. 2. Write the formula for each compound. 3. Write the formula for each compound. 4. Name each compound. 5. Name each compound. 6. Draw the structure of each compound. 7. Draw the structure of each compound. 6. 7. a. b. c. d. e. | 10,561 | 3,602 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Corrosion/Sacrificial_Anode |
Sacrificial Anodes are highly active metals that are used to prevent a less active material surface from corroding. Sacrificial Anodes are created from a metal alloy with a more negative electrochemical potential than the other metal it will be used to protect. The sacrificial anode will be consumed in place of the metal it is protecting, which is why it is referred to as a "sacrificial" anode. When metal surfaces come into contact with electrolytes, they undergo an electrochemical reaction known as corrosion. Corrosion is the process of returning a metal to its natural state as an ore and in this process, causing the metal to disintegrate and its structure to grow weak. These metal surfaces are used all around us -- from pipelines to buildings to ships. It is important to ensure that these metals last as long as they can and thus necessitates what is known as cathode protection. Sacrificial anodes are among several forms of cathode protection. Other forms of cathode protection are Metal in seawater is one such example with the iron metal coming into contact with electrolytes. Under normal circumstances, the iron metal would react with the electrolytes and begin to corrode, growing weaker in structure and disintegrating. The addition of zinc, a sacrificial anode, would prevent the iron metal from "corroding". According to the table of Standard Reduction Potentials, the standard reduction potential of zinc is about -0.76 volts. The standard reduction potential of iron is about -0.44 volts. This difference in reduction potential means that Zinc would oxidize much faster than iron would. In fact, zinc would oxidize completely before iron would begin to react. The materials used for sacrificial anodes are either relatively pure active metals, such as zinc or magnesium, or are magnesium or aluminum alloys that have been specifically developed for use as sacrificial anodes. In applications where the anodes are buried, a special backfill material surrounds the anode in order to insure that the anode will produce the desired output. Since the sacrificial anode works by introducing another metal surface with a more negative electronegative and much more anodic surface. The current will flow from the newly introduced anode and the protected metal becomes cathodic creating a galvanic cell. The oxidation reactions are transferred from the metal surface to the galvanic anode and will be sacrificed in favor of the protected metal structure. Sacrificial anodes are normally supplied with either lead wires or cast-m straps to facilitate their connection to the structure being protected. The lead wires may be attached to the structure by welding or mechanical connections. These should have a low resistance and should be insulated to prevent increased resistance or damage due to corrosion. When anodes with cast-in straps are used, the straps can either be welded directly to the structure or the straps can be used as locations for attachment. A low resistance mechanically adequate attachment is required for good protection and resistance to mechanical damage. In the process of providing electrons for the cathodic protection of a less active metal the more active metal corrodes. The more active metal (anode) is sacrificed to protect the less active metal (cathode). The amount of corrosion depends on the metal being used as an anode but is directly proportional to the amount of current supplied. Sacrificial Anodes are used to protect the hulls of ships, water heaters, pipelines, distribution systems, above-ground tanks, underground tanks, and refineries. The anodes in sacrificial anode cathodic protection systems must be periodically inspected and replaced when consumed. | 3,744 | 3,603 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Nomenclature_of_Alkanes/Nomenclature_of_Cycloalkanes |
Cycloalkanes are cyclic , meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings. Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane. If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this: In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures. (polycyclic) Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in . Cyclic hydrocarbons have the prefix "cyclo-". For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens. The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. Because cyclopropane is a substituent, it would be named a cyclopropyl-substituted alkane. (ex: 1-chlorocyclohexane or cholorocyclohexane is acceptable) Notice that "f" of fluoro alphabetically precedes the "m" of methyl. Although "di" alphabetically precedes "f", it is not used in determining the alphabetical order. 8) If the substituents of the cycloalkane are related by the cis or trans configuration, then indicate the configuration by placing "cis-" or "trans-" in front of the name of the structure. Notice that chlorine and the methyl group are both pointed in the same direction on the axis of the molecule; therefore, they are cis. 9) After all the functional groups and substituents have been mentioned with their corresponding numbers, the name of the cycloalkane can follow. Cycloalkanes are very similar to the alkanes in reactivity, except for the very small ones, especially cyclopropane. Cyclopropane is significantly more reactive than what is expected because of the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are approximately 109.5°. In cyclopropane, the bond angles are 60°. With the electron pairs this close together, there is a significant amount of repulsion between the bonding pairs joining the carbon atoms, making the bonds easier to break. Alcohol (-OH) substituents take the highest priority for carbon atom numbering in IUPAC nomenclature. The carbon atom with the alcohol substituent must be labeled as 1. Molecules containing an alcohol group have an ending " ", indicating the presence of an alcohol group. If there are two alcohol groups, the molecule will have a "di-" prefix before "-ol" (diol). If there are three alcohol groups, the molecule will have a "tri-" prefix before "-ol" (triol), etc. The alcohol substituent is given the lowest number even though the two methyl groups are on the same carbon atom and labeling 1 on that carbon atom would give the lowest possible numbers. Numbering the location of the alcohol substituent is unnecessary because the ending "-ol" indicates the presence of one alcohol group on carbon atom number 1. There are many other like alcohol, which are later covered in an organic chemistry course, and they determine the ending name of a molecule. The naming of these functional groups will be explained in depth later as their chemical properties are explained. Although alkynes determine the name ending of a molecule, alkyne as a substituent on a cycloalkane is not possible because alkynes are planar and would require that the carbon that is part of the ring form 5 bonds, giving the carbon atom a negative charge. However, a cycloalkane with a triple bond-containing substituent is possible if the triple bond is not directly attached to the ring. Name the following structures. 1) 2) 3) 4) 5) 6) 7) 8) 1,1-dibromo-5-fluoro-3-butyl-7-methylcyclooctane 9) trans-1-bromo-2-chlorocyclopentane 10) 1,1-dibromo-2,3-dichloro-4-propylcyclobutane 11) 2-methyl-1-ethyl-1,3-dipropylcyclopentane 12) cycloheptane-1,3,5-triol 14) 15) 16) 17) 18) 19) 1) cyclodecane 2) chlorocyclopentane or 1-chlorocyclopentane 3) trans-1-chloro-2-methylcycloheptane 4) 3-cyclopropyl-6-methyldecane 5) cyclopentylcyclodecane or 1-cyclopentylcyclodecane 6) 1,3-dibromo-1-chloro-2-fluorocycloheptane 7) 1-cyclobutyl-4-isopropylcyclohexane 13) cyclohexane 14) cyclohexanol 15) chlorocyclohexane 16) cyclopentylcyclohexane 17) 1-chloro-3-methylcyclobutane 18) 2,3-dimethylcyclohexanol 19) cis-1-propyl-2-methylcyclopentane | 6,034 | 3,604 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/05%3A_Electrons_in_Atoms/5.10%3A_Heisenberg_Uncertainty_Principle |
Lasers have numerous applications outside the physics lab. These devices can be employed to measure distances accurately, and many commercial instruments are used in construction for laying out a building site. When the light beam strikes a solid object, it is reflected back, and the device determines how far away the object is. There is such a significant difference between the mass of the light beam (photons) and the mass of the object, that the beam does not disturb the object at all. A historically significant measurement of interest was the use of a laser to measure the distance from the Earth to the moon. The impact of the photons from the laser on the moon had absolutely no effect on the moon's orbit. Another feature that is unique to quantum mechanics is the uncertainty principle. The states that it is impossible to simultaneously determine both the position and the velocity of a particle. The detection of an electron, for example, would be made by way of its interaction with photons of light. Since photons and electrons have nearly the same energy, any attempt to locate an electron with a photon will knock the electron off course, resulting in uncertainty about where the electron is located (see below). We do not have to worry about the uncertainty principle with large everyday objects because of their mass. If you are looking for something with a flashlight, the photons coming from the flashlight are not going to cause the thing that you are looking for to move. However, this is not the case with atomic-sized particles, and has led scientists to a new understanding about how to envision the location of the electrons within atoms. | 1,690 | 3,605 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Lipids/Properties_and_Classification_of_Lipids/Phospholipids |
Phospholipids are the main constituents of cell membranes. They resemble the triglycerides in being ester or amide derivatives of glycerol or sphingosine with fatty acids and phosphoric acid. The phosphate moiety of the resulting phosphatidic acid is further esterified with ethanolamine, choline or serine in the phospholipid itself. The following diagram shows the structures of some of these components. ing on the diagram will change it to display structures for two representative phospholipids. Note that the fatty acid components (R & R') may be saturated or unsaturated. As ionic amphiphiles, phospholipids aggregate or self-assemble when mixed with water, but in a different manner than the soaps and detergents. Because of the two pendant alkyl chains present in phospholipids and the unusual mixed charges in their head groups, micelle formation is unfavorable relative to a bilayer structure. If a phospholipid is smeared over a small hole in a thin piece of plastic immersed in water, a stable planar bilayer of phospholipid molecules is created at the hole. As shown in the following diagram, the polar head groups on the faces of the bilayer contact water, and the hydrophobic alkyl chains form a nonpolar interior. The phospholipid molecules can move about in their half the bilayer, but there is a significant energy barrier preventing migration to the other side of the bilayer. This bilayer membrane structure is also found in aggregate structures called . Liposomes are microscopic vesicles consisting of an aqueous core enclosed in one or more phospholipid layers. They are formed when phospholipids are vigorously mixed with water. Unlike micelles, liposomes have both aqueous interiors and exteriors. A cell may be considered a very complex liposome. The bilayer membrane that separates the interior of a cell from the surrounding fluids is largely composed of phospholipids, but it incorporates many other components, such as cholesterol, that contribute to its structural integrity. Protein channels that permit the transport of various kinds of chemical species in and out of the cell are also important components of cell membranes. The interior of a cell contains a variety of structures (organelles) that conduct chemical operations vital to the cells existence. Molecules bonded to the surfaces of cells serve to identify specific cells and facilitate interaction with external chemical entities. The are also membrane lipids. They are the major component of the myelin sheath surrounding nerve fibers. Multiple Sclerosis is a devastating disease in which the myelin sheath is lost, causing eventual paralysis. ), | 2,661 | 3,606 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.09%3A_Buffers |
Buffer solutions, which are of enormous importance in controlling pH in various processes, can be understood in terms of acid/base equilibrium. A buffer is created in a solution which contains both a weak acid and its conjugate base. This creates to absorb excess H or supply H to replace what is lost due to neutralization. The calculation of the pH of a buffer is straightforward using an ICE table approach. What is the pH of a solution that is 0.150 M in KF and 0.250 M in HF? The reaction of interest is \[HF \rightleftharpoons H^+ + F^- \nonumber \] Let’s use an ICE table! \[ K_a = \dfrac{[H^+,F^-]}{[HF]} \nonumber \] \[ 10^{-3.17} M = \dfrac{x(0.150 \,M + x)}{0.250 \,M - x} \nonumber \] This expression results in a quadratic relationship, leading to two values of \(x\) that will make it true. Rejecting the negative root, the remaining root of the equation indicates \[[ H^+]= 0.00111\,M \nonumber \] So the pH is given by \[ pH = -log_{10} (0.00111) = 2.95 \nonumber \] For buffers made from acids with sufficiently large values of pK the buffer problem can be simplified since the concentration of the acid and its conjugate base will be determined by their pre-equilibrium values. In these cases, the pH can be calculated using the Henderson-Hasselbalch approximation. If one considers the expression for \(K_a\) \[ K_a = \dfrac{[H^+,A^-]}{[HA]} = [H^+]\dfrac{[H^-]}{[HA]} \nonumber \] Taking the log of both sides and multiplying by -1 yields \[ pK_a= pH - \log_{10} \dfrac{[A^-]}{[HA]} \nonumber \] An rearrangement produces the form of the Henderson-Hasselbalch approxmimation. \[ pH= pK_a - \log_{10} \dfrac{[A^-]}{[HA]} \nonumber \] It should be noted that this approximation will fail if: since the equilibrium concentration will deviate wildly from the pre-equilibrium values under these conditions. | 1,837 | 3,607 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Physical_Properties_of_Alkenes |
Alkenes contains a carbon-carbon double bond. This carbon-carbon double bond changes the physicals properties of alkenes. At room temperatue, alkenes exist in all three phases, solid, liquids, and gases. Melting and boiling points of alkenes are similar to that of alkanes, however, isomers of cis alkenes have lower melting points than that of trans isomers. Alkenes display a weak dipole-dipole interactions due to the electron-attracting sp carbon. Alkenes are a family of hydrocarbons (compounds containing carbon and hydrogen only) containing a carbon-carbon double bond. The first two are: You can work out the formula of any of them using: \(C_nH_{2n}\). The list is limited to the first two, because after that there are isomers which affect the names. All the alkenes with 4 or more carbon atoms in them show structural isomerism. This means that there are two or more different structural formulae that you can draw for each molecular formula. For example, with C4H8, it isn't too difficult to come up with these three structural isomers: There is, however, another isomer. But-2-ene also exhibits geometric isomerism. The carbon-carbon double bond doesn't allow any rotation about it. That means that it is possible to have the CH3 groups on either end of the molecule locked either on one side of the molecule or opposite each other. These are called cis-but-2-ene (where the groups are on the same side) or trans-but-2-ene (where they are on opposite sides). Cis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. For an explanation of the two ways of naming these two compounds. Ethene, Propene, and Butene exists as colorless gases. Members of the 5 or more carbons such as Pentene, Hexene, and Heptene are liquid, and members of the 15 carbons or more are solids. Alkenes are lighter than water and are insoluble in water due to their non-polar characteristics. Alkenes are only soluble in nonpolar solvents. Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes. The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Ethene, propene and the various butenes are gases at room temperature. All the rest that you are likely to come across are liquids. Boiling points of alkenes depends on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. Each alkene has 2 fewer electrons than the alkane with the same number of carbons. Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers. Chemical structure and fuctional groups can affect the polarity of alkenes compounds. The sp carbon is much more electron-withdrawing than the sp3 hybridize orbitals, therefore, creates a weak dipole along the substituent weak alkenly carbon bond. The two individual dipoles together form a net molecular dipole. In trans-subsituted alkenes, the dipole cancel each other out. In cis-subsituted alkenes there is a net dipole, therefore contributing to higher boiling in cis-isomers than trans-isomers. 1. Compound containing carbon-carbon double bonds are called: 2. Cis-alkenes exhibit a weak net dipole-dipole interactions: 3. Which has a lower melting poin Jim Clark ( ) | 3,891 | 3,608 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./06.E%3A_The_Structure_of_Atoms_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact The lamps in street lights use emission of light from excited states of atoms to produce a characteristic glow. Light is generated by electron bombardment of a metal vapor. Of calcium and strontium, which metal vapor would you use to produce yellow light? Which metal would you use to produce red light? Calculate the energy associated with each transition and propose an explanation for the colors of the emitted light. Lasers have useful medical applications because their light is directional (permitting tight focus of the laser beam for precise cutting), monochromatic, and intense. Carbon dioxide lasers, emitting at a wavelength of 1.06 × 10 nm, are typically used in surgery. An excimer (meaning “excited dimer”) laser emits light in the ultraviolet region of the spectrum. An example of such a laser is krypton fluoride (KrF), which emits light at a wavelength of 248 nm. What is the energy in joules of a mole of photons emitted from this laser? How much more energetic is a single photon of this wavelength than a photon from a carbon dioxide laser used in surgery (10,600 nm)? Wavelengths less than 10 nm are needed to “see” objects on an atomic or molecular scale. Such imaging can be accomplished with an electron microscope, which uses electric and magnetic fields to focus and accelerate a beam of electrons to a high velocity. Electron microscopy is now a powerful tool in chemical research. What electron velocity is needed to produce electrons with a wavelength of 4 × 10 nm, which is sufficient to produce an image of an atom? If electromagnetic radiation were used, what region of the electromagnetic spectrum would this correspond to? Microwave ovens operate by emitting microwave radiation, which is primarily absorbed by water molecules in food. The absorbed radiation is converted to heat through rapid oscillations of polar water molecules, which cooks the food and warms beverages. If 7.2 × 10 photons are needed to heat 150.0 g of water from 20.0°C to 100.0°C in a microwave oven, what is the frequency of the microwaves? Metal objects should not be placed in a microwave oven because they cause sparks. Why does this cause sparks? The magnitude of the energy gap between an excited state and a ground state determines the color of visible light that is absorbed. The observed color of an object is not the color of the light it absorbs but rather the complement of that color. The accompanying rosette, first developed by Isaac Newton, shows the colors increasing in energy from red to violet. Any two colors that are opposite each other are said to be (e.g., red and green are complementary). Given the absorption spectra and following table, what are the colors of the objects that produce spectra A, B, and C? Photodegradation of atmospheric ozone occurs via the reaction O + ν → O + O; the maximum absorption occurs at approximately 255 nm. In what region of the electromagnetic spectrum does this occur? Based on this information, what would be the effect of depleting the ozone layer of Earth’s atmosphere? A microscope’s resolution (its ability to distinguish between two points separated by a given distance) depends on the wavelength of light used to illuminate an object. The resolution is given by the equation = λ/2 , where is a constant related to the aperture. If a microscope has an aperture constant of 0.25, what is the smallest distance between two objects that can be resolved using the following light sources? Silver bromide is the photosensitive material in 35 mm photographic film. When monochromatic light falls on film, the photons are recorded if they contain sufficient energy to react with silver bromide in the film. Given that the minimum energy needed to do this is approximately 57.9 kJ/mol, explain why red light is used to light a darkroom. What happens when the door to the darkroom is opened, allowing yellow light to enter? A lighting system has recently been developed that uses a quartz bulb the size of a golf ball filled with an inert gas and a small amount of sulfur. When irradiated by microwaves, the bulb puts out as much light as hundreds of high-intensity mercury vapor lamps. Because 1000 kJ/mol is needed to ionize sulfur, can this process occur simply by irradiating sulfur atoms with microwaves? Explain your answer. The following table lists the ionization energies of some common atmospheric species: An artist used a pigment that has a significant absorption peak at 450 nm, with a trace absorption at 530 nm. Based on the color chart and table in Problem 6, what was the color of the paint? Draw the absorption pattern. What would the absorption spectrum have looked like if the artist had wanted green? Using absorption spectra, explain why an equal combination of red and yellow paints produces orange. You live in a universe where an electron has four different spins ( = +½, +¼, −½, −¼) and the periodic table has only 36 elements. Which elements would be noble gases? What would the periodic table look like? (Assume that the Pauli exclusion principle is still valid.) If you were living on a planet where there were three quantum numbers ( , , ) instead of four, what would be the allowed combinations for an electron in a 3 orbital? How many electrons would this orbital contain assuming the Pauli exclusion principle were still in effect? How does this compare with the actual number of allowed combinations found on Earth? X-rays are frequently used to project images of the human body. Recently, however, a superior technique called magnetic resonance imaging (MRI) has been developed that uses spin to image tissues in spectacular detail. In MRI, spinning hydrogen nuclei in an organic material are irradiated with photons that contain enough energy to flip the protons to the opposite orientation. If 33.121 kJ/mol of energy is needed to flip a proton, what is the resonance frequency required to produce an MRI spectrum? Suggest why this frequency of electromagnetic radiation would be preferred over x-rays. Vanadium has been found to be a key component in a biological catalyst that reduces nitrogen to ammonia. What is the valence electron configuration of vanadium? What are the quantum numbers for each valence electron? How many unpaired electrons does vanadium have? Tellurium, a metal used in semiconductor devices, is also used as a coloring agent in porcelains and enamels. Illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule using tellurium metal. A new element is believed to have been discovered by a team of Russian and American scientists, although its existence is yet to be independently confirmed. Six atoms of element 117, temporarily named ununseptium, were created by smashing together isotopes of calcium with the element berkelium. Give the following: | 6,960 | 3,609 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.11%3A_Vapor-Liquid_Equilibrium |
When a liquid such as water or alcohol is exposed to air in an open container, the liquid evaporates. This happens because the distribution of speeds (and hence kinetic energies) among molecules in a liquid is similar to that illustrated for gases, shown again below. At any given instant a small fraction of the molecules in the liquid phase will be moving quite fast. If one of these is close to the surface and is traveling upward, it can escape the attraction of its fellow molecules entirely and pass into the gas phase. As the higher energy molecules depart, the average energy of the molecules in the liquid decreases and the temperature of the liquid falls. Heat energy will be absorbed from the surroundings, an effect which you can feel if you let water or alcohol evaporate from your skin. Absorption of heat maintains the average molecular speed in the liquid, so that, given enough time, all the liquid can evaporate. The heat absorbed during the entire process corresponds to the enthalpy of vaporization. If the liquid is placed in a closed, rather than an open, container, we no longer find that it evaporates completely. Once a certain partial pressure of gas has been built up by the evaporation of liquid, no more change occurs, and the amount of liquid remains constant. The partial pressure attained in this way is called the vapor pressure of the liquid. It is different for different liquids and increases with temperature for a given liquid. So long as some liquid is present, the vapor pressure is always the same, regardless of the size of the container or the quantity of liquid. For example, we find that any size sample of water held at 25°C will produce a vapor pressure of 23.8 mmHg (3.168 kPa) in any closed container, provided only that all the water does not evaporate. On the macroscopic level, once the vapor pressure has been attained in a closed container, evaporation appears to stop, as seen in the water bottle below. On the microscopic level, though, molecules are still escaping from the liquid surface into the vapor above, as shown in the accompanying figure. The amount of vapor remains the same only because molecules are reentering the liquid just as fast as they are escaping from it. The molecules of the vapor behave like any other gas: they bounce around colliding with each other and the walls of the container. However, one of these “walls” is the surface of the liquid. In most cases a molecule colliding with the liquid surface will enter the body of the liquid, not have enough energy to escape, and be recaptured. When the liquid is first introduced into the container, there are very few molecules of vapor and the rate of recapture will be quite low, but as more and more molecules evaporate, the chances of a recapture will become proportionately larger. Eventually the vapor pressure will be attained, and the rate of recapture will exactly balance the rate of escape. There will then be no net evaporation of liquid or condensation of gas. Once the vapor-liquid system has attained this state, it will appear on the macroscopic level not to be undergoing any change in its properties. The amount, the volume, the pressure, the temperature, the density, etc., of both liquid and gas will all remain constant with time. When this happens to a system, it is said to be in an or to have attained . Later, we will encounter many other quite different examples of equilibrium, but they all have one property in common. The lack of change on the macroscopic level is always the result of two opposing microscopic processes whose rates are equal. The effect of each process is to nu1lify the effect of the other. Since both microscopic processes are still in motion, such a situation is often referred to as The magnitude of the vapor pressure of a liquid depends mainly on two factors: the strength of the forces holding the molecules together and the temperature. It is easy to see that if the intermolecular forces are weak, the vapor pressure will be high. Weak intermolecular forces will permit molecules to escape relatively easily from the liquid. The rate at which molecules escape will thus be high. Quite a large concentration of molecules will have to build up in the gas phase before the rate of reentry can balance the escape rate. Consequently the vapor pressure will be large. By contrast, strong intermolecular forces result in a low escape rate, and only a small concentration of molecules in the vapor is needed to balance it. The vapor pressure of a liquid is quite a sensitive indicator of small differences in intermolecular forces. In the case of the alkanes, for example, we find that the vapor pressure of normal pentane at 25°C is 512.3 mmHg (68.3 kPa) while that of normal hexane is 150.0 mmHg (20.0 kPa) and that of normal heptane only 45.7 mmHg (6.1 kPa), despite the fact that the intermolecular forces for the three substances differ by less than 10 percent. The following video compares the vapor pressures of pentane, hexane, and heptane, by injecting each into a column of mercury. If you look closely, you can see the small amount of each liquid on top of the mercury. This serves to demonstrate the described difference in vapor pressure: stronger intermolecular forces result in lower vapor pressure. The other major factor governing the magnitude of the vapor pressure of a liquid is temperature. At a low temperature only a minute fraction of the molecules have enough energy to escape from the liquid. As the temperature is raised, this fraction increases very rapidly and the vapor pressure increases with it, which makes sense given our previous discussion on . Moreover, the higher the temperature, the more rapid the rate of increase of the energetic fraction of molecules. The result is a variation of vapor pressure with temperature like that shown for four liquids alkanes. Note from this figure how the vapor-pressure increase for a 10°C increase in temperature is larger at higher temperatures. The following video also showcases the effect of heat on vapor pressure. The subject is one of the novelty "drinking bird" devices. These birds operate due to vapor pressure. There are two distinct spaces above the liquid, one in the base of the bird, and one in the head of the bird which is connected to the liquid by a long glass tube. By heating the head of the bird, the temperature of the vapor there increase, thus the vapor pressure increases, and it pushes the liquid back down into the base. When the base is heated, the gas in the base increases in vapor pressure compared to the head, and thus pushes liquid back up the neck of the bird. | 6,663 | 3,610 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Case_Study%3A_Battery_Types |
Ranging from the very crude to the highly sophisticated, batteries come in a plethora of variety. Batteries in short are electrochemical cells that produce a current of electricity via chemical reactions. More specifically, batteries produce electrical energy from oxidation-reduction reactions. A collection of electrochemical cells wired in series is properly called a battery. A flashlight battery is really a single electrochemical cell, while a car battery is really a battery since it is three electrochemical cells in series. About 2000 years later the Voltaic Pile, a stack of individual cells of zinc and copper disks immersed in sulfuric acid, was created by the Italian Count Volta and effectively replaced the use of the Leyden Jar, an instrument that stored static electricity for future use. Volta's battery is considered the first electrochemical cell and the reaction for which is as follows: : \(Zn \rightarrow Zn^{2+} + 2e^-\) : \(2H^+ + 2e^- \rightarrow H_2\) Because zinc is higher in the electrochemical series, the zinc anode reacts with sulfate anions and is oxidized whilst protons are reduced to hydrogen gas. The copper cathode remains unchanged and acts only as electrode for the chemical reaction. Because the Voltaic pile was unafe to use and the cell power diminished over time, it was abandoned. Electrochemical cells typically consist of an anode (the negative electrode where oxidation occurs), a cathode (the positive electrode where reduction occurs), and an electrolyte (the medium conducting anions and cations within a reaction) all contained within a cell. Electrons flow in a closed circuit from the anode to the cathode. Depending on the configuration of the cell and the electrolyte used, a salt bridge may be necessary to conduct ions from one half cell to another as an electric charge is created when electrons move from electrode to another. The difference created would keep electrons from flowing any further. Because a salt bridge permits the flux of ions, a balance in charge is kept between the half cells whilst keeping them separate. The two main categories of batteries are primary and secondary. Essentially, primary cells are batteries which cannot be recharged while secondary cells are rechargeable. The distinction begs the question as to why primary cells are still in use today, and the reason being is that primary cells have lower self-discharge rates meaning that they can be stored for longer periods of time than rechargeable batteries and maintain nearly the same capacity as before. Reserve and backup batteries present a unique example of this advantage of primary cells. In reserve, or stand-by, batteries components of the battery containing active chemicals are separated until the battery is needed, thus greatly decreasing self-discharge. An excellent example is the Water-Activated Battery. As opposed to inert reserve batteries, backup batteries are already activated and functional but not producing any current until the main power supply fails. Devices that generate electric energy via the digestion of carbohydrates, fats, and protiens by enzymes. The most common biobatteries are the lemon or potato battery and the frog or ox-head battery better described as a "muscular pile". In a lemon cell, the energy for the battery is not produced by the lemon but by the metal electrodes. Usually zinc and copper electrodes are inserted into a lemon (the electrolyte being citric acid) and connected by a circuit. The zinc is oxidized in the lemon in order to reach a preferred lower energy level and the electrons discharged provide the energy. Using zinc and copper electrodes, a lemon can produce about 0.9 Volts. While not technically a biobattery, an Earth battery is comprised of two different electrodes which are either buried underground or immersed in natural bodies of water which tap into Telluric currents to produce electric energy. During the 1860s, a French man named George Lelanche developed the Lelanche cell also known today as the dry-cell battery. A dry-cell battery is a battery with a paste electrolyte (as opposed to a wet-cell battery with a liquid electrolyte) in the the middle of its cylinder and attached are metal electrodes. A dry-cell battery is a primary cell that cannot be reused. In order to function, each dry-cell battery has a cathode and an anode. Some examples of dry-cell batteries used in everyday objects today are remote controls, clocks, and calculators. Figure: Modern dry cell batteries. from Types of dry-cell batteries are zinc-carbon batteries, alkaline-cell batteries, and mercury batteries. Before zinc-carbon batteries were used, mercury batteries were the main resource. It was not until mercury was known to become harmful that zinc-carbon batteries replaced it. Batteries may produce the following potential problems or hazards: Dry-cell batteries are the most common battery type used today. Essentially, the battery is comprised of a metal electrode (or graphite rod) surrounded by a moist electrolyte paste that is enclosed in a metal cylinder. 1.5 volts is the most commonly used voltage for dry-cell batteries. The sizes of dry-cell batteries vary, however, it does not change the voltage of the battery. Zinc-carbon cells were the first really portable energy source. These cells have a short lifetime and the zinc casings become porous as the zinc is converted to zinc chloride. The substances in the cell that leak out are corrosive to metal and can terminally destroy electronic equipment or flashlights. Zinc-carbon cells produce 1.5 volts. For a dry-cell battery to operate, oxidation will occur from the zinc anode and reduction will take place in the cathode. The most common type of cathode is a carbon graphite. Once reactants have been turned into products, the dry-cell battery will work to produce electricity. For example, in a dry-cell battery, once \(Zn^{2+}\) has been oxidized to react with \(NH_3\), it will produce chloride salt to insure that too much \(NH_3\) will not block the current of the cathode. \[Zn^{2+}_{(aq)} + 2NH_{3\;(g)} + 2Cl^-_{(aq)} \rightarrow [Zn(NH_3)_2]Cl_{2\; (s)}\] How does the reaction work? While the zinc anode is being oxidized, it is producing electrons that will be captured by reducing Maganese from an oxidation state of +4 to a +3. The electrons produced by Zinc will then connect to the cathode to produce it's product. Recently, the most popular dry-cell battery to be used has been the alkaline-cell battery. In the zinc-carbon battery shown above, the zinc is not easily dissolved in basic solutions. Though it is fairly cheap to construct a zinc-carbon battery, the alkaline-cell battery is favored because it can last much longer. Instead of using \(NH_4Cl\) as an electrolyte, the alkaline-cell battery will use \(NaOH\) or \(KOH\) instead. The reaction will occur the same where zinc is oxidized and it will react with \(OH^-\) instead. \[Zn^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Zn(OH)_{2\; (s)}\] Once the chemicals in the dry-cell battery can no longer react together, the dry-cell battery is dead and be recharged. Alkaline electrochemical cells have a much longer lifetime but the zinc case still becomes porous as the cell is discharged and the substances inside the cell are still corrosive. Alkaline cells produce 1.54 volts. Mercury cells offer a long lifetime in a small size but the mercury produced as the cell discharges is very toxic. This mercury is released into the atmosphere if the cells are incinerated in the trash. About 90% of the 1.4 million pounds of mercury in our garbage comes from mercury cells. Mercury cells only produce 1.3 Volts. \[HgO + Zn + H_2O \rightarrow Hg + Zn(OH)_2\] Mercury batteries utilize either pure mercuric oxide or a mix of mercuric oxide with manganese dioxide as the cathode. The anode is made with zinc and is separated from the cathode with a piece of paper or other porous substance that has been soaked in the electrolyte (which is generally either sodium or potassium oxide). In the past, these batteries were widely used because of their long shelf life of about 10 years, and also because of their stable, steady voltage output. Also, they had the highest capacity per size. They were popular for use in button-type battery applications, such as watches or hearing aids. However, the environmental impact for the amount of mercury present in the batteries became an issue, and the mercury batteries were discontinued from public sale. The lead-acid battery used in cars and trucks consists of six electrochemical cells joined in series. Each cell in a lead-acid battery produces 2 volts. The electrodes are composed of lead and are immersed in sulfuric acid. The negative electrodes are spongy lead metal and the positive electrodes are lead impregnated with lead oxide. As the battery is discharged, metallic lead is oxidized to lead sulfate at the negative electrodes and lead oxide is reduced to lead sulfate at the positive electrodes. When a lead-acid battery is recharged by an alternator, electrons are forced to flow in the opposite direction which reverses the reaction. \[Pb + PbO_2 + 2 H_2SO_4 \rightarrow 2 PbSO_4 + 2 H_2O\] Nickel-cadmium cells can also be regenerated by reversing the flow of the electrons in a battery charger. The cadmium is oxidized in these cells to cadmium hydroxide and the nickel is reduced. Nickel-cadmium cells generate 1.46 Volts. \[Cd + NiO_2 + 2 H_2O \rightarrow Cd(OH)_2 + Ni(OH)_2\] A Nickel-metal Hyride battery is a secondary cell very similar to the nickel-cadmium cell except that it uses a hydrogen-absorbing alloy in place of cadmium. The Nickel-metal Hyride battery has 2-3 times the capacity of a nickel-cadmium cell. | 9,758 | 3,611 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.08%3A_Sigma_and_Pi_Bonds |
A description of the double bond is the model shown in Figure 1. In this case only two of the orbitals on each C atom are involved in the formation of hybrids. Consequently hybrids are formed, separated by an angle of 120°. Two of these hybrids from each C atom overlap with H 1 orbitals, while the third overlaps with an hybrid on the other C atom. This overlap directly between the two C atoms is called a , and is abbreviated by the Greek letter σ. This orbital has no nodes: electron density exists continuously from around one atom to the other atom. The sp2 hybrid orbitals on each carbon atom involve the 2 and two of the 2 orbitals, leaving a single 2 orbital on each carbon atom. A second carbon-carbon bond is formed by the overlap of these two remaining orbitals. This is called a , Greek letter π. The pi bond (π bond) has two halves—one above the plane of the molecule, and the other below it. Each of the two electrons in the pi bond (π bond) exists both above and below the plane of the four H atoms and the two C atoms. The pi bond can be thought of as a standing wave with a single node in the plane of the molecule. If your workstation is enabled for JCE Software, you will see two videos below which compare the behavior of a standing wave with zero nodes versus a standing wave with one node (otherwise, see the drum animation below). The wave with a single node has higher energy. The sigma bond between the two carbon atoms does not have a node in the plane of the molecule. The pi bond between the two carbon atoms has one node in the plane of the molecule. Thus the pi molecular orbital is higher in energy and is the highest occupied molecular orbital (the HOMO). Alternatively, we can envision the molecular orbitals with the Drum Model described earlier. Imagine the two atoms opposite one another where a diagonal meets the edge of the drum at extreme left and right points. The m mode has no nodes, so the maximum amplitude of the standing wave is between the atoms, representing a high electron density sigma bond. The m mode has a linear node between the atoms, and maximum amplitude in front of, and behind, the node, representing the pi bond. In 3D, this linear node would be a plane, separating the two lobes of high electron density that constitute the pi bond. Because the pi bond has less electron density between the atoms, it is of in the MO diagram and is than the sigma bond. Overall this sigma-pi picture of the double bond is reminiscent of a hot dog in a bun. The sigma bond (σ bond) corresponds to the frankfurter, while the pi bond corresponds to the bun on either side of it. Although the sigma-pi picture is more complex than the bent-bond picture of the double bond, it is much used by organic chemists (those chemists interested in carbon compounds). The sigma-pi model is especially helpful in understanding what happens when visible light or other radiation is absorbed by a molecule. Further discussion on this topic is found in the sections on Spectra and Structure of Atoms and Molecules. In actual fact the difference between the two models of the double bond (the first model described here and the second found in the section on Orbital Descriptions of Multiple Bonds) is more apparent than real. They are related to each other in much the same way as and orbitals are related to hybrids. Figure 2 shows two dot-density diagrams for a carbon-carbon double bond in a plane through both carbon nuclei but at right angles to the plane of the molecule. Figure 2 corresponds to a sigma-pi model with the sigma bond (σ bond) in color and the pi bond in gray. Figure 2 shows two bent bonds. Careful inspection reveals that both diagrams are dot-for-dot the same. Only the color coding of the dots is different. Thus the bent-bond and sigma-pi models of the double bond are just two different ways of dividing up the same overall electron density. A similar situation applies to triple bonds, such as that found in a molecule of ethyne (acetylene), . As shown in Figure 3 ,we can regard this triple bond as being the result of three overlaps of hybrids on different carbon atoms forming three bent bonds. Alternatively we can regard it as being composed of one sigma bond and two pi bonds, the sigma bond being due to the overlap of an hybrid from each carbon atom. Again both pictures of the bond correspond to the same overall electron density, and hence both are describing the same physical reality. We can use whichever one seems more convenient for the problem under consideration. | 4,586 | 3,612 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.07%3A_Experimental_methods_of_chemical_kinetics |
Studies of the dynamics of chemical processes impinge on almost every area of chemistry and biochemistry. It is useful for students even at the general chemistry level to have some understanding of the experimental techniques that have informed what you have already learned about kinetics. Kinetics deals with rates of change, so as a preface to this section, it is useful to consider the range of times within which the study of chemical processes takes place. Prior to about 1945, the practical range for kinetic investigations began at about 10 sec and extended out to around 10 sec, this upper limit being governed by the length of time that a graduate student (who usually does the actual work) can expect to get a Ph.D. degree. The development of flash photolysis and flow methods in the mid-1940s moved the lower time limit down to the millisecond range. The most dramatic changes began to occur in the 1970s as the availablity of lasers capable of very short-lived pulses and fast electronics gradually opened up the study of reactions that are completed in nanoseconds, picoseconds, and even femtoseconds (10 sec). The basic way of obtaining the information needed to determine rate constants and reaction orders is to bring the reactants together and then measure successive changes in concentration of one of the components as a function of time. Two important requirements are Measuring the concentration of a reactant or product directly — that is by chemical analysis, is awkward and seldom necessary. When it cannot be avoided, the reaction sample must usually be in some way in order to stop any further change until its composition can be analyzed. This may be accomplished in various ways, depending on the particular reaction. For reactions carried out in solution, especially enzyme-catalyzed ones, it is sometimes practical to add a known quantity of acid or base to change the pH, or to add some other inhibitory agent. More commonly, however, the preferred approach is to observe some physical property whose magnitude is to the extent of the reaction. These methods are applicable to reactions that are not excessively fast, typically requiring a few minutes or hours to run to completion. They were about the only methods available before 1950. Most first-year laboratory courses students will include at least one experiment based on one of these methods. is perhaps the most widely-used technique. If either a reactant or a product is colored, the reaction is easily followed by recording the change in transmission of an appropriate wavelength after the reaction is started. When a beam of light passes through a solution containing a colored substance, the fraction that is absorbed is directly proportional to the concentration of that substance and to the length of the light's path through the solution. The latter can be controlled by employing a cell or cuvette having a fixed path length. If is the intensity of the light incident on the cell and is the intensity that emerges on the other side, then the is just 100 × / . Because a limited 1-100 scale of light absorption is often inadequate to express the many orders of magnitude frequently encountered, a logarithmic term optical density is often employed. The relation between this, the cell path length, concentration, and innate absorption ability of the colored substance is expressed by . The simplest absorbance measuring device is a in which a beam of white light from an incandescent lamp is passed through a cell (often just a test tube) and onto a photodetector whose electrical output is directly proportional to the light intensity. Before beginning the experiment, a sets the meter to zero when the light path is blocked off, and a sets it to 100 when a cell containing an uncolored solution (a "blank") is inserted. The sensitivity and selectivity of such an arrangement is greatly enhanced by adjusting the wavelength of the light to match the absorption spectrum of the substance being measured. Thus if the substance has a yellow color, it is because blue light is being absorbed, so a blue color filter is placed in the light path. More sophisticated employ two cells (one for the sample and another for the blank. They also allow one to select the particular wavelength range that is most strongly absorbed by the substance under investigation, which can often extend into the near-ultraviolet region. measurements (made with a ) can be useful for reactions that lead to the formation a fine precipitate/ A very simple student laboratory experiment of this kind can be carried out by placing a conical flask containing the reaction mixture on top of a marked piece of paper. The effects of changing the temperature or reactant concentrations can be made by observing how long it takes for precipitate formation to obscure the mark on the paper. such a fluorescence and polarimetry (measurement of the degree to which a solution rotates the plane of polarized light) are also employed when applicable. The traditional experimental methods described above all assume that we can follow the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can we do if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity. If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined. Although this method is very simple in principle, it can be rather complicated in practice, as the illustration shows. Owing to the rather large volumes required, his method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred. These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates. The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed. Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the actvity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner. works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution. In order to investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated . Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium. Many reactions are known which do not take place in the absence of light whose wavelength is sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl with H , which procedes explosively when the system is illuminated with visible light. In , a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means. refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner", meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly ; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present. All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species producd by photolysis. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy. In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Flash durations of around 1 millisecond permitted one to follow processes having lifetimes in the microsecond range, but the advent of fast lasers gradually extended this to picoseconds and femtoseconds. Flash photolysis revolutionized the study of organic photochemistry, especially that relating to the chemistry of free radicals and other reactive species that cannot be isolated or stored, but which can easily be produced by photolysis of a suitable precursor. It has proven invaluable for understanding the complicated kinetics relating to atmospheric chemistry and smog formation. More recently, flash photolysis has become an important tool in biochemistry and cellular physiology. Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water \[2 H_2O → H_3O^+ + OH^–\] and the formation of the triiodide ion in aqueous solution \[I^– + I_2 → I_3^–\] Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system. For example, if the reaction A B is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process. For the general case illustrated here, the quantity " " being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a , will vary with time according to = e After the abrupt perturbation at time , the relaxation time is defined as the half-time for the return to equilibrium — that is, as the time required for xo to decrease by Δ /e = Δ /2.718. The derivation of and the relations highlighted in yellow can be found in most standard kinetics textbooks. are probably most commonly used. They can be brought about in various ways: This is the method that Manfred Eigen pioneered when, in the early 1960's, he measured the rate constant of what was then the fastest reaction ever observed: H + OH → H O = 1.3 × 10 M sec A detailed description of the kinetics of this process can be seen here. Eigen shared the 1967 Nobel Prize in Chemistry with Porter and Norrish, who developed flash photolysis. Eigen's Nobel lecture, in keeping with his legendary sense of humor, was titled "immeasurably fast reactions". His later work has focused on self-organizing systems and the origin of life, molecular genetics, and neurology. According to the Le Chatelier principle, a change in the applied pressure will shift the equilibrium state of any reaction which involves a change in the volume of a system. Aside from the obvious examples associated with changes in the number of moles of gases, there are many more subtle cases involving formation of complexes, hydration shells and surface adsorption, and phase changes. One area of considerable interest is the study of protein folding, which has implications in diseases such as Parkinson's and Alzheimer's. The pressure-jump is applied to the cell through a flexible membrane that is activated by a high-pressure gas supply, or through an electrically-actuated piezoelectric crystal. The latter method is employed in the device shown here, which can produce P-jumps of around 1GPa over sub-millisecond time intervals. More on P-jumps: this 1999 review covers many non-biochemical applications. See also this article on protein folding and aggregation. When a change in pressure propagates through a gas at a rate greater than the ordinary compressions and rarefactions associated with the travel of sound, a moving front (a ) of very high pressure forms. This in turn generates an almost instantaneous rise in the temperature that can approach several thousand degrees in magnitude. A shock tube is an apparatus in which shock waves can be generated and used to study the kinetics of gas-phase reactions that are otherwise inaccessible to kinetic measurements. Since all molecules tend to dissociate at high temperatures, shock tubes are widely used to study dissociation processes and the chemistry of the resulting fragments. For example, the shock-induced decomposition of carbon suboxide provides an efficient means of investigating carbon atom reactions: \[C_3O_2 → C + CO\] Shock tube techniques are also useful for studying combustion reactions, including those that proceed explosively. The shock tube itself consists of two sections separated by a breakable diaphragm of metal or plastic. One section is filled with a "driver" gas at a very high pressure, commonly helium, but often mixed with other inert gases to adjust the properties of the shock. The other, longer section of the tube contains the "driven" gas — the reactants — at a low pressure, usually less than 1 atmosphere. The reaction is initiated by causing the diaphragm to rupture, either by means of a mechanical plunger or by raising the pressure beyond its bursting point. The kinetics of the reaction are monitored by means of an absorption or other optical monitoring device that is positioned at a location along the reaction tube that is appropriate to the time course of the reaction. | 16,199 | 3,613 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.10%3A_Heats_of_Fusion_and_Solidification |
Suppose that you are holding an ice cube in your hand. It feels cold because heat energy leaves your hand and enters the ice cube. What happens to the ice cube? It melts. However, the temperature during a phase change remains constant. So the heat that is being lost by your hand does not raise the temperature of the ice above its melting temperature of \(0^\text{o} \text{C}\). Rather, all the heat goes into the change of state. Energy is absorbed during the process of changing ice into water. The water that is produced also remains at \(0^\text{o} \text{C}\) until all of the ice is melted. All solids absorb heat as they melt to become liquids. The gain of heat in this endothermic process goes into changing the state, rather than changing the temperature. The of a substance is the heat absorbed by one mole of that substance as it is converted from a solid to a liquid. Since the melting of any substance absorbs heat, it follows that the freezing of any substance releases heat. The of a substance is the heat released by one mole of that substance as it is converted from a liquid to a solid. Since fusion and solidification of a given substance are the exact opposite processes, the numerical value of the molar heat of fusion is the same as the numerical value of the molar heat of solidification, but opposite in sign. In other words, \(\Delta H_\text{fus} = - \Delta H_\text{solid}\). The figure below shows all of the possible changes of state along with the direction of heat flow during each process. Every substance has a unique value for its molar heat of fusion, depending on the amount of energy required to disrupt the intermolecular forces present in the solid. When \(1 \: \text{mol}\) of ice at \(0^\text{o} \text{C}\) is converted to \(1 \: \text{mol}\) of liquid water at \(0^\text{o} \text{C}\), \(6.01 \: \text{kJ}\) of heat are absorbed from the surroundings. When \(1 \: \text{mol}\) of water at \(0^\text{o} \text{C}\) freezes to ice at \(0^\text{o} \text{C}\), \(6.01 \: \text{kJ}\) of heat is released into the surroundings. \[\begin{array}{ll} \ce{H_2O} \left( s \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{fus} = 6.01 \: \text{kJ/mol} \\ \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( s \right) & \Delta H_\text{solid} = -6.01 \: \text{kJ/mol} \end{array}\nonumber \] The molar heats of fusion and solidification of a given substance can be used to calculate the heat absorbed or released when various amounts are melted or frozen. Calculate the heat absorbed when \(31.6 \: \text{g}\) of ice at \(0^\text{o} \text{C}\) is completely melted. The mass of ice is first converted to moles. This is then multiplied by the conversion factor of \(\left( \frac{6.01 \: \text{kJ}}{1 \: \text{mol}} \right)\) in order to find the \(\text{kJ}\) of heat absorbed. \[31.6 \: \text{g ice} \times \frac{1 \: \text{mol ice}}{18.02 \: \text{g ice}} \times \frac{6.01 \: \text{kJ}}{1 \: \text{mol ice}} = 10.5 \: \text{kJ}\nonumber \] The given quantity is a bit less than 2 moles of ice, and so just less than \(12 \: \text{kJ}\) of heat is absorbed by the melting process. | 3,149 | 3,614 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/03%3A_The_Langmuir_Isotherm/3.01%3A_Introduction |
Whenever a gas is in contact with a solid there will be an equilibrium established between the molecules in the gas phase and the corresponding adsorbed species (molecules or atoms) which are bound to the surface of the solid. As with all chemical equilibria, the position of equilibrium will depend upon a number of factors: In general, factors (2) and (3) exert opposite effects on the concentration of adsorbed species - that is to say that the surface coverage may be increased by raising the gas pressure but will be reduced if the surface temperature is raised. The Langmuir isotherm was developed by Irving Langmuir in 1916 to describe the dependence of the surface coverage of an adsorbed gas on the pressure of the gas above the surface at a fixed temperature. There are many other types of isotherm (Temkin, Freundlich ...) which differ in one or more of the assumptions made in deriving the expression for the surface coverage; in particular, on how they treat the surface coverage dependence of the enthalpy of adsorption. Whilst the Langmuir isotherm is one of the simplest, it still provides a useful insight into the pressure dependence of the extent of surface adsorption. When considering adsorption isotherms it is conventional to adopt a definition of surface coverage (θ) which defines the maximum (saturation) surface coverage of a particular adsorbate on a given surface always to be unity, i.e. \(θ_{max} = 1\). This way of defining the surface coverage differs from that usually adopted in surface science where the more common practice is to equate \(θ\) with the ratio of adsorbate species to surface substrate atoms (which leads to saturation coverages which are almost invariably less than unity). | 1,737 | 3,615 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Physical_Properties_of_Alkanes |
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless. The boiling points shown are for the "straight chain" isomers of which there is more than one. The first four are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers! Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane. There is not a significant difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size. Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules. For example, the boiling points of the three isomers of \(C_5H_{12}\) are: The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"! Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds. When a molecular substance dissolves in water, the following must occur: Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water. As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water.; the alkane does not dissolve. The energy only description of solvation is an oversimplification because entropic effects are also important when things dissolve. In most organic solvents, the primary forces of attraction between the solvent molecules are - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility. Jim Clark ( ) | 3,307 | 3,617 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/20%3A_Organic_Chemistry/20.3%3A_Aldehydes_Ketones_Carboxylic_Acids_and_Esters |
Another class of organic molecules contains a carbon atom connected to an oxygen atom by a double bond, commonly called a carbonyl group. The trigonal planar carbon in the carbonyl group can attach to two other substituents leading to several subfamilies (aldehydes, ketones, carboxylic acids and esters) described in this section. Both and contain a , a functional group with a carbon-oxygen double bond. The names for aldehyde and ketone compounds are derived using similar nomenclature rules as for alkanes and alcohols, and include the class-identifying suffixes and , respectively. In an aldehyde, the carbonyl group is bonded to at least one hydrogen atom. In a ketone, the carbonyl group is bonded to two carbon atoms. As text, an aldehyde group is represented as –CHO; a ketone is represented as –C(O)– or –CO–. In both aldehydes and ketones, the geometry around the carbon atom in the carbonyl group is trigonal planar; the carbon atom exhibits hybridization. Two of the orbitals on the carbon atom in the carbonyl group are used to form σ bonds to the other carbon or hydrogen atoms in a molecule. The remaining hybrid orbital forms a σ bond to the oxygen atom. The unhybridized orbital on the carbon atom in the carbonyl group overlaps a orbital on the oxygen atom to form the π bond in the double bond. Like the \(\mathrm{C=O}\) bond in carbon dioxide, the \(\mathrm{C=O}\) bond of a carbonyl group is polar (recall that oxygen is significantly more electronegative than carbon, and the shared electrons are pulled toward the oxygen atom and away from the carbon atom). Many of the reactions of aldehydes and ketones start with the reaction between a Lewis base and the carbon atom at the positive end of the polar \(\mathrm{C=O}\) bond to yield an unstable intermediate that subsequently undergoes one or more structural rearrangements to form the final product (Figure \(\Page {1}\)). The importance of molecular structure in the reactivity of organic compounds is illustrated by the reactions that produce aldehydes and ketones. We can prepare a carbonyl group by oxidation of an alcohol—for organic molecules, oxidation of a carbon atom is said to occur when a carbon-hydrogen bond is replaced by a carbon-oxygen bond. The reverse reaction—replacing a carbon-oxygen bond by a carbon-hydrogen bond—is a reduction of that carbon atom. Recall that oxygen is generally assigned a –2 oxidation number unless it is elemental or attached to a fluorine. Hydrogen is generally assigned an oxidation number of +1 unless it is attached to a metal. Since carbon does not have a specific rule, its oxidation number is determined algebraically by factoring the atoms it is attached to and the overall charge of the molecule or ion. In general, a carbon atom attached to an oxygen atom will have a more positive oxidation number and a carbon atom attached to a hydrogen atom will have a more negative oxidation number. This should fit nicely with your understanding of the polarity of C–O and C–H bonds. The other reagents and possible products of these reactions are beyond the scope of this chapter, so we will focus only on the changes to the carbon atoms: Methane represents the completely reduced form of an organic molecule that contains one carbon atom. Sequentially replacing each of the carbon-hydrogen bonds with a carbon-oxygen bond would lead to an alcohol, then an aldehyde, then a carboxylic acid (discussed later), and, finally, carbon dioxide: \[\ce{CH4⟶CH3OH⟶CH2O⟶HCO2H⟶CO2} \nonumber \] What are the oxidation numbers for the carbon atoms in the molecules shown here? In this example, we can calculate the oxidation number (review the chapter on oxidation-reduction reactions if necessary) for the carbon atom in each case (note how this would become difficult for larger molecules with additional carbon atoms and hydrogen atoms, which is why organic chemists use the definition dealing with replacing C–H bonds with C–O bonds described). Indicate whether the marked carbon atoms in the three molecules here are oxidized or reduced relative to the marked carbon atom in ethanol: There is no need to calculate oxidation states in this case; instead, just compare the types of atoms bonded to the marked carbon atoms: reduced (bond to oxygen atom replaced by bond to hydrogen atom); oxidized (one bond to hydrogen atom replaced by one bond to oxygen atom); oxidized (2 bonds to hydrogen atoms have been replaced by bonds to an oxygen atom) Aldehydes are commonly prepared by the oxidation of alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Alcohols that have their –OH groups in the middle of the chain are necessary to synthesize a ketone, which requires the carbonyl group to be bonded to two other carbon atoms: An alcohol with its –OH group bonded to a carbon atom that is bonded to no or one other carbon atom will form an aldehyde. An alcohol with its –OH group attached to two other carbon atoms will form a ketone. If three carbons are attached to the carbon bonded to the –OH, the molecule will not have a C–H bond to be replaced, so it will not be susceptible to oxidation. Formaldehyde, an aldehyde with the formula , is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by weight. Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance. Dimethyl ketone, CH COCH , commonly called acetone, is the simplest ketone. It is made commercially by fermenting corn or molasses, or by oxidation of 2-propanol. Acetone is a colorless liquid. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals. The odor of vinegar is caused by the presence of acetic acid, a carboxylic acid, in the vinegar. The odor of ripe bananas and many other fruits is due to the presence of esters, compounds that can be prepared by the reaction of a carboxylic acid with an alcohol. Because esters do not have hydrogen bonds between molecules, they have lower vapor pressures than the alcohols and carboxylic acids from which they are derived (Figure \(\Page {2}\)). Both and contain a carbonyl group with a second oxygen atom bonded to the carbon atom in the carbonyl group by a single bond. In a carboxylic acid, the second oxygen atom also bonds to a hydrogen atom. In an ester, the second oxygen atom bonds to another carbon atom. The names for carboxylic acids and esters include prefixes that denote the lengths of the carbon chains in the molecules and are derived following nomenclature rules similar to those for inorganic acids and salts (see these examples): The functional groups for an acid and for an ester are shown in red in these formulas. The hydrogen atom in the functional group of a carboxylic acid will react with a base to form an ionic salt: Carboxylic acids are weak acids, meaning they are not 100% ionized in water. Generally only about 1% of the molecules of a carboxylic acid dissolved in water are ionized at any given time. The remaining molecules are undissociated in solution. We prepare carboxylic acids by the oxidation of aldehydes or alcohols whose –OH functional group is located on the carbon atom at the end of the chain of carbon atoms in the alcohol: Esters are produced by the reaction of acids with alcohols. For example, the ester ethyl acetate, CH CO CH CH , is formed when acetic acid reacts with ethanol: The simplest carboxylic acid is formic acid, HCO H, known since 1670. Its name comes from the Latin word , which means “ant”; it was first isolated by the distillation of red ants. It is partially responsible for the pain and irritation of ant and wasp stings, and is responsible for a characteristic odor of ants that can be sometimes detected in their nests. Acetic acid, CH CO H, constitutes 3–6% vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present. Yeast cells present in the juice carry out the fermentation reactions. The fermentation reactions change the sugar present in the juice to ethanol, then to acetic acid. Pure acetic acid has a penetrating odor and produces painful burns. It is an excellent solvent for many organic and some inorganic compounds, and it is essential in the production of cellulose acetate, a component of many synthetic fibers such as rayon. The distinctive and attractive odors and flavors of many flowers, perfumes, and ripe fruits are due to the presence of one or more esters (Figure \(\Page {3}\)). Among the most important of the natural esters are fats (such as lard, tallow, and butter) and oils (such as linseed, cottonseed, and olive oils), which are esters of the trihydroxyl alcohol glycerine, C H (OH) , with large carboxylic acids, such as palmitic acid, CH (CH ) CO H, stearic acid, CH (CH ) CO H, and oleic acid, \(\mathrm{CH_3(CH_2)_7CH=CH(CH_2)_7CO_2H}\). Oleic acid is an unsaturated acid; it contains a \(\mathrm{C=C}\) double bond. Palmitic and stearic acids are saturated acids that contain no double or triple bonds. Functional groups related to the carbonyl group include the –CHO group of an aldehyde, the –CO– group of a ketone, the –CO H group of a carboxylic acid, and the –CO R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group. | 10,387 | 3,618 |
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What are some of the things that chemists do? Like most scientists, they observe and measure components of the natural world. Based on these observations they try to place things into useful, appropriate categories and to formulate scientific laws which summarize the results of a great many observations. Indeed, it is a fundamental belief of all science that natural events do not occur in a completely unpredictable fashion. Instead, they obey natural laws. Therefore observations and made on one occasion can be duplicated by the same or another person on another occasion. Communication of such results, another important activity, affords an opportunity for the entire scientific community to test an individual’s work. Eventually a consensus is reached, and there is agreement on a new law. Like other scientists, chemists try to explain their observations and laws by means of theories or models. They constantly make use of , and other very small particles. Using such theories as their guides, chemists synthesize new materials. Well over 3 million compounds are now known, and more than 9000 are in large-scale commercial production. Even a backpacker going “back to nature” takes along synthetic materials such as nylon, aluminum, and aspirin. Chemists also analyze the substances they make and those found in nature. of a substance is the first step in understanding its chemical properties, and of some materials in the natural world is essential in controlling air and water pollution. Another role that chemists play is in studying the processes (chemical reactions) by which one substance can be transformed into another. Will the reactions occur without prodding? If so, how quickly? Is energy given off? Can the reactions be controlled - made to occur only when we want them to? These and many other problems that interest chemists are the subject matter of this online textbook. We will return to each several times and in increasing detail. Do not forget, however, that chemistry is more than just what chemists do. Many persons in other sciences as well as in daily life are constantly doing chemistry, whether they call it by name or not. Indeed, each of us is an intricate combination of chemicals, and everything we do depends on chemical reactions. Although space limitations will prevent us from exploring all but a fraction of the applications of chemistry to other fields, we have included such excursions as often as seemed appropriate. From them we hope that you will be able to learn how to apply chemical facts and principles to the problems you will face in the future. Many of the problems are probably not even known yet, and we could not possibly anticipate them. If you have learned how to think “chemically” or “scientifically,” however, we believe that you will be better prepared to face them. Chemists have a unique way of characterizing the objects of their study. To a chemist, matter containing one kind of atom or molecule is considered a , so water (H O) and aluminum metal (Al) are pure substances. To a chemist, aluminum is an , the simplest kind of pure substance which contains only one kind of atom (a physicist might consider aluminum a very complex "mixture" of things like leptons, protons, quarks). To a chemist, the pure substance water is a (it contains two kinds of atoms bound to one another in just one kind of molecule). An environmentalist might consider water "pure" even if it contains the normal amount of dissolved oxygen and carbon dioxide, but no other "pollutants". To a chemist, water containing oxygen is no longer a pure substance, but a . In pure water, the ratio of hydrogen to oxygen atoms is always 2:1, but as soon as oxygen is dissolved in the water, the ratio is no longer fixed (because oxygen does not form a new molecule by reacting with H O), a sure sign of impurity to a chemist. Because the dissolved oxygen does not react with water to form a new compound, we consider "dissolving" to be a process (one that is in terms of atoms or molecules and bond formation or breaking). Changes like dissolution, evaporation, and condensation are considered "physical changes" because, while they are clearly interesting changes, they might be studied in with no reference to atomic recombinations. If, however, the dissolved oxygen is used by a fish to convert food to carbon dioxide (CO ) and water (H O), a has occurred, because the process is best at the level of atoms, molecules, and bonds between them. The process by which the dissolved oxygen is absorbed by fish might be considered a "biological process" because knowledge of the structure of the fish gills is important in its understanding. Once more, the is more important than the in determining whether a process is considered "chemical", "physical", or "biological". Later in this book, we will use atomic and molecular concepts like non-covalent bond formation to understand dissolution, and then it should probably be considered a chemical process, and it will become clear that no absolute distinction between physical and chemical processes is possible (or necessary). It is interesting that water containing ice cubes is a pure substance to a chemist (only H O molecules are present), but it is . To a chemist, "heterogeneous" does not indicate "impurity", nor does "homogeneous" indicate purity. Oxygen (or salt) dissolved in water is a , because every part of it looks like every other part. Of course, when we say "homogeneous" we usually assume that the substance is not being observed with anything with higher magnification than a crude microscope. | 5,642 | 3,620 |
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The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine , widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it." As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed. Recall that the density of a gas is its mass to volume ratio, \(ρ=\dfrac{m}{V}\). Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, = , provides us with a means of deriving such a mathematical formula to relate the density of a gas to its volume in the proof shown in Example \(\Page {1}\). Use = to derive a formula for the density of gas in g/L \[PV = nRT \nonumber \] Rearrange to get (mol/L): \[\dfrac{n}{v}=\dfrac{P}{RT} \nonumber \] Multiply each side of the equation by the molar mass, ℳ. When moles are multiplied by ℳ in g/mol, g are obtained: \[(ℳ)\left(\dfrac{n}{V}\right)=\left(\dfrac{P}{RT}\right)(ℳ) \nonumber \] \[ℳ/V=ρ=\dfrac{Pℳ}{RT} \nonumber \] A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas? \[ρ=\dfrac{Pℳ}{RT} \nonumber \] \[\mathrm{0.0847\:g/L=760\cancel{torr}×\dfrac{1\cancel{atm}}{760\cancel{torr}}×\dfrac{\mathit{ℳ}}{0.0821\: L\cancel{atm}/mol\: K}×290\: K} \nonumber \] ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H , 2.02 g/mol) We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or pressure. Gas densities are often reported at . Using the Ideal Gas Law and Density of a Gas Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane? Strategy: First solve the empirical formula problem using methods discussed earlier. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula: \[\mathrm{85.7\: g\: C×\dfrac{1\: mol\: C}{12.01\: g\: C}=7.136\: mol\: C\hspace{20px}\dfrac{7.136}{7.136}=1.00\: mol\: C} \nonumber \] \[\mathrm{14.3\: g\: H×\dfrac{1\: mol\: H}{1.01\: g\: H}=14.158\: mol\: H\hspace{20px}\dfrac{14.158}{7.136}=1.98\: mol\: H} \nonumber \] Empirical formula is CH [empirical mass (EM) of 14.03 g/empirical unit]. Next, use the density equation related to the ideal gas law to determine the molar mass: \[d=\dfrac{Pℳ}{RT}\hspace{20px}\mathrm{\dfrac{1.56\: g}{1.00\: L}=0.984\: atm×\dfrac{ℳ}{0.0821\: L\: atm/mol\: K}×323\: K} \nonumber \] ℳ = 42.0 g/mol, \(\dfrac{ℳ}{Eℳ}=\dfrac{42.0}{14.03}=2.99\), so (3)(CH ) = C H (molecular formula) Acetylene, a fuel used welding torches, is comprised of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene? Empirical formula, CH; Molecular formula, C H Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, , to its amount in moles, : \[ℳ=\mathrm{\dfrac{grams\: of\: substance}{moles\: of\: substance}}=\dfrac{m}{n} \nonumber \] The ideal gas equation can be rearranged to isolate n: \[n=\dfrac{PV}{RT} \nonumber \] and then combined with the molar mass equation to yield: \[ℳ=\dfrac{mRT}{PV} \nonumber \] This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass. The approximate molar mass of a volatile liquid can be determined by: Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform? Since \[ℳ=\dfrac{m}{n} \nonumber \] and \[n=\dfrac{PV}{RT} \nonumber \] substituting and rearranging gives \[ℳ=\dfrac{mRT }{PV} \nonumber \] then \[ℳ=\dfrac{mRT}{PV}=\mathrm{\dfrac{(0.494\: g)×0.08206\: L⋅atm/mol\: K×372.8\: K}{0.976\: atm×0.129\: L}=120\:g/mol} \nonumber \] A sample of phosphorus that weighs 3.243 × 10 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor? 124 g/mol P Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container ( Figure \(\Page {2}\)). The pressure exerted by each individual gas in a mixture is called its . This observation is summarized by : : \[P_{Total}=P_A+P_B+P_C+...=\sum_iP_i \nonumber \] In the equation is the total pressure of a mixture of gases, is the partial pressure of gas A; is the partial pressure of gas B; is the partial pressure of gas C; and so on. The partial pressure of gas A is related to the total pressure of the gas mixture via its , a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components: \[P_A=X_A×P_{Total}\hspace{20px}\ce{where}\hspace{20px}X_A=\dfrac{n_A}{n_{Total}} \nonumber \] where , , and are the partial pressure, mole fraction, and number of moles of gas A, respectively, and is the number of moles of all components in the mixture. A 10.0-L vessel contains 2.50 × 10 mol of H , 1.00 × 10 mol of He, and 3.00 × 10 mol of Ne at 35 °C. The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using \(P=\dfrac{nRT}{V}\): \[P_\mathrm{H_2}=\mathrm{\dfrac{(2.50×10^{−3}\:mol)(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=6.32×10^{−3}\:atm} \nonumber \] \[P_\ce{He}=\mathrm{\dfrac{(1.00×10^{−3}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=2.53×10^{−3}\:atm} \nonumber \] \[P_\ce{Ne}=\mathrm{\dfrac{(3.00×10^{−4}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=7.58×10^{−4}\:atm} \nonumber \] The total pressure is given by the sum of the partial pressures: \[P_\ce{T}=P_\mathrm{H_2}+P_\ce{He}+P_\ce{Ne}=\mathrm{(0.00632+0.00253+0.00076)\:atm=9.61×10^{−3}\:atm} \nonumber \] A 5.73-L flask at 25 °C contains 0.0388 mol of N , 0.147 mol of CO, and 0.0803 mol of H . What is the total pressure in the flask in atmospheres? 1.137 atm Here is another example of this concept, but dealing with mole fraction calculations. A gas mixture used for anesthesia contains 2.83 mol oxygen, O , and 8.41 mol nitrous oxide, N O. The total pressure of the mixture is 192 kPa. The mole fraction is given by \[X_A=\dfrac{n_A}{n_{Total}} \nonumber \] and the partial pressure is \[P_A = X_A \times P_{Total} \nonumber \] For O , \[X_{O_2}=\dfrac{n_{O_2}}{n_{Total}}=\mathrm{\dfrac{2.83 mol}{(2.83+8.41)\:mol}=0.252} \nonumber \] and \[P_{O_2}=X_{O_2}×P_{Total}=\mathrm{0.252×192\: kPa=48.4\: kPa} \nonumber \] For N O, \[X_{N_2O}=\dfrac{n_{N_2O}}{n_{Total}}=\mathrm{\dfrac{8.41\: mol}{(2.83+8.41)\:mol}=0.748} \nonumber \] and \[P_{N_2O}=X_{N_2O}×P_{Total}=\mathrm{(0.748)×192\: kPa = 143.6 \: kPa} \nonumber \] What is the pressure of a mixture of 0.200 g of H , 1.00 g of N , and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C? 1.87 atm A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure \(\Page {3}\)), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer. However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The , which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure \(\Page {4}\)); more detailed information on the temperature dependence of water vapor can be found in Table \(\Page {1}\), and vapor pressure will be discussed in more detail in the next chapter on liquids. If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure \(\Page {3}\), what is the partial pressure of argon? According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water: \[P_\ce{T}=P_\ce{Ar}+P_\mathrm{H_2O} \nonumber \] Rearranging this equation to solve for the pressure of argon gives: \[P_\ce{Ar}=P_\ce{T}−P_\mathrm{H_2O} \nonumber \] The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so: \[P_\ce{Ar}=\mathrm{750\:torr−25.2\:torr=725\:torr} \nonumber \] A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure? 0.583 L Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions. We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure. Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure. We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to \[\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber \] a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant. The explanation for this is illustrated in Figure \(\Page {4}\). According to Avogadro’s law, equal volumes of gaseous N , H , and NH , at the same temperature and pressure, contain the same number of molecules. Because one molecule of N reacts with three molecules of H to produce two molecules of NH , the volume of H required is three times the volume of N , and the volume of NH produced is two times the volume of N . Propane, C H ( ), is used in gas grills to provide the heat for cooking. What volume of O ( ) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion. The ratio of the volumes of C H and O will be equal to the ratio of their coefficients in the balanced equation for the reaction: \[\begin{align}
&\ce{C3H8}(g)+\ce{5O2}(g) ⟶ &&\ce{3CO2}(g)+\ce{4H2O}(l)\\
\ce{&1\: volume + 5\: volumes &&3\: volumes + 4\: volumes}
\end{align} \nonumber \] From the equation, we see that one volume of C H will react with five volumes of O : \[\mathrm{2.7\cancel{L\:C_3H_8}×\dfrac{5\: L\:\ce{O2}}{1\cancel{L\:C_3H_8}}=13.5\: L\:\ce{O2}} \nonumber \] A volume of 13.5 L of O will be required to react with 2.7 L of C H . An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C H , at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 10 L of O at 0 °C and 1 atm, will be required to burn the acetylene? \[\ce{2C2H2 + 5O2⟶4CO2 + 2H2O} \nonumber \] 3.34 tanks (2.34 × 10 L) Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H ( ), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N ? \[\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \nonumber \] Because equal volumes of H and NH contain equal numbers of molecules and each three molecules of H that react produce two molecules of NH , the ratio of the volumes of H and NH will be equal to 3:2. Two volumes of NH , in this case in units of billion ft , will be formed from three volumes of H : \[\mathrm{683\cancel{billion\:ft^3\:NH_3}×\dfrac{3\: billion\:ft^3\:H_2}{2\cancel{billion\:ft^3\:NH_3}}=1.02×10^3\:billion\:ft^3\:H_2} \nonumber \] The manufacture of 683 billion ft of NH required 1020 billion ft of H . (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.) What volume of O ( ) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C H ( ), measured under the same conditions of temperature and pressure? The products are CO and water vapor. 51.0 L What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid? \[\ce{2Ga}(s)+\ce{6HCl}(aq)⟶\ce{2GaCl3}(aq)+\ce{3H2}(g) \nonumber \] To convert from the mass of gallium to the volume of H ( ), we need to do something like this: The first two conversions are: \[\mathrm{8.88\cancel{g\: Ga}×\dfrac{1\cancel{mol\: Ga}}{69.723\cancel{g\: Ga}}×\dfrac{3\: mol\:H_2}{2\cancel{mol\: Ga}}=0.191\:mol\: H_2} \nonumber \] Finally, we can use the ideal gas law: \[V_\mathrm{H_2}=\left(\dfrac{nRT}{P}\right)_\mathrm{H_2}=\mathrm{\dfrac{0.191\cancel{mol}×0.08206\: L\cancel{atm\:mol^{−1}\:K^{−1}}×300\: K}{0.951\:atm}=4.94\: L} \nonumber \] Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen? 1.30 × 10 L The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost \(\dfrac{1}{3}\) is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this radiation is absorbed by certain substances in the atmosphere, known as greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14 °C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure \(\Page {6}\)). There is strong evidence from multiple sources that higher atmospheric levels of CO are caused by human activity, with fossil fuel burning accounting for about \(\dfrac{3}{4}\) of the recent increase in CO . Reliable data from ice cores reveals that CO concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure \(\Page {7}\)). The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products. | 19,066 | 3,622 |
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Previously, we saw that it is the of the entropy changes of the system and surroundings that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to focus our attention on the system rather than the surroundings, and would like to avoid having to calculate the entropy change of the surroundings explicitly. In this unit we introduce a new thermodynamic function, the , which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. However, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as "Gibbs energy." The free energy enables us to do this for changes that occur at a constant temperature and pressure (the ) or constant temperature and volume (the .) The Gibbs energy (also known as the or ) is defined as \[G = H – T S \label{23.4.1}\] in which \(S\) refers to the entropy of the . Since \(H\), \(T\) and \(S\) are all state functions, so is \(G\). Thus for any change in state (under constant temperature), we can write the extremely important relation \[ΔG = ΔH – T ΔS \label{23.4.2}\] How does this simple equation encompass the entropy change of the world \(ΔS_{total}\), which we already know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition \[ΔS_{total} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}\] we would first like to get rid of \(ΔS_{surr}\). How can a chemical reaction (a change in the ) affect the entropy of the ? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat across the system boundary. The enthalpy change of the reaction \(ΔH\) is defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be \(–q_p\) which will cause the entropy of the surroundings to change by \(–q_p / T = –ΔH/T\). We can therefore rewrite Equation \(\ref{23.4.3}\) as \[ΔS_{total} = \dfrac{- ΔH}{T} + ΔS_{sys} \label{23.4.4}\] Multiplying each side by \(-T\), we obtain \[-TΔS_{total} = ΔH - TΔS_{sys} \label{23.4.5}\] which expresses the entropy change of the world in terms of thermodynamic properties of the exclusively. If \(-TΔS_{total}\) is denoted by \(ΔG\), then we have Equation \(\ref{23.4.2}\) which defines the change for the process. From the foregoing, you should convince yourself that \(G\) will decrease in any process occurring at constant temperature and pressure which is accompanied by an overall increase in the entropy. The constant temperature is a consequence of the temperature and the enthalpy appearing in the preceding Equation \(\ref{23.4.5}\). Since most chemical and phase changes of interest to chemists take place under such conditions, the Gibbs energy is the most useful of all the thermodynamic properties of a substance, and (as we shall see in the lesson that follows this one) it is closely linked to the equilibrium constant. Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether \(ΔH\) or the \(TΔS\) term dominates. This is technically correct, but misleading because it disguises the important fact that \(ΔS_{total}\), which this equation expresses in an indirect way, is the criterion of spontaneous change. We will deal only with the Gibbs energy in this course. The is of interest mainly to chemical engineers (whose industrial-scale processes are often confined to tanks and reactors of fixed volume) and some geochemists whose interest is centered on the chemistry that occurs deep within the earth's surface. Remember that \(ΔG\) is meaningful only for changes in which the . These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction. \(ΔG\) serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Moreover, it determines the direction and extent of chemical change. In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the \(TΔS\) term). \[\ce{H_2O(l) \rightarrow H2O(s)} \label{23.5.6}\] water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H O diminishes, and the process proceeds spontaneously. In a spontaneous change, Gibbs energy decreases and never increases. An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows: This might seem strange, given the key importance \(ΔG\) in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate \(ΔG\). As we will show in the lesson that follows this one, it is far more convenient to work with the equilibrium constant of a reaction, within which \(ΔG\) is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of \(ΔG\) depends on the of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with \(ΔH\). Recalling the condition for spontaneous change \[ΔG = ΔH – TΔS < 0\] it is apparent that the temperature dependence of Δ depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of \(ΔH\) and \(ΔS\) are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation For any given reaction, the sign of \(ΔH\) can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process: Both enthalpic \(\Delta H\) and entropic \(-T\Delta S\) terms will be negative, so \(ΔG\) will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures. If the reaction is sufficiently exothermic it can force \(ΔG\) negative only at temperatures below which \(|TΔS| < |ΔH|\). This means that there is a temperature \(T = ΔH / ΔS\) at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition. This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that \(TΔS > ΔH\). Since the effect of the temperature is to "magnify" the influence of a positive \(ΔS\), the process will be spontaneous at temperatures above \(T = ΔH / ΔS\). (Think of melting and boiling.) With both \(ΔH\) and \(ΔS\) working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form. The plots above are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of Δ and Δ . The other two plots on each diagram are only for the chemistry-committed. You have already been introduced to the terms such as \(ΔU^o\) and \(ΔH^o\) in which the \(^o\) sign indicates that all components (reactants and products) are in their . This concept of standard states is especially important in the case of the free energy, so let's take a few moments to review it. More exact definitions of the conventional standard states can be found in most physical chemistry textbooks. In specialized fields such as biochemistry and oceanography, alternative definitions may apply. For example, the "standard pH" of zero (corresponding to \([H^{+}] = 1\,M\)) is impractical in biochemistry, so pH = 7 is commonly employed. For most practical purposes, the following definitions are good enough: To make use of Gibbs energies to predict chemical changes, we need to know the free energies of the individual components of the reaction. For this purpose we can combine the standard enthalpy of formation and the standard entropy of a substance to get its \[ΔG_f^o = ΔH_f^o – TΔS_f^o \label{23.4.7}\] Recall that the symbol ° refers to the of a substance measured under the conditions of 1 atm pressure or an effective concentration of 1 mol L and a temperature of 298 K. Then determine the standard Gibbs energy of the reaction according to \[ ΔG^o = \sum ΔG_f^o \;(\text{products})– \sum ΔG_f^o \;(\text{reactants}) \label{24.4.8}\] As with standard heats of formation, the standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 K. Standard Gibbs free energies of formation are normally found directly from tables. Once the values for all the reactants and products are known, the standard Gibbs energy change for the reaction is found by Equation \(\ref{23.4.7}\). Most tables of thermodynamic values list \(ΔG_f^o\) values for common substances (e.g., ), which can, of course, always be found from values of and . Find the standard Gibbs energy change for the reaction \[\ce{CaCO3(s) \rightarrow CaO (s) + CO2(g)} \nonumber\] The \(ΔG_f^o°\) values for the three components of this reaction system are \(\ce{CaCO3(s)}\): –1128 kJ mol , CaO : –603.5 kJ mol , CO : –137.2 kJ mol . Substituting into Equation \(\ref{23.4.7}\), we have \[ΔG^o = (–603.5 –137.2) – (–1128) kJ\, mol^{–1} = +130.9\, kJ\, mol^{–1} \nonumber \] This indicates that the process is not spontaneous under standard conditions (i.e., solid calcium carbone will not form solid calcium oxide and CO at 1 atm partial pressure at 25° C). This reaction is carried out on a huge scale to manufacture cement, so it is obvious that the process can be spontaneous under different conditions. The practical importance of the Gibbs energy is that it allows us to make predictions based on the properties (Δ values) of the reactants and products themselves, eliminating the need to experiment. But bear in mind that while thermodynamics always correctly predicts whether a given process take place (is spontaneous in the thermodynamic sense), it is unable to tell us if it take place at an observable rate. When thermodynamics says "no", it means exactly that. When it says "yes", it means "maybe". The reaction \[\ce{ 1/2 O2(g) + H2(g) → H2O(l)} \nonumber\] is used in fuel cells to produce an electrical current. The reaction can also be carried out by direct combustion. : molar entropies in J mol K : O (g) 205.0; H (g)130.6; H O(l) 70.0; H O(l) Δ ° = –285.9 kJ mol . Use this information to find First, we need to find \(ΔH^o\) and \(ΔS^o\) for the process. Recalling that the standard enthalpy of formation of the elements is zero, \[\begin{align*} ΔH^o &= ΔH^p_f(\text{products}) – ΔH^°_f(\text{reactants}) \\[4pt] &= –285.9\, kJ\, mol^{–1} – 0 \\[4pt] &= –285.9 \,kJ \,mol^{–1} \end{align*}.\] Similarly, \[\begin{align*} ΔS^o &= S^o_f(\text{products}) – S^o_f(\text{reactants}) \\[4pt] &= (70.0) – (½ \times 205.0 + 130.6) \\[4pt] &= –163\, J\, K^{–1}mol^{–1} \end{align*}\] The foregoing example illustrates an important advantage of fuel cells. Although direct combustion of a mole of hydrogen gas yields more energy than is produced by the same net reaction within the fuel cell, the latter, in the form of electrical energy, can be utilized at nearly 100-percent energy efficiency by a motor or some other electrical device. If the thermal energy released by direct combustion were supplied to a heat engine, second-law considerations would require that at least half of this energy be "wasted" to the surroundings. Δ refer to in which all components (reactants and products) are in their . The \(ΔG_f^o\) of a substance, like \(ΔH_f^o\), refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. Both of these terms are by definition zero for the elements in their standard states. There are only a few common cases in which this might create some ambiguity: Ions in aqueous solution are a special case; their standard free energies are relative to the hydrated hydrogen ion \(\ce{H^{+}(aq)}\) which is assigned \(ΔG_f^o = 0\). \(ΔG\) is very different from ΔG°. The distinction is nicely illustrated in Figure \(\Page {5}\) in which Δ is plotted on a vertical axis for two hypothetical reactions having opposite signs of Δ . The horizontal axis schematically expresses the relative concentrations of reactants and products at any point of the process. Note that the origin corresponds to the composition at which half of the reactants have been converted into products. Take careful note of the following: The important principle you should understand from this is that a negative Δ does not mean that the reactants will be completely transformed into products. By the same token, a positive Δ does not mean that no products are formed at all. It should now be clear from the discussion above that a given reaction carried out under standard conditions is characterized by a of . The reason for the Gibbs energy minimum at equilibrium relates to the increase in entropy when products and reactants coexist in the same phase. As seen in the plot, even a minute amount of "contamination" of products by reactants reduces the free energy below that of the pure products. In contrast, composition of a chemical reaction system undergoes continual change until the equilibrium state is reached. So the a single reaction can have an infinite number of Δ values, reflecting the infinite possible compositions between the extremes of pure reactants (zero extent of reaction) and pure products (unity extent of reaction). In the example of a reaction A → B, depicted in the above diagram, the standard free energy of the products is smaller than that of the reactants , so the reaction will take place spontaneously. T . For reactions in which products and reactants occupy a single phase (gas or solution), the meaning of "spontaneous" is that the equilibrium composition will correspond to an extent of reaction greater than 0.5 but smaller than unity.Note, however, that for values in excess of about ±50 kJ mol , the equilibrium composition will be negligibly different from zero or unity extent-of-reaction. The physical meaning of Δ is that it tells us how far the free energy of the system has changed from ° of the pure reactants . As the reaction proceeds to the right, the composition changes, and Δ begins to fall. When the composition reaches , Δ reaches its minimum value and further reaction would cause it to rise. But because free energy can only decrease but never increase, this does not happen. The composition of the system remains permanently at its equilibrium value. A . extent-of-reaction diagram for a non-spontaneous reaction can be interpreted in a similar way; the equilibrium composition will correspond to an extent of reaction greater than zero but less than 0.5. In this case, the minimum at reflects the increase in entropy when the reactants are "contaminated" by a small quantity of products. If all this detail about Δ seems a bit overwhelming, do not worry: it all gets hidden in the equilibrium constant and reaction quotient that we discuss in the next lesson! Although it is \(ΔG\) \(ΔG^o\) that serves as a criterion for spontaneous change at constant temperature and pressure, \(ΔG^o\) values are so readily available that they are often used to get a rough idea of whether a given chemical change is possible. This is practical to do in some cases, but not in others: It generally works for reactions such as \[\ce{4 NH_3(g) + 5 O_2(g) → 4 NO(g) + 6 H_2O(g)} \nonumber\] with \(ΔG^o = –1,010\, kJ\). (industrially important for the manufacture of nitric acid) because \(ΔG^o\) is so negative that the reaction will be spontaneous and virtually complete under just about any reasonable set of conditions. The following reaction expresses the fact that the water molecule is thermodynamically stable: \[\ce{2 H_2(g) + 1/2 O_2(g)→ H_2O(l)} \nonumber\] with \(ΔG^o = –237.2 \,kJ\). Note that this refers to water (the standard state of H O at 25°). If you think about it, a negative standard Gibbs energy of formation (of which this is an example) can in fact be considered a definition of molecular stability. Similarly, dissociation of dihydrogen into its atoms is highly unlikely under standard conditions: \[\ce{H_2O(g) → 2 H(g) + O(g)} \nonumber\] with \(ΔG^o = +406.6\, kJ\). Again, an analogous situation would apply to any stable molecule. Now consider the dissociation of dinitrogen tetroxide \[\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\] with \(ΔG^o = +2.8 kJ\). in which the positive value of tells us that N O at 1 atm pressure will not change into two moles of NO at the same pressure, but owing to the small absolute value of , we can expect the spontaneity of the process to be quite sensitive to both the temperature (as shown in the table below) and to the pressure in exactly the way the Le Chatelier principle predicts. For reactions involving dissolved ions, one has to be quite careful. Thus for the dissociation of the weak hydrofluoric acid \[\ce{HF(aq) → H^+(aq) + F^–(aq)} \nonumber\] with \(ΔG^o = –317 \,kJ\). it is clear that a 1 mol/L solution of HF will not dissociate into 1M ions, but this fact is not very useful because if the HF is added to water, the initial concentration of the fluoride ion will be zero (and that of H very close to zero), and the Le Chatelier principle again predicts that dissociation will be spontaneous. It is common knowledge that dissociation of water into hydrogen- and hydroxyl ions occurs only very sparingly: \[\ce{H_2O(l) → H^+(aq) + OH^–(aq) } \nonumber\] with \(ΔG^o = 79.9 \,kJ\). which correctly predicts that the water will not form 1M (effective concentration) of the ions, but this is hardly news if you already know that the product of these ion concentrations can never exceed 10 at 298K. Finally, consider this most familiar of all phase change processes, the vaporization of liquid water: \[\ce{H_2O(l) → H_2O(g)} \nonumber\] with \(ΔG^o = 8.58 \,kJ \). Conversion of liquid water to its vapor at 1 atm partial pressure does not take place at 25° C, at which temperature the equilibrium partial pressure of the vapor (the "vapor pressure") is only 0.031 atm (23.8 torr.) Gaseous H O at a pressure of 1 atm can only exist at 100° C. Of course, water left in an open container at room temperature will spontaneously evaporate if the partial pressure of water vapor in the air is less than 0.031 atm, corresponding to a relative humidity of under 100% A reaction is in its equilibrium state when \[ΔG = ΔH – TΔS = 0 \label{23.4.1a}\] The temperature at which this occurs is given by \[T = \dfrac{ΔH}{TΔS} \label{23.4.1b}\] If we approximate \(ΔH\) by \(ΔH^o\) and \(ΔS\) by \(ΔS^o\), so Equation \ref{23.4.1a} would be \[ΔG \approx ΔH^o – TΔS^o = 0 \label{23.4.1aa}\] We can then estimate the normal boiling point of a liquid. From the following thermodynamic data for water: Because ΔH° values are normally expressed in kilojoules while ΔS° is given in joules, a very common student error is to overlook the need to express both in the same units. We find that liquid water is in equilibrium with water vapor at a partial pressure of 1 atm when the temperature is \[T = \dfrac{44,100\, J}{118.7\, J\, K^{–1}} = 371.5\, K\] But " ", you say? Very true. The reason we are off here is that both Δ ° and Δ ° have their own temperature dependencies; we are using the "standard" 25° values without correcting them to 100° C. Nevertheless, if you think about it, the fact that we can estimate the boiling point of a liquid from a table of thermodynamic data should be rather impressive! Of course, the farther one gets from 298 K, the more unreliable will be the result. Thus for the dissociation of dihydrogen into its atoms, All one can say here is that H will break down at something over 3000 K or so. (You may already know that molecules will dissociate into their atoms at high temperatures.) We tend to think of high temperatures as somehow "forcing" molecules to dissociate into their atoms, but this is wrong. In order to get the H–H bond to vibrate so violently through purely thermal excitation that the atoms would fly apart, a temperature more like 30,000 K would be required. The proper interpretation is at the temperature corresponding to Δ sufficient to overcome the H-H bond strength. The \(T\Delta S\) term interacts with the \(ΔH\) term in \(\Delta G\) to determine whether the reaction can take place at a given temperature. This can be more clearly understood by examining plots of \(TΔS^o\) and \(ΔH^o\) as functions of the temperature for some actual reactions. Of course these parameters refer to standard states that generally do not correspond to the temperatures, pressures, or concentrations that might be of interest in an actual case. Nevertheless, these quantities are easily found and they can usefully predict the way that temperature affects these systems. \[\ce{C(graphite) + O_2(g) → CO_2(g)} \nonumber \] This , like most such reactions, is . The positive entropy change is due mainly to the greater mass of \(\ce{CO2}\) molecules compared to those of \(\ce{O2}\). \[\ce{3 H_2 + N_2 → 2 NH_3(g) } \nonumber\] The decrease in moles of gas in the drives the entropy change negative, making the reaction . Thus higher T, which speeds up the reaction, also reduces its extent. \[ \ce{N_2O_4(g) → 2 NO_2(g)} \nonumber\] are typically endothermic with positive entropy change, and are therefore . \[\ce{ 1/2 N_2 (g) + O_2 (g)→ NO_2(g)} \nonumber\] This reaction is , meaning that . But because the reverse reaction is kinetically inhibited, NO can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable. The appellation “free energy” for \(G\) has led to so much confusion that many scientists now refer to it simply as the . The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: \(ΔG\) is the maximum amount of energy, which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible. A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although \(G\) has the of energy (joules, or in its intensive form, J mol ), it lacks one of the most important attributes of energy in that it is not . Thus, although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating in energy anywhere else. Referring to \(G\) as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the of energy. he increase in the entropy. The quotient \(–ΔG/T\) is in fact identical with \(ΔS_{total}\), the entropy change of the world, whose increase is the primary criterion for an \(G\) differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas \(H\) and \(S\) can be related to the quantity and distribution of energy in a collection of molecules. The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier. | 25,464 | 3,623 |
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Compounds in which two or more benzene rings are fused together were described in an , and they present interesting insights into aromaticity and reactivity. The smallest such hydrocarbon is naphthalene. Naphthalene is stabilized by resonance. Three canonical resonance contributors may be drawn, and are displayed in the following diagram. The two structures on the left have one discrete benzene ring each, but may also be viewed as 10-pi-electron annulenes having a bridging single bond. The structure on the right has two benzene rings which share a common double bond. From heats of hydrogenation or combustion, the resonance energy of naphthalene is calculated to be 61 kcal/mole, 11 kcal/mole less than that of two benzene rings (2 * 36). As expected from an average of the three resonance contributors, the carbon-carbon bonds in naphthalene show variation in length, suggesting some localization of the double bonds. The C –C bond is 1.36 Å long, whereas the C –C bond length is 1.42 Å. This contrasts with the structure of benzene, in which all the C–C bonds have a common length, 1.39 Å. Naphthalene is more reactive than benzene, both in substitution and addition reactions, and these reactions tend to proceed in a manner that maintains one intact benzene ring. The following diagram shows three oxidation and reduction reactions that illustrate this feature. In the last example, catalytic hydrogenation of one ring takes place under milder conditions than those required for complete saturation (the decalin product exists as cis/trans isomers). Electrophilic substitution reactions take place more rapidly at C , although the C product is more stable and predominates at equilibrium. Examples of these reactions will be displayed by clicking on the diagram. The kinetically favored C orientation reflects a preference for generating a cationic intermediate that maintains one intact benzene ring. By clicking on the diagram a second time, the two naphthenonium intermediates created by attack at C and C will be displayed. The structure and chemistry of more highly fused benzene ring compounds, such as nthracene and phenanthreneshow many of the same characteristics described above. ), | 2,226 | 3,624 |
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Proteins which occur in nature differ from each other primarily because their side chains are different. Partly this is a matter of composition. In wool, for example, 11 percent of the side chains are cysteine, while no cysteine occurs in silk at all. To a much larger extent though, the differences between different proteins is a matter of the sequence in which the different side chains occur. This is especially true of globular proteins like enzymes. The sequence of side chains along the backbone of peptide bonds in a polypeptide is said to constitute its . The 20 different amino acids permit construction of a tremendous variety of primary structures. Consider, for example, how many tripeptides similar to that shown in Eq. 1 on the page on polypeptide chains can be constructed from the 20 amino acids. In the example shown, the first amino acid in the chain is glycine, but it might just as well be proline or any other of the 20 amino acids. Thus there are 20 possibilities for the first place in the chain. Similarly there are 20 possibilities for the second place in the chain, making a total of 20 × 20 = 400 possible combinations. For each of these 400 structures we can again choose from among 20 amino acids for the third place in the chain, giving a grand total of 400 × 20 = 20 = 8000 possible structures for the tripeptide. A general formula for the number of primary structures for a polypeptide containing amino acid units is 20 —a very large number indeed when you consider that most proteins contain at least 50 amino acid residues. [20 = (2 × 10) = 2 × 10 = 10 × 10 = 10 ] Primary structure is conventionally specified by writing the three-letter abbreviations for each amino acid, starting with the —NH end of the polymer. In some cases, this is even shortened to the 1-letter abbreviation sequence. For example, the structure which reading from the -NH end is alanine, glycine, glycine would be specified as or Note that Ala-Gly-Gly is not the same as Gly-Gly-Ala. In the latter case glycine rather than alanine has the free —NH group. Because its ends are different, there is a directional character in the polypeptide chain. Both three letter and single letter abbreviations are shown in Figure \(\Page {1}\) : Determination of the primary structure of a protein is a difficult and complicated problem. It also is a rather important one—the sequence of amino acids governs the three-dimensional shape and ultimately the biological function of the protein. Consequently much effort has gone into methods by which primary structure can be elucidated. Insulin was the first protein whose amino acid sequence was determined. This pioneering work, completed in 1953 after some 10 years of effort, earned a Nobel Prize for British biochemist Frederick Sanger (born 1918). He found the primary structure to be
Note how there are two chains in this structure, one with 21 side chains and the other with 30. These two chains are linked in two places with disulfide (—S—S—) bridges, each connecting two cysteine residues in different chains. In order to determine the primary structure of a protein, a known mass of pure sample is first boiled in acid or base until it is completely hydrolyzed to individual amino acids. The amino acid mixture is then separated chromatographically and the exact amount of each amino acid determined. In this way, one can find that for every 3 mol serine in the insulin molecule, there are 6 mol leucine. The next step is to break down the protein into smaller fragments. Disulfide bridges are broken by oxidation after which the protein is selectively hydrolyzed by enzymes, called proteases, such as trypsin or chymotrypsin. In a favorable case one will then have several fragments each containing 10 or 20 amino acid residues. These can then be separated and analyzed individually. Using Edman degradation, so named for Pehr Edman, the sequence of amino acids in one of these polypeptide fragments is usually determined using phenylisothiocyanate, , which selectively attacks the —NH end of the polypeptide chain. This reaction is carried out under basic conditions. Addition of acid then splits off the terminal amino acid, and it can be identified. Since the rest of the polypeptide chain is left intact, this process can be repeated, and each amino acid in the sequence can be attacked, removed, and identified. By snipping off amino acids one at a time, one eventually finds the complete sequence for the fragment. This whole process can be automated and hence sped up considerably. Once the fragments have been sequenced, it becomes a matter of ordering them correctly. Since different proteases hydrolyze peptide bonds at different places in the amino acid sequence, different fragmentation patterns can be used to determine the sequence for the whole protein. The end of a fragment from a trypsin digest will be in the middle of a fragment from a chymotrypsin digest, for instance. This provides a relatively quick means to sequence unknown protein sequences. Edman degradation is not the only method for which proteins are sequenced nowadays. Mass spectrometry can sequence polypeptides of 20 to 30 amino acids in length, using a technique called tandem mass spectrometry. In this method, a polypeptide is sent through one mass spectrometer, which ionizes the polypeptide. The charged peptide then enters a collision chamber, causing the peptide to fragment at different peptide bonds. The resulting fragments are then measured by a second mass spectrometer. The resultant spectrum can determine the peptide sequence by differences in mass of the fragments. With the advent of DNA sequencing methods and the success of the Human Genome Project, many protein sequences are now determined indirectly, through the genetic code. When the DNA sequence for a protein is known, this can be used to determine the protein sequence. By the same token, a known protein sequence can be used to determine the gene coding for that protein. Thus there are many different ways to determine protein sequences, all of which compliment each other. As methods for determining primary structure have become more advanced, a great many proteins have been sequenced, and some interesting comparisons can be made. A particularly intriguing example is that of cytochrome c, an electron carrier which is found in all organisms that use oxygen for respiration. When samples of cytochrome c from different organisms are compared, it is found that the amino acid sequence is usually different in each case. Moreover, the more widely separated two species are in their macroscopic features, the greater the degree of difference in their protein sequences. When horse cytochrome c is compared with that of yeast, 45 out of 104 residues are different. Only two substitutions are found between chicken and duck, and cytochrome c is identical in the pig, cow, and sheep. The magnitudes of these changes coincide quite well with biological taxonomy based on macroscopically observable differences. Cytochrome c can be used to trace biological evolution from unicellular organisms to today’s diverse species, and even to estimate times at which branching occurred in the family tree of life. This makes molecular methods a powerful tool for the evolutionary biologist as well. | 7,339 | 3,625 |
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Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time. The progress of a simple reaction (A → B) is shown in the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in . The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time. \[\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1} \] Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation \(\ref{Eq1}\) is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first. Reaction rates generally decrease with time as reactant concentrations decrease. A Discussing Average Reaction Rates. Link: We can use Equation \(\ref{Eq1}\) to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure \(\Page {2}\). Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table \(\Page {1}\) and are shown in the graph in . The data in Table \(\Page {1}\) were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). The for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t ) and at the end of the interval (t ). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows: The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: If the reaction rate is calculated during the last interval given in Table \(\Page {1}\)(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h): In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide: \[\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2} \] The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in , the volume of CO gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed: \[\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3} \] The concentration of the reactant—in this case sucrose— with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration with time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction ( ) corresponds to sucrose, so the reaction rate is generally defined as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4} \] Consider the thermal decomposition of gaseous N O to NO and O via the following equation: Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. balanced chemical equation reaction rate expressions Because O has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O and write that expression. The balanced chemical equation shows that 2 mol of N O must decompose for each 1 mol of O produced and that 4 mol of NO are produced for every 1 mol of O produced. The molar ratios of O to N O and to NO are thus 1:2 and 1:4, respectively. This means that the rate of change of [N O ] and [NO ] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO is produced at four times the rate of O , the rate of production of NO is divided by 4. The reaction rate expressions are as follows: \(\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}\) The is used in the manufacture of sulfuric acid. A key step in this process is the reaction of \(SO_2\) with \(O_2\) to produce \(SO_3\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. The of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time. The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time. Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0. Using the reaction shown in Example \(\Page {1}\), calculate the reaction rate from the following data taken at 56°C: \[2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber \] balanced chemical equation and concentrations at specific times reaction rate Calculate the reaction rate in the interval between t = 240 s and t = 600 s. From Example \(\Page {1}\), the reaction rate can be evaluated using any of three expressions: Subtracting the initial concentration from the final concentration of N O and inserting the corresponding time interval into the rate expression for N O , Substituting actual values into the expression, Similarly, NO can be used to calculate the reaction rate: Allowing for experimental error, this is the same rate obtained using the data for N O . The data for O can also be used: Again, this is the same value obtained from the N O and NO data. Thus, the reaction rate does not depend on which reactant or product is used to measure it. Using the data in the following table, calculate the reaction rate of \(SO_2(g)\) with \(O_2(g)\) to give \(SO_3(g)\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time. | 10,775 | 3,626 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.19%3A_Common_Reducing_Agents |
A good reducing agent must be able to donate electrons readily, meaning it must not have a high electronegativity. Among the elements, low electronegativity is characteristic of good reducing agents. Molecules and ions which contain relatively electropositive elements which have low oxidation numbers are also good reducing agents. Bear these general rules in mind as we examine examples of common reducing agents in the following paragraphs. All metals have low and are relatively electropositive, and so they lose electrons fairly easily. Therefore, most metals are good reducing agents. on the left of the periodic table exhibit this property to the greatest extent, and some of them, such as Li or Na, can even reduce H O: \[\ce{2Li(s) + 2H2O(l) -> 2Li^+(aq) + 2OH^{–}(aq) + H2(g)} \nonumber \] Skip to the 1 minute mark to get directly to the reaction. Other metals, such as Fe or Zn, cannot reduce H O but can reduce hydronium ions, and so they dissolve in acid solution: | 994 | 3,627 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.09%3A_Strong_Acids_and_Bases |
The most straight-forward examples involving and deal with acids and bases. , like HCl or HNO , are such good proton donors that none of their own molecules can remain in aqueous solution. All HCl molecules, for example, transfer their protons to H O molecules, and so the solution contains only H O ( ) and Cl ( ) ions. Similarly, the ions of , like BaO or NaH, are such good proton acceptors that they cannot remain in aqueous solution. All O ions, for example, are converted to OH ions by accepting protons from H O molecules, and the H O molecules are also converted to OH . Therefore a solution of BaO contains only Ba ( ) and OH ( ) ions. Table \(\Page {1}\) lists molecules and ions which act as strong acids and bases in aqueous solution. In addition to those which react completely with H O to form H O and OH , any compound which itself contains these ions will serve as a strong acid or base. Note that the strength of an acid refers only to its ability to donate protons to H O molecules and the strength of a base to its ability to accept protons from H O molecules. The acidity or basicity of a solution, on the other hand, depends on the concentration as well as the strength of the dissolved acid or base. H O (Only a few compounds like H OCl and H ONO are known to contain hydronium ions.) OH [Only LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) , Sr(OH) and Ba(OH) are sufficiently soluble to produce large concentrations of OH (aq).] HCl, HBr, HI HClO O (Li O, Na O, K O, Rb O, Cs O, CaO, SrO, and BaO are soluble Notice that the cations of the strong bases are also soluble in water, as seen in the table from . For example, all the Group 1 cations form strong bases and all Group 1 cations are soluble in water. Therefore, you can use your prior understanding of solubility rules to guide your understanding of strong bases. As a general rules, if a cation is soluble in water, it can form a strong base. As a general rule, strong proton donors are molecules in which a hydrogen is attached to a rather electronegative atom, such as oxygen or a halogen. Considerable electron density is shifted away from hydrogen in such a molecule, making it possible for hydrogen ions to depart without taking along any electrons. The strong acids in Table \(\Page {1}\) fit this rule nicely. They are either hydrogen halides (HCl, HBr, HI) or oxyacids (whose general formula is H XO ). Below are the resonance structures for oxoacids after they donate a proton. The reason for the strength of the following acids is the stability of the anion, which is shown by the number of resonance structures and the distribution of the negative charge amongst all of the oxygen atoms. The distribution of the negative charge can be visualized in the 3D structure, with red being representing negative charge and blue representing positive charge. The 3D structure represents the average of the resonance structures shown to the left. Perchloric Acid Nitric Acid The Lewis structures indicate a proton bonded to oxygen in each of the oxyacids, hence their general name. Note that for a strong oxyacid the number of oxygens is always larger by two or more than the number of hydrogens. That is, in the general formula H XO , ≥ + 2. The strength of a base depends on its ability to attract and hold a proton. Therefore bases often have negative charges, and they invariably have at least one lone pair of electrons which can form a coordinate covalent bond to a proton. The strong bases in Table 1 might be thought of as being derived from neutral molecules by successive removal of protons. For example, OH can be obtained by removing H from H O, and O can be obtained by removing H from OH . When the strong bases are considered this way, it is not surprising that they are good proton acceptors. | 3,828 | 3,628 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.15%3A_Maximum_Useful_Work |
The Gibbs free energy has another very useful property. When a spontaneous chemical reaction occurs, ,–Δ , , . Symbolically, \[-\Delta G = w_{max} \nonumber \] For a reaction which is not spontaneous Δ is positive and is negative. This means that work must be done on the system (through some outside intervention) to force the nonspontaneous reaction to occur. The minimum work that must be done is given by Δ . As an example of the utility of this interpretation of Δ , consider the recovery of Al from Al O ore: \[\ce{Al2O3→2Al + \frac{3}{2}O2}\qquad \Delta G^{\circ}_{m}(298 \text{ K}) = 1576.4 \frac{\text{ kJ}}{\text{ mol}} \nonumber \] The positive Δ tells us that at least 1576.4 kJ of work must be done on 1 mol Al O to effect this change. In a modern aluminum manufacturing plant this work is supplied electrically, and the electricity is often provided by burning coal. Assuming coal to be mainly carbon, we can write \[\ce{C(s) + O2(g)→CO2(g)}\qquad \Delta G^{\circ}_{m}(298 \text{ K}) = -394.4\frac{\text{ kJ}}{\text{ mol}} \nonumber \] Thus 1 mol C can do almost exactly one-quarter the work required to decompose 1 mol Al O and we must burn at least 4 mol C to process each 1 mol Al O ore. (In practice the aluminum smelting process is only 17 percent efficient, so it is necessary to burn nearly 6 times the theoretical 4 mol C.) In the context we have just described, energy is energy that is , not energy that we can get for nothing. When a spontaneous process occurs and there is a free energy decrease, it is the which decreases. According to the first law of thermodynamics, energy be consumed in any process, but according to the second law, free (or available) energy is consumed in a spontaneous process. When we talk about consuming energy resources by burning fossil fuels, it is the availability of energy that is used up. The energy originally stored in a fuel is converted to heat energy and dispersed to the surroundings. Once this has happened its usefulness is lost. There is no way of abstracting this energy from the surroundings and using it to lift a weight or do other useful work, because that would correspond to the reversal of a spontaneous process. The second law thus adds a very important qualification to the first law. While the first law tells us that we cannot destroy energy, the second law tells as that we cannot recycle it either. | 2,417 | 3,629 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.5%3A_Importance_of_Nomenclature |
The primary function of chemical nomenclature is to ensure that a spoken or written chemical name leaves no ambiguity concerning which chemical compound the name refers to: each chemical name should refer to a single substance. A less important aim is to ensure that each substance has a single name (although a limited number of alternative names is acceptable in some cases). The form of nomenclature used depends on the audience to which it is addressed. As such, no single form exists, but rather there are different forms that are more or less appropriate in different circumstances. A common name will often suffice to identify a chemical compound in a particular set of circumstances. However, in a few specific circumstances (such as the construction of large indices), it becomes necessary to ensure that each compound has a unique name. In discussing chemistry nomenclatures is it necessary to identify the type of compound including stoichiometry and type of constituent atoms. The first separation of importance is to distinguish between and compounds. Unfortunately, this separation is not always clear. even to experienced chemists. Historically (two centuries ago) the definitions were that inorganic compounds were synthesized by geological systems and organic compounds were found in biological systems. The modern definitions argues that organic compounds are any molecule containing carbon and by default this means that inorganic chemistry deals with molecules lacking carbon. The confusion arise when chemists that study carbon based materials like diamond or graphite, but that argument is beyond the scope of this text. The following sections presents the current nomenclature rules for inorganic and organic molecules. | 1,764 | 3,630 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/13%3A_Properties_of_Solutions/13.06%3A_Colloids |
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table \(\Page {1}\). Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols, which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible. Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure \(\Page {1}\). Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic. Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid. In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure \(\Page {2}\)). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin. Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure \(\Page {3}\). Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels. Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”: Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate \(\ce{[CH_3(CH_2)_{16}CO_2−Na^{+}]}\), and detergents, such as sodium dodecyl sulfate \(\ce{[CH_3(CH_2)_{11}OSO_3−Na^{+}]}\), whose structures are as follows: When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise. A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats. Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles, which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (Figure \(\Page {4a}\)). A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers, extended sheets consisting of a double layer of molecules. As shown in Figure \(\Page {4b}\), the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution. A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks. A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves. | 11,122 | 3,632 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.02%3A_Reaction_Rates_Typically_Change_with_Time |
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic. On this page we extend the concept of differential rate laws introduced on the previous page to integral rate laws and reaction that are of great importance in most practical applications of kinetics. Measuring instantaneous rates as we have described on the previous page of this unit is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision. The ordinary rate law (more precisely known as the or rate law) tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants ( of ) change with time. For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products. This is the purpose of an . This is easy to do, but only some courses expect you to know how to do it. For a quick run-through, . If you have had even a bit of calculus, here is an opportunity to put it to use! How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case, A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing. If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion. If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%. The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially imporant in industrial processes in which the balances of these costs affect the profitability of the operation. Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the (or half-time) of the reaction. The rate at which a reactant is consumed in a first-order process is proportional to its concentration at that time. This general relationship, , is said to follow an . Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function so efficiently describes such changes stems from the remarkable property that = ; that is, is its own derivative, so the rate of change of is identical to its value at any point. A nice discussion of this property can be found here . The integrated rate law for a first-order reaction A → products is a common example of the law of exponential change. For a reactant A, its concentration [A] at time is given by [A] = [A] × e in which [A] is its initial concentration and is the first-order rate constant. The "e" in the exponential term is of course the base of the natural logarithms, and the negative sign in its exponent means that the value of this term diminishes as increases, as we would expect for any kind of a decay process. A more convenient form of the integrated rate law is obtained by taking the natural logarithm of both sides: ln [A] = – + ln [A] This has the form of an equation for a straight line = + in which the slope corresponds to the rate constant . After a period of one half-life, = and we can write (in which we express the exponential as a function in order to make it stand out more prominently.) Taking logarithms of both sides (remember that
ln e = ) yields Solving for the half-life, we obtain the simple relation which tells us that . This means that 100,000 molecules of a reactant will be reduced to 50,000 in the same time interval needed for ten molcules to be reduced to five. It should be clear that the rate constant and the half life of a first-order process are inversely related. The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds? From the above equation, = –0.693/(600 s) = 0.00115 s The decay of radioactive nuclei is always a first-order process. The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years. a) What fraction of the Am in a smoke detector will have decayed after 50 years? How long will it take for the activity to decline to 80% of its initial value? What would be the "seventh-life" of Am ? Integration of the second-order rate law yields which is easily rearranged into a form of the equation for a straight line (try showing this yourself!) and yields plots similar to the one shown on the left below. The half-life is given by (see for details) Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to its constancy for a first-order reaction. For this reason, the concept of half-life for a second-order reaction is far less useful. In some reactions, the rate is apparently independent of the reactant concentraton, in which case rate = [A] = Note the word "apparently" in the preceding sentence; For this reason, reactions that follow zero-order kinetics are often referred to as reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left. There are two general conditions that can give rise to zero-order rates: This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface ( ) or to an enzyme. For example, the decomposition of nitrous oxide in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional kinetics when carried out entirely in the gas phase. In this case, the N O molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site. Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an . If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order. This is most often seen when two or more reactants are involved. Thus if the reaction A + B → products is first-order in both reactants so that rate = [A,B] then if B is present in great excess, the reaction will appear to be zero order in B (and first order overall). This commonly happens when B is H O and the reaction is carried out in aqueous solution. The following table compares the rate parameters of zero-, first-, and second-order reactions of the form . See also from Purdue University that shows plots of [A], ln [A], and 1/[A] for each reaction order. | 8,281 | 3,633 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Chemical_Properties_of_Alkanes |
Alkanes are when compared with other chemical species. This is because the backbone carbon atoms in alkanes have attained their octet of electrons through forming four covalent bonds (the maximum allowed number of bonds under the octet rule; which is why carbon's valence number is 4). These four bonds formed by carbon in alkanes are sigma bonds, which are more stable than other types of bond because of the greater overlap of carbon's atomic orbitals with neighboring atoms' atomic orbitals. To make alkanes react, the input of additional energy is needed; either through heat or radiation. Gasoline is a mixture of the alkanes and unlike many chemicals, can be stored for long periods and transported without problem. It is only when ignited that it has enough energy to continue reacting. This property makes it difficult for alkanes to be converted into other types of organic molecules. (There are only a few ways to do this). Alkanes are also , as one can observe, oil, an alkane, floats on water. Alkanes are . Since only C and H atoms are present, alkanes are nonpolar. Alkanes are immiscible in water but freely miscible in other non-polar solvents. Alkanes consisting of weak dipole dipole bonds can not break the strong hydrogen bond between water molecules hence it is not miscible in water. The same character is also shown by alkenes. Because alkanes contain only carbon and hydrogen, combustion produces compounds that contain only carbon, hydrogen, and/or oxygen. Like other hydrocarbons, combustion under most circumstances produces mainly carbon dioxide and water. However, alkanes require more heat to combust and do not release as much heat when they combust as other classes of hydrocarbons. Therefore, combustion of alkanes produces higher concentrations of organic compounds containing oxygen, such as aldehydes and ketones, when combusting at the same temperature as other hydrocarbons. The general formula for alkanes is C H ; the simplest possible alkane is therefore methane, CH . The next simplest is ethane, C H ; the series continues indefinitely. Each carbon atom in an alkane has sp³ hybridization. Alkanes are also known as paraffins, or collectively as the paraffin series. These terms are also used for alkanes whose carbon atoms form a single, unbranched chain. Branched-chain alkanes are called isoparaffins. through are very flammable gases at standard temperature and pressure (STP). is an extremely flammable liquid boiling at 36 °C and boiling points and melting points steadily increase from there; octadecane is the first alkane which is solid at room temperature. Longer alkanes are waxy solids; candle wax generally has between C and C chains. As chain length increases ultimately we reach polyethylene, which consists of carbon chains of indefinite length, which is generally a hard white solid. Alkanes react only very poorly with ionic or other polar substances. The pKa values of all alkanes are above 50, and so they are practically inert to acids and bases. This inertness is the source of the term paraffins (Latin para + affinis, with the meaning here of "lacking affinity"). In crude oil the alkane molecules have remained chemically unchanged for millions of years. However redox reactions of alkanes, in particular with oxygen and the halogens, are possible as the carbon atoms are in a strongly reduced condition; in the case of methane, the lowest possible oxidation state for carbon (−4) is reached. Reaction with oxygen leads to combustion without any smoke; with halogens, substitution. In addition, alkanes have been shown to interact with, and bind to, certain transition metal complexes. Free radicals, molecules with unpaired electrons, play a large role in most reactions of alkanes, such as cracking and reformation where long-chain alkanes are converted into shorter-chain alkanes and straight-chain alkanes into branched-chain isomers. In highly branched alkanes and cycloalkanes, the bond angles may differ significantly from the optimal value (109.5°) in order to allow the different groups sufficient space. This causes a tension in the molecule, known as steric hinderance, and can substantially increase the reactivity. The same is preferred for alkenes too. | 4,263 | 3,634 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Polymers/Hydrogen_Bromide_and_Aklenes%3A_The_Peroxide_Effect |
This page gives you the facts and simple uncluttered mechanisms for the free radical addition of hydrogen bromide to alkenes - often known as the "peroxide effect". If you want the mechanisms explained to you in more detail, there is a link at the bottom of the page. A symmetrical alkene is one like ethene where the groups at both ends of the carbon-carbon double bond are the same. The reaction happens at room temperature in the presence of organic peroxides or some oxygen from the air. Alkenes react very slowly with oxygen to produce traces of organic peroxides - so the two possible conditions are equivalent to each other. The reaction is a simple addition of the hydrogen bromide. For example, with ethene: With a symmetrical alkene you get exactly the same product in the absence of the organic peroxides or oxygen - but the mechanism is different. Hydrogen halides (hydrogen chloride, hydrogen bromide and the rest) usually react with alkenes using an electrophilic addition mechanism. However, in the presence of organic peroxides, hydrogen bromide adds by a different mechanism. If you are interested, you will find the for the addition of hydrogen bromide and other hydrogen halides to alkenes if you follow this link. You may need to explore several pages in this section. With the organic peroxides present you get a The chain is initiated by free radicals produced by an oxygen-oxygen bond in the organic peroxide breaking. These free radicals extract a hydrogen atom from a hydrogen bromide molecule to produce bromine radicals. A bromine radical joins to the ethene using one of the electrons in the pi bond. That creates a new radical with the single electron on the other carbon atom. That radical reacts with another HBr molecule to produce bromoethane and another bromine radical to continue the process. etc Eventually two free radicals hit each other and produce a molecule of some sort. The process stops here because no new free radicals are formed. An unsymmetrical alkene is one like propene where the groups at either end of the carbon-carbon double bond are different.The reaction happens under the same conditions as with a symmetrical alkene, but there is a complication because the hydrogen and the bromine can add in two different ways. Which way they add depends on whether there are organic peroxides (or oxygen) present or not. Normally, when a molecule HX adds to a carbon-carbon double bond, the hydrogen becomes attached to the carbon with the more hydrogens on already. This is known as . Because the HBr adds on the "wrong way around " in the presence of organic peroxides, this is often known as the or . In the absence of peroxides, hydrogen bromide adds to propene via an electrophilic addition mechanism. That gives the product predicted by Markovnikov's Rule. This is exactly the same as in the ethene case above. When the bromine radical joins to the propene, it attaches so that a secondary radical is formed. This is more stable (and so easier to form) than the primary radical which would be formed if it attached to the other carbon atom. That radical reacts with another HBr molecule to produce 1-bromopropane and another bromine radical to continue the process. etc Eventually two free radicals hit each other and produce a molecule of some sort. The process stops here because no new free radicals are formed. The reason that hydrogen bromide adds in an anti-Markovnikov fashion in the presence of organic peroxides is simply a question of reaction rates. The free radical mechanism is much faster than the alternative electrophilic addition mechanism. Both mechanisms happen, but most of the product is the one from the free radical mechanism because that is working faster. With the other hydrogen halides, the opposite is true. This is due to the relatively high hydrogen-chlorine bond strength. In the case of hydrogen bromide, both steps of the propagation stage are exothermic. You will find questions on this on the Help! page (see below). Jim Clark ( ) | 4,042 | 3,635 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.02%3A_AcidBase_Titrations |
Before 1800, most used H SO , HCl, or HNO as acidic titrants, and K CO or Na CO as basic titrants. A titration’s end point was determined using litmus as an indicator, which is red in acidic solutions and blue in basic solutions, or by the cessation of CO effervescence when neutralizing \(\text{CO}_3^{2-}\). Early examples of acid–base titrimetry include determining the acidity or alkalinity of solutions, and determining the purity of carbonates and alkaline earth oxides. The determination of acidity and alkalinity continue to be important applications of acid–base titrimetry. We will take a closer look at these applications later in this section. Three limitations slowed the development of acid–base titrimetry: the lack of a strong base titrant for the analysis of weak acids, the lack of suitable indicators, and the absence of a theory of acid–base reactivity. The introduction, in 1846, of NaOH as a strong base titrant extended acid–base titrimetry to the determination of weak acids. The synthesis of organic dyes provided many new indicators. Phenolphthalein, for example, was first synthesized by Bayer in 1871 and used as an indicator for acid–base titrations in 1877. Despite the increased availability of indicators, the absence of a theory of acid–base reactivity made it difficult to select an indicator. The development of equilibrium theory in the late 19th century led to significant improvements in the theoretical understanding of acid–base chemistry, and, in turn, of acid–base titrimetry. Sørenson’s establishment of the pH scale in 1909 provided a rigorous means to compare indicators. The determination of acid–base dissociation constants made it possible to calculate a theoretical titration curve, as outlined by Bjerrum in 1914. For the first time analytical chemists had a rational method for selecting an indicator, making acid–base titrimetry a useful alternative to gravimetry. In the overview to this chapter we noted that a titration’s end point should coincide with its equivalence point. To understand the relationship between an acid–base titration’s end point and its equivalence point we must know how the titrand’s pH changes during a titration. In this section we will learn how to calculate a titration curve using the equilibrium calculations from . We also will learn how to sketch a good approximation of any acid–base titration curve using a limited number of simple calculations. For our first titration curve, let’s consider the titration of 50.0 mL of 0.100 M HCl using a titrant of 0.200 M NaOH. When a strong base and a strong acid react the only reaction of importance is \[\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \label{9.1}\] Although we have not written reaction \ref{9.1} as an equilibrium reaction, it is at equilibrium; however, because its equilibrium constant is large—it is ( ) or \(1.00 \times 10^{14}\)—we can treat reaction \ref{9.1} as though it goes to completion. The first task is to calculate the volume of NaOH needed to reach the equivalence point, . At the equivalence point we know from reaction \ref{9.1} that \[\begin{aligned} \text { moles } \mathrm{HCl}=& \text { moles } \mathrm{NaOH} \\ M_{a} \times V_{a} &=M_{b} \times V_{b} \end{aligned} \nonumber\] where the subscript ‘ ’ indicates the acid, HCl, and the subscript ‘ ’ indicates the base, NaOH. The volume of NaOH needed to reach the equivalence point is \[V_{e q}=V_{b}=\frac{M_{a} V_{a}}{M_{b}}=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{(0.200 \ \mathrm{M})}=25.0 \ \mathrm{mL} \nonumber\] Before the equivalence point, HCl is present in excess and the pH is determined by the concentration of unreacted HCl. At the start of the titration the solution is 0.100 M in HCl, which, because HCl is a strong acid, means the pH is \[\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log \left[\text{HCl} \right] = -\log (0.100)=1.00 \nonumber\] After adding 10.0 mL of NaOH the concentration of excess HCl is \[[\text{HCl}] = \frac {(\text{mol HCl})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] \[[\mathrm{HCl}]=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})-(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0500 \ \mathrm{M} \nonumber\] and the pH increases to 1.30. At the equivalence point the moles of HCl and the moles of NaOH are equal. Since neither the acid nor the base is in excess, the pH is determined by the dissociation of water. \[\begin{array}{c}{K_{w}=1.00 \times 10^{-14}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.00 \times 10^{-7}}\end{array} \nonumber\] Thus, the pH at the equivalence point is 7.00. For volumes of NaOH greater than the equivalence point, the pH is determined by the concentration of excess OH . For example, after adding 30.0 mL of titrant the concentration of OH is \[[\text{OH}^-] = \frac {(\text{mol NaOH})_\text{added} - (\text{mol HCl})_\text{initial}} {\text{total volume}} = \frac {M_b V_b - M_a V_a} {V_a + V_b} \nonumber\] \[\left[\mathrm{OH}^{-}\right]=\frac{(0.200 \ \mathrm{M})(30.0 \ \mathrm{mL})-(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{30.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0125 \ \mathrm{M} \nonumber\] To find the concentration of H O we use the expression \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0125}=8.00 \times 10^{-13} \ \mathrm{M} \nonumber\] to find that the pH is 12.10. Table 9.2.1
and Figure 9.2.1
show additional results for this titration curve. You can use this same approach to calculate the titration curve for the titration of a strong base with a strong acid, except the strong base is in excess before the equivalence point and the strong acid is in excess after the equivalence point. Construct a titration curve for the titration of 25.0 mL of 0.125 M NaOH with 0.0625 M HCl. The volume of HCl needed to reach the equivalence point is \[V_{e q}=V_{a}=\frac{M_{b} V_{b}}{M_{a}}=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{(0.0625 \ \mathrm{M})}=50.0 \ \mathrm{mL} \nonumber\] Before the equivalence point, NaOH is present in excess and the pH is determined by the concentration of unreacted OH . For example, after adding 10.0 mL of HCl \[\begin{array}{c}{\left[\mathrm{OH}^{-}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})-(0.0625 \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0714 \ \mathrm{M}} \\ {\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0714 \ \mathrm{M}}=1.40 \times 10^{-13} \ \mathrm{M}}\end{array} \nonumber\] the pH is 12.85. For the titration of a strong base with a strong acid the pH at the equivalence point is 7.00. For volumes of HCl greater than the equivalence point, the pH is determined by the concentration of excess HCl. For example, after adding 70.0 mL of titrant the concentration of HCl is \[[\mathrm{HCl}]=\frac{(0.0625 \ \mathrm{M})(70.0 \ \mathrm{mL})-(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{70.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0132 \ \mathrm{M} \nonumber\] giving a pH of 1.88. Some additional results are shown here. For this example, let’s consider the titration of 50.0 mL of 0.100 M acetic acid, CH COOH, with 0.200 M NaOH. Again, we start by calculating the volume of NaOH needed to reach the equivalence point; thus \[\operatorname{mol} \ \mathrm{CH}_{3} \mathrm{COOH}=\mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[M_{a} \times V_{a}=M_{b} \times V_{b} \nonumber\] \[V_{e q}=V_{b}=\frac{M_{a} V_{a}}{M_{b}}=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{(0.200 \ \mathrm{M})}=25.0 \ \mathrm{mL} \nonumber\] Before we begin the titration the pH is that for a solution of 0.100 M acetic acid. Because acetic acid is a weak acid, we calculate the pH using the method outlined in \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] \[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^-\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=\frac{(x)(x)}{0.100-x}=1.75 \times 10^{-5} \nonumber\] finding that the pH is 2.88. Adding NaOH converts a portion of the acetic acid to its conjugate base, CH COO . \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \label{9.2}\] Because the equilibrium constant for reaction \ref{9.2} is quite large \[K=K_{\mathrm{a}} / K_{\mathrm{w}}=1.75 \times 10^{9} \nonumber\] we can treat the reaction as if it goes to completion. Any solution that contains comparable amounts of a weak acid, HA, and its conjugate weak base, A , is a buffer. As we learned in , we can calculate the pH of a buffer using the Henderson–Hasselbalch equation. \[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \nonumber\] Before the equivalence point the concentration of unreacted acetic acid is \[\left[\text{CH}_3\text{COOH}\right] = \frac {(\text{mol CH}_3\text{COOH})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] and the concentration of acetate is \[[\text{CH}_3\text{COO}^-] = \frac {(\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_b V_b} {V_a + V_b} \nonumber\] For example, after adding 10.0 mL of NaOH the concentrations of CH COOH and CH COO are \[\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})-(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}} = 0.0500 \text{ M} \nonumber\] \[\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=\frac{(0.200 \ \mathrm{M})(10.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0333 \ \mathrm{M} \nonumber\] which gives us a pH of \[\mathrm{pH}=4.76+\log \frac{0.0333 \ \mathrm{M}}{0.0500 \ \mathrm{M}}=4.58 \nonumber\] At the equivalence point the moles of acetic acid initially present and the moles of NaOH added are identical. Because their reaction effectively proceeds to completion, the predominate ion in solution is CH COO , which is a weak base. To calculate the pH we first determine the concentration of CH COO \[\left[\mathrm{CH}_{3} \mathrm{COO}^-\right]=\frac{(\mathrm{mol} \ \mathrm{NaOH})_{\mathrm{added}}}{\text { total volume }}= \frac{(0.200 \ \mathrm{M})(25.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0667 \ \mathrm{M} \nonumber\] Alternatively, we can calculate acetate’s concentration using the initial moles of acetic acid; thus \[\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=\frac{\left(\mathrm{mol} \ \mathrm{CH}_{3} \mathrm{COOH}\right)_{\mathrm{initial}}}{\text { total volume }} = \frac{(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}} = 0.0667 \text{ M} \nonumber\] Next, we calculate the pH of the weak base as shown earlier in \[\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q) \nonumber\] \[K_{\mathrm{b}}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}=\frac{(x)(x)}{0.0667-x}=5.71 \times 10^{-10} \nonumber\] \[x=\left[\mathrm{OH}^{-}\right]=6.17 \times 10^{-6} \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{6.17 \times 10^{-6}}=1.62 \times 10^{-9} \ \mathrm{M} \nonumber\] finding that the pH at the equivalence point is 8.79. After the equivalence point, the titrant is in excess and the titration mixture is a dilute solution of NaOH. We can calculate the pH using the same strategy as in the titration of a strong acid with a strong base. For example, after adding 30.0 mL of NaOH the concentration of OH is \[\left[\mathrm{OH}^{-}\right]=\frac{(0.200 \ \mathrm{M})(30.0 \ \mathrm{mL})-(0.100 \ \mathrm{M})(50.0 \ \mathrm{mL})}{30.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0125 \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{0.0125}=8.00 \times 10^{-13} \ \mathrm{M} \nonumber\] giving a pH of 12.10. Table 9.2.2
and Figure 9.2.2
show additional results for this titration. You can use this same approach to calculate the titration curve for the titration of a weak base with a strong acid, except the initial pH is determined by the weak base, the pH at the equivalence point by its conjugate weak acid, and the pH after the equivalence point by excess strong acid. Construct a titration curve for the titration of 25.0 mL of 0.125 M NH with 0.0625 M HCl. The volume of HCl needed to reach the equivalence point is \[V_{a q}=V_{a}=\frac{M_{b} V_{b}}{M_{a}}=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{(0.0625 \ \mathrm{M})}=50.0 \ \mathrm{mL} \nonumber\] Before adding HCl the pH is that for a solution of 0.100 M NH . \[K_{\mathrm{b}}=\frac{[\mathrm{OH}^-]\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]}=\frac{(x)(x)}{0.125-x}=1.75 \times 10^{-5} \nonumber\] \[x=\left[\mathrm{OH}^{-}\right]=1.48 \times 10^{-3} \ \mathrm{M} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{[\mathrm{OH}^-]}=\frac{1.00 \times 10^{-14}}{1.48 \times 10^{-3} \ \mathrm{M}}=6.76 \times 10^{-12} \ \mathrm{M} \nonumber\] The pH at the beginning of the titration, therefore, is 11.17. Before the equivalence point the pH is determined by an \(\text{NH}_3/\text{NH}_4^+\) buffer. For example, after adding 10.0 mL of HCl \[\left[\mathrm{NH}_{3}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})-(0.0625 \ \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0714 \ \mathrm{M} \nonumber\] \[\left[\mathrm{NH}_{4}^{+}\right]=\frac{(0.0625 \ \mathrm{M})(10.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+10.0 \ \mathrm{mL}}=0.0179 \ \mathrm{M} \nonumber\] \[\mathrm{pH}=9.244+\log \frac{0.0714 \ \mathrm{M}}{0.0179 \ \mathrm{M}}=9.84 \nonumber\] At the equivalence point the predominate ion in solution is \(\text{NH}_4^+\). To calculate the pH we first determine the concentration of \(\text{NH}_4^+\) \[\left[\mathrm{NH}_{4}^{+}\right]=\frac{(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{25.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}}=0.0417 \ \mathrm{M} \nonumber\] and then calculate the pH \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^{+}\right]}=\frac{(x)(x)}{0.0417-x}=5.70 \times 10^{-10} \nonumber\] obtaining a value of 5.31. After the equivalence point, the pH is determined by the excess HCl. For example, after adding 70.0 mL of HCl \[[\mathrm{HCl}]=\frac{(0.0625 \ \mathrm{M})(70.0 \ \mathrm{mL})-(0.125 \ \mathrm{M})(25.0 \ \mathrm{mL})}{70.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=0.0132 \ \mathrm{M} \nonumber\] and the pH is 1.88. Some additional results are shown here. We can extend this approach for calculating a weak acid–strong base titration curve to reactions that involve multiprotic acids or bases, and mixtures of acids or bases. As the complexity of the titration increases, however, the necessary calculations become more time consuming. Not surprisingly, a variety of algebraic and spreadsheet approaches are available to aid in constructing titration curves. The following papers provide information on algebraic approaches to calculating titration curves: (a) Willis, C. J. , , 659–663; (b) Nakagawa, K. , , 673–676; (c) Gordus, A. A. , , 759–761; (d) de Levie, R. , , 209–217; (e) Chaston, S. , , 878–880; (f) de Levie, R. , , 585–590. The following papers provide information on the use of spreadsheets to generate titration curves: (a) Currie, J. O.; Whiteley, R. V. , , 923–926; (b) Breneman, G. L.; Parker, O. J. , , 46–47; (c) Carter, D. R.; Frye, M. S.; Mattson, W. A. , , 67–71; (d) Freiser, H. , CRC Press: Boca Raton, 1992. To evaluate the relationship between a titration’s equivalence point and its end point we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching an acid–base titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of 0.100 M CH COOH with 0.200 M NaOH to illustrate our approach. This is the same example that we used to develop the calculations for a weak acid–strong base titration curve. You can review the results of that calculation in and in . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 25.0 mL. Next we draw our axes, placing pH on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point volume, we draw a vertical line that intersects the -axis at 25.0 mL of NaOH. Figure 9.2.3
a shows the first step in our sketch. Before the equivalence point the titrand’s pH is determined by a buffer of acetic acid, CH COOH, and acetate, CH COO . Although we can calculate a buffer’s pH using the Henderson–Hasselbalch equation, we can avoid this calculation by making a simple assumption. You may recall from that a buffer operates over a pH range that extends approximately ±1 pH unit on either side of the weak acid’s p value. The pH is at the lower end of this range, pH = p – 1, when the weak acid’s concentration is \(10 \times\) greater than that of its conjugate weak base. The buffer reaches its upper pH limit, pH = p + 1, when the weak acid’s concentration is \(10 \times\) smaller than that of its conjugate weak base. When we titrate a weak acid or a weak base, the buffer spans a range of volumes from approximately 10% of the equivalence point volume to approximately 90% of the equivalence point volume. The actual values are 9.09% and 90.9%, but for our purpose, using 10% and 90% is more convenient; that is, after all, one advantage of an approximation! Figure 9.2.3
b shows the second step in our sketch. First, we superimpose acetic acid’s ladder diagram on the -axis, including its buffer range, using its p value of 4.76. Next, we add two points, one for the pH at 10% of the equivalence point volume (a pH of 3.76 at 2.5 mL) and one for the pH at 90% of the equivalence point volume (a pH of 5.76 at 22.5 mL). The third step is to add two points after the equivalence point. The pH after the equivalence point is fixed by the concentration of excess titrant, NaOH. Calculating the pH of a strong base is straightforward, as we saw earlier. Figure 9.2.3
c includes points (see ) for the pH after adding 30.0 mL and after adding 40.0 mL of NaOH. Next, we draw a straight line through each pair of points, extending each line through the vertical line that represents the equivalence point’s volume (Figure 9.2.3
d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.2.3
e). A comparison of our sketch to the exact titration curve (Figure 9.2.3
f) shows that they are in close agreement. Sketch a titration curve for the titration of 25.0 mL of 0.125 M NH with 0.0625 M HCl and compare to the result from . The figure below shows a sketch of the titration curve. The black dots and curve are the approximate sketch of the titration curve. The points in are the calculations from . The two black points before the equivalence point ( = 5 mL, pH = 10.24 and = 45 mL, pH= 8.24) are plotted using the p of 9.244 for \(\text{NH}_4^+\). The two black points after the equivalence point ( = 60 mL, pH = 2.13 and = 80 mL, pH= 1.75 ) are from the answer to . As shown in the following example, we can adapt this approach to any acid–base titration, including those where exact calculations are more challenging, including the titration of polyprotic weak acids and bases, and the titration of mixtures of weak acids or weak bases. Sketch titration curves for the following two systems: (a) the titration of 50.0 mL of 0.050 M H A, a diprotic weak acid with a p of 3 and a p of 7; and (b) the titration of a 50.0 mL mixture that contains 0.075 M HA, a weak acid with a p of 3, and 0.025 M HB, a weak acid with a p of 7. For both titrations, assume that the titrant is 0.10 M NaOH. Figure 9.2.4
a shows the titration curve for H A, including the ladder diagram for H A on the -axis, the two equivalence points at 25.0 mL and at 50.0 mL, two points before each equivalence point, two points after the last equivalence point, and the straight-lines used to sketch the final titration curve. Before the first equivalence point the pH is controlled by a buffer of H A and HA . An HA /A buffer controls the pH between the two equivalence points. After the second equivalence point the pH reflects the concentration of excess NaOH. Figure 9.2.4
b shows the titration curve for the mixture of HA and HB. Again, there are two equivalence points; however, in this case the equivalence points are not equally spaced because the concentration of HA is greater than that for HB. Because HA is the stronger of the two weak acids it reacts first; thus, the pH before the first equivalence point is controlled by a buffer of HA and A . Between the two equivalence points the pH reflects the titration of HB and is determined by a buffer of HB and B . After the second equivalence point excess NaOH determines the pH. Sketch the titration curve for 50.0 mL of 0.050 M H A, a diprotic weak acid with a p of 3 and a p of 4, using 0.100 M NaOH as the titrant. The fact that p falls within the buffer range of p presents a challenge that you will need to consider. The figure below shows a sketch of the titration curve. The titration curve has two equivalence points, one at 25.0 mL \((\text{H}_2\text{A} \rightarrow \text{HA}^-)\) and one at 50.0 mL (\(\text{HA}^- \rightarrow \text{A}^{2-}\)). In sketching the curve, we plot two points before the first equivalence point using the p of 3 for H A \[V_{\mathrm{HCl}}=2.5 \ \mathrm{mL}, \mathrm{pH}=2 \text { and } V_{\mathrm{HCl}}=22.5 \ \mathrm{mL}, \mathrm{pH}=4 \nonumber\] two points between the equivalence points using the p of 5 for HA \[V_{\mathrm{HCl}}=27.5 \ \mathrm{mL}, \mathrm{pH}=3, \text { and } V_{\mathrm{HCl}}=47.5 \ \mathrm{mL}, \mathrm{pH}=5 \nonumber\] and two points after the second equivalence point \[V_{\mathrm{HCl}}=70 \ \mathrm{mL}, \mathrm{pH}=12.22 \text { and } V_{\mathrm{HCl}}=90 \ \mathrm{mL}, \mathrm{pH}=12.46 \nonumber\] Drawing a smooth curve through these points presents us with the following dilemma—the pH appears to increase as the titrant’s volume approaches the first equivalence point and then appears to decrease as it passes through the first equivalence point. This is, of course, absurd; as we add NaOH the pH cannot decrease. Instead, we model the titration curve before the second equivalence point by drawing a straight line from the first point ( = 2.5 mL, pH = 2) to the fourth point ( = 47.5 mL, pH= 5), ignoring the second and third points. The results is a reasonable approximation of the exact titration curve. Earlier we made an important distinction between a titration’s end point and its equivalence point. The difference between these two terms is important and deserves repeating. An equivalence point, which occurs when we react stoichiometrically equal amounts of the analyte and the titrant, is a theoretical not an experimental value. A titration’s end point is an experimental result that represents our best estimate of the equivalence point. Any difference between a titration’s equivalence point and its corresponding end point is a source of determinate error. Earlier we learned how to calculate the pH at the equivalence point for the titration of a strong acid with a strong base, and for the titration of a weak acid with a strong base. We also learned how to sketch a titration curve with only a minimum of calculations. Can we also locate the equivalence point without performing any calculations. The answer, as you might guess, often is yes! For most acid–base titrations the inflection point—the point on a titration curve that has the greatest slope—very nearly coincides with the titration’s equivalence point. The red arrows in , for example, identify the equivalence points for the titration curves in . An inflection point actually precedes its corresponding equivalence point by a small amount, with the error approaching 0.1% for weak acids and weak bases with dissociation constants smaller than 10 , or for very dilute solutions [Meites, L.; Goldman, J. A. , , 472–479]. The principal limitation of an inflection point is that it must be present and easy to identify. For some titrations the inflection point is missing or difficult to find. Figure 9.2.5
, for example, demonstrates the affect of a weak acid’s dissociation constant, , on the shape of its titration curve. An inflection point is visible, even if barely so, for acid dissociation constants larger than 10 , but is missing when is 10 . An inflection point also may be missing or difficult to see if the analyte is a multiprotic weak acid or weak base with successive dissociation constants that are similar in magnitude. To appreciate why this is true let’s consider the titration of a diprotic weak acid, H A, with NaOH. During the titration the following two reactions occur. \[\mathrm{H}_{2} \mathrm{A}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HA}^{-}(a q) \label{9.3}\] \[\mathrm{HA}^{-}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{A}^{2-}(a q) \label{9.4}\] To see two distinct inflection points, reaction \ref{9.3} must essentially be complete before reaction \ref{9.4} begins. Figure 9.2.6
shows titration curves for three diprotic weak acids. The titration curve for maleic acid, for which is approximately \(20000 \times\) larger than , has two distinct inflection points. Malonic acid, on the other hand, has acid dissociation constants that differ by a factor of approximately 690. Although malonic acid’s titration curve shows two inflection points, the first is not as distinct as the second. Finally, the titration curve for succinic acid, for which the two values differ by a factor of only \(27 \times\), has only a single inflection point that corresponds to the neutralization of \(\text{HC}_2\text{H}_4\text{O}_4^-\) to \(\text{C}_2\text{H}_4\text{O}_4^{2-}\). In general, we can detect separate inflection points when successive acid dissociation constants differ by a factor of at least 500 (a \(\Delta\) of at least 2.7). The same holds true for mixtures of weak acids or mixtures of weak bases. To detect separate inflection points when titrating a mixture of weak acids, their p values must differ by at least a factor of 500. One interesting group of weak acids and weak bases are organic dyes. Because an organic dye has at least one highly colored conjugate acid–base species, its titration results in a change in both its pH and its color. We can use this change in color to indicate the end point of a titration provided that it occurs at or near the titration’s equivalence point. As an example, let’s consider an indicator for which the acid form, HIn, is yellow and the base form, In , is red. The color of the indicator’s solution depends on the relative concentrations of HIn and In . To understand the relationship between pH and color we use the indicator’s acid dissociation reaction \[\mathrm{HIn}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{In}^{-}(a q) \nonumber\] and its equilibrium constant expression. \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \label{9.5}\] Taking the negative log of each side of Equation \ref{9.5}, and rearranging to solve for pH leaves us with a equation that relates the solution’s pH to the relative concentrations of HIn and In . \[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{[\mathrm{In}^-]}{[\mathrm{HIn}]} \label{9.6}\] If we can detect HIn and In with equal ease, then the transition from yellow-to-red (or from red-to-yellow) reaches its midpoint, which is orange, when the concentrations of HIn and In are equal, or when the pH is equal to the indicator’s p . If the indicator’s p and the pH at the equivalence point are identical, then titrating until the indicator turns orange is a suitable end point. Unfortunately, we rarely know the exact pH at the equivalence point. In addition, determining when the concentrations of HIn and In are equal is difficult if the indicator’s change in color is subtle. We can establish the range of pHs over which the average analyst observes a change in the indicator’s color by making two assumptions: that the indicator’s color is yellow if the concentration of HIn is \(10 \times\) greater than that of In and that its color is red if the concentration of HIn is \(10 \times\) smaller than that of In . Substituting these inequalities into Equation \ref{9.6} \[\begin{array}{l}{\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{1}{10}=\mathrm{p} K_{\mathrm{a}}-1} \\ {\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{10}{1}=\mathrm{p} K_{\mathrm{a}}+1}\end{array} \nonumber\] shows that the indicator changes color over a pH range that extends ±1 unit on either side of its p . As shown in Figure 9.2.7
, the indicator is yel-ow when the pH is less than p – 1 and it is red when the pH is greater than p + 1. For pH values between p – 1 and p + 1 the indicator’s color passes through various shades of orange. The properties of several common acid–base indicators are listed in Table 9.2.3
. You may wonder why an indicator’s pH range, such as that for phenolphthalein, is not equally distributed around its p value. The explanation is simple. Figure 9.2.7
presents an idealized view in which our sensitivity to the indicator’s two colors is equal. For some indicators only the weak acid or the weak base is colored. For other indicators both the weak acid and the weak base are colored, but one form is easier to see. In either case, the indicator’s pH range is skewed in the direction of the indicator’s less colored form. Thus, phenolphthalein’s pH range is skewed in the direction of its colorless form, shifting the pH range to values lower than those suggested by Figure 9.2.7
. The relatively broad range of pHs over which an indicator changes color places additional limitations on its ability to signal a titration’s end point. To minimize a determinate titration error, the indicator’s entire pH range must fall within the rapid change in pH near the equivalence point. For example, in Figure 9.2.8
we see that phenolphthalein is an appropriate indicator for the titration of 50.0 mL of 0.050 M acetic acid with 0.10 M NaOH. Bromothymol blue, on the other hand, is an inappropriate indicator because its change in color begins well before the initial sharp rise in pH, and, as a result, spans a relatively large range of volumes. The early change in color increases the probability of obtaining an inaccurate result, and the range of possible end point volumes increases the probability of obtaining imprecise results. Suggest a suitable indicator for the titration of 25.0 mL of 0.125 M NH with 0.0625 M NaOH. You constructed a titration curve for this titration in and . The pH at the equivalence point is 5.31 (see ) and the sharp part of the titration curve extends from a pH of approximately 7 to a pH of approximately 4. Of the indicators in , methyl red is the best choice because its p value of 5.0 is closest to the equivalence point’s pH and because the pH range of 4.2–6.3 for its change in color will not produce a significant titration error. An alternative approach for locating a titration’s end point is to monitor the titration’s progress using a sensor whose signal is a function of the analyte’s concentration. The result is a plot of the entire titration curve, which we can use to locate the end point with a minimal error. A pH electrode is the obvious sensor for monitoring an acid–base titration and the result is a . For example, Figure 9.2.9
a shows a small portion of the potentiometric titration curve for the titration of 50.0 mL of 0.050 M CH COOH with 0.10 M NaOH, which focuses on the region that contains the equivalence point. The simplest method for finding the end point is to locate the titration curve’s inflection point, which is shown by the arrow. This is also the least accurate method, particularly if the titration curve has a shallow slope at the equivalence point. See for more details about pH electrodes. . Titration curves for the titration of 50.0 mL of 0.050 M CH COOH with 0.10 M NaOH: (a) normal titration curve; (b) first derivative titration curve; (c) second derivative titration curve; (d) Gran plot. The arrows show the location of each titration’s end point. Another method for locating the end point is to plot the first derivative of the titration curve, which gives its slope at each point along the -axis. Examine Figure 9.2.9
a and consider how the titration curve’s slope changes as we approach, reach, and pass the equivalence point. Because the slope reaches its maximum value at the inflection point, the first derivative shows a spike at the equivalence point (Figure 9.2.9
b). The second derivative of a titration curve can be more useful than the first derivative because the equivalence point intersects the volume axis. Figure 9.2.9
c shows the resulting titration curve. Suppose we have the following three points on our titration curve: Mathematically, we can approximate the first derivative as \(\Delta \text{pH} / \Delta V\), where \(\Delta \text{pH}\) is the change in pH between successive additions of titrant. Using the first two points, the first derivative is \[\frac{\Delta \mathrm{pH}}{\Delta V}=\frac{6.10-6.00}{23.91-23.65}=0.385 \nonumber\] which we assign to the average of the two volumes, or 23.78 mL. For the second and third points, the first derivative is 0.455 and the average volume is 24.02 mL. We can approximate the second derivative as \(\Delta (\Delta \text{pH} / \Delta V) / \Delta V\), or \(\Delta^2 \text{pH} / \Delta V^2\). Using the two points from our calculation of the first derivative, the second derivative is \[\frac{\Delta^{2} \mathrm{p} \mathrm{H}}{\Delta V^{2}}=\frac{0.455-0.385}{24.02-23.78}=0.292 \nonumber\] which we assign to the average of the two volumes, or 23.90 mL. Note that calculating the first derivative comes at the expense of losing one piece of information (three points become two points), and calculating the second derivative comes at the expense of losing two pieces of information. Derivative methods are particularly useful when titrating a sample that contains more than one analyte. If we rely on indicators to locate the end points, then we usually must complete separate titrations for each analyte so that we can see the change in color for each end point. If we record the titration curve, however, then a single titration is sufficient. The precision with which we can locate the end point also makes derivative methods attractive for an analyte that has a poorly defined normal titration curve. Derivative methods work well only if we record sufficient data during the rapid increase in pH near the equivalence point. This usually is not a problem if we use an automatic titrator, such as the one seen earlier in . Because the pH changes so rapidly near the equivalence point—a change of several pH units over a span of several drops of titrant is not unusual—a manual titration does not provide enough data for a useful derivative titration curve. A manual titration does contain an abundance of data during the more gently rising portions of the titration curve before and after the equivalence point. This data also contains information about the titration curve’s equivalence point. Consider again the titration of acetic acid, CH COOH, with NaOH. At any point during the titration acetic acid is in equilibrium with H O and CH COO \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] for which the equilibrium constant is \[K_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\] Before the equivalence point the concentrations of CH COOH and CH COO are \[[\text{CH}_3\text{COOH}] = \frac {(\text{mol CH}_3\text{COOH})_\text{initial} - (\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_a V_a - M_b V_b} {V_a + V_b} \nonumber\] \[[\text{CH}_3\text{COO}^-] = \frac {(\text{mol NaOH})_\text{added}} {\text{total volume}} = \frac {M_b V_b} {V_a + V_b} \nonumber\] Substituting these equations into the expression and rearranging leaves us with \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(M_{b} V_{b}\right) /\left(V_{a}+V_{b}\right)}{\left\{M_{a} V_{a}-M_{b} V_{b}\right\} /\left(V_{a}+V_{b}\right)} \nonumber\] \[K_{a} M_{a} V_{a}-K_{a} M_{b} V_{b}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(M_{b} V_{b}\right) \nonumber\] \[\frac{K_{a} M_{a} V_{a}}{M_{b}}-K_{a} V_{b}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] V_{b} \nonumber\] Finally, recognizing that the equivalence point volume is \[V_{eq}=\frac{M_{a} V_{a}}{M_{b}} \nonumber\] leaves us with the following equation. \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times V_{b}=K_{\mathrm{a}} V_{eq}-K_{\mathrm{a}} V_{b} \nonumber\] For volumes of titrant before the equivalence point, a plot of \(V_b \times [\text{H}_3\text{O}^+]\) versus is a straight-line with an -intercept of and a slope of – . shows a typical result. This method of data analysis, which converts a portion of a titration curve into a straight-line, is a . Values of determined by this method may have a substantial error if the effect of activity is ignored. See for a discussion of activity. The reaction between an acid and a base is exothermic. Heat generated by the reaction is absorbed by the titrand, which increases its temperature. Monitoring the titrand’s temperature as we add the titrant provides us with another method for recording a titration curve and identifying the titration’s end point (Figure 9.2.10
). Before we add the titrant, any change in the titrand’s temperature is the result of warming or cooling as it equilibrates with the surroundings. Adding titrant initiates the exothermic acid–base reaction and increases the titrand’s temperature. This part of a thermometric titration curve is called the titration branch. The temperature continues to rise with each addition of titrant until we reach the equivalence point. After the equivalence point, any change in temperature is due to the titrant’s enthalpy of dilution and the difference between the temperatures of the titrant and titrand. Ideally, the equivalence point is a distinct intersection of the titration branch and the excess titrant branch. As shown in Figure 9.2.10
, however, a thermometric titration curve usually shows curvature near the equivalence point due to an incomplete neutralization reaction or to the excessive dilution of the titrand and the titrant during the titration. The latter problem is minimized by using a titrant that is 10–100 times more concentrated than the analyte, although this results in a very small end point volume and a larger relative error. If necessary, the end point is found by extrapolation. Although not a common method for monitoring an acid–base titration, a thermometric titration has one distinct advantage over the direct or indirect monitoring of pH. As discussed earlier, the use of an indicator or the monitoring of pH is limited by the magnitude of the relevant equilibrium constants. For example, titrating boric acid, H BO , with NaOH does not provide a sharp end point when monitoring pH because boric acid’s of \(5.8 \times 10^{-10}\) is too small (Figure 9.2.11
a). Because boric acid’s enthalpy of neutralization is fairly large, –42.7 kJ/mole, its thermometric titration curve provides a useful endpoint (Figure 9.2.11
b). Thus far we have assumed that the titrant and the titrand are aqueous solutions. Although water is the most common solvent for acid–base titrimetry, switching to a nonaqueous solvent can improve a titration’s feasibility. For an amphoteric solvent, SH, the autoprotolysis constant, , relates the concentration of its protonated form, \(\text{SH}_2^+\), to its deprotonated form, S \[\begin{aligned} 2 \mathrm{SH} &\rightleftharpoons\mathrm{SH}_{2}^{+}+\mathrm{S}^{-} \\ K_{\mathrm{s}} &=\left[\mathrm{SH}_{2}^{+}\right,\mathrm{S}^-] \end{aligned} \nonumber\] and the solvent’s pH and pOH are \[\begin{array}{l}{\mathrm{pH}=-\log \left[\mathrm{SH}_{2}^{+}\right]} \\ {\mathrm{pOH}=-\log \left[\mathrm{S}^{-}\right]}\end{array} \nonumber\] You should recognize that is just specific form of when the solvent is water. The most important limitation imposed by is the change in pH during a titration. To understand why this is true, let’s consider the titration of 50.0 mL of \(1.0 \times 10^{-4}\) M HCl using \(1.0 \times 10^{-4}\) M NaOH as the titrant. Before the equivalence point, the pH is determined by the untitrated strong acid. For example, when the volume of NaOH is 90% of , the concentration of H O is \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{M_{a} V_{a}-M_{b} V_{b}}{V_{a}+V_{b}} = \frac{\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})-\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(45.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+45.0 \ \mathrm{mL}} = 5.3 \times 10^{-6} \ \mathrm{M} \nonumber\] and the pH is 5.3. When the volume of NaOH is 110% of , the concentration of OH is \[\left[\mathrm{OH}^{-}\right]=\frac{M_{b} V_{b}-M_{a} V_{a}}{V_{a}+V_{b}} = \frac{\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(55.0 \ \mathrm{mL})-\left(1.0 \times 10^{-4} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{55.0 \ \mathrm{mL}+50.0 \ \mathrm{mL}} = 4.8 \times 10^{-6} \ \mathrm{M} \nonumber\] and the pOH is 5.3. The titrand’s pH is \[\mathrm{pH}=\mathrm{p} K_{w}-\mathrm{pOH}=14.0-5.3=8.7 \nonumber\] and the change in the titrand’s pH as the titration goes from 90% to 110% of is \[\Delta \mathrm{pH}=8.7-5.3=3.4 \nonumber\] If we carry out the same titration in a nonaqueous amphiprotic solvent that has a of \(1.0 \times 10^{-20}\), the pH after adding 45.0 mL of NaOH is still 5.3. However, the pH after adding 55.0 mL of NaOH is \[\mathrm{pH}=\mathrm{p} K_{s}-\mathrm{pOH}=20.0-5.3=14.7 \nonumber\] In this case the change in pH \[\Delta \mathrm{pH}=14.7-5.3=9.4 \nonumber\] is significantly greater than that obtained when the titration is carried out in water. Figure 9.2.12
shows the titration curves in both the aqueous and the nonaqueous solvents. Another parameter that affects the feasibility of an acid–base titration is the titrand’s dissociation constant. Here, too, the solvent plays an important role. The strength of an acid or a base is a relative measure of how easy it is to transfer a proton from the acid to the solvent or from the solvent to the base. For example, HF, with a of \(6.8 \times 10^{-4}\), is a better proton donor than CH COOH, for which is \(1.75 \times 10^{-5}\). The strongest acid that can exist in water is the hydronium ion, H O . HCl and HNO are strong acids because they are better proton donors than H O and essentially donate all their protons to H O, leveling their acid strength to that of H O . In a different solvent HCl and HNO may not behave as strong acids. If we place acetic acid in water the dissociation reaction \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] does not proceed to a significant extent because CH COO is a stronger base than H O and H O is a stronger acid than CH COOH. If we place acetic acid in a solvent that is a stronger base than water, such as ammonia, then the reaction \[\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3}\rightleftharpoons\mathrm{NH}_{4}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-} \nonumber\] proceeds to a greater extent. In fact, both HCl and CH COOH are strong acids in ammonia. All other things being equal, the strength of a weak acid increases if we place it in a solvent that is more basic than water, and the strength of a weak base increases if we place it in a solvent that is more acidic than water. In some cases, however, the opposite effect is observed. For example, the p for NH is 4.75 in water and it is 6.40 in the more acidic glacial acetic acid. In contradiction to our expectations, NH is a weaker base in the more acidic solvent. A full description of the solvent’s effect on the p of weak acid or the p of a weak base is beyond the scope of this text. You should be aware, however, that a titration that is not feasible in water may be feasible in a different solvent. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical acid–base titrimetric method. Although each method is unique, the following description of the determination of protein in bread provides an instructive example of a typical procedure. The description here is based on Method 13.86 as published in , 8th Ed., Association of Official Agricultural Chemists: Washington, D. C., 1955. This method is based on a determination of %w/w nitrogen using the Kjeldahl method. The protein in a sample of bread is oxidized to \(\text{NH}_4^+\) using hot concentrated H SO . After making the solution alkaline, which converts \(\text{NH}_4^+\) to NH , the ammonia is distilled into a flask that contains a known amount of HCl. The amount of unreacted HCl is determined by a back titration using a standard strong base titrant. Because different cereal proteins contain similar amounts of nitrogen—on average there are 5.7 g protein for every gram of nitrogen—we multiply the experimentally determined %w/w N by a factor of 5.7 gives the %w/w protein in the sample. Transfer a 2.0-g sample of bread, which previously has been air-dried and ground into a powder, to a suitable digestion flask along with 0.7 g of a HgO catalyst, 10 g of K SO , and 25 mL of concentrated H SO . Bring the solution to a boil. Continue boiling until the solution turns clear and then boil for at least an additional 30 minutes. After cooling the solution below room temperature, remove the Hg catalyst by adding 200 mL of H O and 25 mL of 4% w/v K S. Add a few Zn granules to serve as boiling stones and 25 g of NaOH. Quickly connect the flask to a distillation apparatus and distill the NH into a collecting flask that contains a known amount of standardized HCl. The tip of the condenser must be placed below the surface of the strong acid. After the distillation is complete, titrate the excess strong acid with a standard solution of NaOH using methyl red as an indicator (Figure 9.2.13
). 1. Oxidizing the protein converts all of its nitrogen to \(\text{NH}_4^+\). Why is the amount of nitrogen not determined by directly titrating the \(\text{NH}_4^+\) with a strong base? There are two reasons for not directly titrating the ammonium ion. First, because \(\text{NH}_4^+\) is a very weak acid (its is \(5.6 \times 10^{-10}\)), its titration with NaOH has a poorly-defined end point. Second, even if we can determine the end point with acceptable accuracy and precision, the solution also contains a substantial concentration of unreacted H SO . The presence of two acids that differ greatly in concentration makes for a difficult analysis. If the titrant’s concentration is similar to that of H SO , then the equivalence point volume for the titration of \(\text{NH}_4^+\) is too small to measure reliably. On the other hand, if the titrant’s concentration is similar to that of \(\text{NH}_4^+\), the volume needed to neutralize the H SO is unreasonably large. 2. Ammonia is a volatile compound as evidenced by the strong smell of even dilute solutions. This volatility is a potential source of determinate error. Is this determinate error negative or positive? Any loss of NH is loss of nitrogen and, therefore, a loss of protein. The result is a negative determinate error. 3. Identify the steps in this procedure that minimize the determinate error from the possible loss of NH . Three specific steps minimize the loss of ammonia: (1) the solution is cooled below room temperature before we add NaOH; (2) after we add NaOH, the digestion flask is quickly connected to the distillation apparatus; and (3) we place the condenser’s tip below the surface of the HCl to ensure that the NH reacts with the HCl before it is lost through volatilization. 4. How does K S remove Hg , and why is its removal important? Adding sulfide precipitates Hg as HgS. This is important because NH forms stable complexes with many metal ions, including Hg . Any NH that reacts with Hg is not collected during distillation, providing another source of determinate error. Although many quantitative applications of acid–base titrimetry have been replaced by other analytical methods, a few important applications continue to find use. In this section we review the general application of acid–base titrimetry to the analysis of inorganic and organic compounds, with an emphasis on applications in environmental and clinical analysis. First, however, we discuss the selection and standardization of acidic and basic titrants. The most common strong acid titrants are HCl, HClO , and H SO . Solutions of these titrants usually are prepared by diluting a commercially available concentrated stock solution. Because the concentration of a concentrated acid is known only approximately, the titrant’s concentration is determined by standardizing against one of the primary standard weak bases listed in Table 9.2.4
. The nominal concentrations of the concentrated stock solutions are 12.1 M HCl, 11.7 M HClO , and 18.0 M H SO . The actual concentrations of these acids are given as %w/v and vary slightly from lot-to-lot. (a) The end point for this titration is improved by titrating to the second equivalence point, boiling the solution to expel CO2, and retitrating to the second equivalence point. The reaction in this case is \[\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow \mathrm{CO}_{2}+2 \mathrm{Na}^{+}+3 \mathrm{H}_{2} \mathrm{O} \nonumber\] (b) -(hydroxymethyl)aminomethane often goes by the shorter name of TRIS or THAM. (c) Potassium hydrogen phthalate often goes by the shorter name of KHP. (d) Because it is not very soluble in water, dissolve benzoic acid in a small amount of ethanol before diluting with water. The most common strong base titrant is NaOH, which is available both as an impure solid and as an approximately 50% w/v solution. Solutions of NaOH are standardized against any of the primary weak acid standards listed in Table \(\Page [4|\). Using NaOH as a titrant is complicated by potential contamination from the following reaction between dissolved CO and OH . \[\mathrm{CO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}( l) \label{9.7}\] Any solution in contact with the atmosphere contains a small amount of CO from the equilibrium \[\mathrm{CO}_{2}(g)\rightleftharpoons\mathrm{CO}_{2}(a q) \nonumber\] During the titration, NaOH reacts both with the titrand and with CO , which increases the volume of NaOH needed to reach the titration’s end point. This is not a problem if the end point pH is less than 6. Below this pH the \(\text{CO}_3^{2-}\) from reaction \ref{9.7} reacts with H O to form carbonic acid. \[\mathrm{CO}_{3}^{2-}(a q)+2 \mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q). \label{9.8}\] Combining reaction \ref{9.7} and reaction \ref{9.8} gives an overall reaction that does not include OH . \[\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l ) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber\] Under these conditions the presence of CO does not affect the quantity of OH used in the titration and is not a source of determinate error. If the end point pH is between 6 and 10, however, the neutralization of \(\text{CO}_3^{2-}\) requires one proton \[\mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{HCO}_{3}^{-}(a q) \nonumber\] and the net reaction between CO and OH is \[\mathrm{CO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{HCO}_{3}^{-}(a q) \nonumber\] Under these conditions some OH is consumed in neutralizing CO , which results in a determinate error. We can avoid the determinate error if we use the same end point pH for both the standardization of NaOH and the analysis of our analyte, although this is not always practical. Solid NaOH is always contaminated with carbonate due to its contact with the atmosphere, and we cannot use it to prepare a carbonate-free solution of NaOH. Solutions of carbonate-free NaOH are prepared from 50% w/v NaOH because Na CO is insoluble in concentrated NaOH. When CO is absorbed, Na CO precipitates and settles to the bottom of the container, which allow access to the carbonate-free NaOH. When pre- paring a solution of NaOH, be sure to use water that is free from dissolved CO . Briefly boiling the water expels CO ; after it cools, the water is used to prepare carbonate-free solutions of NaOH. A solution of carbonate-free NaOH is relatively stable if we limit its contact with the atmosphere. Standard solutions of sodium hydroxide are not stored in glass bottles as NaOH reacts with glass to form silicate; instead, store such solutions in polyethylene bottles. Acid–base titrimetry is a standard method for the quantitative analysis of many inorganic acids and bases. A standard solution of NaOH is used to determine the concentration of inorganic acids, such as H PO or H AsO , and inorganic bases, such as Na CO are analyzed using a standard solution of HCl. If an inorganic acid or base that is too weak to be analyzed by an aqueous acid–base titration, it may be possible to complete the analysis by adjusting the solvent or by an indirect analysis. For example, when analyzing boric acid, H BO , by titrating with NaOH, accuracy is limited by boric acid’s small acid dissociation constant of \(5.8 \times 10^{-10}\). Boric acid’s value increases to \(1.5 \times 10^{-4}\) in the presence of mannitol, because it forms a stable complex with the borate ion, which results is a sharper end point and a more accurate titration. Similarly, the analysis of ammonium salts is limited by the ammonium ion’ small acid dissociation constant of \(5.7 \times 10^{-10}\). We can determine \(\text{NH}_4^+\) indirectly by using a strong base to convert it to NH , which is removed by distillation and titrated with HCl. Because NH is a stronger weak base than \(\text{NH}_4^+\) is a weak acid (its is \(1.58 \times 10^{-5}\)), the titration has a sharper end point. We can analyze a neutral inorganic analyte if we can first convert it into an acid or a base. For example, we can determine the concentration of \(\text{NO}_3^-\) by reducing it to NH in a strongly alkaline solution using Devarda’s alloy, a mixture of 50% w/w Cu, 45% w/w Al, and 5% w/w Zn. \[3 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{Al}(s)+5 \mathrm{OH}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 8 \mathrm{AlO}_{2}^{-}(a q)+3 \mathrm{NH}_{3}(a q) \nonumber\] The NH is removed by distillation and titrated with HCl. Alternatively, we can titrate \(\text{NO}_3^-\) as a weak base by placing it in an acidic nonaqueous solvent, such as anhydrous acetic acid, and using HClO as a titrant. Acid–base titrimetry continues to be listed as a standard method for the determination of alkalinity, acidity, and free CO in waters and wastewaters. is a measure of a sample’s capacity to neutralize acids. The most important sources of alkalinity are OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), although other weak bases, such as phosphate, may contribute to the overall alkalinity. Total alkalinity is determined by titrating to a fixed end point pH of 4.5 (or to the bromocresol green end point) using a standard solution of HCl or H SO . Results are reported as mg CaCO /L. Although a variety of strong bases and weak bases may contribute to a sample’s alkalinity, a single titration cannot distinguish between the possible sources. Reporting the total alkalinity as if CaCO is the only source provides a means for comparing the acid-neutralizing capacities of different samples. When the sources of alkalinity are limited to OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), separate titrations to a pH of 4.5 (or the bromocresol green end point) and a pH of 8.3 (or the phenolphthalein end point) allow us to determine which species are present and their respective concentrations. Titration curves for OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\)are shown in Figure 9.2.14
. For a solution that contains OH alkalinity only, the volume of strong acid needed to reach each of the two end points is identical (Figure 9.2.14
a). When the only source of alkalinity is \(\text{CO}_3^{2-}\), the volume of strong acid needed to reach the end point at a pH of 4.5 is exactly twice that needed to reach the end point at a pH of 8.3 (Figure 9.2.14
b). If a solution contains \(\text{HCO}_3^-\) alkalinity only, the volume of strong acid needed to reach the end point at a pH of 8.3 is zero, but that for the pH 4.5 end point is greater than zero (Figure 9.2.14
c). A mixture of OH and \(\text{CO}_3^{2-}\) or a mixture of \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\) also is possible. Consider, for example, a mixture of OH and \(\text{CO}_3^{2-}\). The volume of strong acid to titrate OH is the same whether we titrate to a pH of 8.3 or a pH of 4.5. Titrating \(\text{CO}_3^{2-}\) to a pH of 4.5, however, requires twice as much strong acid as titrating to a pH of 8.3. Consequently, when we titrate a mixture of these two ions, the volume of strong acid needed to reach a pH of 4.5 is less than twice that needed to reach a pH of 8.3. For a mixture of \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\) the volume of strong acid needed to reach a pH of 4.5 is more than twice that needed to reach a pH of 8.3. Table 9.2.5
summarizes the relationship between the sources of alkalinity and the volumes of titrant needed to reach the two end points. A mixture of OH and \(\text{HCO}_3^-\) is unstable with respect to the formation of \(\text{CO}_3^{2-}\). Problem 15 in the end-of-chapter problems asks you to explain why this is true. is a measure of a water sample’s capacity to neutralize base and is divided into strong acid and weak acid acidity. Strong acid acidity from inorganic acids such as HCl, HNO , and H SO is common in industrial effluents and in acid mine drainage. Weak acid acidity usually is dominated by the formation of H CO from dissolved CO , but also includes contributions from hydrolyzable metal ions such as Fe , Al , and Mn . In addition, weak acid acidity may include a contribution from organic acids. Acidity is determined by titrating with a standard solution of NaOH to a fixed pH of 3.7 (or the bromothymol blue end point) and to a fixed pH of 8.3 (or the phenolphthalein end point). Titrating to a pH of 3.7 provides a measure of strong acid acidity, and titrating to a pH of 8.3 provides a measure of total acidity. Weak acid acidity is the difference between the total acidity and the strong acid acidity. Results are expressed as the amount of CaCO that can be neutralized by the sample’s acidity. An alternative approach for determining strong acid and weak acid acidity is to obtain a potentiometric titration curve and use a Gran plot to determine the two equivalence points. This approach has been used, for example, to determine the forms of acidity in atmospheric aerosols [Ferek, R. J.; Lazrus, A. L.; Haagenson, P. L.; Winchester, J. W. , , 315–324]. As is the case with alkalinity, acidity is reported as mg CaCO /L. Water in contact with either the atmosphere or with carbonate-bearing sediments contains free CO in equilibrium with CO ( ) and with aqueous H CO , \(\text{HCO}_3^-\) and \(\text{CO}_3^{2-}\). The concentration of free CO is determined by titrating with a standard solution of NaOH to the phenolphthalein end point, or to a pH of 8.3, with results reported as mg CO /L. This analysis essentially is the same as that for the determination of total acidity and is used only for water samples that do not contain strong acid acidity. Free CO is the same thing as CO . Acid–base titrimetry continues to have a small, but important role for the analysis of organic compounds in pharmaceutical, biochemical, agricultur- al, and environmental laboratories. Perhaps the most widely employed acid–base titration is the for organic nitrogen. Examples of analytes determined by a Kjeldahl analysis include caffeine and saccharin in pharmaceutical products, proteins in foods, and the analysis of nitrogen in fertilizers, sludges, and sediments. Any nitrogen present in a –3 oxidation state is oxidized quantitatively to \(\text{NH}_4^+\). Because some aromatic heterocyclic compounds, such as pyridine, are difficult to oxidize, a catalyst is used to ensure a quantitative oxidation. Nitrogen in other oxidation states, such as nitro and azo nitrogens, are oxidized to N , which results in a negative determinate error. Including a reducing agent, such as salicylic acid, converts this nitrogen to a –3 oxidation state, eliminating this source of error. Table 9.2.6
provides additional examples in which an element is converted quantitatively into a titratable acid or base. Several organic functional groups are weak acids or weak bases. Carboxylic (–COOH), sulfonic (–SO H) and phenolic (–C H OH) functional groups are weak acids that are titrated successfully in either aqueous or non-aqueous solvents. Sodium hydroxide is the titrant of choice for aqueous solutions. Nonaqueous titrations often are carried out in a basic solvent, such as ethylenediamine, using tetrabutylammonium hydroxide, (C H ) NOH, as the titrant. Aliphatic and aromatic amines are weak bases that are titrated using HCl in aqueous solutions, or HClO in glacial acetic acid. Other functional groups are analyzed indirectly following a reaction that produces or consumes an acid or base. Typical examples are shown in Table 9.2.7
. [1]: (CH CO) O + ROH \(\rightarrow\) CH COOR + [2]: (CH CO) ) + H O \(\rightarrow\) 2 the species that is titrated is shown in for alcohols, reaction [1] is carried out in pyridine to prevent the hydrolysis of acetic anhydride by water. After reaction [1] is complete, water is added to covert any unreacted acetic anhydride to acetic acid (reaction [2]) Many pharmaceutical compounds are weak acids or weak bases that are analyzed by an aqueous or a nonaqueous acid–base titration; examples include salicylic acid, phenobarbital, caffeine, and sulfanilamide. Amino acids and proteins are analyzed in glacial acetic acid using HClO as the titrant. For example, a procedure for determining the amount of nutritionally available protein uses an acid–base titration of lysine residues [(a) Molnár-Perl, I.; Pintée-Szakács, M. , , 159–166; (b) Barbosa, J.; Bosch, E.; Cortina, J. L.; Rosés, M. , , 177–181]. The quantitative relationship between the titrand and the titrant is determined by the titration reaction’s stoichiometry. If the titrand is polyprotic, then we must know to which equivalence point we are titrating. The following example illustrates how we can use a ladder diagram to determine a titration reaction’s stoichiometry. A 50.00-mL sample of a citrus drink requires 17.62 mL of 0.04166 M NaOH to reach the phenolphthalein end point. Express the sample’s acidity as grams of citric acid, C H O , per 100 mL. Because citric acid is a triprotic weak acid, we first must determine if the phenolphthalein end point corresponds to the first, second, or third equivalence point. Citric acid’s ladder diagram is shown in Figure 9.2.15
a. Based on this ladder diagram, the first equivalence point is between a pH of 3.13 and a pH of 4.76, the second equivalence point is between a pH of 4.76 and a pH of 6.40, and the third equivalence point is greater than a pH of 6.40. Because phenolphthalein’s end point pH is 8.3–10.0 (see ), the titration must proceed to the third equivalence point and the titration reaction is \[ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q)+3 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) \nonumber\] To reach the equivalence point, each mole of citric acid consumes three moles of NaOH; thus \[(0.04166 \ \mathrm{M} \ \mathrm{NaOH})(0.01762 \ \mathrm{L} \ \mathrm{NaOH})=7.3405 \times 10^{-4} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[7.3405 \times 10^{-4} \ \mathrm{mol} \ \mathrm{NaOH} \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{3 \ \mathrm{mol} \ \mathrm{NaOH}}= 2.4468 \times 10^{-4} \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \nonumber\] \[2.4468 \times 10^{-4} \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \times \frac{192.1 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}=0.04700 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} \nonumber\] Because this is the amount of citric acid in a 50.00 mL sample, the concentration of citric acid in the citrus drink is 0.09400 g/100 mL. The complete titration curve is shown in Figure 9.2.15
b. Your company recently received a shipment of salicylic acid, C H O , for use in the production of acetylsalicylic acid (aspirin). You can accept the shipment only if the salicylic acid is more than 99% pure. To evaluate the shipment’s purity, you dissolve a 0.4208-g sample in water and titrate to the phenolphthalein end point, using 21.92 mL of 0.1354 M NaOH. Report the shipment’s purity as %w/w C H O . Salicylic acid is a diprotic weak acid with p values of 2.97 and 13.74. Because salicylic acid is a diprotic weak acid, we must first determine to which equivalence point it is being titrated. Using salicylic acid’s p values as a guide, the pH at the first equivalence point is between 2.97 and 13.74, and the second equivalence points is at a pH greater than 13.74. From , phenolphthalein’s end point is in the pH range 8.3–10.0. The titration, therefore, is to the first equivalence point for which the moles of NaOH equal the moles of salicylic acid; thus \[(0.1354 \ \mathrm{M})(0.02192 \ \mathrm{L})=2.968 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[2.968 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{138.12 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{\mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}=0.4099 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \nonumber\] \[\frac{0.4099 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}}{0.4208 \ \mathrm{g} \text { sample }} \times 100=97.41 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3} \nonumber\] Because the purity of the sample is less than 99%, we reject the shipment. In an indirect analysis the analyte participates in one or more preliminary reactions, one of which produces or consumes acid or base. Despite the additional complexity, the calculations are straightforward. The purity of a pharmaceutical preparation of sulfanilamide, C H N O S, is determined by oxidizing the sulfur to SO and bubbling it through H O to produce H SO . The acid is titrated to the bromothymol blue end point using a standard solution of NaOH. Calculate the purity of the preparation given that a 0.5136-g sample requires 48.13 mL of 0.1251 M NaOH. The bromothymol blue end point has a pH range of 6.0–7.6. Sulfuric acid is a diprotic acid, with a p of 1.99 (the first value is very large and the acid dissociation reaction goes to completion, which is why H SO4 is a strong acid). The titration, therefore, proceeds to the second equivalence point and the titration reaction is \[\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{4}^{2-}(a q) \nonumber\] Using the titration results, there are \[(0.1251 \ \mathrm{M} \ \mathrm{NaOH})(0.04813 \ \mathrm{L} \ \mathrm{NaOH})=6.021 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] \[6.012 \times 10^{-3} \text{ mol NaOH} \times \frac{1 \text{ mol} \mathrm{H}_{2} \mathrm{SO}_{4}} {2 \text{ mol NaOH}} = 3.010 \times 10^{-3} \text{ mol} \mathrm{H}_{2} \mathrm{SO}_{4} \nonumber\] \[3.010 \times 10^{-3} \ \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{4} \times \frac{1 \ \mathrm{mol} \text{ S}}{\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{4}} \times \ \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{\mathrm{mol} \text{ S}} \times \frac{168.17 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}= 0.5062 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S} \nonumber\] produced when the SO is bubbled through H O . Because all the sulfur in H SO comes from the sulfanilamide, we can use a conservation of mass to determine the amount of sulfanilamide in the sample. \[\frac{0.5062 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S}}{0.5136 \ \mathrm{g} \text { sample }} \times 100=98.56 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{6} \mathrm{H}_{4} \mathrm{N}_{2} \mathrm{O}_{2} \mathrm{S} \nonumber\] The concentration of NO in air is determined by passing the sample through a solution of H O , which oxidizes NO to HNO , and titrating the HNO with NaOH. What is the concentration of NO , in mg/L, if a 5.0 L sample of air requires 9.14 mL of 0.01012 M NaOH to reach the methyl red end point The moles of HNO produced by pulling the sample through H O is \[(0.01012 \ \mathrm{M})(0.00914 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HNO}_{3}}{\mathrm{mol} \ \mathrm{NaOH}}=9.25 \times 10^{-5} \ \mathrm{mol} \ \mathrm{HNO}_{3} \nonumber\] A conservation of mass on nitrogen requires that each mole of NO produces one mole of HNO ; thus, the mass of NO in the sample is \[9.25 \times 10^{-5} \ \mathrm{mol} \ \mathrm{HNO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{NO}_{2}}{\mathrm{mol} \ \mathrm{HNO}_{3}} \times \frac{46.01 \ \mathrm{g} \ \mathrm{NO}_{2}}{\mathrm{mol} \ \mathrm{NO}_{2}}=4.26 \times 10^{-3} \ \mathrm{g} \ \mathrm{NO}_{2} \nonumber\] and the concentration of NO is \[\frac{4.26 \times 10^{-3} \ \mathrm{g} \ \mathrm{NO}_{2}}{5 \ \mathrm{L} \text { air }} \times \frac{1000 \ \mathrm{mg}}{\mathrm{g}}=0.852 \ \mathrm{mg} \ \mathrm{NO}_{2} \ \mathrm{L} \text { air } \nonumber\] For a back titration we must consider two acid–base reactions. Again, the calculations are straightforward. The amount of protein in a sample of cheese is determined by a Kjeldahl analysis for nitrogen. After digesting a 0.9814-g sample of cheese, the nitrogen is oxidized to \(\text{NH}_4^+\), converted to NH with NaOH, and the NH distilled into a collection flask that contains 50.00 mL of 0.1047 M HCl. The excess HCl is back titrated with 0.1183 M NaOH, requiring 22.84 mL to reach the bromothymol blue end point. Report the %w/w protein in the cheese assuming there are 6.38 grams of protein for every gram of nitrogen in most dairy products. The HCl in the collection flask reacts with two bases \[\mathrm{HCl}(a q)+\mathrm{NH}_{3}(a q) \rightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{Cl}^{-}(a q) \nonumber\] \[\mathrm{HCl}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}^{-}(a q) \nonumber\] The collection flask originally contains \[(0.1047 \ \mathrm{M \ HCl})(0.05000 \ \mathrm{L \ HCl})=5.235 \times 10^{-3} \mathrm{mol} \ \mathrm{HCl} \nonumber\] of which \[(0.1183 \ \mathrm{M} \ \mathrm{NaOH})(0.02284 \ \mathrm{L} \ \mathrm{NaOH}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{NaOH}}=2.702 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] react with NaOH. The difference between the total moles of HCl and the moles of HCl that react with NaOH is the moles of HCl that react with NH . \[5.235 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl}-2.702 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} =2.533 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] Because all the nitrogen in NH comes from the sample of cheese, we use a conservation of mass to determine the grams of nitrogen in the sample. \[2.533 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \times \frac{1 \ \mathrm{mol} \ \mathrm{NH}_{3}}{\mathrm{mol} \ \mathrm{HCl}} \times \frac{14.01 \ \mathrm{g} \ \mathrm{N}}{\mathrm{mol} \ \mathrm{NH}_{3}}=0.03549 \ \mathrm{g} \ \mathrm{N} \nonumber\] The mass of protein, therefore, is \[0.03549 \ \mathrm{g} \ \mathrm{N} \times \frac{6.38 \ \mathrm{g} \text { protein }}{\mathrm{g} \ \mathrm{N}}=0.2264 \ \mathrm{g} \text { protein } \nonumber\] and the % w/w protein is \[\frac{0.2264 \ \mathrm{g} \text { protein }}{0.9814 \ \mathrm{g} \text { sample }} \times 100=23.1 \ \% \mathrm{w} / \mathrm{w} \text { protein } \nonumber\] Limestone consists mainly of CaCO , with traces of iron oxides and other metal oxides. To determine the purity of a limestone, a 0.5413-g sample is dissolved using 10.00 mL of 1.396 M HCl. After heating to expel CO , the excess HCl was titrated to the phenolphthalein end point, requiring 39.96 mL of 0.1004 M NaOH. Report the sample’s purity as %w/w CaCO . The total moles of HCl used in this analysis is \[(1.396 \ \mathrm{M})(0.01000 \ \mathrm{L})=1.396 \times 10^{-2} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] Of the total moles of HCl \[(0.1004 \ \mathrm{M} \ \mathrm{NaOH})(0.03996 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{NaOH}} =4.012 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] are consumed in the back titration with NaOH, which means that \[ 1.396 \times 10^{-2} \ \mathrm{mol} \ \mathrm{HCl}-4.012 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \\ =9.95 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \nonumber\] react with the CaCO . Because \(\text{CO}_3^{2-}\) is dibasic, each mole of CaCO consumes two moles of HCl; thus \[9.95 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HCl} \times \frac{1 \ \mathrm{mol} \ \mathrm{CaCO}_{3}}{2 \ \mathrm{mol} \ \mathrm{HCl}} \times \\ \frac{100.09 \ \mathrm{g} \ \mathrm{CaCO}_{3}}{\mathrm{mol} \ \mathrm{CaCO}_{3}}=0.498 \ \mathrm{g} \ \mathrm{CaCO}_{3} \nonumber\] \[\frac{0.498 \ \mathrm{g} \ \mathrm{CaCO}_{3}}{0.5143 \ \mathrm{g} \text { sample }} \times 100=96.8 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{CaCO}_{3} \nonumber\] Earlier we noted that we can use an acid–base titration to analyze a mixture of acids or bases by titrating to more than one equivalence point. The concentration of each analyte is determined by accounting for its contribution to each equivalence point. The alkalinity of natural waters usually is controlled by OH , \(\text{HCO}_3^-\), and \(\text{CO}_3^{2-}\), present singularly or in combination. Titrating a 100.0-mL sample to a pH of 8.3 requires 18.67 mL of 0.02812 M HCl. A second 100.0-mL aliquot requires 48.12 mL of the same titrant to reach a pH of 4.5. Identify the sources of alkalinity and their concentrations in milligrams per liter. Because the volume of titrant to reach a pH of 4.5 is more than twice that needed to reach a pH of 8.3, we know from , that the sample’s alkalinity is controlled by \(\text{CO}_3^{2-}\) and \(\text{HCO}_3^-\). Titrating to a pH of 8.3 neutralizes \(\text{CO}_3^{2-}\) to \(\text{HCO}_3^-\) \[\mathrm{CO}_{3}^{2-}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q) \nonumber\] but there is no reaction between the titrant and \(\text{HCO}_3^-\) (see ). The concentration of \(\text{CO}_3^{2-}\) in the sample, therefore, is \[{(0.02812 \ \mathrm{M \ HCl})(0.01867 \ \mathrm{L \ HCl}) \times} {\frac{1 \ \mathrm{mol} \ \mathrm{CO}_3^{2-}}{\mathrm{mol} \ \mathrm{HCl}}=5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \nonumber\] \[\frac{5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-}}{0.1000 \ \mathrm{L}} \times \frac{60.01 \ \mathrm{g} \ \mathrm{CO}_{3}^{2-}}{\mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \times \frac{1000 \ \mathrm{mg}}{\mathrm{g}}=315.1 \ \mathrm{mg} / \mathrm{L} \nonumber\] Titrating to a pH of 4.5 neutralizes \(\text{CO}_3^{2-}\) to H CO and neutralizes \(\text{HCO}_3^-\) to H CO (see ). \[\begin{array}{l}{\mathrm{CO}_{3}^{2-}(a q)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+2 \mathrm{Cl}^{-}(a q)} \\ {\mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCl}(a q) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{Cl}^{-}(a q)}\end{array} \nonumber\] Because we know how many moles of \(\text{CO}_3^{2-}\) are in the sample, we can calculate the volume of HCl it consumes. \[{5.250 \times 10^{-4} \ \mathrm{mol} \ \mathrm{CO}_{3}^{2-} \times \frac{2 \ \mathrm{mol} \ \mathrm{HCl}}{\mathrm{mol} \ \mathrm{CO}_{3}^{2-}} \times} {\frac{1 \ \mathrm{L} \ \mathrm{HCl}}{0.02812 \ \mathrm{mol} \ \mathrm{HCl}} \times \frac{1000 \ \mathrm{mL}}{\mathrm{L}}=37.34 \ \mathrm{mL} \ \mathrm{HCl}} \nonumber\] This leaves 48.12 mL–37.34 mL, or 10.78 mL of HCl to react with \(\text{HCO}_3^-\). The amount of \(\text{HCO}_3^-\) in the sample is \[{(0.02812 \ \mathrm{M \ HCl})(0.01078 \ \mathrm{L} \ \mathrm{HCl}) \times} {\frac{1 \ \mathrm{mol} \ \mathrm{H} \mathrm{CO}_{3}^{-}}{\mathrm{mol} \ \mathrm{HCl}}=3.031 \times 10^{-4} \ \mathrm{mol} \ \mathrm{HCO}_{3}^{-}} \nonumber\] The sample contains 315.1 mg \(\text{CO}_3^{2-}\)/L and 185.0 mg \(\text{HCO}_3^-\)/L Samples that contain a mixture of the monoprotic weak acids 2–methylanilinium chloride (C H NCl, p = 4.447) and 3–nitrophenol (C H NO , p = 8.39) can be analyzed by titrating with NaOH. A 2.006-g sample requires 19.65 mL of 0.200 M NaOH to reach the bromocresol purple end point and 48.41 mL of 0.200 M NaOH to reach the phenolphthalein end point. Report the %w/w of each compound in the sample. Of the two analytes, 2-methylanilinium is the stronger acid and is the first to react with the titrant. Titrating to the bromocresol purple end point, therefore, provides information about the amount of 2-methylanilinium in the sample. \[(0.200\ \mathrm{M} \ \mathrm{NaOH} )(0.01965 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{143.61 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{\mathrm{mol} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}=0.564 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl} \nonumber\] \[\frac{0.564 \ \mathrm{g} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl}}{2.006 \ \mathrm{g} \text { sample }} \times 100=28.1 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{7} \mathrm{H}_{10} \mathrm{NCl} \nonumber\] Titrating from the bromocresol purple end point to the phenolphthalein end point, a total of 48.41 mL – 19.65 mL = 28.76 mL, gives the amount of NaOH that reacts with 3-nitrophenol. The amount of 3-nitrophenol in the sample, therefore, is \[(0.200 \ \mathrm{M} \ \mathrm{NaOH}) (0.02876 \ \mathrm{L}) \times \frac{1 \ \mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{\mathrm{mol} \ \mathrm{NaOH}} \times \frac{139.11 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{\mathrm{mol} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}=0.800 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3} \nonumber\] \[\frac{0.800 \ \mathrm{g} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3}}{2.006 \ \mathrm{g} \text { sample }} \times 100=39.8 \ \% \mathrm{w} / \mathrm{w} \ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{3} \nonumber\] shows how we can use an acid–base titration to determine the forms of alkalinity in waters and their concentrations. We can extend this approach to other systems. For example, if we titrate a sample to the methyl orange end point and the phenolphthalein end point using either a strong acid or a strong base, we can determine which of the following species are present and their concentrations: H PO , \(\text{H}_2\text{PO}_4^-\), \(\text{HPO}_4^{2-}\), \(\text{PO}_4^{3-}\), HCl, and NaOH. As outlined in Table 9.2.8
, each species or mixture of species has a unique relationship between the volumes of titrant needed to reach these two end points. Note that mixtures containing three or more these species are not possible. Use a ladder diagram to convince yourself that mixtures containing three or more of these species are unstable. and are, respectively, the volume of titrant at the phenolphthalein and methyl orange end points when no information is provided, the volume at each end point is zero In addition to a quantitative analysis and a qualitative analysis, we also can use an acid–base titration to characterize the chemical and physical properties of matter. Two useful characterization applications are the determination of a compound’s equivalent weight and the determination of its acid dissociation constant or its base dissociation constant. Suppose we titrate a sample of an impure weak acid to a well-defined end point using a monoprotic strong base as the titrant. If we assume the titration involves the transfer of protons, then the moles of titrant needed to reach the end point is \[\text { moles titrant }=\frac{n \text { moles titrant }}{\text { moles analyte }} \times \text { moles analyte } \nonumber\] If we know the analyte’s identity, we can use this equation to determine the amount of analyte in the sample \[\text { grams analyte }=\text { moles titrant } \times \frac{1 \text { mole analyte }}{n \text { moles analyte }} \times F W \text { analyte } \nonumber\] where is the analyte’s formula weight. But what if we do not know the analyte’s identify? If we titrate a pure sample of the analyte, we can obtain some useful information that may help us establish its identity. Because we do not know the number of protons that are titrated, we let = 1 and replace the analyte’s formula weight with its equivalent weight ( ) \[\text { grams analyte }=\text { moles titrant } \times \frac{1 \text { equivalent analyte }}{1 \text { mole analyte }}=E W \text { analyte } \nonumber\] where \[F W=n \times E W \nonumber\] A 0.2521-g sample of an unknown weak acid is titrated with 0.1005 M NaOH, requiring 42.68 mL to reach the phenolphthalein end point. Determine the compound’s equivalent weight. Which of the following compounds is most likely to be the unknown weak acid? The moles of NaOH needed to reach the end point is \[(0.1005 \ \mathrm{M} \ \mathrm{NaOH})(0.04268 \ \mathrm{L} \ \mathrm{NaOH})=4.289 \times 10^{-3} \ \mathrm{mol} \ \mathrm{NaOH} \nonumber\] The equivalents of weak acid are the same as the moles of NaOH used in the titration; thus, he analyte’s equivalent weight is \[E W=\frac{0.2521 \ \mathrm{g}}{4.289 \times 10^{-3} \text { equivalents }}=58.78 \ \mathrm{g} / \mathrm{equivalent} \nonumber\] The possible formula weights for the weak acid are 58.78 g/mol ( = 1), 117.6 g/mol ( = 2), and 176.3 g/mol ( = 3). If the analyte is a monoprotic weak acid, then its formula weight is 58.78 g/mol, eliminating ascorbic acid as a possibility. If it is a diprotic weak acid, then the analyte’s formula weight is either 58.78 g/mol or 117.6 g/mol, depending on whether the weak acid was titrated to its first or its second equivalence point. Succinic acid, with a formula weight of 118.1 g/mole is a possibility, but malonic acid is not. If the analyte is a triprotic weak acid, then its formula weight is 58.78 g/mol, 117.6 g/mol, or 176.3 g/mol. None of these values is close to the formula weight for citric acid, eliminating it as a possibility. Only succinic acid provides a possible match. Figure 9.2.16
shows the potentiometric titration curve for the titration of a 0.500-g sample an unknown weak acid. The titrant is 0.1032 M NaOH. What is the weak acid’s equivalent weight? The first of the two visible end points is approximately 37 mL of NaOH. The analyte’s equivalent weight, therefore, is \[(0.1032 \ \mathrm{M} \ \mathrm{NaOH})(0.037 \ \mathrm{L}) \times \frac{1 \text { equivalent }}{\mathrm{mol} \ \mathrm{NaOH}}=3.8 \times 10^{-3} \text { equivalents } \nonumber\] \[E W=\frac{0.5000 \ \mathrm{g}}{3.8 \times 10^{-3} \text { equivalents }}=1.3 \times 10^{2} \ \mathrm{g} / \mathrm{equivalent} \nonumber\] Another application of acid–base titrimetry is the determination of a weak acid’s or a weak base’s dissociation constant. Consider, for example, a solution of acetic acid, CH COOH, for which the dissociation constant is \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\] When the concentrations of CH COOH and CH COO are equal, the expression reduces to = [H O ], or pH = p . If we titrate a solution of acetic acid with NaOH, the pH equals the p when the volume of NaOH is approximately 1⁄2 . As shown in Figure 9.2.17
, a potentiometric titration curve provides a reasonable estimate of acetic acid’s p . Recall that pH = p is a step on a ladder diagram, which divides the pH axis into two regions, one where the weak acid is the predominate species, and one where its conjugate weak base is the predominate species. This method provides a reasonable estimate for a weak acid’s p if the acid is neither too strong nor too weak. These limitations are easy to appreciate if we consider two limiting cases. For the first limiting case, let’s assume the weak acid, HA, is more than 50% dissociated before the titration begins (a relatively large value); in this case the concentration of HA before the equivalence point is always less than the concentration of A and there is no point on the titration curve where [HA] = [A ]. At the other extreme, if the acid is too weak, then less than 50% of the weak acid reacts with the titrant at the equivalence point. In this case the concentration of HA before the equivalence point is always greater than that of A . Determining the p by the half-equivalence point method overestimates its value if the acid is too strong and underestimates its value if the acid is too weak. Use the potentiometric titration curve in to estimate the p values for the weak acid in . At 1⁄2 , or approximately 18.5 mL, the pH is approximately 2.2; thus, we estimate that the analyte’s p is 2.2. A second approach for determining a weak acid’s p is to use a Gran plot. For example, earlier in this chapter we derived the following equation for the titration of a weak acid with a strong base. \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \times V_{b}=K_{a} V_{e q}-K_{a} V_{b} \nonumber\] A plot of [H O ] \(\times\) versus for volumes less than the equivalence point yields a straight line with a slope of – . Other linearizations have been developed that use the entire titration curve or that require no assumptions [(a) Gonzalez, A. G.; Asuero, A. G. , , 29–33; (b) Papanastasiou, G.; Ziogas, I.; Kokkindis, G. , , 119–135]. This approach to determining an acidity constant has been used to study the acid–base properties of humic acids, which are naturally occurring, large molecular weight organic acids with multiple acidic sites. In one study a humic acid was found to have six titratable sites, three which were identified as carboxylic acids, two which were believed to be secondary or tertiary amines, and one which was identified as a phenolic group [Alexio, L. M.; Godinho, O. E. S.; da Costa, W. F. , , 35–39]. Values of determined by this method may have a substantial error if the effect of activity is ignored. See for a discussion of activity. In an acid–base titration, the volume of titrant needed to reach the equivalence point is proportional to the moles of titrand. Because the pH of the titrand or the titrant is a function of its concentration, the change in pH at the equivalence point—and thus the feasibility of an acid–base titration—depends on their respective concentrations. Figure 9.2.18
, for example, shows a series of titration curves for the titration of several concentrations of HCl with equimolar solutions NaOH. For titrand and titrant concentrations smaller than 10 M, the change in pH at the end point is too small to provide an accurate and a precise result. Acid–base titrimetry is an example of a total analysis technique in which the signal is proportional to the absolute amount of analyte. See for a discussion of the difference between total analysis techniques and concentration techniques. A minimum concentration of 10 M places limits on the smallest amount of analyte we can analyze successfully. For example, suppose our analyte has a formula weight of 120 g/mol. To successfully monitor the titration’s end point using an indicator or a pH probe, the titrand needs an initial volume of approximately 25 mL. If we assume the analyte’s formula weight is 120 g/mol, then each sample must contain at least 3 mg of analyte. For this reason, acid–base titrations generally are limited to major and minor analytes. We can extend the analysis of gases to trace analytes by pulling a large volume of the gas through a suitable collection solution. We need a volume of titrand sufficient to cover the tip of the pH probe or to allow for an easy observation of the indicator’s color. A volume of 25 mL is not an unreasonable estimate of the minimum volume. One goal of analytical chemistry is to extend analyses to smaller samples. Here we describe two interesting approaches to titrating μL and pL samples. In one experimental design (Figure 9.2.19
), samples of 20–100 μL are held by capillary action between a flat-surface pH electrode and a stainless steel sample stage [Steele, A.; Hieftje, G. M. , , 2884–2888]. The titrant is added using the oscillations of a piezoelectric ceramic device to move an angled glass rod in and out of a tube connected to a reservoir that contains the titrant. Each time the glass tube is withdrawn an approximately 2 nL microdroplet of titrant is released. The microdroplets are allowed to fall onto the sample, with mixing accomplished by spinning the sample stage at 120 rpm. A total of 450 microdroplets, with a combined volume of 0.81–0.84 μL, is dispensed between each pH measurement. In this fashion a titration curve is constructed. This method has been used to titrate solutions of 0.1 M HCl and 0.1 M CH COOH with 0.1 M NaOH. Absolute errors ranged from a minimum of +0.1% to a maximum of –4.1%, with relative standard deviations from 0.15% to 4.7%. Samples as small as 20 μL were titrated successfully. Another approach carries out the acid–base titration in a single drop of solution [(a) Gratzl, M.; Yi, C. , , 2085–2088; (b) Yi, C.; Gratzl, M. , , 1976–1982; (c) Hui, K. Y.; Gratzl, M. , , 695–698; (d) Yi, C.; Huang, D.; Gratzl, M. , , 1580–1584; (e) Xie, H.; Gratzl, M. , , 3665–3669]. The titrant is delivered using a microburet fashioned from a glass capillary micropipet (Figure 9.2.20
). The microburet has a 1-2 μm tip filled with an agar gel membrane. The tip of the microburet is placed within a drop of the sample solution, which is suspended in heptane, and the titrant is allowed to diffuse into the sample. The titration’s progress is monitored using an acid–base indicator and the time needed to reach the end point is measured. The rate of the titrant’s diffusion from the microburet is determined by a prior calibration. Once calibrated the end point time is converted to an end point volume. Samples usually consist of picoliter volumes (10 liters), with the smallest sample being 0.7 pL. The precision of the titrations is about 2%. Titrations conducted with microliter or picoliter sample volumes require a smaller absolute amount of analyte. For example, diffusional titrations have been conducted on as little as 29 femtomoles (10 moles) of nitric acid. Nevertheless, the analyte must be present in the sample at a major or minor level for the titration to give accurate and precise results. When working with a macro–major or a macro–minor sample, an acid–base titration can achieve a relative error of 0.1–0.2%. The principal limitation to accuracy is the difference between the end point and the equivalence point. An acid–base titration’s relative precision depends primarily on the precision with which we can measure the end point volume and the precision in detecting the end point. Under optimum conditions, an acid–base titration has a relative precision of 0.1–0.2%. We can improve the relative precision by using the largest possible buret and by ensuring we use most of its capacity in reaching the end point. A smaller volume buret is a better choice when using costly reagents, when waste disposal is a concern, or when we must complete the titration quickly to avoid competing chemical reactions. An automatic titrator is particularly useful for titrations that require small volumes of titrant because it provides significantly better precision (typically about ±0.05% of the buret’s volume). The precision of detecting the end point depends on how it is measured and the slope of the titration curve at the end point. With an indicator the precision of the end point signal usually is ±0.03–0.10 mL. Potentiometric end points usually are more precise. For an acid–base titration we can write the following general analytical equation to express the titrant’s volume in terms of the amount of titrand \[\text { volume of titrant }=k \times \text { moles of titrand } \nonumber\] where , the sensitivity, is determined by the stoichiometry between the titrand and the titrant. Consider, for example, the determination of sulfurous acid, H SO , by titrating with NaOH to the first equivalence point \[\mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{OH}^{-}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l )+\mathrm{HSO}_{3}^{-}(a q) \nonumber\] At the equivalence point the relationship between the moles of NaOH and the moles of H SO is \[\mathrm{mol} \ \mathrm{NaOH}=\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] Substituting the titrant’s molarity and volume for the moles of NaOH and rearranging \[M_{\mathrm{NaOH}} \times V_{\mathrm{NNOH}}=\mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] \[V_{\mathrm{NaOH}}=\frac{1}{M_{\mathrm{NaOH}}} \times \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] we find that is \[k=\frac{1}{M_{\mathrm{NaOH}}} \nonumber\] There are two ways in which we can improve a titration’s sensitivity. The first, and most obvious, is to decrease the titrant’s concentration because it is inversely proportional to the sensitivity, . The second approach, which applies only if the titrand is multiprotic, is to titrate to a later equivalence point. If we titrate H SO to its second equivalence point \[ \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+2 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{3}^{2-}(a q)\nonumber\] then each mole of H SO consumes two moles of NaOH \[\mathrm{mol} \ \mathrm{NaOH}=2 \times \mathrm{mol} \ \mathrm{H}_{2} \mathrm{SO}_{3} \nonumber\] and the sensitivity becomes \[k=\frac{2}{M_{\mathrm{NaOH}}} \nonumber\] In practice, however, any improvement in sensitivity is offset by a decrease in the end point’s precision if a larger volume of titrant requires us to refill the buret. For this reason, standard acid–base titrimetric procedures are written to ensure that a titration uses 60–100% of the buret’s volume. Acid–base titrants are not selective. A strong base titrant, for example, reacts with all acids in a sample, regardless of their individual strengths. If the titrand contains an analyte and an interferent, then selectivity depends on their relative acid strengths. Let’s consider two limiting situations. If the analyte is a stronger acid than the interferent, then the titrant will react with the analyte before it begins reacting with the interferent. The feasibility of the analysis depends on whether the titrant’s reaction with the interferent affects the accurate location of the analyte’s equivalence point. If the acid dissociation constants are substantially different, the end point for the analyte can be determined accurately. Conversely, if the acid dissociation constants for the analyte and interferent are similar, then there may not be an accurate end point for the analyte. In the latter case a quantitative analysis for the analyte is not possible. In the second limiting situation the analyte is a weaker acid than the interferent. In this case the volume of titrant needed to reach the analyte’s equivalence point is determined by the concentration of both the analyte and the interferent. To account for the interferent’s contribution to the end point, an end point for the interferent must be available. Again, if the acid dissociation constants for the analyte and interferent are significantly different, then the analyte’s determination is possible. If the acid dissociation constants are similar, however, there is only a single equivalence point and we cannot separate the analyte’s and the interferent’s contributions to the equivalence point volume. Acid–base titrations require less time than most gravimetric procedures, but more time than many instrumental methods of analysis, particularly when analyzing many samples. With an automatic titrator, however, concerns about analysis time are less significant. When performing a titration manually our equipment needs—a buret and, perhaps, a pH meter—are few in number, inexpensive, routinely available, and easy to maintain. Automatic titrators are available for between $3000 and $10 000. | 98,786 | 3,636 |
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Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation. . Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure \(\Page {1}\)). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down. Waves have characteristic properties (Figure \(\Page {2}\)). As you may have noticed in Figure \(\Page {1}\), waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (λ), distance between two corresponding points in a wave—between the midpoints of two peaks or two troughs. \(\lambda\) is the lowercase Greek lambda, and \( u\) is the lowercase Greek nu. Wavelengths are described by a unit of distance, typically meters. The frequency (ν), the number of oscillations (i.e., of a wave) that pass a particular point in a given period of time. of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations persecond (1/s = s ), which in the SI system is called the hertz (Hz). \[\begin{align} \text{(wavelength)(frequency)} &= speed \\[4pt] \lambda u &=v \label{eq1a} \\[4pt] \left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{wave}}{second} \right ) &=\dfrac{meters}{second} \label{eq1b} \end{align}\] Be careful not to confuse the symbols for the speed, \(v\), with the frequency, \( u\). Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs. Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation, which is energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields. (Figure \(\Page {3}\)). Some forms of electromagnetic radiation are shown in Figure \(\Page {4}\). In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light (c), which is the speed with which all forms of electromagnetic , a fundamental physical constant with a value of 2.99792458 × 10 m/s (which is about 3.00 ×10 m/s or 1.86 × 10 mi/s). This is about a million times faster than the speed of sound. Because the various kinds of electromagnetic radiation all have the same speed ( ), they differ in only wavelength and frequency. As shown in Figure \(\Page {4}\) and Table \(\Page {1}\), the wavelengths of familiar electromagnetic radiation range from 10 m for radio waves to 10 m for gamma rays, which are emitted by nuclear reactions. By replacing with in Equation \(\Page {1}\), we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength: \[ \begin{array}{cc} c=\lambda u \\ u =\dfrac{c}{\lambda } \end{array} \label{eq2} \] For example, the frequency of radio waves is about 10 Hz, whereas the frequency of gamma rays is about 10 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10 m (700 nm, or 4.3 × 10 Hz) and 4 × 10 m (400 nm, or 7.5 × 10 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly when frequency decreases, the wavelength increases. Within this visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in Figure \(\Page {4}\). A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms. As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength: \[ E\; \propto\; u \label{\(\Page {3}\)}\] \[ E\; \propto\; \dfrac{1}{\lambda } \label{\(\Page {4}\)}\] Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. The energy of electromagnetic radiation with frequency and wavelength. Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation? frequency wavelength Substitute the value for the speed of light in meters per second into Equation \(\Page {2}\) to calculate the wavelength in meters. From Equation \ref{eq2}, we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 10 m/s: \[λν = c = 2.998 \times 10^8 m/s\] Thus the wavelength λ is given by \[ \lambda =\dfrac{c}{ u }=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right )=2.965\; m \] As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun? 8.45 mm Electromagnetic Radiation: Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A basic knowledge of the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A is a periodic oscillation by which energy is transmitted through space. All waves are , repeating regularly in both space and time. Waves are characterized by several interrelated properties: , the distance between successive waves; , the number of waves that pass a fixed point per unit time; , the rate at which the wave propagates through space; and , the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. consists of two perpendicular waves, one electric and one magnetic, propagating at the . Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. ( ) | 7,772 | 3,637 |
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To date, about 115 different elements have been discovered; by definition, each is chemically unique. To understand why they are unique, you need to understand the structure of the atom (the fundamental, individual particle of an element) and the characteristics of its components. In most cases, the symbols for the elements are derived directly from each element’s name, such as C for carbon, U for uranium, Ca for calcium, and Po for polonium. Elements have also been named for their properties [such as radium (Ra) for its radioactivity], for the native country of the scientist(s) who discovered them [polonium (Po) for Poland], for eminent scientists [curium (Cm) for the Curies], for gods and goddesses [selenium (Se) for the Greek goddess of the moon, Selene], and for other poetic or historical reasons. Some of the symbols used for elements that have been known since antiquity are derived from historical names that are no longer in use; only the symbols remain to indicate their origin. Examples are Fe for iron, from the Latin ; Na for sodium, from the Latin ; and W for tungsten, from the German . Examples are in What single parameter uniquely characterizes the atom of a given element? It is not the atom's relative mass, as we will see in the section on isotopes below. It is, rather, the number of protons in the nucleus, which we call the and denote by the symbol . Each proton carries an electric charge of +1, so the atomic number also specifies the electric charge of the nucleus. In the neutral atom, the within the nucleus are balanced by outside it. Atomic numbers were first worked out in 1913 by Henry Moseley, a young member of Rutherford's research group in Manchester. Moseley searched for a measurable property of each element that increases linearly with atomic number. He found this in a class of X-rays emitted by an element when it is bombarded with electrons. The frequencies of these X-rays are unique to each element, and they increase uniformly in successive elements. Moseley found that the square roots of these frequencies give a straight line when plotted against Z; this enabled him to sort the elements in order of increasing atomic number. You can think of the atomic number as a kind of serial number of an element, commencing at 1 for hydrogen and increasing by one for each successive element. The chemical name of the element and its symbol are uniquely tied to the atomic number; thus the symbol "Sr" stands for strontium, whose atoms all have = 38. This is just the sum of the numbers of protons and neutrons in the nucleus. It is sometimes represented by the symbol , so \[A = Z + N \label{2.4.1}\] in which is the atomic number and is the . The term simply refers to any particular kind of nucleus. For example, a nucleus of atomic number 7 is a nuclide of nitrogen. Any nuclide is characterized by the pair of numbers ( ). The element symbol depends on alone, so the symbol Mg is used to specify the mass-26 nuclide of manganese, whose name implies =12. A more explicit way of denoting a particular kind of nucleus is to add the atomic number as a subscript. Of course, this is somewhat redundant, since the symbol Mg implies =12, but it is sometimes a convenience when discussing several nuclides. Figure \(\Page {2}\): Formalism used for identifying specific nuclide (any particular kind of nucleus) The element carbon (C) has an atomic number of 6, which means that all neutral carbon atoms contain 6 protons and 6 electrons. In a typical sample of carbon-containing material, 98.89% of the carbon atoms also contain 6 neutrons, so each has a mass number of 12. An isotope of any element can be uniquely represented as \(^A_Z X\), where X is the atomic symbol of the element. The isotope of carbon that has 6 neutrons is therefore \(_6^{12} C\). The subscript indicating the atomic number is actually because the atomic symbol already uniquely specifies Z. Consequently, \(_6^{12} C\) is more often written as C, which is read as “carbon-12.” Nevertheless, the value of Z is commonly included in the notation for nuclear reactions because these reactions involve changes in Z. Recall that the nuclei of most atoms contain neutrons as well as protons. Unlike protons, the number of neutrons is not absolutely fixed for most elements. Atoms that have the same number of protons, and hence the same atomic number, but different numbers of neutrons are called isotopes. All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons. Two nuclides having the atomic number but different mass numbers are known as . Most elements occur in nature as mixtures of isotopes, but twenty-three of them (including beryllium and fluorine, shown in the table) are monoisotopic. For example, there are three of magnesium: Mg (79% of all Mg atoms), Mg (10%), and Mg (11%); all three are present in all compounds of magnesium in about these same proportions. Approximately 290 isotopes occur in nature. The two heavy isotopes of hydrogen are especially important— so much so that they have names and symbols of their own: \[\underset{\text{protium}}{\ce{^1_1H} } \label{2.4.2a}\] \[\underset{\text{deuterium}}{\ce{^2_1H \equiv D}} \label{2.4.2b}\] \[\underset{\text{tritium}}{\ce{^2_1H \equiv T}} \label{2.4.2c}\] Deuterium accounts for only about 15 out of every one million atoms of hydrogen. Tritium, which is radioactive, is even less abundant. All the tritium on the earth is a by-product of the decay of other radioactive elements. For carbon, in addition to \(^{12}C\), a typical sample of carbon contains 1.11% \(_6^{13} C\) ( C), with 7 neutrons and 6 protons, and a trace of \(_6^{14} C\) ( C), with 8 neutrons and 6 protons. The nucleus of C is not stable, however, but undergoes a slow radioactive decay that is the basis of the carbon-14 dating technique used in archeology. Many elements other than carbon have more than one stable isotope; tin, for example, has 10 isotopes. The properties of some common isotopes are in Sources of isotope data: G. Audi et al., Nuclear Physics A 729 (2003): 337–676; J. C. Kotz and K. F. Purcell, Chemistry and Chemical Reactivity, 2nd ed., 1991. How Elements Are Represented on the Periodic Table: An element with three stable isotopes has 82 protons. The separate isotopes contain 124, 125, and 126 neutrons. Identify the element and write symbols for the isotopes. : number of protons and neutrons : element and atomic symbol : : The element with 82 protons (atomic number of 82) is lead: Pb. For the first isotope, A = 82 protons + 124 neutrons = 206. Similarly, A = 82 + 125 = 207 and A = 82 + 126 = 208 for the second and third isotopes, respectively. The symbols for these isotopes are \(^{206}_{82}Pb\), \(^{207}_{82}Pb\), and \(^{208}_{82}Pb\), which are usually abbreviated as \(^{206}Pb\), \(^{207}Pb\), and \(^{208}Pb\). Identify the element with 35 protons and write the symbols for its isotopes with 44 and 46 neutrons. : \(\ce{^{79}_{35}Br}\) and \(\ce{^{81}_{35}Br}\) or, more commonly, \(\ce{^{79}Br}\) and \(\ce{^{81}Br}\). Each atom of an element contains the same number of protons, which is the ( ). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called . Each isotope of a given element has the same atomic number but a different ( ), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the ( ), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. | 7,826 | 3,638 |
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In the years following Kekulé brilliant proposal for the structure of benzene, organic chemistry underwent a tremendous expansion, and in the process a wide variety of paradigms or working hypotheses were developed about what kinds of compounds could "exist" and what kinds of reactions could occur. In many cases, acceptance of these hypotheses appeared to stifle many possible lines of investigation and caused contrary evidence to be pigeonholed as "interesting but not conclusive." As one example, the paradigm of angle strain was believed to wholly preclude substances that we know now are either stable or important reaction intermediates, such as cubane ( ), cyclopropanone ( ), and benzyne ( and ). No paradigm did more to retard the development of organic chemistry than the notion that, with a "few" exceptions, compounds with bonds between carbon and transition metals (Fe, Co, Ni, Ti, and so on) are inherently unstable. This idea was swept away in 1951 with the discovery of ferrocene (\(C_5H_5)_2Fe\)) by P. L. Pauson. Ferrocene has unheard of properties for an organoiron compound, stable to more than 500 and able to be dissolved in, and recovered from, concentrated sulfuric acid! Pauson's work started an avalanche of research on transition metals in the general area between organic and inorganic chemistry, which has flourished ever since and has led to an improved understanding of important biochemical processes.
and (1977) | 1,471 | 3,639 |
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. In addition to these individual basis; please contact Describe the differences in behavior between NaOH and CH OH in aqueous solution. Which solution would be a better conductor of electricity? Explain your reasoning. What is the relationship between the strength of the electrostatic attraction between oppositely charged ions and the distance between the ions? How does the strength of the electrostatic interactions change as the size of the ions increases? Which will result in the release of more energy: the interaction of a gaseous sodium ion with a gaseous oxide ion or the interaction of a gaseous sodium ion with a gaseous bromide ion? Why? Which will result in the release of more energy: the interaction of a gaseous chloride ion with a gaseous sodium ion or a gaseous potassium ion? Explain your answer. What are the predominant interactions when oppositely charged ions are Several factors contribute to the stability of ionic compounds. Describe one type of interaction that ionic compounds. Describe the interactions that ionic compounds. What is the relationship between the electrostatic attractive energy between charged particles and the distance between the particles? The interaction of a sodium ion and an oxide ion. The electrostatic attraction energy between ions of opposite charge is directly proportional to the charge on each ion ( and in Equation 9.1). Thus, more energy is released as the charge on the ions increases (assuming the internuclear distance does not increase substantially). A sodium ion has a +1 charge; an oxide ion, a −2 charge; and a bromide ion, a −1 charge. For the interaction of a sodium ion with an oxide ion, = +1 and = −2, whereas for the interaction of a sodium ion with a bromide ion, = +1 and = −1. The larger value of × for the sodium ion–oxide ion interaction means it will release more energy. How does the energy of the electrostatic interaction between ions with charges +1 and −1 compare to the interaction between ions with charges +3 and −1 if the distance between the ions is the same in both cases? How does this compare with the magnitude of the interaction between ions with +3 and −3 charges? How many grams of gaseous MgCl are needed to give the same electrostatic attractive energy as 0.5 mol of gaseous LiCl? The ionic radii are Li = 76 pm, Mg = 72 pm, and Cl = 181 pm. Sketch a diagram showing the relationship between potential energy and internuclear distance (from = ∞ to = 0) for the interaction of a bromide ion and a potassium ion to form gaseous KBr. Explain why the energy of the system increases as the distance between the ions decreases from = to = 0. Calculate the magnitude of the electrostatic attractive energy ( , in kilojoules) for 85.0 g of gaseous SrS ion pairs. The observed internuclear distance in the gas phase is 244.05 pm. What is the electrostatic attractive energy ( , in kilojoules) for 130 g of gaseous HgI ? The internuclear distance is 255.3 pm. According to Equation 9.1, in the first case = (+1)(−1) = −1; in the second case, = (+3)(−1) = −3. Thus, will be three times larger for the +3/−1 ions. For +3/−3 ions, = (+3)(−3) = −9, so will be nine times larger than for the +1/−1 ions. At , the energy of the system increases due to electron–electron repulsions between the overlapping electron distributions on ions. At very short internuclear distances, electrostatic repulsions between adjacent also become important. What is the bond order about the central atom(s) of hydrazine (N H ), nitrogen, and diimide (N H )? Draw Lewis electron structures for each compound and then arrange these compounds in order of increasing N–N bond distance. Which of these compounds would you expect to have the largest N–N bond energy? Explain your answer. What is the carbon–carbon bond order in ethylene (C H ), BrH CCH Br, and FCCH? Arrange the compounds in order of increasing C–C bond distance. Which would you expect to have the largest C–C bond energy? Why? From each pair of elements, select the one with the greater bond strength? Explain your choice in each case. From each pair of elements, select the one with the greater bond strength? Explain your choice in each case. Approximately how much energy per mole is required to completely dissociate acetone [(CH ) CO] and urea [(NH ) CO] into their constituent atoms? Approximately how much energy per mole is required to completely dissociate ethanol, formaldehyde, and hydrazine into their constituent atoms? Is the reaction of diimine (N H ) with oxygen to produce nitrogen and water exothermic or endothermic? Quantify your answer. N H , bond order 1; N H , bond order 2; N , bond order 3; N–N bond distance: N < N H < N H ; Largest bond energy: N ; Highest bond order correlates with strongest and shortest bond. Why do ionic compounds such as KI exhibit substantially less than 100% ionic character in the gas phase? Of the compounds LiI and LiF, which would you expect to behave more like a classical ionic compound? Which would have the greater dipole moment in the gas phase? Explain your answers. Predict whether each compound is purely covalent, purely ionic, or polar covalent. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Based on relative electronegativities, classify the bonding in each compound as ionic, covalent, or polar covalent. Indicate the direction of the bond dipole for each polar covalent bond. Classify each species as having 0%–40% ionic character, 40%–60% ionic character, or 60%–100% ionic character based on the type of bonding you would expect. Justify your reasoning. If the bond distance in HCl (dipole moment = 1.109 D) were double the actual value of 127.46 pm, what would be the effect on the charge localized on each atom? What would be the percent negative charge on Cl? At the actual bond distance, how would doubling the charge on each atom affect the dipole moment? Would this represent more ionic or covalent character? Calculate the percent ionic character of HF (dipole moment = 1.826 D) if the H–F bond distance is 92 pm. Calculate the percent ionic character of CO (dipole moment = 0.110 D) if the C–O distance is 113 pm. Calculate the percent ionic character of PbS and PbO in the gas phase, given the following information: for PbS, = 228.69 pm and µ = 3.59 D; for PbO, = 192.18 pm and µ = 4.64 D. Would you classify these compounds as having covalent or polar covalent bonds in the solid state? Compare and contrast covalent and ionic compounds with regard to What are the similarities between plots of the overall energy versus internuclear distance for an ionic compound and a covalent compound? Why are the plots so similar? Which atom do you expect to be the central atom in each of the following species? Which atom is the central atom in each of the following species? What is the relationship between the number of bonds typically formed by the period 2 elements in groups 14, 15, and 16 and their Lewis electron structures? Although formal charges do not represent actual charges on atoms in molecules or ions, they are still useful. Why? Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate? Give the electron configuration and the Lewis dot symbol for the following. How many more electrons can each atom accommodate? Based on Lewis dot symbols, predict the preferred oxidation state of Be, F, B, and Cs. Based on Lewis dot symbols, predict the preferred oxidation state of Br, Rb, O, Si, and Sr. Based on Lewis dot symbols, predict how many bonds gallium, silicon, and selenium will form in their neutral compounds. Determine the total number of valence electrons in the following. Determine the total number of valence electrons in the following. Draw Lewis electron structures for the following. Draw Lewis electron structures for the following. Draw Lewis electron structures for CO , NO , SO , and NO . From your diagram, predict which pair(s) of compounds have similar electronic structures. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed. Write Lewis dot symbols for each pair of elements. For a reaction between each pair of elements, predict which element is the oxidant, which element is the reductant, and the final stoichiometry of the compound formed. Use Lewis dot symbols to predict whether ICl and NO are chemically reasonable formulas. Draw a plausible Lewis electron structure for a compound with the molecular formula Cl PO. Draw a plausible Lewis electron structure for a compound with the molecular formula CH O. While reviewing her notes, a student noticed that she had drawn the following structure in her notebook for acetic acid: Why is this structure not feasible? Draw an acceptable Lewis structure for acetic acid. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures. A student proposed the following Lewis structure shown for acetaldehyde. Why is this structure not feasible? Draw an acceptable Lewis structure for acetaldehyde. Show the formal charges of all nonhydrogen atoms in both the correct and incorrect structures. Draw the most likely structure for HCN based on formal charges, showing the formal charge on each atom in your structure. Does this compound have any plausible resonance structures? If so, draw one. Draw the most plausible Lewis structure for NO . Does this ion have any other resonance structures? Draw at least one other Lewis structure for the nitrate ion that is not plausible based on formal charges. At least two Lewis structures can be drawn for BCl . Using arguments based on formal charges, explain why the most likely structure is the one with three B–Cl single bonds. Using arguments based on formal charges, explain why the most feasible Lewis structure for SO has two sulfur–oxygen double bonds. At least two distinct Lewis structures can be drawn for N . Use arguments based on formal charges to explain why the most likely structure contains a nitrogen–nitrogen double bond. Is H–O–N=O a reasonable structure for the compound HNO ? Justify your answer using Lewis electron dot structures. Is H–O=C–H a reasonable structure for a compound with the formula CH O? Use Lewis electron dot structures to justify your answer. Explain why the following Lewis structure for SO is or is not reasonable. [Ar]4 3 4 Selenium can accommodate two more electrons, giving the Se ion. [Ar]4 3 4 Krypton has a closed shell electron configuration, so it cannot accommodate any additional electrons. 1 2 Lithium can accommodate one additional electron in its 2 orbital, giving the Li ion. [Kr]5 Strontium has a filled 5 subshell, and additional electrons would have to be placed in an orbital with a higher energy. Thus strontium has no tendency to accept an additional electron. 1 Hydrogen can accommodate one additional electron in its 1 orbital, giving the H ion. Be , F , B , Cs K is the reductant; S is the oxidant. The final stoichiometry is K S. Sr is the reductant; Br is the oxidant. The final stoichiometry is SrBr . Al is the reductant; O is the oxidant. The final stoichiometry is Al O . Mg is the reductant; Cl is the oxidant. The final stoichiometry is MgCl . The only structure that gives both oxygen and carbon an octet of electrons is the following: The student’s proposed structure has two flaws: the hydrogen atom with the double bond has four valence electrons (H can only accommodate two electrons), and the carbon bound to oxygen only has six valence electrons (it should have an octet). An acceptable Lewis structure is The formal charges on the correct and incorrect structures are as follows: The most plausible Lewis structure for NO is: There are three equivalent resonance structures for nitrate (only one is shown), in which nitrogen is doubly bonded to one of the three oxygens. In each resonance structure, the formal charge of N is +1; for each singly bonded O, it is −1; and for the doubly bonded oxygen, it is 0. The following is an example of a Lewis structure that is plausible: This structure nitrogen has six bonds (nitrogen can form only four bonds) and a formal charge of –1. With four S–O single bonds, each oxygen in SO has a formal charge of −1, and the central sulfur has a formal charge of +2. With two S=O double bonds, only two oxygens have a formal charge of –1, and sulfur has a formal charge of zero. Lewis structures that minimize formal charges tend to be lowest in energy, making the Lewis structure with two S=O double bonds the most probable. Yes. This is a reasonable Lewis structure, because the formal charge on all atoms is zero, and each atom (except H) has an octet of electrons. 3. Below are the all Lewis dot structure with formal charges (in red) for Sulfate ( ). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms. 4. Below is the resonance for , formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge. 5. The resonance for , and the formal charges (in red). 6. The resonance for , and the formal charges (in red). 7. The resonance hybrid for , hybrid bonds are in red. 8. The resonance hybrid for , hybrid bonds are in red. What is the major weakness of the Lewis system in predicting the electron structures of PCl and other species containing atoms from period 3 and beyond? The compound aluminum trichloride consists of Al Cl molecules with the following structure (lone pairs of electrons removed for clarity): Does this structure satisfy the octet rule? What is the formal charge on each atom? Given the chemical similarity between aluminum and boron, what is a plausible explanation for the fact that aluminum trichloride forms a dimeric structure rather than the monomeric trigonal planar structure of BCl ? Draw Lewis electron structures for ClO , IF , SeCl , and SbF . Draw Lewis electron structures for ICl , Cl PO, Cl SO, and AsF . Draw plausible Lewis structures for the phosphate ion, including resonance structures. What is the formal charge on each atom in your structures? Draw an acceptable Lewis structure for PCl , a compound used in manufacturing a form of cellulose. What is the formal charge of the central atom? What is the oxidation number of the central atom? Using Lewis structures, draw all of the resonance structures for the BrO ion. Draw an acceptable Lewis structure for xenon trioxide (XeO ), including all resonance structures. ClO (one of four equivalent resonance structures) The formal charge on phosphorus is 0, while three oxygen atoms have a formal charge of −1 and one has a formal charge of zero. or 1. 2. 3. 10 (Sodium and higher) 4. 5. | 15,201 | 3,641 |
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Ladder diagrams are a useful tool for evaluating chemical reactivity and for providing a reasonable estimate of a chemical system’s composition at equilibrium. If we need a more exact quantitative description of the equilibrium condition, then a ladder diagram is insufficient; instead, we need to find an algebraic solution. In this section we will learn how to set‐up and solve equilibrium problems. We will start with a simple problem and work toward more complex problems. If we place an insoluble compound such as Pb(IO ) in deionized water, the solid dissolves until the concentrations of Pb and \(\text{IO}_3^-\) satisfy the solubility product for Pb(IO ) . At equilibrium the solution is saturated with Pb(IO ) , which means simply that no more solid can dissolve. How do we determine the equilibrium concentrations of Pb and \(\text{IO}_3^-\), and what is the molar solubility of Pb(IO ) in this saturated solution? When we first add solid Pb(IO ) to water, the concentrations of Pb and \(\text{IO}_3^-\) are zero and the reaction quotient, , is \[Q_r = \left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_{3}^{-}\right]^{2}=0 \nonumber\] As the solid dissolves, the concentrations of these ions increase, but remains smaller than . We reach equilibrium and “satisfy the solubility product” when = . We begin by writing the equilibrium reaction and the solubility product expression for Pb(IO ) . \[\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{IO}_{3}^{-}(a q) \nonumber\] e \[K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_{3}^{-}\right]^{2}=2.5 \times 10^{-13} \label{6.1}\] As Pb(IO ) dissolves, two \(\text{IO}_3^-\) ions form for each ion of Pb . If we assume that the change in the molar concentration of Pb at equilibrium is , then the change in the molar concentration of \(\text{IO}_3^-\) is 2 . The following table helps us keep track of the initial concentrations, the change in con‐ centrations, and the equilibrium concentrations of Pb2+ and \(\text{IO}_3^-\). Because a solid, such as Pb(IO ) , does not appear in the solubility product expression, we do not need to keep track of its concentration. Remember, however, that the value applies only if there is some solid Pb(IO ) present at equilibrium. Substituting the equilibrium concentrations into Equation \ref{6.1} and solving gives \[(x)(2 x)^{2}=4 x^{3}=2.5 \times 10^{-13} \nonumber\] \[x=3.97 \times 10^{-5} \nonumber\] Substituting this value of back into the equilibrium concentration expressions for Pb and \(\text{IO}_3^-\) gives their concentrations as \[\left[\mathrm{Pb}^{2+}\right]=x=4.0 \times 10^{-5} \mathrm{M} \text { and }\left[\mathrm{IO}_{3}^{-}\right]=2 x=7.9 \times 10^{-5} \nonumber\] Because one mole of Pb(IO ) contains one mole of Pb , the molar solubility of Pb(IO ) is equal to the concentration of Pb , or \(4.0 \times 10^{-5}\) M. We can express a compound’s solubility in two ways: as its molar solubility (mol/L) or as its mass solubility (g/L). Be sure to express your answer clearly. Calculate the molar solubility and the mass solubility for Hg Cl , given the following solubility reaction and value. \[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad K_{\mathrm{sp}}=1.2 \times 10^{-8} \nonumber\] When Hg Cl dissolves, two Cl are produced for each ion of \(\text{Hg}_2^{2+}\). If we assume is the change in the molar concentration of \(\text{Hg}_2^{2+}\), then the change in the molar concentration of Cl is 2 . The following table helps us keep track of our solution to this problem. Substituting the equilibrium concentrations into the expression forHg Cl gives \[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(2 \mathrm{x})^{2}=4 x^{3}=1.2 \times 10^{-18} \nonumber\] \[x=6.69 \times 10^{-7} \nonumber\] Substituting back into the equilibrium expressions for \(\text{Hg}_2^{2+}\) and Cl gives their concentrations as \[\left[\mathrm{Hg}_{2}^{2+}\right]=x=6.7 \times 10^{-7} \ \mathrm{M} \quad\left[\mathrm{Cl}^{-}\right]=2 x=1.3 \times 10^{-6} \ \mathrm{M} \nonumber\] The molar solubility is equal to [\(\text{Hg}_2^{2+}\)], or \(6.7 \times 10^{-7}\) mol/L. Calculating the solubility of Pb(IO ) in deionized water is a straightforward problem because the solid’s dissolution is the only source of Pb and \(\text{IO}_3^-\). But what if we add Pb(IO ) to a solution of 0.10 M Pb(NO ) ? Before we set‐up and solve this problem algebraically, think about the system’s chemistry and decide whether the solubility of Pb(IO ) will increase, decrease, or remain the same. Beginning a problem by thinking about the likely answer is a good habit to develop. Knowing what answers are reasonable will help you spot errors in your calculations and give you more confidence that your solution to a problem is correct. Because the solution already contains a source of Pb , we can use Le Châtelier’s principle to predict that the solubility of Pb(IO ) is smaller than that in our previous problem. We begin by setting up a table to help us keep track of the concentrations of Pb and \(\text{IO}_3^-\) as this system moves toward and reaches equilibrium. Substituting the equilibrium concentrations into Equation \ref{6.1} \[(0.10+x)(2 x)^{2}=2.5 \times 10^{-13} \nonumber\] and multiplying out the terms on the equation’s left side leaves us with \[4 x^{3}+0.40 x^{2}=2.5 \times 10^{-13} \label{6.2}\] This is a more difficult equation to solve than that for the solubility of Pb(IO ) in deionized water, and its solution is not immediately obvious. We can find a rigorous solution to Equation \ref{6.2} using computational software packages and spreadsheets, some of which are described in . There are several approaches to solving cubic equations, but none are computationally easy using just paper and pencil. How might we solve Equation \ref{6.2} if we do not have access to a computer? One approach is to use our understanding of chemistry to simplify the problem. From Le Châtelier’s principle we know that a large initial concentration of Pb will decrease significantly the solubility of Pb(IO ) . One reasonable assumption is that the initial concentration of Pb is very close to its equilibrium concentration. If this assumption is correct, then the following approximation is reasonable \[\left[\mathrm{Pb}^{2+}\right]=0.10+x \approx 0.10 \nonumber\] Substituting this approximation into Equation \ref{6.1} and solving for gives \[(0.10)(2 x)^{2}=0.4 x^{2}=2.5 \times 10^{-13} \nonumber\] \[x=7.91 \times 10^{-7} \nonumber\] Before we accept this answer, we must verify that our approximation is reasonable. The difference between the actual concentration of Pb , which is 0.10 + M, and our assumption that the concentration of Pb is 0.10 M is \(7.9 \times 10^{-7}\), or \(7.9 \times 10^{-4}\) % of the assumed concentration. This is a negligible error. If we accept the result of our calculation, we find that the equilibrium concentrations of Pb and \(\text{IO}_3^-\) are \[\left[\mathrm{Pb}^{2+}\right]=0.10+x \approx 0.10 \ \mathrm{M} \text { and }\left[\mathrm{IO}_{3}^{-}\right]=2 x=1.6 \times 10^{-6} \ \mathrm{M} \nonumber\] \[\begin{aligned} \% \text { error } &=\frac{\text { actual }-\text { assumed }}{\text { assumed }} \times 100 \\ &=\frac{(0.10+x)-0.10}{0.10} \times 100 \\ &=\frac{7.91 \times 10^{-7}}{0.10} \times 100 \\ &=7.91 \times 10^{-4} \% \end{aligned} \nonumber\] The molar solubility of Pb(IO ) is equal to the additional concentration of Pb in solution, or \(7.9 \times 10^{-4}\) mol/L. As expected, we find that Pb(IO ) is less soluble in the presence of a solution that already contains one of its ions. This is known as the . As outlined in the following example, if an approximation leads to an error that is unacceptably large, then we can extend the process of making and evaluating approximations. One “rule of thumb” when making an approximation is that it should not introduce an error of more than ±5%. Although this is not an unreasonable choice, what matters is that the error makes sense within the context of the problem you are solving. Calculate the solubility of Pb(IO ) in \(1.0 \times 10^{-4}\) M Pb(NO ) . If we let equal the change in the concentration of Pb , then the equilibrium concentrations of Pb and \(\text{IO}_3^-\) are \[\left[\mathrm{P} \mathrm{b}^{2+}\right]=1.0 \times 10^{-4}+ \ x \text { and }\left[\mathrm{IO}_{3}^-\right]=2 x \nonumber\] Substituting these concentrations into Equation \ref{6.1} leaves us with \[\left(1.0 \times 10^{-4}+ \ x\right)(2 x)^{2}=2.5 \times 10^{-13} \nonumber\] To solve this equation for , let’s make the following assumption \[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x \approx 1.0 \times 10^{-4} \ \mathrm{M} \nonumber\] Solving for gives its value as \(2.50 \times 10^{-5}\); however, when we substitute this value for back, we find that the calculated concentration of Pb at equilibrium \[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x=1.0 \times 10^{-4}+ \ 2.50 \times 10^{-5}=1.25 \times 10^{-4} \ \mathrm{M} \nonumber\] is 25% greater than our assumption of \(1.0 \times 10^{-4}\) M. This error is unreasonably large. Rather than shouting in frustration, let’s make a new assumption. Our first assumption—that the concentration of Pb is \(1.0 \times 10^{-4}\) M—was too small. The calculated concentration of \(1.25 \times 10^{-4}\) M, therefore, probably is a too large, but closer to the correct concentration than was our first assumption. For our second approximation, let’s assume that \[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x=1.25 \times 10^{-4} \mathrm{M} \nonumber\] Substituting into Equation \ref{6.1} and solving for gives its value as \(2.24 \times 10^{-5}\). The resulting concentration of Pb is \[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ 2.24 \times 10^{-5}=1.22 \times 10^{-4} \ \mathrm{M} \nonumber\] which differs from our assumption of \(1.25 \times 10^{-4}\) M by 2.4%. Because the original concentration of Pb is given to two significant figure, this is a more reasonable error. Our final solution, to two significant figures, is \[\left[\mathrm{Pb}^{2+}\right]=1.2 \times 10^{-4} \ \mathrm{M} \text { and }\left[\mathrm{IO}_{3}\right]=4.5 \times 10^{-5} \ \mathrm{M} \nonumber\] and the molar solubility of Pb(IO ) is \(2.2 \times 10^{-5}\) mol/L. This iterative approach to solving the problems is known as the . Calculate the molar solubility for Hg Cl in 0.10 M NaCl and compare your answer to its molar solubility in deionized water (see ). We begin by setting up a table to help us keep track of the concentrations \(\text{Hg}_2^{2+}\) and Cl as this system moves toward and reaches equilibrium. Substituting the equilibrium concentrations into the expression for Hg Cl leaves us with a difficult to solve cubic equation. \[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(0.10+2 x)^{2}=4 x^{3}+0.40 x^{2}+0.010 x \nonumber\] Let’s make an assumption to simplify this problem. Because we expect the value of to be small, let’s assume that \[\left[\mathrm{Cl}^{-}\right]=0.10+2 x \approx 0.10 \nonumber\] This simplifies our problem to \[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(0.10)^{2}=0.010 x=1.2 \times 10^{-18} \nonumber\] which gives the value of as \(1.2 \times 10^{-16}\) M. The difference between the actual concentration of Cl , which is (0.10 + 2 ) M, and our assumption that it is 0.10 M introduces an error of \(2.4 \times 10^{-13}\) %. This is a negligible error. The molar solubility of Hg Cl is the same as the concentration of \(\text{Hg}_2^{2+}\), or \(1.2 \times 10^{-16}\) M. As expected, the molar solubility in 0.10 M NaCl is less than \(6.7 \times 10^{-7}\) mol/L, which is its solubility in water (see solution to ). Calculating the solubility of Pb(IO ) in a solution of Pb(NO ) is more complicated than calculating its solubility in deionized water. The calculation, however, is still relatively easy to organize and the simplifying assumptions are fairly obvious. This problem is reasonably straightforward because it involves only one equilibrium reaction and one equilibrium constant. Determining the equilibrium composition of a system with multiple equilibrium reactions is more complicated. In this section we introduce a systematic approach to setting‐up and solving equilibrium problems. As shown in Table 6.7.1
, this approach involves four steps. Step 1 Write all relevant equilibrium reactions and equilibrium constant expressions. Step 2 Count the unique species that appear in the equilibrium constant expressions; these are your unknowns. You have enough information to solve the problem if the number of unknowns equals the number of equilibrium constant expressions Step 3 Combine your equations and solve for one unknown. Whenever possible, simplify the algebra by making appropriate assumptions. If you make an assumption, set a limit for its error. This decision influences your evaluation of the assumption. Step 4 Check your assumptions. If any assumption proves invalid, return to the previous step and continue solving. The problem is complete when you have an answer that does not violate any of your assumptions. In addition to equilibrium constant expressions, two other equations are important to this systematic approach to solving an equilibrium problem. The first of these equations is a , which simply is a statement that matter is conserved during a chemical reaction. In a solution of acetic acid, for example, the combined concentrations of the conjugate weak acid, CH COOH, and the conjugate weak base, CH COO , must equal acetic acid’s initial concentration, \(C_{\text{CH}_3\text{COOH}}\). \[C_{\mathrm{CH}_{\mathrm{3}} \mathrm{COOH}}=\left[\mathrm{CH}_{3} \mathrm{COOH}\right]+\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \nonumber\] You may recall from that this is the difference between a formal concentration and a molar concentration. The variable represents a formal concentration. The second equation is a , which requires that the total positive charge from the cations equal the total negative charge from the anions. Mathematically, the charge balance equation is \[\sum_{i=1}^{n}\left(z^{+}\right)_{i}\left[{C^{z}}^+\right]_{i} = -\sum_{j=1}^{m}(z^-)_{j}\left[{A^{z}}^-\right]_{j} \nonumber\] where [ ] and [ ] are, respectively, the concentrations of the cation and the anion, and ( ) and ( ) are the charges for the cation and the anion. Every ion in solution, even if it does not appear in an equilibrium reaction, must appear in the charge balance equation. For example, the charge balance equation for an aqueous solution of Ca(NO ) is \[2 \times\left[\mathrm{Ca}^{2+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{+}\right]+\left[\mathrm{NO}_{3}^-\right] \nonumber\] Note that we multiply the concentration of Ca by two and that we include the concentrations of H O and OH . A charge balance is a conservation of a charge. The minus sign in front of the summation term on the right side of the charge balance equation ensures that both summations are positive. There are situations where it is impossible to write a charge balance equation because we do not have enough information about the solution’s composition. For example, suppose we fix a solution’s pH using a buffer. If the buffer’s composition is not specified, then we cannot write a charge balance equation. Write mass balance equations and a charge balance equation for a 0.10 M solution of NaHCO . It is easier to keep track of the species in solution if we write down the reactions that define the solution’s composition. These reactions are the dissolution of a soluble salt \[\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \nonumber\] and the acid–base dissociation reactions of \(\text{HCO}_3^-\) and H O \[\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \nonumber\] \[\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber\] \[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\] The mass balance equations are \[0.10 \mathrm{M}=\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]+\left[\mathrm{HCO}_{3}^{-}\right]+\left[\mathrm{CO}_{3}^{2-}\right] \nonumber\] \[0.10 \ \mathrm{M}=\left[\mathrm{Na}^{+}\right] \nonumber\] and the charge balance equation is \[\left[\mathrm{Na}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{HCO}_{3}^-\right]+2 \times\left[\mathrm{CO}_{3}^{2-}\right] \nonumber\] Write appropriate mass balance and charge balance equations for a solution containing 0.10 M KH PO and 0.050 M Na HPO . To help us determine what ions are in solution, let’s write down all the reaction needed to prepare the solutions and the equilibrium reactions that take place within these solutions. These reactions are the dissolution of two soluble salts \[\mathrm{KH}_{2} \mathrm{PO}_{4}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \nonumber\] \[\mathrm{NaHPO}_{4}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) \nonumber\] and the acid–base dissociation reactions for \(\text{H}_2\text{PO}_4^-\), \(\text{HPO}_4^{2-}\). and H O. \[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) \nonumber\] \[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \nonumber\] \[\mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{PO}_{4}^{3-}(a q) \nonumber\] \[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\] Note that we did not include the base dissociation reaction for \(\text{HPO}_4^{2-}\) because we already accounted for its product, \(\text{H}_2\text{PO}_4^-\), in another reaction. The mass balance equations for K and Na are straightforward \[\left[\mathrm{K}^{+}\right]=0.10 \ \mathrm{M} \text { and }\left[\mathrm{Na}^{+}\right]=0.10 \ \mathrm{M} \nonumber\] but the mass balance equation for phosphate takes a bit more thought. Both \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) produce the same ions in solution. We can, therefore, imagine that the solution initially contains 0.15 M KH PO , which gives the following mass balance equation. \[\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]+\left[\mathrm{H}_{2} \mathrm{PO}_{4}^-\right]+\left[\mathrm{HPO}_{4}^{2-}\right]+\left[\mathrm{PO}_{4}^{3-}\right]=0.15 \ \mathrm{M} \nonumber\] The charge balance equation is \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{K}^{+}\right]+\left[\mathrm{Na}^{+}\right] =\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]+2 \times\left[\mathrm{HPO}_{4}^{2-}\right] +3 \times\left[\mathrm{PO}_{4}^{3-}\right]+\left[\mathrm{OH}^{-}\right] \nonumber\] To illustrate the systematic approach to solving equilibrium problems, let’s calculate the pH of 1.0 M HF. Two equilibrium reactions affect the pH. The first, and most obvious, is the acid dissociation reaction for HF \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \nonumber\] for which the equilibrium constant expression is \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=6.8 \times 10^{-4} \label{6.3}\] The second equilibrium reaction is the dissociation of water, which is an obvious yet easily neglected reaction \[2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\] \[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.00 \times 10^{-14} \label{6.4}\] Counting unknowns, we find four: [HF], [F ], [H O ], and [OH ]. To solve this problem we need two additional equations. These equations are a mass balance equation on hydrofluoric acid \[C_{\mathrm{HF}}=[\mathrm{HF}]+\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M} \label{6.5}\] and a charge balance equation \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{F}^{-}\right] \label{6.6}\] With four equations and four unknowns, we are ready to solve the problem. Before doing so, let’s simplify the algebra by making two assumptions. . Because HF is a weak acid, we know that the solution is acidic. For an acidic solution it is reasonable to assume that \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]>>\left[\mathrm{OH}^{-}\right] \nonumber\] which simplifies the charge balance equation to \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{F}^{-}\right] \label{6.7}\] . Because HF is a weak acid, very little of it dissociates to form F . Most of the HF remains in its conjugate weak acid form and it is reasonable to assume that \[[\mathrm{HF}]>>\left[\mathrm{F}^{-}\right] \nonumber\] which simplifies the mass balance equation to \[C_{\mathrm{HF}}=[\mathrm{HF}]=1.0 \ \mathrm{M} \label{6.8}\] For this exercise let’s accept an assumption if it introduces an error of less than ±5%. Substituting Equation \ref{6.7} and Equation \ref{6.8} into Equation \ref{6.3}, and solving for the concentration of H O gives us \[\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\mathrm{C}_{\mathrm{HF}}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}}{\mathrm{C}_{\mathrm{HF}}}=6.8 \times 10^{-4} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{K_{\mathrm{a}} C_{\mathrm{HF}}}=\sqrt{\left(6.8 \times 10^{-4}\right)(1.0)}=2.6 \times 10^{-2} \nonumber\] Before accepting this answer, we must verify our assumptions. The first assumption is that [OH ] is significantly smaller than [H O ]. Using Equation \ref{6.4}, we find that \[\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{1.00 \times 10^{-14}}{2.6 \times 10^{-2}}=3.8 \times 10^{-13} \nonumber\] Clearly this assumption is acceptable. The second assumption is that [F ] is significantly smaller than [HF]. From Equation \ref{6.7} we have \[\left[\mathrm{F}^{-}\right]=2.6 \times 10^{-2} \ \mathrm{M} \nonumber\] Because [F ] is 2.60% of , this assumption also is acceptable. Given that [H O ] is \(2.6 \times 10^{-2}\) M, the pH of 1.0 M HF is 1.59. How does the calculation change if we require that the error introduced in our assumptions be less than ±1%? In this case we no longer can assume that [HF] >> [F ] and we cannot simplify the mass balance equation. Solving the mass balance equation for [HF] \[[\mathrm{HF}]=C_{\mathrm{HF}}-\left[\mathrm{F}^{-}\right]=C_{\mathrm{HF}}-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber\] and substituting into the expression along with Equation \ref{6.7} gives \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}}{C_{\mathrm{HF}}-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\] Rearranging this equation leaves us with a quadratic equation \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}+K_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-K_{\mathrm{a}} C_{\mathrm{HF}}=0 \nonumber\] which we solve using the quadratic formula \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \nonumber\] where , , and are the coefficients in the quadratic equation \[a x^{2}+b x+c=0 \nonumber\] Solving a quadratic equation gives two roots, only one of which has chemical significance. For our problem, the equation’s roots are \[x=\frac{-6.8 \times 10^{-4} \pm \sqrt{\left(6.8 \times 10^{-4}\right)^{2}-(4)(1)\left(-6.8 \times 10^{-4}\right)}}{(2)(1)} \nonumber\] \[x=\frac{-6.8 \times 10^{-4} \pm 5.22 \times 10^{-2}}{2} \nonumber\] \[x=2.57 \times 10^{-2} \text { or }-2.64 \times 10^{-2} \nonumber\] Only the positive root is chemically significant because the negative root gives a negative concentration for H O . Thus, [H O ] is \(2.57 \times 10^{-2}\) M and the pH is 1.59. You can extend this approach to calculating the pH of a monoprotic weak base by replacing with , replacing with the weak base’s concentration, and solving for [OH ] in place of [H O ]. Calculate the pH of 0.050 M NH . State any assumptions you make in solving the problem, limiting the error for any assumption to ±5%. The value for NH is \(1.75 \times 10^{-5}\). To determine the pH of 0.050 M NH , we need to consider two equilibrium reactions: the base dissociation reaction for NH \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] and water’s dissociation reaction. \[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\] These two reactions contain four species whose concentrations we need to consider: NH , \(\text{NH}_4^+\), H O , and OH . We need four equations to solve the problem—these equations are the equation for NH \[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=1.75 \times 10^{-5} \nonumber\] the equation for H O \[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\] a mass balance equation on ammonia \[C_{\mathrm{NH}_{3}}=0.050 \ \mathrm{M}=\left[\mathrm{NH}_{3}\right]+\left[\mathrm{NH}_{4}^{+}\right] \nonumber\] and a charge balance equation \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right] \nonumber\] To solve this problem, we will make two assumptions. Because NH is a base, our first assumption is \[\left[\mathrm{OH}^{-}\right]>>\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber\] which simplifies the charge balance equation to \[\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right] \nonumber\] Because NH is a weak base, our second assumption is \[\left[\mathrm{NH}_{3}\right]>>\left[\mathrm{NH}_{4}^{+}\right] \nonumber\] which simplifies the mass balance equation to \[C_{\mathrm{NH}_{3}}=0.050 \ \mathrm{M}=\left[\mathrm{NH}_{3}\right] \nonumber\] Substituting the simplified charge balance equation and mass balance equation into the equation leave us with \[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{OH}^{-}\right]}{C_{\mathrm{NH}_3}}=\frac{\left[\mathrm{OH}^{-}\right]^{2}}{C_{\mathrm{NH_3}}}=1.75 \times 10^{-5} \nonumber\] \[\left[\mathrm{OH}^{-}\right]=\sqrt{K_{\mathrm{b}} C_{\mathrm{NH_3}}}=\sqrt{\left(1.75 \times 10^{-5}\right)(0.050)}=9.35 \times 10^{-4} \nonumber\] Before we accept this answer, we must verify our two assumptions. The first assumption is that the concentration of OH is significantly greater than the concentration of H O . Using , we find that \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{9.35 \times 10^{-4}}=1.07 \times 10^{-11} \nonumber\] Clearly this assumption is acceptable. Our second assumption is that the concentration of NH is significantly greater than the concentration of \(\text{NH}_4^+\). Using our simplified charge balance equation, we find that \[\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right]=9.35 \times 10^{-4} \nonumber\] Because the concentration of \(\text{NH}_4^+\) is 1.9% of \(C_{\text{NH}_3}\), our second assumption also is reasonable. Given that [H O ] is \(1.07 \times 10^{-11}\), the pH is 10.97. A more challenging problem is to find the pH of a solution that contains a polyprotic weak acid or one of its conjugate species. As an example, consider the amino acid alanine, whose structure is shown in Figure 6.7.1
. The ladder diagram in Figure 6.7.2
shows alanine’s three acid–base forms and their respective areas of predominance. For simplicity, we identify these species as H L , HL, and L .
. Ladder diagram for alanine. Alanine hydrochloride is the salt of the diprotic weak acid H L and Cl . Because H L has two acid dissociation reactions, a complete systematic solution to this problem is more complicated than that for a monoprotic weak acid. The ladder diagram in Figure 6.7.2
helps us simplify the problem. Because the areas of predominance for H L and L are so far apart, we can assume that a solution of H L will not contain a significant amount of L . As a result, we can treat H L as though it is a monoprotic weak acid. Calculating the pH of 0.10 M alanine hydrochloride, which is 1.72, is left to the reader as an exercise. The alaninate ion is a diprotic weak base. Because L has two base dissociation reactions, a complete systematic solution to this problem is more complicated than that for a monoprotic weak base. Once again, the ladder diagram in Figure 6.7.2
helps us simplify the problem. Because the areas of predominance for H L and L are so far apart, we can assume that a solution of L will not contain a significant amount of H L . As a result, we can treat L as though it is a monoprotic weak base. Calculating the pH of 0.10 M sodium alaninate, which is 11.42, is left to the reader as an exercise. Finding the pH of a solution of alanine is more complicated than our previous two examples because we cannot ignore the presence of either H L or L . To calculate the solution’s pH we must consider alanine’s acid dissociation reaction \[\mathrm{HL}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{L}^{-}(a q) \nonumber\] and its base dissociation reaction \[\mathrm{HL}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{L}^{+}(a q) \nonumber\] and, as always, we must also consider the dissociation of water \[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\] This leaves us with five unknowns—[H L ], [HL], [L ], [H O ], and [OH ]—for which we need five equations. These equations are and for alanine \[K_{\mathrm{a} 2}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{L}^{-}\right]}{[\mathrm{HL}]} \nonumber\] \[K_{\mathrm{b} 2}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a1}}}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]}{[\mathrm{HL}]} \nonumber\] the equation \[K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\] a mass balance equation for alanine \[C_{\mathrm{HL}}=\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+[\mathrm{HL}]+[\mathrm{L}^{-}] \nonumber\] and a charge balance equation \[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{OH}^-]+[\mathrm{L^-}] \nonumber\] Because HL is a weak acid and a weak base, it seems reasonable to assume that little of it will dissociate and that \[[\mathrm{HL}]>>\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+[\mathrm{L}^-] \nonumber\] which allows us to simplify the mass balance equation to \[C_{\mathrm{HL}}=[\mathrm{HL}] \nonumber\] Next we solve for [H L ] \[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]=\frac{K_{\mathrm{w}}[\mathrm{HL}]}{K_{\mathrm{a1}}\left[\mathrm{OH}^{-}\right]}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right,\mathrm{HL}]}{K_{\mathrm{a1}}}=\frac{C_{\mathrm{HL}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a1}}} \nonumber\] and solve for [L ] \[[\mathrm{L^-}]=\frac{K_{a2}[\mathrm{HL}]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{K_{a2} C_{\mathrm{HL}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\] Substituting these equations for [H L ] and [L ], and the equation for , into the charge balance equation give us \[\frac{C_{\mathrm{HL}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a1}}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\frac{K_{a2} C_{\mathrm{HL}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\] which we simplify to \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(\frac{C_{\mathrm{HL}}}{K_{\mathrm{a1}}}+1\right)=\frac{1}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\left(K_{\mathrm{w}}+K_{a2} C_{\mathrm{HL}}\right) \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}=\frac{\left(K_{\mathrm{a} 2} C_{\mathrm{HL}}+K_{\mathrm{w}}\right)}{\frac{C_{\mathrm{HL}}}{K_{\mathrm{a1}}}+1}=\frac{K_{\mathrm{a1}}\left(K_{\mathrm{a2}} C_{\mathrm{HL}}+K_{\mathrm{w}}\right)}{C_{\mathrm{HL}}+K_{\mathrm{a1}}} \nonumber\] \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{\frac{\left(K_{\mathrm{a1}} K_{a2} C_{\mathrm{HL}}+K_{\mathrm{a1}} K_{\mathrm{w}}\right)}{C_{\mathrm{HL}}+K_{\mathrm{a1}}}} \nonumber\] We can further simplify this equation if << , and if << , leaving us with \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{K_{\mathrm{a1}} K_{\mathrm{a} 2}} \nonumber\] For a solution of 0.10 M alanine the [H O ] is \[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{\left(4.487 \times 10^{-3}\right)\left(1.358 \times 10^{-10}\right)}=7.806 \times 10^{-7} \ \mathrm{M} \nonumber\] or a pH of 6.11. Verify that each assumption in our solution for the pH of 0.10 M alanine is reasonable, using ±5% as the limit for the acceptable error. In solving for the pH of 0.10 M alanine, we made the following three assumptions: (a) [HL] >> [H L ] + [L ]; (b) << ; and (c) << . Assumptions (b) and (c) are easy to check. The value of (\(4.487 \times 10^{-3}\)) is 4.5% of (0.10), and w (\(4.487 \times 10^{-17}\)) is 0.074% of (\(6.093 \times 10^{-14}\)). Each of these assumptions introduces an error of less than ±5%. To test assumption (a) we need to calculate the concentrations of H L and L , which we accomplish using the equations for and . \[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right,\mathrm{HL}]}{K_{\mathrm{a1}}}=\frac{\left(7.807 \times 10^{-7}\right)(0.10)}{4.487 \times 10^{-3}}=1.74 \times 10^{-5} \nonumber\] \[\left[\mathrm{L}^{-}\right]=\frac{K_{a 2}[\mathrm{HL}]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{\left(1.358 \times 10^{-10}\right)(0.10)}{7.807 \times 10^{-7}}=1.74 \times 10^{-5} \nonumber\] Because these concentrations are less than ±5% of HL, the first assumption also is acceptable. One method for increasing a precipitate’s solubility is to add a ligand that forms soluble complexes with one of the precipitate’s ions. For example, the solubility of AgI increases in the presence of NH due to the formation of the soluble \(\text{Ag(NH}_3)_2^+\) complex. As a final illustration of the systematic approach to solving equilibrium problems, let’s calculate the molar solubility of AgI in 0.10 M NH . We begin by writing the relevant equilibrium reactions, which includes the solubility of AgI, the acid–base chemistry of NH and H O, and the metal‐ligand complexation chemistry between Ag and NH . \[\begin{array}{c}{\operatorname{AgI}(s)\rightleftharpoons\operatorname{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)} \\ {\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q)} \\ {2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)} \\ {\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)}\end{array} \nonumber\] This leaves us with seven unknowns—[Ag ], [I ], [NH ], [\(\text{NH}_4^+\) ], [OH ], [H O ], and [\(\text{Ag(NH}_3)_2^+\)]—and a need for seven equations. Four of the equations we need to solve this problem are the equilibrium constant expressions \[K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]=8.3 \times 10^{-17} \label{6.9}\] \[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=1.75 \times 10^{-5} \label{6.10}\] \[K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.00 \times 10^{-14} \label{6.11}\] \[\beta_{2}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_{3}\right]^{2}}=1.7 \times 10^{7} \label{6.12}\] We still need three additional equations. The first of these equations is a mass balance for NH . \[C_{\mathrm{NH}_{3}}=\left[\mathrm{NH}_{3}\right]+\left[\mathrm{NH}_{4}^{+}\right]+2 \times\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.13}\] In writing this mass balance equation we multiply the concentration of \(\text{Ag(NH}_3)_2^+\) by two since there are two moles of NH per mole of \(\text{Ag(NH}_3)_2^+\). The second additional equation is a mass balance between iodide and silver. Because AgI is the only source of I and Ag , each iodide in solution must have an associated silver ion, which may be Ag or \(\text{Ag(NH}_3)_2^+\) ; thus \[\left[\mathrm{I}^{-}\right]=\left[\mathrm{Ag}^{+}\right]+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.14}\] Finally, we include a charge balance equation. \[\left[\mathrm{Ag}^{+}\right]+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]+\left[\mathrm{NH}_{4}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{OH}^-]+[\mathrm{I}^-] \label{6.15}\] Although the problem looks challenging, three assumptions greatly simplify the algebra. Because the formation of the \(\text{Ag(NH}_3)_2^+\) complex is so favorable (\(\beta_2\) is \(1.7 \times 10^7\)), there is very little free Ag in solution and it is reasonable to assume that \[\left[\mathrm{Ag}^{+}\right]<<\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \nonumber\] . Because NH is a weak base we may reasonably assume that most uncomplexed ammonia remains as NH ; thus \[\left[\mathrm{NH}_{4}^{+}\right]<<\left[\mathrm{NH}_{3}\right] \nonumber\] Because for AgI is significantly smaller than \(\beta_2\) for \(\text{Ag(NH}_3)_2^+\), the solubility of AgI probably is small enough that very little ammonia is needed to form the metal–ligand complex; thus \[\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]<<\left[\mathrm{NH}_{3}\right] \nonumber\] As we use these assumptions to simplify the algebra, let’s set ±5% as the limit for error. Assumption two and assumption three suggest that the concentration of NH is much larger than the concentrations of either \(\text{NH}_4^+\) or \(\text{Ag(NH}_3)_2^+\), which allows us to simplify the mass balance equation for NH to \[C_{\mathrm{NH}_{3}}=\left[\mathrm{NH}_{3}\right] \label{6.16}\] Finally, using assumption one, which suggests that the concentration of \(\text{Ag(NH}_3)_2^+\) is much larger than the concentration of Ag , we simplify the mass balance equation for I to \[\left[\mathrm{I}^{-}\right]=\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.17}\] Now we are ready to combine equations and to solve the problem. We begin by solving Equation \ref{6.9} for [Ag ] and substitute it into \(\beta_2\) (Equation \ref{6.12}), which leaves us with \[\beta_{2}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right,\mathrm{I}^-]}{K_{\mathrm{sp}}\left[\mathrm{NH}_{3}\right]^{2}} \label{6.18}\] Next we substitute Equation \ref{6.16} and Equation \ref{6.17} into Equation \ref{6.18}, obtaining \[\beta_{2}=\frac{\left[\mathrm{I}^{-}\right]^{2}}{K_{\mathrm{sp}}\left(C_{\mathrm{NH}_3}\right)^{2}} \label{6.19}\] Solving Equation \ref{6.19} for [I ] gives \[\left[\mathrm{I}^{-}\right]=C_{\mathrm{NH}_3} \sqrt{\beta_{2} K_{s p}} = \\ (0.10) \sqrt{\left(1.7 \times 10^{7}\right)\left(8.3 \times 10^{-17}\right)}=3.76 \times 10^{-6} \ \mathrm{M} \nonumber\] Because one mole of AgI produces one mole of I , the molar solubility of AgI is the same as the [I ], or \(3.8 \times 10^{-6}\) mol/L. Before we accept this answer we need to check our assumptions. Substituting [I ] into Equation \ref{6.9}, we find that the concentration of Ag is \[\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{p}}}{[\mathrm{I}^-]}=\frac{8.3 \times 10^{-17}}{3.76 \times 10^{-6}}=2.2 \times 10^{-11} \ \mathrm{M} \nonumber\] Substituting the concentrations of I and Ag+ into the mass balance equation for iodide (Equation \ref{6.14}), gives the concentration of \(\text{Ag(NH}_3)_2^+\) as \[\left[\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]=[\mathrm{I}^-]-\left[\mathrm{Ag}^{+}\right]=3.76 \times 10^{-6}-2.2 \times 10^{-11}=3.76 \times 10^{-6} \ \mathrm{M} \nonumber\] Our first assumption that [Ag ] is significantly smaller than the [\(\text{Ag(NH}_3)_2^+\)] is reasonable. Substituting the concentrations of Ag and \(\text{Ag(NH}_3)_2^+\) into Equation \ref{6.12} and solving for [NH ], gives \[\left[\mathrm{NH}_{3}\right]=\sqrt{\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right] \beta_{2}}}=\sqrt{\frac{3.76 \times 10^{-6}}{\left(2.2 \times 10^{-11}\right)\left(1.7 \times 10^{7}\right)}}=0.10 \ \mathrm{M} \nonumber\] From the mass balance equation for NH3 (Equation \ref{6.12}) we see that [\(\text{NH}_4^+\)] is negligible, verifying our second assumption that \([\text{NH}_4^+]\) is significantly smaller than [NH ]. Our third assumption that [\(\text{Ag(NH}_3)_2^+\)] is significantly smaller than [NH ] also is reasonable. Did you notice that our solution to this problem did not make use of Equation \ref{6.15}, the charge balance equation? The reason for this is that we did not try to solve for the concentration of all seven species. If we need to know the reaction mixture’s complete composition at equilibrium, then we will need to incorporate the charge balance equation into our solution. | 42,132 | 3,643 |
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. In addition to these individual basis; please contact 4Fe(s) + 3O (g) → 2Fe O (s) Why does the entropy of the universe increase in this reaction even though gaseous molecules, which have a high entropy, are consumed? Based on this table, can you conclude that entropy is related to the nature of functional groups? Explain your reasoning. The text states that the magnitude of ΔS tends to be similar for a wide variety of compounds. Based on the values in the table, do you agree? Ba(NO ) (aq) + 2NaI(aq) → BaI (aq) + 2NaNO (aq) You want to determine the absolute entropy of BaI , but that information is not listed in your tables. However, you have been able to obtain the following information: You know that ΔG° for the reaction at 25°C is 22.64 kJ/mol. What is ΔH° for this reaction? What is S° for BaI ? | 844 | 3,644 |
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. In addition to these individual basis; please contact 7. London dispersion forces increase with increasing atomic mass. Iodine is a solid while bromine is a liquid due to the greater intermolecular interactions between the heavier iodine atoms. Iodine is less soluble than bromine in virtually all solvents because it requires more energy to separate \(I_2\) molecules than \(Br_2\) molecules. 11. 15. In dental amalgam, the mercury atoms are locked in a solid phase that does not undergo corrosion under physiological conditions; hence, the mercury atoms cannot readily diffuse to the surface where they could vaporize or undergo chemical reaction. 21. Dissolve the mixture of A and B in a solvent in which they are both soluble when hot and relatively insoluble when cold, filter off any undissolved B, and cool slowly. Pure A should crystallize, while B stays in solution. If B were less soluble, it would be impossible to obtain pure A by this method in a single step, because some of the less soluble compound (B) will always be present in the solid that crystallizes from solution. hexane and methanol 1. \(6.7 \times 10^4\; amu\) 3. 9.24 atm 5. The \(CaCl_2\) solution will have a lower vapor pressure, because it contains three times as many particles as the glucose solution. 7. 0.36 m \(NaCl\), 2.6 g \(NaCl\) 9. 60 g NaBr 11. 700 g \(NaCl\) 13. \(MgCl_2\) produces three particles in solution versus two for \(NaCl\), so the same molal concentration of \(MgCl_2\) will produce a 50% greater freezing point depression than for \(NaCl\). Nonetheless, the molar mass of \(MgCl_2\) is 95.3 g/mol versus 48.45 g/mol for \(NaCl\). Consequently, a solution containing 1 g \(NaCl\) per 1000 g \(H_2O\) will produce a freezing point depression of 0.064°C versus 0.059°C for a solution containing 1 g \(MgCl_2\) per 1000 g \(H_2O\). Thus, given equal cost per gram, \(NaCl\) is more effective. Yes, \(MgCl_2\) would be effective at −8°C; a 1.43 m solution (136 g per 1000 g H2O) would be required. 16. k = 1.81(°C•kg)/mol, molecular mass of urea = 60.0 g/mol | 2,095 | 3,647 |
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Aldehydes and ketones contain the carbonyl group. Aldehydes are considered the most important functional group. They are often called the formyl or methanoyl group. Aldehydes derive their name from the ration of cohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms. Aldehydes and ketones are organic compounds which incorporate a , C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an . If neither is hydrogen, the compound is a . The IUPAC system of nomenclature assigns a characteristic suffix to aldehydes. For example, H C=O is methan , more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC. Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on. If the aldehyde moiety (-CHO) is attached to a ring the suffix is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring. The IUPAC system names are given on top while the common name is given on the bottom in parentheses. The IUPAC system of nomenclature assigns a characteristic suffix of to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function. ketones The IUPAC system names are given on top while the common name is given on the bottom in parentheses. As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone. For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending is added to the end of the parent chain name. diketones In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix is included before the parent chain name to indicate that it is in a ring. As with other ketones the ending is replaced with the to indicate the presence of a ketone. With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an is not removed from the end, but the suffix is added. When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the is named as a substituent. However, the in hydroxyl is generally removed. When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. When carbonyls are included with an alkene the following order is followed: Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary. A) butanal B) 2-hydroxycyclopentanone C) 2,3-pentanedione D) 1,3-cyclohexanedione E) 4-hydoxy-3-methyl-2-butanone F) ( ) 3-methyl-2-hepten-4-one G) 3-oxobutanal H) -3-bromocyclohexanecarboaldehyde I) butanedial J) -2-methyl-3-hexenal
| 4,843 | 3,648 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.11%3A_Electronegativity |
The ability of an atom in a molecule to attract a shared electron pair to itself, forming a , is called its . The negative side of a polar covalent bond corresponds to the element. Furthermore the more polar a bond, the larger the in electronegativity of the two atoms forming it. Unfortunately there is no direct way of measuring electronegativity. Dipole-moment measurements tell us about the electrical behavior of electron pairs in the molecule, not just the bonding pair in which we are interested. Also, the polarity of a bond depends on whether the bond is a single, double, or triple bond and on what the other atoms and electron pairs in a molecule are. Therefore the dipole moment cannot tell us quantitatively the difference between the electronegativities of two bonded atoms. Various attempts have been made over the years to derive a scale of electronegativities for the elements, none of which is entirely satisfactory. Nevertheless most of these attempts agree in large measure in telling us which elements are more electronegative than others. The best-known of these scales was devised by the Nobel prize-winning California chemist Linus Pauling (1901 to 1994) and is shown in the periodic table found below. In this scale a value of 4.0 is arbitrarily given to the most electronegative element, fluorine, and the other electronegativities are scaled relative to this value. As can be seen from this table, elements with electronegativities of 2.5 or more are all nonmetals in the top right-hand comer of the periodic table. These have been color-coded dark red. By contrast, elements with negativities of 1.3 or less are all metals on the lower left of the table. These elements have been coded in dark gray. They are often referred to as the most elements, and they are the metals which invariably form . Between these two extremes we notice that most of the remaining metals (largely transition metals) have electronegativities between 1.4 and 1.9 (light gray), while most of the remaining nonmetals have electronegativities between 2.0 and 2.4 (light red). Another feature worth noting is the very large differences in electronegativities in the top right-hand comer of the table. Fluorine, with an electronegativity of 4, is by far the most electronegative element. At 3.5 oxygen is a distant second, while chlorine and nitrogen are tied for third place at 3.0. If the electronegativity values of two atoms are very different, the bond between those atoms is largely ionic. In most of the typical ionic compounds discussed in the previous chapter, the difference is greater than 1.5, although it is dangerous to attach too much significance to this figure since electronegativity is only a semiquantitative concept. As the electronegativity difference becomes smaller, the bond becomes more covalent. An important example of an almost completely covalent bond between two different atoms is that between carbon (2.5) and hydrogen (2.1). The properties of numerous compounds of hydrogen and carbon (hydrocarbons) are described in . These properties indicate that the C―H bond has almost no polar character. Without consulting the table of electronegativities (use the ), arrange the following bonds in order of decreasing polarity: B—Cl, Ba—Cl, Be—Cl, Br—Cl, Cl—Cl. We first need to arrange the elements in order of increasing electronegativity. Since the electronegativity increases in going up a column of the periodic table, we have the following relationships: Also since the electronegativity increases across the periodic table, we have Since B is a group III element on the borderline between metals and non-metals, we easily guess that which gives us the complete order Among the bonds listed, therefore, the Ba—Cl bond corresponds to the largest difference in electronegativity, i.e., to the most nearly ionic bond. The order of bond polarity is thus where the final bond, Cl—Cl,is, of course, purely covalent. | 3,966 | 3,650 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/04%3A_Alkanes/4.01%3A_Prelude_to_Alkanes |
Although this chapter is concerned with the chemistry of only one class of compounds, saturated hydrocarbons or alkanes, several fundamental principles are developed that we shall use extensively in later chapters. The study of some of these principles has been associated traditionally more with physical chemistry than with organic chemistry. We include them here, at the beginning of out discussion of organic reactions, because they provide a sound basis for understanding the key questions concerning the practical use of organic reactions. Is the equilibrium point of a given reaction far enough toward the desired products to be useful? Can conditions be found in which the reaction will take place at a practical rate? How can unwanted side reactions be suppressed? Initially, we will be concerned with the physical properties of alkanes and how these properties can be correlated by the important concept of homology. This will be followed by a brief survey of the occurrence and uses of hydrocarbons, with special reference to the petroleum industry. Chemical reactions of alkanes then will be discussed, with special emphasis on combustion and substitution reactions, These reactions are employed to illustrate how we can predict and use energy changes - particularly \(\Delta H\), the heat evolved or absorbed by a reacting system, which often can be estimated from bond energies. Then we consider some of the problems involved in predicting reaction rates in the context of a specific reaction, the chlorination of methane. The example is complex, but it has the virtue that we are able to break the overall reaction into quite simple steps. Before proceeding further, it will be well to reiterate what an alkane is, lest you be confused as to the difference between alkanes and alkenes. are compounds of carbon and hydrogen only, without double bonds, triple bonds, or rings. They all conform to the general formula \(\ce{C_nH_{2n+2}}\) and sometimes are called hydrocarbons, open-chain saturated hydrocarbons, or hydrocarbons. The nomenclature of alkanes has been discussed in , and you may find it well to review before proceeding. and (1977) | 2,186 | 3,653 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Catalyst_Examples/Hydroamination |
In general, alkynes are found to produce higher TOFs than alkenes, which can be explained by easier This indicates the same sensitivity to catalyst design as that demonstrated by aminoalkenes. Aminodienes respond to such changes in ligation environment more readily than aminoalkenes. This is best explained by the ability for the diene to datively coordinate to the metal center and form an McGrane and Livinghouse synthesized a titanium catalyst to convert a substituted alkyne into a cyclic organotitanium compound , which was then used to synthesize the final product monomorine . Scheme 20 shows the synthetic pathway undertaken to produce monomorine. The mechanism proceeds through a [2+2] reaction, presented in Scheme 14.
Scheme 31 illustrates usage of Gooßen’s catalyst to afford the Z stereoisomer of , also known as lansiumamide A, a natural product. | 971 | 3,656 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Case_Studies%3A_Proteins/Permanent_Hair_Wave |
The formation of disulfide bonds has a direct application in producing curls in hair by the permanent wave process. Hair keratin consists of many protein alpha-helices. Three alpha-helices are interwoven into a left-handed coil called a protofibril. Eleven protofibrils are bonded and coiled together to make a microfibril. Hundreds of these microfibrils are cemented into an irregular bundle called a macrofibril. These in turn are mixed with dead and living cells to make a complete strand of hair. Although it may seem incredible, in order for hair to grow 6 inches in one year, 9-1/2 turns of a -helix must be produced every second. The alpha-helices are extensively cross-linked with disulfide bonds from cysteine. These bonds enable keratin to have a somewhat elastic nature. If the alpha -helices stretch unevenly past each other, the disulfide cross-links return them to the original position when the tension is released. Disulfide bonds are formed by oxidation of the sulfhydryl groups on cysteine. Different protein chains or loops within a single chain are held together by the strong covalent disulfide bonds. The alpha-helices in the hair strands are bonded by disulfide links. In the permanent wave process, a basic reducing substance (usually ammonium thioglycolate) is first added to reduce and rupture some of the disulfide cross-links. When the hair gets wet, water molecules intrude into the keratin strands. The sheer numbers of water molecules are able to disrupt some of the hydrogen bonds which also help to keep the alpha-helices aligned. The helices are able to slip past each other and will retain a new shape in the hair drying process as new hydrogen bonds are formed. The hair strands are able for a short time to maintain the new curl in the hair. For a permanent wave, we will continue the discussion from the use of the reducing agent. The hair is put on rollers or curlers. Since the alpha-helices are no longer tightly cross-linked to each other, the alpha-helices can shift positions in relation to each other. An oxidizing agent, usually a dilute solution of hydrogen peroxide, (also called the neutralizer) is added to reform the disulfide bonds in their new positions. The permanent will hold these new disulfide bond positions until the hair grows out, since new hair growth is of course not treated. | 2,362 | 3,657 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/10%3A_Energetics_of_Chemical_Reactions |
We begin our study of the energetics of chemical reactions with our understanding of mass relationships, determined by the stoichiometry of balanced reactions and the relative atomic masses of the elements. We will assume a conceptual understanding of energy based on the physics of mechanics, and in particular, we will assume the law of conservation of energy. In developing a molecular understanding of the reaction energetics, we will further assume our understanding of chemical bonding via valence shell electron pair sharing and molecular orbital theory. The heat released or consumed in a chemical reaction is typically amongst the most easily observed and most readily appreciated consequences of the reaction. Many chemical reactions are performed routinely specifically for the purpose of utilizing the heat released by the reaction. We are interested here in an understanding of the energetics of chemical reactions. Specifically, we wish to know what factors determine whether heat is absorbed or released during a chemical reaction. With that knowledge, we seek to quantify and predict the amount of heat anticipated in a chemical reaction. We expect to find that the quantity of heat absorbed or released during a reaction is related to the bonding of the molecules involved in the reaction. Prior to answering these questions, we must first answer a few questions regarding the nature of heat. Despite our common familiarity with heat (particularly in Houston), the concept of heat is somewhat elusive to define. We recognize heat as "whatever it is that makes things hot", but this definition is too imprecise to permit measurement or any other conceptual progress. Exactly how to we define and measure heat? We can define in a variety of ways a temperature scale which permits quantitative measurement of "how hot" an object is. Such scales are typically based on the expansion and contraction of materials, particularly of liquid mercury, or on variation of resistance in wires or thermocouples. Using such scales, we can easily show that heating an object causes its temperature to rise. It is important, however, to distinguish between heat and temperature. These two concepts are not one and the same. To illustrate the difference, we begin by measuring the temperature rise produced by a given amount of heat, focusing on the temperature rise in \(1000 \: \text{g}\) of water produced by burning \(1.0 \: \text{g}\) of methane gas. We discover by performing this experiment repeatedly that the temperature of this quantity of water always rises by exactly \(13.3^\text{o} \text{C}\). Therefore, the same quantity of heat must always be produced by reaction of this quantity of methane. If we burn \(1.0 \: \text{g}\) of methane to heat \(500 \: \text{g}\) of water instead, we observe a temperature rise of \(26.6^\text{o} \text{C}\). If we burn \(1.0 \: \text{g}\) of methane to heat \(1000 \: \text{g}\) of iron, we observe a temperature rise of \(123^\text{o} \text{C}\). Therefore, the temperature rise observed as a function of the quantity of material heated as well as the nature of the material heated. Consequently, \(13.3^\text{o} \text{C}\) is not an approximate measure of this quantity of heat, since we cannot say that the burning of \(1.0 \: \text{g}\) of methane "produces" \(13.3^\text{o} \text{C}\) of heat. Such a statement is clearly revealed to be nonsense, so the concepts of temperature and heat must be kept distinct. Our observations do reveal that we can relate the temperature rise produced in a substance to a fixed quantity of heat, provided that we specify the type and amount of the substance. Therefore, we define a property for each substance, called the , which relates the temperature rise to the quantity of heat absorbed. We define \(q\) to be the quantity of heat, and \(\Delta T\) to be the temperature rise produced by this heat. The heat capacity \(C\) is defined by \[q = C \Delta T\] This equation, however, is only a definition and does not help us calculate either \(q\) or \(C\), since we know neither one. Next, however, we observe that we can also elevate the temperature of a substance , that is, by doing work on it. As simple examples, we can warm water by stirring it, or warm metal by rubbing or scraping it. (As a historical note, these observations were crucial in establishing that heat is equivalent to work in its effect on matter, demonstrating that heat is therefore a form of energy.) Although it is difficult to do, we can measure the amount of work required to elevate the temperature of \(1 \: \text{g}\) of water by \(1^\text{o} \text{C}\). We find that the amount of work required is invariably equal to \(4.184 \: \text{J}\). Consequently, adding \(4.184 \: \text{J}\) of energy to \(1 \: \text{g}\) of water must elevate the energy of the water molecules by an amount measured by \(1^\text{o} \text{C}\). By conservation of energy, the energy of the water molecules does not depend on how that energy was acquired. Therefore, the increase in energy measured by a \(1^\text{o} \text{C}\) temperature increase is the same regardless of whether the water was heated or stirred. As such, \(4.184 \: \text{J}\) must also be the amount of energy added to the water molecules when they are by \(1^\text{o} \text{C}\) rather than stirred. We have therefore effectively measured the heat \(q\) required to elevate the temperature of \(1 \: \text{g}\) of water by \(1^\text{o} \text{C}\). Referring back to the definition of heat capacity, we now can calculate that the heat capacity of \(1 \: \text{g}\) of water must be \(4.184 \: \frac{\text{J}}{^\text{o} \text{C}}\). The heat capacity of a substance is referred to as the of the substance, usually indicated by the symbol \(c_s\). The specific heat of water is \(4.184 \: \frac{\text{J}}{^\text{o} \text{C}}\). Determining the heat capacity (or specific heat) of water is an extremely important measurement for two reasons. First, from the heat capacity of water we can determine the heat capacity of any other substance very simply. Imagine taking a hot \(5.0 \: \text{g}\) iron weight at \(100^\text{o} \text{C}\) and placing it in \(10.0 \: \text{g}\) of water at \(25^\text{o} \text{C}\). We know from experience that the iron bar will be cooled and the water will be heated until both have achieved the same temperature. This is an easy experiment to perform, and we find that the final temperature of the iron and water is \(28.8^\text{o} \text{C}\). Clearly, the temperature of the water has been raised by \(3.8^\text{o} \text{C}\). From the definition of heat capacity and the specific heat of water, we can calculate that the water must have absorbed an amount of heat \(q = \left( 10.0 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) \left( 3.8^\text{o} \text{C} \right) = 1.59 \: \text{J}\). By conservation of energy, this must be the amount of heat by the \(1 \: \text{g}\) iron weight, whose temperature was lowered by \(71.2^\text{o} \text{C}\). Again referring to the definition of heat capacity, we can calculate the specific heat of the iron bar to be \(c_s = \frac{-159 \: \text{J}}{\left( -71.2^\text{o} \text{C} \right) \left( 5.0 \: \text{g} \right)} = 0.45 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}}\). Following this procedure, we can easily produce extensive tables of heat capacities for many substances. Second, and perhaps more importantly for our purposes, we can use the known specific heat of water to measure the heat released in any chemical reaction. To analyze a previous example, we observed that the combustion of \(1.0 \: \text{g}\) of methane gas released sufficient heat to increase the temperature of \(1000 \: \text{g}\) of water by \(13.3^\text{o} \text{C}\). The heat capacity of \(1000 \: \text{g}\) of water must be \(\left( 1000 \: \text{g} \right) \left( 4.184 \: \frac{\text{J}}{\text{g} ^\text{o} \text{C}} \right) = 4184 \: \text{J}{^\text{o} \text{C}}\). Therefore, by the definition of heat capacity, elevating the temperature of \(1000 \: \text{g}\) of water by \(13.3^\text{o} \text{C}\) must require \(55,650 \: \text{J} = 55.65 \: \text{kJ}\) of heat. The method of measuring reaction energies by capturing the heat evolved in a water bath and measuring the temperature rise produced in that water bath is called . This method is dependent on the equivalence of heat and work as transfers of energy, and on the law of conservation of energy. Following this procedure, we can straightforwardly measure the heat released or absorbed in any easily performed chemical reaction. For reactions which are difficult to initiate or which occur only under restricted conditions or which are exceedingly slow, we will require alternative methods. Hydrogen gas, which is of potential interest nationally as a clean fuel, can be generated by the reaction of carbon (coal) and water: \[\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)\] Calorimetry reveals that this reaction requires the of \(90.1 \: \text{kJ}\) of heat for every mole of \(\ce{C} \left( s \right)\) consumed. By convention, when heat is absorbed during a reaction, we consider the quantity of heat to be a positive number: in chemical terms, \(q > 0\) for an reaction. When heat is evolved, the reaction is and \(q < 0\) by convention. It is interesting to ask where this input energy goes when the reaction occurs. One way to answer this question is to consider the fact that the reaction converts one fuel, \(\ce{C} \left( s \right)\), into another, \(\ce{H_2} \left( g \right)\). To compare the energy available in each fuel, we can measure the heat evolved in the combustion of each fuel with one mole of oxygen gas. We observe that \[\ce{C} \left( s \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right)\] produces \(393.5 \: \text{kJ}\) for one mole of carbon burned; hence \(q = -393.5 \: \text{kJ}\). The reaction \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\] produces \(483.6 \: \text{kJ}\) for two moles of hydrogen gas burned, so \(q = -483.6 \: \text{kJ}\). It is evident that more energy is available from combustion of the hydrogen fuel than from combustion of the carbon fuel, so it is not surprising that conversion of the carbon fuel to hydrogen fuel requires the input of energy. Of considerable importance is the observation that the heat input in the reaction of coal and water, \(90.1 \: \text{kJ}\) is exactly equal to the between the heat evolved, \(-393.5 \: \text{kJ}\), in the combustion of carbon and the heat evolved, \(-483.6 \: \text{kJ}\), in the combustion of hydrogen. This is not a coincidence: if we take the combustion of carbon and add it to the of the combustion of hydrogen, we get \[\begin{align} \ce{C} \left( s \right) + \ce{O_2} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) \\ 2 \ce{H_2O} \left( g \right) &\rightarrow 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \\ \ce{C} \left( s \right) + \ce{O_2} \left( g \right) + 2 \ce{H_2O} \left( g \right) &\rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \end{align}\] Canceling the \(\ce{O_2} \left( g \right)\) from both sides, since it is net neither a reactant nor product. The final equation is equivalent to the reaction of carbon and water presented earlier. Thus, taking the combustion of carbon and "subtracting" the combustion of hydrogen (or more accurately, adding the reverse of the combustion of hydrogen) yields the final equation above. And, the heat of the combustion of carbon the heat of the combustion of hydrogen equals the heat of the reaction between carbon and water. By studying many chemical reactions in this way, we discover that this result, known as , is general. (Although we have not considered the restriction, applicability of this law requires that all reactions considered proceed under similar conditions: we will consider all reactions to occur at constant pressure.) A pictorial view of Hess' Law as applied to the heat of the reaction of carbon and water is illustrative. In Figure 10.1, the reactants \(\ce{C} \left( s \right) + 2 \ce{H_2O} \left( g \right)\) are placed together in a box, representing the of the materials involved in the reaction prior to the reaction. The products \(\ce{CO_2} \left( g \right) + 2 \ce{H_2} \left( g \right)\) are placed together in a second box representing the state of the materials involved after the reaction. The reaction arrow connecting these boxes is labeled with the heat of this reaction. Now we take these same materials and place them in a third box containing \(\ce{C} \left( s \right)\), \(\ce{O_2} \left( g \right)\), and \(2 \ce{H_2} \left( g \right)\). This box is connected to the reactant and product boxes with reaction arrows, labeled by the heats of reaction in the combustion of carbon and the combustion of hydrogen. This picture of Hess' Law reveals that the heat of reaction along the "path" directly connecting the reactant state to the product state is exactly equal to the total heat of reaction along the alternative "path" connecting reactants to products via the intermediate state containing \(\ce{C} \left( s \right)\), \(\ce{O_2} \left( g \right)\), and \(2 \ce{H_2} \left( g \right)\). A consequence of our observation of Hess' Law is therefore that the net heat evolved or absorbed during a reaction is independent of the path connecting the reactant to product. (This statement is again subject to our restriction that all reactions in the alternative path must occur under constant pressure conditions.) A slightly different view of Figure 10.1 results from beginning at the reactant box and following a complete circuit through the other boxes leading back to the reactant box, summing the net heats of reaction as we go. We discover that the net heat transferred (again provided that all reactions occur under constant pressure) is exactly zero. This is a statement of the conservation of energy: the energy in the reactant state does not depend upon the processes which produced that state. Therefore, we cannot extract any energy from the reactants by a process which simply recreates the reactants. Were this not the case, we could endlessly produce unlimited quantities of energy by following the circuitous path which continually reproduces the initial reactants. By this reasoning, we can define an energy function whose value for the reactants is independent of how the reactant state was prepared. Likewise, the value of this energy function in the product state is independent of how the products are prepared. We choose this function, \(H\), so that the change in the function, \(\Delta H = H_\text{products} - H_\text{reactants}\), is equal to the heat of reaction \(q\) under constant pressure conditions. \(H\), which we call the , is a , since its value depends only on the state of the materials under consideration, that is, the temperature, pressure, and composition of these materials. The concept of a state function is somewhat analogous to the idea of elevation. Consider the difference in elevation between the first floor and the third floor of a building. This difference is independent of the path we choose to get from the first floor to the third floor. We can simply climb up two flights of stairs, or we can climb one flight of stairs, walk the length of the building, then walk a second flight of stairs. Or we can ride the elevator. We could even walk outside and have a crane lift us to the roof of the building, from which we can climb down to the third floor. Each path produces exactly the same elevation gain, even though the distance traveled is significantly different from one path to the next. This is simply because the elevation is a "state function". Our elevation, standing on the third floor, is independent of how we got to the third floor, and the same is true of the first floor. Since the elevation thus is a state function, the elevation gain is independent of the path. Now, the existence of an energy state function \(H\) is of considerable importance in calculating heats of reaction. Consider the prototypical reaction in Figure 10.2a, with reactants \(\ce{R}\) being converted to products \(\ce{P}\). We wish to calculate the heat absorbed or released in this reaction, which is \(\Delta H\). Since \(H\) is a state function, we can follow any path from \(\ce{R}\) to \(\ce{P}\) and calculate \(\Delta H\) along that path. In Figure 10.2b, we consider one such possible path, consisting of two reactions passing through an intermediate state containing all the atoms involved in the reaction, each in elemental form. This is a useful intermediate state since it can be used for any possible chemical reaction. For example, in Figure 10.1, the atoms involved in the reaction are \(\ce{C}\), \(\ce{H}\), and \(\ce{O}\), each of which are represented in the intermediate state in elemental form. We can see in Figure 10.2b that the \(\Delta H\) for the overall reaction is now the difference between the \(\Delta H\) in the formation of the products \(\ce{P}\) from the elements and the \(\Delta H\) in the formation of the reactants \(\ce{R}\) from the elements. a. b. The \(\Delta H\) values for formation of each material from the elements are thus of general utility in calculating \(\Delta H\) for any reaction of interest. We therefore define the for reactant \(\ce{R}\), as \[\text{elements in standard state} \rightarrow \ce{R}\] and the heat involved in this reaction is the , designated by \(\Delta H_f^\text{o}\). The subscript \(f\), standing for "formation", indicates that the \(\Delta H\) is for the reaction creating the material from the elements in standard state. The superscript \(^\text{o}\) indicates that the reactions occur under constant standard pressure conditions of \(1 \: \text{atm}\). From Figure 10.2b, we see that the heat of any reaction can be calculated from \[Delta H_f^\text{o} = \Delta H_{f, \: \text{products}}^\text{o} - \Delta H_{f, \: \text{reactants}}^\text{o}\] Extensive tables of \(\Delta H_f^\text{o}\) have been compiled and published. This allows us to calculate with complete confidence the heat of reaction for any reaction of interest, even including hypothetical reactions which may be difficult to perform or impossibly slow to react. The for a molecule is the energy required to separate the two bonded atoms to great distance. We recall that the total energy of the bonding electrons is lower when the two atoms are separated by the bond distance than when they are separated by a great distance. As such, the energy input required to separate the atoms elevates the energy of the electrons when the bond is broken. We can use diatomic bond energies to calculate the heat of reaction \(\Delta H\) for any reaction involving only diatomic molecules. We consider two simple examples. First, the reaction \[\ce{H_2} \left( g \right) + \ce{Br} \left( g \right) \rightarrow \ce{H} \left( g \right) + \ce{HBr} \left( g \right)\] is observed to be endothermic with heat of reaction \(70 \: \frac{\text{kJ}}{\text{mol}}\). Note that this reaction can be viewed as consisting entirely of the breaking of the \(\ce{H_2}\) bond followed by the formation of the \(\ce{HBr}\) bond. Consequently, we must input energy equal to the bond energy of \(\ce{H_2}\) \(\left( 436 \: \frac{\text{kJ}}{\text{mol}} \right)\), but in forming the \(\ce{HBr}\) bond we recover output energy equal to the bond energy of \(\ce{HBr}\) \(\left( 366 \: \frac{\text{kJ}}{\text{mol}} \right)\). Therefore the heat of the overall equation at constant pressure must be equal to the difference in these bond energies, \(70 \: \frac{\text{kJ}}{\text{mol}}\). Now we can answer the question, at least for this reaction, of where the energy "goes" during the reaction. The reason this reaction absorbs energy is that the bond which must be broken, \(\ce{H_2}\), is stronger than the bond which is formed, \(\ce{HBr}\). Note that energy is released when the \(\ce{HBr}\) bond is formed, but the amount of energy released is less than the amount of energy required to break the \(\ce{H_2}\) bond in the first place. The second example is similar: \[\ce{H_2} \left( g \right) + \ce{Br_2} \left( g \right) \rightarrow 2 \ce{HBr} \left( g \right)\] This reaction is exothermic with \(\Delta H^\text{o} = -103 \: \frac{\text{kJ}}{\text{mol}}\). In this case, we must break an \(\ce{H_2}\) bond, with energy \(436 \: \frac{\text{kJ}}{\text{mol}}\), and a \(\ce{Br_2}\) bond, with energy \(193 \: \frac{\text{kJ}}{\text{mol}}\). Since two \(\ce{HBr}\) molecules are formed, we must form two \(\ce{HBr}\) bonds, each with bond energy \(366 \: \frac{\text{kJ}}{\text{mol}}\). In total, then, breaking the bonds in the reactants requires \(629 \: \frac{\text{kJ}}{\text{mol}}\), and forming the new bonds releases \(732 \: \frac{\text{kJ}}{\text{mol}}\), for a net release of \(103 \: \frac{\text{kJ}}{\text{mol}}\). This calculation reveals that the reaction is exothermic because, although we must break one very strong bond and one weaker bond, we form two strong bonds. There are two items worth reflection in these examples. First, energy is released in a chemical reaction due to the of strong bonds. Breaking a bond, on the other hand, always requires the of energy. Second, the reaction of \(\ce{H_2}\) and \(\ce{Br_2}\) does not actually proceed by the two-step process of breaking both reactant bonds, thus forming four free atoms, followed by making two new bonds. The actual process of the reaction is significantly more complicated. The details of this process are irrelevant to the energetics of the reaction, however, since, as we have shown, the heat of reaction \(\Delta H\) does not depend on the path of the reaction. This is another example of the utility of Hess' Law. We now proceed to apply this bond energy analysis to the energetics of reactions involving polyatomic molecules. A simple example is the combustion of hydrogen gas discussed previously. This is an explosive reaction, producing \(483.6 \: \text{kJ}\) per mole of oxygen. Calculating the heat of reaction from bond energies requires us to know the bond energies in \(\ce{H_2O}\). In this case, we must break not one but two bonds: \[\ce{H_2O} \left( g \right) \rightarrow 2 \ce{H} \left( g \right) + \ce{O} \left( g \right)\] The energy required to perform this reaction is measured to be \(926.0 \: \frac{\text{kJ}}{\text{mol}}\). The reaction of hydrogen and oxygen can proceed by a path in which we first break two \(\ce{H_2}\) bonds and one \(\ce{O_2}\) bond, then we follow the reverse of the decomposition of water twice: \[\begin{align} 2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) &\rightarrow 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) \\ 4 \ce{H} \left( g \right) + 2 \ce{O} \left( g \right) &\rightarrow 2 \ce{H_2O} \left( g \right) \\ 2 \ce{H_2} \left( g \right) + \ce{O_2} &\rightarrow 2 \ce{H_2O} \left( g \right) \end{align}\] Therefore, the energy of the final equation must be the energy required to break two \(\ce{H_2}\) bonds and one \(\ce{O_2}\) bond minus twice the energy of the decomposition of water. We calculate that \(\Delta H^\text{o} = 2 \times \left( 436 \: \frac{\text{kJ}}{\text{mol}} \right) + 498.3 \: \frac{\text{kJ}}{\text{mol}} - 2 \times \left( 926.9 \: \frac{\text{kJ}}{\text{mol}} \right) = -483.5 \: \frac{\text{kJ}}{\text{mol}}\). It is clear from this calculation that the formation of water is strongly exothermic because of the very large amount of energy released when two hydrogen atoms and one oxygen atom form a water molecule. It is tempting to use the heat of the decomposition of water to calculate the energy of an \(\ce{O-H}\) bond. Since breaking the two \(\ce{O-H}\) bonds in water requires \(926.9 \: \frac{\text{kJ}}{\text{mol}}\), then we might infer that breaking a single \(\ce{O-H}\) bond requires \(\frac{926.9}{2} \: \frac{\text{kJ}}{\text{mol}} = 463.5 \: \frac{\text{kJ}}{\text{mol}}\). However, the reaction \[\ce{H_2O} \left( g \right) \rightarrow \ce{OH} \left( g \right) + \ce{H} \left( g \right)\] has \(\Delta H^\text{o} = 492 \: \frac{\text{kJ}}{\text{mol}}\). Therefore, the energy required to break an \(\ce{O-H}\) bond in \(\ce{H_2O}\) is not the same as the energy required to break the \(\ce{O-H}\) bond in the \(\ce{OH}\) diatomic molecule. Stated differently, it requires more energy to break the first \(\ce{O-H}\) bond in water than is required to break the second \(\ce{O-H}\) bond. In general, we find that the energy required to break a bond between any two particular atoms depends upon the molecule those two atoms are in. Considering yet again oxygen and hydrogen, we find that the energy required to break the \(\ce{O-H}\) bond in methanol \(\left( \ce{CH_3OH} \right)\) is \(437 \: \frac{\text{kJ}}{\text{mol}}\), which differs substantially from the energy of the first \(\ce{O-H}\) bond in water. Similarly, the energy required to break a single \(\ce{C-H}\) bond in methane \(\left( \ce{CH_4} \right)\) is \(435 \: \frac{\text{kJ}}{\text{mol}}\), but the energy required to break all four \(\ce{C-H}\) bonds in methane is \(1663 \: \frac{\text{kJ}}{\text{mol}}\), which is not equal to four times the energy of one bond. As another such comparison, the energy required to break a \(\ce{C-H}\) bond is \(400 \: \frac{\text{kJ}}{\text{mol}}\) in trichloromethane \(\left( \ce{HCCl_3} \right)\), \(414 \: \frac{\text{kJ}}{\text{mol}}\) in dichloromethane \(\left( \ce{H_2CCl_2} \right)\), and \(422 \: \frac{\text{kJ}}{\text{mol}}\) in chloromethane \(\left( \ce{H_3CCl} \right)\). These observations are somewhat discouraging, since they reveal that, to use bond energies to calculate the heat of a reaction, we must first measure the bond energies for all bonds for all molecules involved in that reaction. This is almost certainly more difficult than it is desirable. On the other hand, we can note that the bond energies for similar bonds in similar molecules are close to one another. The \(\ce{C-H}\) bond energy in any one of the three chloromethanes above illustrate this quite well. We can estimate the \(\ce{C-H}\) bond energy in any one of these chloromethanes by the average \(\ce{C-H}\) bond energy in the three chloromethane molecules, which is \(412 \: \frac{\text{kJ}}{\text{mol}}\). Likewise, the average of the \(\ce{C-H}\) bond energies in methane is \(\frac{1663}{4} \: \frac{\text{kJ}}{\text{mol}} = 416 \: \frac{\text{kJ}}{\text{mol}}\) and is thus a reasonable approximation to the energy required to break a single \(\ce{C-H}\) bond in methane. By analyzing many bond energies in many molecules, we find that, in general, we can approximate the bond energy in any particular molecule by the average of the energies of similar bonds. These average bond energies can then be used to estimate the heat of a reaction without measuring all of the required bond energies. Consider for example the combustion of methane to form water and carbon dioxide: \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\] We can estimate the heat of this reaction by using average bond energies. We must break four \(\ce{C-H}\) bonds at an energy cost of approximately \(4 \times 412 \: \frac{\text{kJ}}{\text{mol}}\) and two \(\ce{O_2}\) bonds at an energy cost of approximately \(2 \times 496 \: \frac{\text{kJ}}{\text{mol}}\). Forming the bonds in the products releases approximately \(2 \times 742 \: \frac{\text{kJ}}{\text{mol}}\) for the two \(\ce{C=O}\) double bonds and \(4 \times 463 \: \frac{\text{kJ}}{\text{mol}}\) for the \(\ce{O-H}\) bonds. Net, the heat of reaction is thus approximately \(\Delta H^\text{o} = 1648 + 992 - 1486 - 1852 = -698 \: \frac{\text{kJ}}{\text{mol}}\). This is a rather rough approximation to the actual heat of combustion of methane, \(-890 \: \frac{\text{kJ}}{\text{mol}}\). Therefore, we cannot use average bond energies to predict accurately the heat of a reaction. We can get an estimate, which may be sufficiently useful. Moreover, we can use these calculations to gain insight into the energetics of the reaction. For example, the combustion of methane is strongly exothermic, which is why methane gas (the primary component in natural gas) is an excellent fuel. From our calculation, we can see that the reaction involved breaking six bonds and forming six new bonds. The bonds formed are substantially stronger than those broken, thus accounting for the net release of energy during the reaction. Assume you have two samples of two different metals, X and Z. The samples are exactly the same mass. Both samples are heated to the same temperature. Then each sample is placed into separate glasses containing identical quantities of cold water, initially at identical temperatures below that of the metals. The final temperature of the water containing metal X is greater than the final temperature of the water containing metal Z. Which of the two metals has the larger heat capacity? Explain your conclusion. If each sample, initially at the same temperature, is heated with exactly \(100 \: \text{J}\) of energy, which sample has the higher final temperature? Explain how Hess' Law is a consequence of conservation of energy. Consider the reaction \[\ce{N_2O_4} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\] Draw Lewis structures for each of \(\ce{N_2O_4}\) and \(\ce{NO_2}\). On the basis of these structures, predict whether the reaction is endothermic or exothermic, and explain your reasoning. Why is the bond energy of \(\ce{H_2}\) not equal to \(\Delta H_f^\text{o}\) of \(\ce{H_2}\)? For what species is the enthalpy of formation related to the bond energy of \(\ce{H_2}\)? Suggest a reason why \(\Delta H\) for the reaction \[\ce{CO_2} \left( g \right) \rightarrow \ce{CO} \left( g \right) + \ce{O} \left( g \right)\] is not equal to \(\Delta H^\text{o}\) for the reaction \[\ce{CO} \left( g \right) \rightarrow \ce{C} \left( g \right) + \ce{O} \left( g \right)\]. Determine whether the reaction is exothermic or endothermic for each of the following circumstances: The heat of combustion of the products is greater than the heat of combustion of the reactants. The enthalpy of formation of the products is greater than the enthalpy of reaction of the reactants. The total of the bond energies of the products is greater than the total of the bond energies for the reactants. ; Chemistry) | 30,680 | 3,658 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/01%3A_The_Basics/1.05%3A_Work_and_Energy |
Temperature, pressure and volume are important variables in the description of physical systems. They will also be important to describe how energy flows from one system to another. Generally, energy can flow in two important forms: 1) work and 2) heat. The bookkeeping needed to track the flow of energy is what the subject of Thermodynamics is all about, so these topics will be discussed at length in subsequent chapters. However, a little bit of review is in order, just to set the foundation for the discussions that are forthcoming. is an important entity in the modern world. We use energy to light our homes, drive our cars, and power our electronic devices. According to Richard Smalley, co-winner of the 1996 Nobel Prize in Chemistry, energy is one of the (if not the biggest) challenge we face moving into the 21 century : Energy can be measured in a multitude of different units, including joules (J), kilojoules (kJ), calories (cal), kilocalories (kcal), as well as several other set of units such as kJ/mol or kcal/mol. A was once defined as the amount of energy needed to raise the temperature of 1 g of water by 1 °C. This definition suggests a convenient property of water called the : \[C = \dfrac{1\,cal}{g\, °C} \nonumber \] The modern definition of a calorie is 4.184 , \[ 1 \,J = 1\,N \cdot m = \dfrac{1 \,kg \, m^2}{s^2} \nonumber \] where a joule is the energy necessary to move a mass a distance of 1 m against a resisting force of 1 N. A dietary is equal to 1000 cal, or 1 kcal, and is often listed on the labels of food containers to indicate the energy content of the food inside. Energy can take the form of (stored energy) and (realized energy) forms. Kinetic energy is the energy of motion. On the other hand, potential energy can be defined as the energy stored in a system that can be converted to kinetic energy someplace in the universe. Kinetic energy of a particle can be expressed as \[E_{kin} = \dfrac{1}{2} mv^2 \label{transKE} \] where \(m\) is the mass of the particle, and \(v\) is the magnitude of its velocity (or speed). Equation \ref{transKE} describes the kinetic energy associated with translation; other expressions exist for different motions (e.g., rotation or vibration). An example of a system in which energy is converted between kinetic energy and potential energy is a Hooke’s Law oscillator. According to Hooke’s Law, the force acting on an object is proportional in magnitude to the displacement of the object from an equilibrium position, and opposite in sign. \[ F = -kx \label{hook} \] In this equation, \(F\) is the force, \(x\) is the displacement from equilibrium, and \(k\) is the constant of proportionality. The negative sign is necessary to insure that the force acting on the object is one that will tend to restore it to an equilibrium position (\(x = 0\)) irrespective of whether \(x\) is positive or negative. As the object that follows Hooke’s Law moves from its equilibrium position, the kinetic energy of its motion is converted into potential energy until there is no more kinetic energy left. At this point, the change in displacement will change direction, returning the object to the equilibrium position by converting potential energy back into kinetic energy. As the object is displaced in the along the x-axis (in the case shown in Figure 1.5.1, this would be accomplished by stretching or compressing the spring), the potential energy increases. The force acting on the object will also increase as the object is displaced and will be directed opposite of the direction of displacement (Equation \ref{hook}). According to Newtonian physics, the potential energy \(U(x)\) is given by the negative integral of the force with respect to position: \[U (x) = - \int F(x) dx \label{energy} \] Substituting Equation \ref{hook} in Equation \ref{energy} yields \[U(x) = - \int (-k x)dx \nonumber \] \[ = \dfrac{1}{2} kx^2 + constant \nonumber \] With the proper choice of coordinate system and other definitions, the constant of integration can be arbitrarily made to be zero (for example, by choosing it to offset any other forces acting on the object, such as the force due to gravity.) The kinetic energy is then given by the total energy minus the potential energy (since the total energy must be constant due to the conservation of energy in the system!) \[ E_{kin} = E_{tot} -U(x) \nonumber \] is defined as the amount of energy expended to move a mass against a resisting force. For a mass being moved along a surface, the amount of energy expended must be sufficient to overcome the resisting force (perhaps due to friction) and also sufficient to cause motion along the entire path. The energy expended as work in this case (if the force is independent of the position of the object being moved) is given by \[w= - F \,\Delta x \nonumber \] where F is the magnitude of the resisting force, and x is the displacement of the object. The negative sign is necessary since the force is acting in the opposite direction of the motion. A more general expression, and one that can be used if the force is not constant over the entire motion is \[ dw = -F \,dx \nonumber \] This expression can then be integrated, including any dependence \(F\) might have on \(x\) as needed for a given system. Another important way that work can be defined includes that for the expansion of a gas sample against an external pressure. In this case, the displacement is defined by a change in volume for the sample: \[dw = -p_{ext} dV \nonumber \] This is a very convenient expression and will be used quite often when discussing the work expended in the expansion of a gas. The conversion of potential energy into kinetic energy generally is accomplished through work which is done someplace in the universe. As such, the concepts of energy and work are inexorably intertwined. They will be central to the study of thermodynamics. An important case of is that of the . For a change to be reversible there can be no net force pushing the change in one direction or the other. In order for this to be the case, the internal pressure (that of the system) and external pressure (that of the surroundings) must be the same. \[p_{int} = p_{ext} = p \nonumber \] In this case, the work of expansion can be calculated by integrating the expression for dw. Making a simple substitution from the allows for the expression in terms of volume and temperature. If the temperature is constant (so that it can be placed before the integral) the expression becomes \[w= - nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = -nRT \ln \left( \dfrac{V_2}{V_1} \right) \label{workIG} \] where \(V_1\) and \(V_2\) are the initial and final volumes of the expansion respectively. Consider 1.00 mol of an ideal gas, expanding isothermally at 273 K, from an initial volume of 11.2 L to a final volume of 22.4 L. What is the final pressure of the gas? Calculate the work of the expansion if it occurs First, let’s calculate the final pressure via Equation \ref{IGL}: \[ p = \dfrac{nRT}{V} = \dfrac{(1.00\,mole)(0.08206\, atm\,L\,mol^{-1} K^{-1})(273\,K)}{22.4\, L} = 1.00 \, atm \nonumber \] (This may be a relationship you remember from General Chemistry – that 1 mole of an idea gas occupies 22.4 L at 0 C!) Okay – now for the irreversible expansion against a constant external pressure: \[dw = -p_{ext}dV \nonumber \] so \[w = -p_{exp} \int _{V_1}^{V_2} dV = -p_{ext} \Delta V \nonumber \] \[ w = -(1.00 \, atm)(22.4 \, L- 11.2\,L) =-11.2 \,atm\,L \nonumber \] But what the heck is an ? It is actually a fairly simply thing to convert from units of to J by using the . \[w = -(11.2 \,atm\,L) \left( \dfrac{8.314\, \frac{J}{mol\, K}}{0.08206\, \frac{atm\, L}{mol\, K}} \right) = -1130 \,J \nonumber \] Note that the negative sign indicated that the system is expending energy by doing work on the surroundings. (This concept will be vital in the Chapter 3!) Now for the reversible pathway. The work done by the system can be calculated for this change using Equation \ref{workIG}: \[w= -(1.00\, mol) (8.314 J\, mol^{-1} K^{-1}) (273 \,K) \ln \left( \dfrac{22.4\,L}{11.2\,L} \right) = -1570\,J \nonumber \] : There are many cases of “limiting ideal behavior” which we use to derive and/or explore the nature of chemical systems. The most obvious case, perhaps, is that of the . | 8,339 | 3,659 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Peptides_and_Proteins/Hydrolysis_of_Proteins |
This page looks briefly at the hydrolysis of proteins into their constituent amino acids using hydrochloric acid. If you have already studied the hydrolysis of amides under acidic conditions, you will find that this is basically the same reaction. That's not surprising because what biologists and biochemists call a peptide link (in proteins, for example) is what chemists call an amide link. With an amide like ethanamide, the carbon-nitrogen bond in the amide group is broken and you get a carboxylic acid formed: \[CH_3CONH_2 + H_2O + H^+ \rightarrow CH_3COOH + NH_4^+\] Now imagine doing the same thing with a simple dipeptide made of any two amino acids. Instead of ammonium ions, you get positive ions made from the -NH groups reacting with hydrogen ions. You need the extra hydrogen ion in the equation (compared with the amide equation) to react with the -NH group on the left-hand end of the dipeptide - the one not involved in the peptide link. If you scale this up to a polypeptide (a protein chain), each of the peptide links will be broken in exactly the same way. That means that you will end up with a mixture of the amino acids that made up the protein - although in the form of their positive ions because of the presence of the hydrogen ions from the hydrochloric acid. There are two ways of carrying out this reaction - an old, slow method, and a new, fast one. The protein is heated with 6 M hydrochloric acid for about 24 hours at 110°C. (6M hydrochloric acid is slightly more than semi-concentrated.) Protein samples are placed in tubes in a sealed container containing 6 M hydrochloric acid in an atmosphere of nitrogen. The whole container is then placed in a microwave oven for about 5 - 30 minutes (depending on the protein) with temperatures up to 200°C. The hydrochloric acid vaporizes, comes into contact with the protein samples and hydrolyses them. This method is used to hydrolyse small samples of protein during protein analysis. | 1,977 | 3,660 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Potentials/Standard_Potentials |
In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work. Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure 19.3 but instead of copper use a strip of cobalt metal and 1 M Co in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V. The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the (E° ), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions,Concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for nonideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C. Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system. It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured. (This is analogous to measuring absolute enthalpies or free energies. Recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram: \[Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{19.9}\] This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu in solution at the copper cathode. All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances ( ). The standard cell potential (E° ) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum: \[E°_{cell} = E°_{cathode} − E°_{anode} \label{19.10}\] In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation \(\ref{19.10}\), when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell. The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials: \[E°_{cell} = E°_{cathode} − E°_{anode}\] Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H . The [H ] in solution is in equilibrium with H gas at a pressure of 1 atm at the Pt-solution interface (Figure \(\Page {2}\)). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation: \[2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{19.11}\] One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction. Figure \(\Page {3}\) shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn , and H ions are reduced to H in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows: \[Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{19.12}\] The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows: \[E°_{cell}=E°_{cathode}−E°_{anode}=0.76\; V\] Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the for that half-reaction.In this example, the standard reduction potential for Zn (aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn , often called the Zn/Zn redox couple, or the Zn/Zn couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E° from E° to obtain E° : 0 − (−0.76 V) = 0.76 V. Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential. E° values do depend on the stoichiometric coefficients for a half-reaction, because it is an property. To measure the potential of the Cu/Cu couple, we can construct a galvanic cell analogous to the one shown in Figure \(\Page {3}\) but containing a Cu/Cu couple in the sample compartment instead of Zn/Zn . When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of E° indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn couple. Hence the reactions that occur spontaneously, indicated by a positive E° , are the reduction of Cu to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H is oxidized to H at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu /Cu couple on the right: \[Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.16}\] The half-cell reactions and potentials of the spontaneous reaction are as follows: \[E°_{cell} = E°_{cathode}− E°_{anode} = 0.34\; V\] Thus the standard electrode potential for the Cu /Cu couple is 0.34 V. Electrode Potentials and ECell: Previously, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other. We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas: \[Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{19.20}\] In this reaction, \(Al_{(s)}\) is oxidized to Al , and H in water is reduced to H gas, which bubbles through the solution, agitating it and breaking up the clogs. The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed , we find the corresponding half-reactions that describe the reduction of H ions in water to H and the oxidation of Al to Al in basic solution: The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution. In Equation \(\ref{19.21}\), two H ions gain one electron each in the reduction; in Equation \(\ref{19.22}\), the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation \(\ref{19.21}\)) by 3 and the oxidation half-reaction (Equation \(\ref{19.22}\)) by 2 to give the same number of electrons in both half-reactions: \[6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{19.25}\] Simplifying by canceling substances that appear on both sides of the equation, \[6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.26}\] We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced: \[2Al + 8O + 14H = 2Al + 8O + 14H \label{19.27}\] The atoms also balance, so Equation \(\ref{19.26}\) is a balanced chemical equation for the redox reaction depicted in Equation \(\ref{19.20}\). The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction. We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in but instead focus on the atoms whose oxidation states change, as illustrated in the following steps: Write the reduction half-reaction and the oxidation half-reaction. For the reaction shown in Equation \(\ref{19.20}\), hydrogen is reduced from H in OH to H , and aluminum is oxidized from Al° to Al : Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction: Balance the charges in each half-reaction by adding electrons. Two electrons are gained in the reduction of H ions to H , and three electrons are lost during the oxidation of Al° to Al : In this case, we multiply Equation \(\ref{19.34}\) (the reductive half-reaction) by 3 and Equation \(\ref{19.35}\) (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions: Adding and, in this case, canceling 8H , 3H O, and 6e , \[2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.38}\] We have three OH and one H on the left side. Neutralizing the H gives us a total of 5H O + H O = 6H O and leaves 2OH on the left side: \[2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.39}\] Check to make sure that all atoms and charges are balanced. Equation \(\ref{19.39}\) is identical to Equation \(\ref{19.26}\), obtained using the first method, so the charges and numbers of atoms on each side of the equation balance. In acidic solution, the redox reaction of dichromate ion (\(Cr_2O_7^{2−}\)) and iodide (\(I^−\)) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green \(Cr^{3+}_{(aq)}\) complex and brown I (aq) ions (Figure \(\Page {4}\)): \[Cr_2O^{2−}_{7(aq)} + I^−_{(aq)} \rightarrow Cr^{3+}_{(aq)} + I_{2(aq)}\] Balance this equation using half-reactions. redox reaction and balanced chemical equation using half-reactions Follow the steps to balance the redox reaction using the half-reaction method. From the standard electrode potentials listed in we find the half-reactions corresponding to the overall reaction: Balancing the number of electrons by multiplying the oxidation reaction by 3, Adding the two half-reactions and canceling electrons, \[Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} + 6I^−_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} + 3I_{2(aq)}\] We must now check to make sure the charges and atoms on each side of the equation balance: The charges and atoms balance, so our equation is balanced. We can also use the alternative procedure, which does not require the half-reactions listed in . Chromium is reduced from \(Cr^{6+}\) in \(Cr_2O_7^{2−}\) to \(Cr^{3+}\), and \(I^−\) ions are oxidized to \(I_2\). Dividing the reaction into two half-reactions, Balancing the atoms other than oxygen and hydrogen, We now balance the O atoms by adding H O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step. Next we balance the H atoms by adding H to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction. We must now add electrons to balance the charges. The reduction half-reaction (2Cr to 2Cr ) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I ) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3: Adding the two half-reactions and canceling substances that appear in both reactions, This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance. Copper is found as the mineral covellite (\(CuS\)). The first step in extracting the copper is to dissolve the mineral in nitric acid (\(HNO_3\)), which oxidizes sulfide to sulfate and reduces nitric acid to \(NO\): \[CuS_{(s)} + HNO_{3(aq)} \rightarrow NO_{(g)} + CuSO_{4(aq)}\] Balance this equation using the half-reaction method. \[3CuS_{(s)} + 8HNO{3(aq)} \rightarrow 8NO_{(g)} + 3CuSO_{4(aq)} + 4H_2O_{(l)}\] Balancing a Redox Reaction in Acidic Conditions: The standard cell potential for a redox reaction (E° ) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram: \[ Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.40}\] We know the values of E° for the reduction of Zn and E° for the reduction of Cu , so we can calculate E° : \[E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V\] This is the same value that is observed experimentally. If the value of E° is positive, the reaction will occur spontaneously as written. If the value of E° is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example 4 and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples. A positive E° means that the reaction will occur spontaneously as written. A negative E° means that the reaction will proceed spontaneously in the opposite direction. A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl , and the other contains a piece of nickel immersed in a 1 M solution of NiCl . The half-reactions that occur when the compartments are connected are as follows: cathode: Ni (aq) + 2e → Ni(s) anode: Ga(s) → Ga (aq) + 3e If the potential for the oxidation of Ga to Ga is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni ? galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions standard electrode potential of reaction occurring at the cathode We have been given the potential for the oxidation of Ga to Ga under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga (aq) + 3e → Ga(s), E° = −0.55 V. Using the value given for E° and the calculated value of E° , we can calculate the standard potential for the reduction of Ni to Ni from Equation \(\ref{19.10}\): This is the standard electrode potential for the reaction Ni (aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni under standard conditions, we must reverse the sign of E° . Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry. A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate \(Hg(CH_3CO_2)_2\) and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of \(MgCl_2\). When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur: If the potential for the oxidation of Mg to Mg is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the anode? 0.85 V We can use this procedure described to measure the standard potentials for a wide variety of chemical substances, some of which are listed in . These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table \(\Page {1}\). All reactants that lie below the SHE in the table are stronger oxidants than H , and all those that lie above the SHE are weaker. The strongest oxidant in the table is F , with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element. \[Ce^{4+}(aq) + e^− \rightleftharpoons Ce^{3+}(aq)\] Similarly, all species in Table \(\Page {1}\) that lie above H reductants H H reductant ΔH reductant Species in Talbe Table \(\Page {1}\) (or Table ) that lie above H are stronger reducing agents (more easily oxidized) than H . Species that lie below H are stronger oxidizing agents. Because the half-reactions shown in Table \(\Page {1}\) are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E° > 0), as illustrated below. The black tarnish that forms on silver objects is primarily Ag S. The half-reaction for reversing the tarnishing process is as follows: reduction half-reaction, standard electrode potential, and list of possible reductants reductants for Ag S, strongest reductant, and potential reducing agent for removing tarnish From their positions inTable \(\Page {1}\), decide which species can reduce Ag S. Determine which species is the strongest reductant. Use Table \(\Page {1}\) to identify a reductant for Ag S that is a common household product. We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag S/Ag couple in Table \(\Page {1}\) or (2) compare E° for each species with E° for the Ag S/Ag couple (−0.69 V). Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: redox reaction and list of standard electrode potentials ( ) reaction spontaneity Adding the two half-reactions gives the overall reaction: \(\textrm{cathode:} \; \mathrm{Be^{2+}(aq)} +\mathrm{2e^-} \rightarrow \mathrm{Be(s)}\) \(\textrm{anode:} \; \mathrm{Sn(s) \rightarrow \mathrm{Sn^{2+}}(s)} +\mathrm{2e^-} \) \(\textrm{total:} \; \mathrm{Sn(s)+ \mathrm{Be^{2+}(aq)} \rightarrow \mathrm{Sn^{2+}}(aq)} + \mathrm{Be(s)}\) The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn ) by metallic beryllium, which has a positive value of E° , will occur spontaneously. The two half-reactions and their corresponding potentials are as follows The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO , and oxygen gas will evolve from the solution. Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: Although the sign of E° tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics. When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called , whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the , must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode. The potential of any reference electrode should be affected by the properties of the solution to be analyzed, and it should also be physically isolated. There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the , which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows: \[Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{19.44}\] \[AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)}\] If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE. A second common reference electrode is the , which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg Cl ; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure \(\Page {4}\)). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows: \[Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{19.45}\] \[Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{19.46}\] At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential. One of the most common uses of electrochemistry is to measure the H ion concentration of a solution. A is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure \(\Page {5}\)). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H ] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H ] as follows (recall that pH = −log[H ]: \[E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{19.47}\] The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH. are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure \(\Page {5}\)). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped \(LaF_3\) as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table \(\Page {2}\). The Standard Hydrogen Electrode (SHE): The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E° ). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E° = E° − E° ). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase. If E° is positive, the reaction will occur spontaneously under standard conditions. If E° is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°. | 32,275 | 3,661 |
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Chemical bonds form when electrons can be simultaneously close to two or more nuclei, but beyond this, there is no simple, easily understood theory that would not only explain why atoms bind together to form molecules, but would also predict the three-dimensional structures of the resulting compounds as well as the energies and other properties of the bonds themselves. Unfortunately, no one theory exists that accomplishes these goals in a satisfactory way for all of the many categories of compounds that are known. Moreover, it seems likely that if such a theory does ever come into being, it will be far from simple. When we are faced with a scientific problem of this complexity, experience has shown that it is often more useful to concentrate instead on developing . A scientific model is something like a theory in that it should be able to explain observed phenomena and to make useful predictions. But whereas a theory can be discredited by a single contradictory case, a model can be useful even if it does not encompass all instances of the phenomena it attempts to explain. We do not even require that a model be a credible representation of reality; all we ask is that be able to explain the behavior of those cases to which it is applicable in terms that are consistent with the model itself. An example of a model that you may already know about is the kinetic molecular theory of gases. Despite its name, this is really a model (at least at the level that beginning students use it) because it does not even try to explain the observed behavior of real gases. Nevertheless, it serves as a tool for developing our understanding of gases, and as a starting point for more elaborate treatments. Given the extraordinary variety of ways in which atoms combine into aggregates, it should come as no surprise that a number of useful bonding models have been developed. Most of them apply only to certain classes of compounds, or attempt to explain only a restricted range of phenomena. In this section we will provide brief descriptions of some of the bonding models; the more important of these will be treated in much more detail in later parts of this chapter. By classical, we mean models that do not take into account the quantum behavior of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics. Although this completely ignores what has been learned about the nature of the electron since the development of quantum theory in the 1920's, these classical models have not only proven extremely useful, but the major ones also serve as the basis for the chemist's general classification of compounds into "covalent" and "ionic" categories. Ever since the discovery early in the 19th century that solutions of salts and other electrolytes conduct electric current, there has been general agreement that the forces that hold atoms together must be electrical in nature. Electrolytic solutions contain ions having opposite electrical charges, opposite charges attract, so perhaps the substances from which these ions come consist of positive and negatively charged atoms held together by electrostatic attraction. It turns out that this is not true generally, but a model built on this assumption does a fairly good job of explaining a rather small but important class of compounds that are called ionic solids. The most well known example of such a compound is sodium chloride, which consists of two interpenetrating lattices of Na and Cl ions arranged in such as way that every ion of one type is surrounded (in three dimensional space) by six ions of opposite charge. The main limitation of this model is that it applies really well only to the small class of solids composed of Group 1 and 2 elements with highly electronegative elements such as the halogens. Although compounds such as CuCl dissociate into ions when they dissolve in water, the fundamental units making up the solid are more like polymeric chains of covalently-bound CuCl molecules that have little ionic character. This model originated with the theory developed by G.N. Lewis in 1916, and it remains the most widely-used model of chemical bonding. The essential element s of this model can best be understood by examining the simplest possible molecule. This is the H , which consists of two nuclei and one electron. First, however, think what would happen if we tried to make the even simpler molecule H . Since this would consist only of two protons whose electrostatic charges would repel each other at all distances, it is clear that such a molecule cannot exist; something more than two nuclei are required for bonding to occur. In the hydrogen molecule ion H we have a third particle, an electron. The effect of this electron will depend on its location with respect to the two nuclei. If the electron is in the space between the two nuclei, it will attract both protons toward itself, and thus toward each other. If the total attraction energy exceeds the internuclear repulsion, there will be a net bonding effect and the molecule will be stable. If, on the other hand, the electron is off to one side, it will attract both nuclei, but it will attract the closer one much more strongly, owing to the inverse-square nature of Coulomb's law. As a consequence, the electron will now help the electrostatic repulsion to push the two nuclei apart. We see, then, that the electron is an essential component of a chemical bond, but that it must be in the right place: between the two nuclei. Coulomb's law can be used to calculate the forces experienced by the two nuclei for various positions of the electron. This allows us to define two regions of space about the nuclei, as shown in the figure. One region, the binding region, depicts locations at which the electron exerts a net binding effect on the new nuclei. Outside of this, in the antibinding region, the electron will actually work against binding. This simple picture illustrates the number one rule of chemical bonding: chemical bonds form when electrons can be simultaneously close to two or more nuclei. It should be pointed out that this principle applies also to the ionic model; as will be explained later in this chapter, the electron that is "lost" by a positive ion ends up being closer to more nuclei (including the one from whose electron cloud it came) in the compound. Quantum models of bonding take into account the fact that a particle as light a the electron cannot really be said to be in any single location. The best we can do is define a region of space in which the probability of finding the electron has some arbitrary value which will always be less than unity. The shape of this volume of space is called an and is defined by a mathematical function that relates the probability to the (x,y,z) coordinates of the molecule. Like other models of bonding, the quantum models attempt to show how more electrons can be simultaneously close to more nuclei. Instead of doing so through purely geometrical arguments, they attempt this by predicting the nature of the orbitals which the valence electrons occupy in joined atoms. According to this model, the bonding electrons act as a kind of fluid that concentrates in the region of each nucleus (lowering the potential energy) and at the same time is able to freely flow between them (reducing the kinetic energy). A summary of the concept, showing its application to a simple molecule, is shown on the next page. Despite its conceptual simplicity and full acknowledgment of the laws of quantum mechanics, this model is not widely known and is rarely taught. Chemical bonding occurs when one or more electrons can be simultaneously close to two nuclei. But how can this be arranged? The conventional picture of the shared electron bond places the bonding electrons in the region the two nuclei. This makes a nice picture, but it is not consistent with the principle that opposite charges attract. This would imply that the electrons would be "happiest" (at the lowest potential energy) when they are very close to a nucleus, not half a bond-length away from two of them! This plot shows how the potential energy of an electron in the hydrogen atom varies with its distance from the nucleus. Notice how the energy falls without limit as the electron approaches the nucleus, represented here as a proton, \(H^+\). If potential energy were the only consideration, the electron would fall right into the nucleus where its potential energy would be minus infinity. When an electron is added to the proton to make a neutral hydrogen atom, it tries to get as close to the nucleus as possible. The Heisenberg uncertainty principle requires the total energy of the electron energy to increase as the volume of space it occupies diminishes. As the electron gets closer to the nucleus, the nuclear charge confines the electron to such a tiny volume of space that its energy rises, allowing it to "float" slightly away from the nucleus without ever falling into it. The shaded region above shows the range of energies and distances from the nucleus the electron is able to assume within the 1 orbital. The electron can thus be regarded as a fluid that occupies a vessel whose walls conform to the red potential energy curves shown above. Note that as the potential energy falls, the kinetic energy increases, but only half as fast (this is called the ). Thus close to the nucleus, the kinetic energy is large and so is the electron's effective velocity. The top of the shaded area defines the work required to raise its potential energy to zero, thus removing it from the atom; this corresponds, of course, to the ionization energy. A quantum particle can be described by a waveform which is the plot of a mathematical function related to the probability of finding the particle at a given location at any time. If the particle is confined to a box, it turns out that the wave does not fall to zero at the walls of the box, but has a finite probability of being found outside it. This means that a quantum particle is able to penetrate, or "tunnel through" its confining boundaries. This remarkable property is called the . In terms of the electron fluid model introduced above, the fluid is able to "leak out" of the atom if another low-energy location can be found nearby. Suppose we now bring a bare proton up close to a hydrogen atom. Each nucleus has its own potential well, but only that of the hydrogen atom is filled, as indicated by the shading in the leftmost potential well. But the electron fluid is able to tunnel through the potential energy barrier separating the two wells; like any liquid, it will seek a common level in the two sides of the container as shown below. The electron is now "simultaneously close to two nuclei" while never being in between them. Bear in mind that this would be physically impossible for a real liquid composed of real molecules; this is purely a quantum effect that is restricted to a low-mass particle such as the electron. Because the same amount of electron fluid is now shared between the two wells, its level in both is lower. The difference between what it is now and what is was before corresponds to the of the hydrogen molecule ion. Now let's make a molecule of . We start with two hydrogen atoms, each with one electron. But there is a problem here: both potential energy wells are already filled with electron fluid; there is no room for any more without pushing the energy way up. But quantum theory again comes to the rescue! If the two electrons have opposite spins, the two fluids are able to interpenetrate each other, very much as two gases are able to occupy the same container. This is depicted by the double shading in the diagram below. When the two hydrogen atoms are within tunneling distance, half of the electron fluid (really the probability of finding the electron) from each well flows into the other well. Because the two fluids are able to interpenetrate, the level is not much different from what it was in the H ion, but the greater of the electron-fluid between the two nuclei makes H a strongly bound molecule. If we tried to join two helium atoms in this way, we would be in trouble. The electron well of He already contains two electrons of opposite spin. There is no room for more electron fluid (without raising the energy), and thus no way the electrons in either He atom can be simultaneously close to two nuclei. Even before G.N.Lewis developed his theory of the shared electron pair bond, it was believed that bonding in many solid salts could be satisfactorily explained on the basis of simple electrostatic forces between the positive and negative ions which are assumed to be the basic molecular units of these compounds. Lewis himself continued to recognize this distinction, which has continued to be a part of the tradition of chemistry; the shared electron pair bond is known as the covalent bond, while the other type is the ionic or electrovalent bond. The covalent bond is formed when two atoms are able to share electrons: whereas the ionic bond is formed when the "sharing" is so unequal that an electron from atom A is completely lost to atom B, resulting in a pair of ions: The two extremes of electron sharing represented by the covalent and ionic models appear to be generally consistent with the observed properties of molecular and ionic solids and liquids. But does this mean that there are really two kinds of chemical bonds, ionic and covalent? According to the ionic electrostatic model, solids such as NaCl consist of positive and negative ions arranged in a crystal lattice. Each ion is attracted to neighboring ions of opposite charge, and is repelled by ions of like charge; this combination of attractions and repulsions, acting in all directions, causes the ion to be tightly fixed in its own location in the crystal lattice. Since electrostatic forces are nondirectional, the structure of an ionic solid is determined purely by geometry: two kinds of ions, each with its own radius, will fall into whatever repeating pattern will achieve the lowest possible potential energy. Surprisingly, there are only a small number of possible structures. When two elements form an ionic compound, is an electron really lost by one atom and transferred to the other one? In order to deal with this question, consider the data on the ionic solid LiF. The average radius of the neutral Li atom is about 2.52Å. Now if this Li atom reacts with an atom of F to form LiF, what is the average distance between the Li nucleus and the electron it has "lost" to the fluorine atom? The answer is 1.56Å; So the answer to the above question is both yes and no: yes, the electron that was now in the 2 orbital of Li is now within the grasp of a fluorine 2 orbital, but no, the electron is now even to the Li nucleus than before, so how can it be "lost"? The one thing that is inarguably true about LiF is that there are more electrons closer to positive nuclei than there are in the separated Li and F atoms. But this is just the condition that gives rise to all forms of chemical bonding: It is obvious that the electron-pair bond brings about this situation, and this is the reason for the stability of the covalent bond. What is not so obvious (until you look at the numbers such as were quoted for LiF above) is that the "ionic" bond results in the same condition; even in the most highly ionic compounds, both electrons are close to both nuclei, and the resulting mutual attractions bind the nuclei together. This being the case, is there really any fundamental difference between the ionic and covalent bond? The answer, according to modern chemical thinking is probably "no"; in fact, there is some question as to whether it is realistic to consider that these solids consist of "ions" in the usual sense. The preferred picture that seems to be emerging is one in which the electron orbitals of adjacent atom pairs are simply skewed so as to place more electron density around the "negative" element than around the "positive" one. This being said, it must be reiterated that the ionic model of bonding is a useful one for many purposes, and there is nothing wrong with using the term "ionic bond" to describe the interactions between the atoms in "ionic solids" such as LiF and NaCl. If there is no such thing as a "completely ionic" bond, can we have one that is completely covalent? The answer is yes, if the two nuclei have equal electron attracting powers. This situation is guaranteed to be the case with homonuclear diatomic molecules— molecules consisting of two identical atoms. Thus in Cl , O , and H , electron sharing between the two identical atoms must be exactly even; in such molecules, the center of positive charge corresponds exactly to the center of negative charge: halfway between the two nuclei. This term was introduced earlier in the course to denote the relative electron-attracting power of an atom. The electronegativity is not the same as the electron affinity; the latter measures the amount of energy released when an electron from an external source "falls into" a vacancy within the outermost orbital of the atom to yield an isolated negative ion. The products of bond formation, in contrast, are not ions and they are not isolated; the two nuclei are now drawn closely together by attraction to the region of high electron density between them. Any shift of electron density toward one atom takes place at the energetic expense of stealing it from the other atom. It is important to understand that electronegativity is not a measurable property of an atom in the sense that ionization energies and electron affinity are; electronegativity is a property that an atom displays when it is bonded to another. Any measurement one does make must necessarily depend on the properties of both of the atoms. By convention, electronegativities are measured on a scale on which the highest value, 4.0, is arbitrarily assigned to fluorine. A number of electronegativity scales have been proposed, each based on slightly different criteria. The most well known of these is due to Linus Pauling, and is based on a study of bond energies in a variety of compounds. The periodic trends in electronegativity are about what one would expect; the higher the nuclear charge and the smaller the atom, the more strongly attractive will it be to an outer-shell electron of an atom within binding distance. The division between the metallic and nonmetallic elements is largely that between those that have Pauling electronegativies greater than about 1.7, and those that have smaller electronegativities. The greater the electronegativity difference between two elements and , the more will be their molecule . It is important to point out, however, that the pairs having the greatest electronegativity differences, the alkali halides, are solids order ordinary conditions and exist as molecules only in the rarefied conditions of the gas phase. Even these possess a certain amount of covalent character, so, as discussed above, there is no such thing as a "purely ionic" bond. It has become more common to place binary compounds on a scale something like that shown here, in which the degree of shading is a rough indication of the number of compounds at any point on the covalent-ionic scale. The covalent-ionic continuum described above is certainly an improvement over the old covalent - - ionic dichotomy that existed only in the textbook and classroom, but it is still only a one-dimensional view of a multidimensional world, and thus a view that hides more than it reveals. The main thing missing is any allowance for the type of bonding that occurs between more pairs of elements than any other: metallic bonding. Intermetallic compounds are rarely even mentioned in introductory courses, but since most of the elements are metals, there are a lot of them, and many play an important role in metallurgy. In metallic bonding, the valence electrons lose their association with individual atoms; they form what amounts to a mobile "electron fluid" that fills the space between the crystal lattice positions occupied by the atoms, (now essentially positive ions.) The more readily this electron delocalization occurs, the more "metallic" the element. Thus instead of the one-dimension chart shown above, we can construct a triangular diagram whose corners represent the three extremes of "pure" covalent, ionic, and metallic bonding. We can take this a step farther by taking into account collection of weaker binding effects known generally as van der Waals forces. Contrary to what is often implied in introductory textbooks, these are the major binding forces in most of the common salts that are not alkali halides; these include NaOH, CaCl , MgSO . They are also significant in solids such as CuCl and solid SO in which infinite covalently-bound chains are held together by ion-induced dipole and similar forces. The only way to represent this four-dimensional bonding-type space in two dimensions is to draw a projection of a tetrahedron, each of its four corners representing the "pure" case of one type of bonding. Note that some of the entries on this diagram (ice, CH , and the two parts of NH ClO ) are covalently bound units, and their placement refers to the binding between these units. Thus the H O molecules in ice are held together mainly by hydrogen bonding, which is a van der Waals force, with only a small covalent contribution. Note: the triangular and tetrahedral diagrams above were adapted from those in the excellent article by William B. Jensen, "Logic, history and the chemistry textbook", Part II, J. Chemical Education 1998: 817-828. | 21,864 | 3,663 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Lipids/Applications_of_Lipids/Lipid_Bilayer_Membranes |
Every cell is enclosed by a membrane which gives structure to the cell and allows for the passage of nutrients and wastes into and out of the cell. The purpose of the bilayer membrane is to separate the cell contents from the outside environment. The outside of the cell is mostly water and the inside of the cell is mostly water. The cell membrane may be coated with other molecules such as carbohydrates and proteins, which serve as receptor sites for other messenger molecules. Interaction with the cell membrane allows for molecular communication signals to pass from outside to inside of the cell. Cell membranes are composed of two classes of molecules: lipids and proteins. The proteins serve as enzymes, carry molecules, and provide the membrane with distinctive functional properties. Details of proteins and enzyme structures are given elsewhere. The lipids provide the structural integrity for the cell. The lipids found in the membrane consist of two parts: hydrophilic (water soluble) and hydrophobic (water insoluble). The hydrophobic portion of the lipids is the non-polar long hydrocarbon chains of two fatty acids. The fatty acids are present as esters bonded to glycerol. The third-OH group on glycerol is ester bonded to phosphate hence the term phospholipid. The phosphate ester portion of the molecule is polar or even ionic and hence is water soluble. A simple interaction of several phospholipids is shown in the graphic on the left. There are two common found in the bilayer: The arrangement of phospholipids in cell membranes has been deduced by X-Ray diffraction data. The phospholipids are arranged as a bilayer (two molecules thick). The phospholipids are stacked with the non-polar hydrocarbon chains pointed inward while the polar ends act as the external surface as shown in graphic on the left. The structure of the bilayer is another application of the solubility principle of "likes dissolve likes". Most of the fatty acids in the membrane are unsaturated because this allows the membrane to be more flexible (cis bonds are bent) to allow certain molecules through the membrane. However, the interaction of the hydrophobic inside of the layer acts as a barrier for ionic and polar molecules from entering the inside of the cell. In animal cells is inserted between the non-polar chains, and makes up about 20% of the molecules of the membrane. This helps to make the membrane more rigid and adds strength. | 2,459 | 3,664 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Case_Studies%3A_Proteins/Angiotnesin_Peptide |
Hypertension (high blood pressure) is a major human disease and despite much research, the problem is still not completely understood. A major part of the problem involves the action of an octapeptide, angiotensin II. This peptide hormone stimulates the constriction of blood vessels which leads to an increase in blood pressure. Angiotensin II is produced by the removal of two amino acids units on angiotensin I by an enzyme in the blood. Angiotensin I does not constrict blood vessels. QUES. What is the difference between angiotensin I and angiotensin II? This difference is responsible for difference in physiological action. Angiotensin I: asp - arg - val - tyr - ile - his - pro - phe - his - leu
Angiotensin II: asp - arg - val - tyr - ile - his - pro - phe. | 788 | 3,665 |
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When addition polymers are formed, no by-products result. Formation of a condensation polymer, on the other hand, produces H O, HCl, or some other simple molecule which escapes as a gas. A familiar example of a condensation polymer is , which is obtained from the reaction of two monomers These two molecules can link up with each other because each contains a reactive functional group, either an amine or a carboxylic acid which reacts to form an amide linkage. They combine as follows: Below is a video of the reaction to form nylon. This reaction is slightly modified from the one described above, as adipoyl chloride, not adipic acid, is used as a reactant. Thus HCl, not H O is produced. This also means that the chain terminates in an acid chloride, rather than the carboxylic acid shown above. Note that an amide linkage is still formed. A solution of adipoyl chloride in cyclohexane is poured on top of an aqueous solution of 1,6-diaminohexane in a beaker. Nylon (6,6) polyamide is formed at the interface of the two immiscible liquids and is carefully drawn from the solution and placed on a glass rod. The rod is then spun, and the Nylon (6,6) polyamide is spun onto the rod. Well-known condensation polymers other than nylon are Dacron, Bakelite, melamine, and Mylar. Nylon makes extremely strong threads and fibers because its long-chain molecules have stronger intermolecular forces than the London forces of polyethylene. Each N—H group in a nylon chain can hydrogen bond to the O of a C=O group in a neighboring chain, as shown below. Therefore the chains cannot slide past one another easily. If you pull on both ends of a nylon thread, for example, it will only stretch slightly. After that it will strongly resist breaking because a large number of hydrogen bonds are holding overlapping chains together. The same is not true of a polyethylene thread in which only London forces attract overlapping chains together, and this is one reason that polyethylene is not used to make thread. | 2,020 | 3,666 |
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are one of the three classes of compounds found in petroleum. They are less abundant than the and , amounting to only a few percent of the total, but they are quite important commercially. Most aromatic hydrocarbons contain a . You will recall from the discussion on that benzene, C H contains a flat ring of six carbon atoms joined by bonds which are intermediate in character between single and double bonds. The benzene ring is usually indicated by In the latter structure the lines represent C—C bonds, but carbon and hydrogen atoms, as well as C—H bonds, have been omitted. The benzene ring is very stable, surviving unchanged in most chemical reactions. It is very different in reactivity and shape from the puckered six-membered rings found in cycloalkanes. Below are 3D Jmol models of both cyclohexane and benzene. Note that the three xylenes are also isomers. Compounds containing two benzene rings joined together, such as naphthalene, are also found in crude oil, though they are much rarer than benzene-related compounds. Aromatic hydrocarbons are much more common in coal than in petroleum, though in the United States they are mostly manufactured from the latter. In addition to their use in motor fuel, they may be made into dyes, plastics, explosives, detergents, insecticides, medicines, and many other products. In 2000, a total of 6.74 × 10 liters of benzene were produced in the US, after compensation for exportation and importation. Some aromatic compounds, benzene among them, are toxic. The compound 1,2-benzopyrene was the cause of the first demonstrated case of occupational disease. During the eighteenth century chimney sweeps in London were found to have extremely high rates of skin cancer relative to the average person. This was eventually traced to the carcinogenic (cancer-causing) properties of 1,2-benzopyrene in the soot which coated the insides of the chimneys they cleaned. Small quantities of the compound were produced by inefficient combustion of coal in the fireplaces used to heat London houses. | 2,064 | 3,668 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.05%3A_Exceptions_to_the_Periodic_Law |
In the process of constructing the first periodic table, Mendeleev encountered several situations where the properties of elements were incompatible with the positions they would be forced to occupy in order of increasing atomic weight. In such a case, Mendeleev chose to emphasize the properties, because in the 1870s it was difficult to determine atomic weights accurately. He assumed that some atomic weights were in error and that ordering of elements ought to be changed to agree with chemical behavior. One example occurs when looking at the of potassium(K) and argon(Ar). While potassium has a smaller atomic weight than argon(39.10 vs. 39.95) the valence properties of potassium suggest that it should follow argon in the periodic table. Mendeleev did not have to contend with this because argon, one of the noble gases, had not been discovered in 1872, but it illustrates the difficulty nicely. There is a break in the regular sequence of valences of the first 24 elements when we come to K and Ar. The alkali metal has a atomic weight than the noble gas and appears before the noble gas in the sequence of atomic weights. All other alkali metals immediately follow noble gases (they have slightly atomic weights). Unless we make an exception to the order of increasing atomic weight for Ar and K, the periodic table would contain a strange anomaly. One of the elements in the vertical column of noble gases would be the extremely reactive silver metal pictured below, Potassium (K). Likewise, the reactive group of alkali metals would contain the gas Argon, pictured below, which is not a metal and is very unreactive. Mendeleev’s assumption that more accurate atomic weight determinations would eliminate situations such as we have just described has turned out to be incorrect. The atomic weights used for K before Ar are modern, highly accurate values, but they still predict the wrong order for Ar and K. The same problem occurs in the case of Co and Ni and of Te and I. Apparently atomic weight, although related to chemical behavior, is not as fundamental as Mendeleev and other early developers of the periodic table thought. | 2,162 | 3,669 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.6%3A_Heats_of_Reactions_-_U_and_H |
When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (\(ΔH_{rxn}\)), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so \(ΔH_{rxn}\) is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so \(ΔH_{rxn}\) is positive. Thus \(ΔH_{rxn} < 0\) , and \(ΔH_{rxn} > 0\) . In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table: If \(ΔH_{rxn}\) is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, (Figure \(\Page {1a}\)). Conversely, if \(ΔH_{rxn}\) is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, (Figure \(\Page {1b}\)). Bond breaking requires an input of energy; bond making releases energy. Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion. \[ \begin{align} \text{heat} + \ce{H_{2}O(s)} & \ce{ -> H_{2}O(l)} & \Delta H > 0 \label{7.6.7} \\[4pt] \ce{H2O (l)} & \ce{-> H2O(s) + heat} & \Delta H < 0 \end{align} \label{7.6.8} \] In both cases, the of the enthalpy change is the same; only the is different. \[ \ce{ 2Al (s ) + Fe2O3 (s ) -> 2Fe (s) + Al2O3 (s )} + 851.5 \; kJ \label{7.6.9} \] Thus \(ΔH = −851.5 \,kJ/mol\) of \(\ce{Fe2O3}\). We can also describe \(ΔH\) for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for \(ΔH\), in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation \ref{7.6.9}, it is the value of Δ corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation: \[\ce{ 2Al (s) + Fe2O3 (s) -> 2Fe (s) + Al2O3 (s)} \quad\quad \Delta H_{rxn}= - 851.5 \; kJ \label{7.6.10} \] If 4 mol of Al and 2 mol of Fe O react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows: \[ - \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{7.6.6} \] The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example \(\Page {1}\). Certain parts of the world, such as southern California and Saudi Arabia, are short of fresh water for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If \(ΔH\) is 6.01 kJ/mol for the reaction \(\ce{H2O(s) → H2O(l)}\) at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 10 metric tons)? (A metric ton is 1000 kg.) energy per mole of ice and mass of iceberg energy required to melt iceberg Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given Δ for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by Δ (+6.01 kJ/mol): \[ \begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{metric \; tons}\, H_{2}O \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{metric \; ton}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right )\\[4pt] & = 5.55\times 10^{10} \; mol H_{2}O \end{align*}\] The energy needed to melt the iceberg is thus \[ \left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber\] Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below. Possible sources of the approximately 3.34 × 10 kJ needed to melt a 1.00 × 10 metric ton iceberg If 17.3 g of powdered aluminum are allowed to react with excess \(\ce{Fe2O3}\), how much heat is produced? One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following: The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. Enthalpy of Reaction: is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the (\(ΔH\)) can be measured. A negative \(ΔH\) means that heat flows from a system to its surroundings; a positive ΔH means that heat flows into a system from its surroundings. For a chemical reaction, the (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of Δ are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\). The magnitude of ΔH also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the (\(ΔH_{fus}\)) and the (\(ΔH_{vap}\)), respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is . The (\(ΔH_{comb}\)) is the enthalpy change that occurs when a substance is burned in excess oxygen. ( ) | 6,331 | 3,670 |
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The benzylic hydrogens of alkyl substituents on a benzene ring are activated toward free radical attack, as noted earlier. Furthermore, \(S_N1\), \(S_N2\) and E1 reactions ofbenzylic halides, show enhanced reactivity, due to the adjacent aromatic ring. The possibility that these observations reflect a general benzylic activation is supported by the susceptibility of alkyl side-chains to oxidative degradation, as shown in the following examples (the oxidized side chain is colored). Such oxidations are normally effected by hot acidic pemanganate solutions, but for large scale industrial operations catalyzed air-oxidations are preferred. Interestingly, if the benzylic position is completely substituted this oxidative degradation does not occur (second equation, the substituted benzylic carbon is colored blue). C H – + + + KMnO + H O & heat –C H – These equations are not balanced. The permanganate oxidant is reduced, usually to Mn(IV) or Mn(II). Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids. Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation). Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases. The reaction of alkyl and aryl halides with reactive metals (usually Li & Mg) to give nucleophilic reagents has been noted. This provides a powerful tool for the conversion of chloro, bromo or iodo substituents into a variety of other groups. Many reactions of these aryl lithium and Grignard reagents will be discussed in later sections, and the following equations provide typical examples of carboxylation, protonation and Gilman coupling. Metal halogen exchange reactions take place at low temperature, and may be used to introduce iodine at designated locations. An example of this method is displayed below. In this example care must be taken to maintain a low temperature, because elimination to an aryne intermediate takes place on warming. The potential reversibility of the aromatic sulfonation reaction was noted earlier. The following equation illustrates how this characteristic of the sulfonic acids may be used to prepare the 3-bromo derivative of ortho-xylene. Direct bromination would give the 4-bromo derivative. ), | 3,166 | 3,671 |
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Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different. is necessarily planar (flat), with the carbon atoms at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial , since all the carbon-carbon bonds are fully eclipsed. reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible. Rings larger than cyclopentane would have angle strain . However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations, such as those shown below. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring). A planar structure for cyclohexane is clearly improbable. The bond angles would necessarily be 120º, 10.5º larger than the ideal tetrahedral angle. Also, every carbon-carbon bond in such a structure would be eclipsed. The resulting angle and eclipsing strains would severely destabilize this structure. If two carbon atoms on opposite sides of the six-membered ring are lifted out of the plane of the ring, much of the angle strain can be eliminated. This structure still has two eclipsed bonds (colored magenta in the drawing) and severe steric crowding of two hydrogen atoms on the "bow" and "stern" of the boat. This is often called . By twisting the boat conformation, the steric hindrance can be partially relieved, but the conformer still retains some of the strains that characterize the boat conformer. Finally, by lifting one carbon above the ring plane and the other below the plane, a relatively strain-free is formed. This is the predominant structure adopted by molecules of cyclohexane. An energy diagram for these conformational interconversions is drawn below. The activation energy for the chair-chair conversion is due chiefly to a high energy twist-chair form (TC), in which significant angle and eclipsing strain are present. A facile twist-boat (TB)-boat (B) equilibrium intervenes as one chair conformer (C) changes to the other. . Investigations concerning the conformations of cyclohexane were initiated by H. Sachse (1890) and E. Mohr (1918), but it was not until 1950 that a full treatment of the manifold consequences of interconverting chair conformers and the different orientations of pendent bonds was elucidated by D. H. R. Barton (Nobel Prize 1969 together with O. Hassel). The following discussion presents some of the essential features of this . On careful examination of a chair conformation of cyclohexane, we find that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termed . The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed because they are aligned parallel to the symmetry axis of the ring. In the stick model shown on the left below, the equatorial hydrogens are colored blue, and the axial hydrogens are red. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character. Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which large substituents assume equatorial orientation. In the two methylcyclohexane conformers shown above, the methyl carbon is colored blue. When the methyl group occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. This crowding or steric hindrance is associated with the red-colored hydrogens in the structure. A careful examination of the axial conformer shows that this steric hindrance is due to two of the methyl group with ring carbons #3 and #5. The use of models is particularly helpful in recognizing and evaluating these relationships. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the between the conformers, and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location. A table of these free energy values (sometimes referred to as A values) may be examined by . Clearly the apparent "size" of a substituent is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine. | 5,543 | 3,672 |
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This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methylbenzene (previously known as toluene) and chlorine. Methylbenzene has a methyl group attached to a benzene ring. The hexagon with the circle inside is the standard symbol for this ring. There is a carbon atom at each corner of the hexagon, and a hydrogen atom on each carbon apart from the one with the methyl group attached. The reaction we are going to explore happens between methylbenzene and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light. The organic product is (chloromethyl)benzene. The brackets in the name emphasizes 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, so this is a substitution reaction. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page. The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. The over-all process is known as free radical substitution, or as a free radical chain reaction. Cl 2Cl 2Cl Cl | 1,508 | 3,673 |
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Our discussion of the atomic theory has indicated that mass is a very important characteristic of atoms — it does not change as chemical reactions occur. , on the other hand, often does change, because atoms or molecules pack together more tightly in and or become more widely separated in when a reaction takes place. From the time Dalton’s theory was first proposed, chemists realized the importance of the masses of atoms, and they spent much time and effort on experiments to determine how much heavier one kind of atom is than another. Dalton, for example, studied a compound of carbon and oxygen which he called carbonic oxide. He found that a 100-g sample contained 42.9 g C and 57.1 g O. In Dalton’s day there were no simple ways to determine the microscopic nature of a compound, and so he did not know the composition of the molecules (and hence the formula) of carbonic oxide. Faced with this difficulty, he did what most scientists would do — make the simplest possible assumption. This was that the molecules of carbonic oxide contained the minimum number of atoms: one of carbon and one of oxygen. Carbonic oxide was the compound we now know as carbon monoxide, CO, and so in this case Dalton was right. However, erroneous assumptions about the formulas for other compounds led to half a century of confusion about atomic weights. Since the formula was CO, Dalton argued that the ratio of the mass of carbon to the mass of oxygen in the compound must be the same as the ratio of the mass of 1 carbon atom to the mass of 1 oxygen atom: In other words, the mass of a carbon atom is about three-quarters (0.75) as great as the mass of an oxygen atom. Notice that this method involves a of masses and that the units cancel, yielding a pure number. That number (0.751, or approximately ¾) is the of a carbon atom compared with an oxygen atom. It tells nothing about the actual masses of carbon or oxygen atoms – only that carbon is three-quarters as heavy as oxygen. The relative masses of the atoms are usually referred to as . Their values are in the , along with the names and symbols for the elements. The atomic-weight scale was originally based on a relative mass of 1 for the lightest atom, hydrogen. As more accurate methods for determining atomic weight were devised, it proved convenient to shift to oxygen and then carbon, but the scale was adjusted so that hydrogen’s relative mass remained close to 1. Thus nitrogen’s atomic weight of 14.0067 tells us that a nitrogen atom has about 14 times the mass of a hydrogen atom. The fact that atomic weights are ratios of masses and have no units does not detract at all from their usefulness. It is very easy to determine how much heavier one kind of atom is than another. Use the to show that the mass of a mercury atom is 2.510 times the mass of a bromine atom. The actual masses of the atoms will be in the same proportion as their relative masses. The atomic weight of mercury is 200.59 and bromine is 79.904. Therefore \[\frac{\text{Mass of a Hg atom}}{\text{Mass of a Br atom}}=\frac{\text{relative mass of a Hg atom}}{\text{relative mass of a Br atom}}=\frac{\text{200}\text{.59}}{\text{79}\text{0.904}}=\text{2.5104} \nonumber \] or: Mass of a Hg atom = 2.5104 \(\cdot\) Mass of a Br atom The atomic-weight table also permits us to obtain the relative masses of molecules. These are called and are calculated by summing the atomic weights of all atoms in the molecule. How heavy would a mercurous bromide molecule be in comparison to a single bromine atom? First, obtain the relative mass of an Hg Br molecule (the molecular weight): \[ \begin{align*} \text{2 Hg atoms: relative mass}= 2 \cdot200.59 &=& 401.18 \\ \text{2 Br atoms: relative mass}= 2 \cdot79.904 &=& 159.808 \\ \text{1 Hg2Br2 molecule: relative mass} &=& 560.99 \end{align*} \nonumber \] Therefore \[\frac{\text{Mass of a Hg}_{\text{2}}\text{Br}_{\text{2}}\text{ molecule}}{\text{Mass of a Br atom}}=\dfrac{\text{560}\text{0.99}}{\text{79}\text{0.904}}=\text{7.0208} \nonumber \] The Hg Br molecule is about 7 times heavier than a bromine atom. | 4,116 | 3,674 |
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Two examples of an dealt with , namely: \[K_{c}=\frac{[trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]}{[cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]} \nonumber \] for the reaction \[\text{cis-C}_2\text{H}_2\text{F}_2 \rightleftharpoons \text{trans-C}_2\text{H}_2\text{F}_2 \nonumber \] In the equation above, notice that both sides have the same molecular formula, yet in the image below, the elements involved are arranged differently. These molecules are called isomers and to learn more about them, check out the following site: . and \[K_{c}=\frac{[\text{ NO}_{\text{2}}]^{\text{2}}}{[\text{ N}_{\text{2}}\text{O}_{\text{4}}]} \nonumber \] for the reaction \[\text{N}_2\text{O}_4 \rightleftharpoons \text{2NO}_2 \nonumber \] In the equation above, we see the decomposition of N O (the clear gas pictured below) into NO , which is the brownish red gas seen below. As you can see in the molecular structure above, this decomposition of N O occurs when the middle bond holding N O together is broken, leaving two NO molecules. For more information on this specific equilibrium, check out the page on Both the trans-cis transition of C H F and the decomposition of N O are particular examples of a more general law governing chemical equilibrium in gases. If we write an equation for a gaseous equilibrium in general in the form \[aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \label{1} \] then the equilibrium constant defined by the equation \[K_{c}=\frac{[\text{ C }]^{c}[\text{ D }]^{d}}{[\text{ A }]^{a}[\text{ B }]^{b}} \label{2} \] is found to be a constant quantity depending only on the temperature and the nature of the reaction. This general result is called the , or the Write expressions for the equilibrium constant for the following reactions: A mixture containing equal concentrations of methane and steam is passed over a nickel catalyst at 1000 K. The emerging gas has the composition [CO] = 0.1027 mol dm , [H ] = 0.3080 mol dm , and [CH ] = [H O] = 0.8973 mol dm . Assuming this mixture is at equilibrium, calculate the equilibrium constant for the reaction \[\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{3H}_2(g) \nonumber \] The equilibrium constant is given by the following equation: \[K_{c}=\frac{[\text{ CO }]\text{ }[\text{ H}_{\text{2}}]^{\text{3}}}{[\text{ CH}_{\text{4}}]\text{ }[\text{ H}_{\text{2}}\text{O }]}=\frac{\text{0}\text{.1027}^{-\text{3}}\times \text{ (0}\text{.3080}^{-\text{3}}\text{)}^{\text{3}}}{\text{0}\text{.8973}^{-\text{3}}\times \text{ 0}\text{.8973}^{-\text{3}}} \nonumber \] \[=\text{3}\text{.727 }\times \text{ 10}^{-\text{3}} \nonumber \] Note: The yield of H at this temperature is quite poor. In the commercial production of H from natural gas, the reaction is run at a somewhat higher temperature where the value is larger. As the above example shows, the equilibrium constant is a dimensionless quantity. This lack of units is a result of the derivation of from which uses the of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects: Thus, when we apply the equilibrium law to reactions which involve pure and pure , we find in such cases that as long as some solid or liquid is present, the actual amount does not affect the position of equilibrium. Accordingly, only the molar concentrations of gaseous species are explicitly written in the expression for the equilibrium constant. For example, the equilibrium constant for the reaction \[\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \label{3} \] Notice below the molecular representation of the above reaction, with the ionic structure of is given by the expression \[\text{K}_c = [\text{CO}_2] \label{4} \] in which only the concentration of the gas appears explicitly because the two solids each have an activity with a value equal to 1. Equation \(\ref{4}\) suggests that if we heat CaCO to a high temperature so that some of it decomposes, the concentration of CO at equilibrium will depend only on the temperature and will not change if the ratio of amount of solid CaCO to amount of solid CaO is altered. Experimentally this is what is observed. Write expressions for the equilibrium constants for the following reactions: Since only gaseous species need be included, we obtain The equilibrium law can be shown experimentally to apply to dilute liquid solutions as well as to mixtures of gases, and the equilibrium-constant expression for a solution reaction can be obtained in the same way as for a gas-phase reaction. Acetic acid, for example, reacts as follows when it dissolves in water: \[\text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COO}^{-}(aq) + \text{H}_3\text{O}^{+} (aq)\label{5} \] Notice in the figure below how the white hydrogen attached to the red oxygen on the acetic acid molecule is transferred to water, forming acetate (acetic acid minus a hydrogen) and a hydronium ion. Symbolically in the equation above, we can see this same transfer, except hydrogen is represented as an H rather than a white ball. In solution only the concentrations of the solute species are explicitly shown in the equilibrium law. Even though the solvent may be a reactant or product, as long as the solution is dilute, any small amount of water produced or used by the chemical reaction will not change the amount of water to an appreciable extent because it is the solvent. As the solvent, water is assigned an activity equal to 1. Thus the activity of water is conventionally not explicitly written as part of the equilibrium law for a reaction in aqueous solution. Since it applies to a weak acid, is called an . (The stands for acid.) Other equilibrium constants in which the activity of the solvent water has a value of 1 and is thus not explicitly included in the equilibrium law are the , , for ionization of a weak base and the solubility product constant, , for dissolution of a slightly soluble compound. The activity of the solvent in a dilute solution has a defined value of 1. This activity is part of the equilibrium law, but it is conventionally not written in the equilibrium law because its numerical value of 1 does not change the numerical value of the equilibrium constant. It is a common error to claim that the molar concentration of the solvent is in someway involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K = K [H O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K = K (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K = K (1), or K = K . Write out expressions for the equilibrium constants for the following ionic equilibria in dilute aqueous solution: We do not explicitly include the activity of H O in the first three examples and the activity of solid BaSO in the fourth because in all of these cases, the acivity of the specific species is equal to 1, and thus would not change the numerical value of the equilibrium constant.. Measurements of the conductivities of acetic acid solutions indicate that the fraction of acetic acid molecules converted to acetate and hydronium ions is Use these data to calculate the equilibrium constant for Equation (5) at each concentration. Consider first 1 dm of solution . This originally contained 0.02 mol CH COOH of which the fraction 0.0296 has ionized. Thus (1 – 0.0296) × 0.02 mol undissociated CH COOH is left, while 0.0296 × 0.02 mol H O and CH COO have been produced. In tabular form Substituting into the expression for gives \[K_{a}=\frac{[\text{ CH}_{\text{3}}\text{COO}^{-}]\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}]}{[\text{ CH}_{\text{3}}\text{COOH }]}=\frac{\text{(5}\text{.92 }\times \text{ 10}^{-\text{4}}\text{)}^{\text{2}}}{\text{0}\text{.0194}}=\text{1}\text{.81 }\times \text{ 10}^{-\text{5}} \nonumber \] A similar calculation on the second solution yields \[K_{a}=\frac{\text{(1}\text{.5083 }\times \text{ 10}^{-\text{5}}\text{)}^{\text{2}}}{\text{1}\text{.2926 }\times \text{ 10}^{-\text{5}}}=\text{1}\text{.760 }\times \text{ 10}^{-\text{5}} \nonumber \] The two values of the equilibrium constant are only in approximate agreement. In more concentrated solutions the agreement is worse. If the concentration is 1 mol dm , for instance, has the value 1.41 × 10 . This is the reason for our statement that the equilibrium law applies to solutions. molview.org was used to create the molecular representations found on this webpage. Check out the site at the following link: . | 8,997 | 3,675 |
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The regular three-dimensional arrangement of atoms or ions in a crystal is usually described in terms of a space lattice and a unit cell. To see what these two terms mean, let us first consider the two-dimensional patterns shown in Figure \(\Page {1}\). We can think of each of these three structures as a large number of repetitions in two directions of the parallel-sided figure shown immediately below each pattern. This parallel-sided figure is the . It represents the simplest, smallest shape from which the overall structure can be constructed. The pattern of made by the comers of the unit cells when they are packed together is called the space lattice (Figure \(\Page {2}\)). The lines joining the points of the space lattice are shown in color. Without some experience, it is quite easy to pick the wrong unit cell for a given structure. Some incorrect choices are shown immediately below the correct choice in the figure. Note in particular that the unit cell for structure , in which each circle is surrounded by six others at the comers of a hexagon, is a hexagon, but a parallelogram of equal sides (a rhombus) with angles of 60 and 120°. Figure \(\Page {2}\) illustrates the space lattice and the unit cell for a real three-dimensional crystal structure—that of sodium chloride. This is the same structure that was shown for lithium hydride, except that the sizes of the ions are different. A unit cell for this structure is a cube whose comers are all occupied by sodium ions. Alternatively, the unit cell could be chosen with chloride ions at the comers. The unit cell of sodium chloride contains sodium ions and chloride ions. In arriving at such an answer we must bear in mind that many of the ions are shared by several adjacent cells (part of Figure \(\Page {2}\) shows this well). Specifically, the sodium ions at the centers of the square faces of the cell are shared by two cells, so that only half of each lies within the unit cell. Since there are six faces to a cube, this makes a total of three sodium ions. In the middle of each edge of the unit cell is a chloride ion which is shared by four adjacent cells and so counts one-quarter. Since there are twelve edges, this makes three chloride ions. At each comer of the cube, a sodium ion is shared by eight other cells. Since there are eight comers, this totals to one more sodium ion. Finally, there is a chloride ion in the body of the cube unshared by any other cell. The grand total is thus four sodium and four chloride ions. A general formula can be derived from the arguments just presented for counting , the number of atoms or ions in a unit cell. It is \[N=N_{\text{body}}\text{ + }\frac{N_{\text{face}}}{\text{2}}\text{ + }\frac{N_{\text{edge}}}{\text{4}}\text{ + }\frac{N_{\text{corner}}}{\text{8}} \nonumber \] | 2,829 | 3,676 |
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The method for determining a reaction rate is relatively straightforward. Since a reaction rate is based on change over time, it must be determined from tabulated values or found experimentally. With the obtained data, it is possible to calculate the reaction rate either algebraically or graphically. What follows is general guidance and examples of measuring the rates of a reaction. Measuring time change is easy; a stopwatch or any other time device is sufficient. However, determining the change in concentration of the reactants or products involves more complicated processes. The change of concentration in a system can generally be acquired in two ways: It does not matter whether an experimenter monitors the reagents or products because there is no effect on the overall reaction. However, since reagents decrease during reaction, and products increase, there is a sign difference between the two rates. Reagent concentration decreases as the reaction proceeds, giving a negative number for the change in concentration. The products, on the other hand, increase concentration with time, giving a positive number. Since the convention is to express the rate of reaction as a positive number, to solve a problem, set the overall rate of the reaction equal to the negative of a reagent's disappearing rate. The overall rate also depends on stoichiometric coefficients. It is worth noting that the process of measuring the concentration can be greatly simplified by taking advantage of the different physical or chemical properties (ie: phase difference, reduction potential, etc.) of the reagents or products involved in the reaction by using the above methods. We have emphasized the importance of taking the sign of the reaction into account to get a positive reaction rate. Now, we will turn our attention to the importance of stoichiometric coefficients. A reaction rate can be reported quite differently depending on which product or reagent selected to be monitored. Given a reaction: \[ aA+bB \rightarrow cC + dD \] the general rate for this reaction is defined as \[rate = - \dfrac{1}{a}\dfrac{ \Delta [A]}{ \Delta t} = - \dfrac{1}{b} \dfrac{\Delta [B]}{\Delta t} = \dfrac{1}{c}\dfrac{ \Delta [C]}{\Delta t} = \dfrac{1}{d}\dfrac{ \Delta [D]}{\Delta t} \label{rate1}\] Equation \(\ref{rate1}\) can also be written as: rate of reaction = \( - \dfrac{1}{a} \) (rate of disappearance of A) = \( - \dfrac{1}{b} \) (rate of disappearance of B) = \( \dfrac{1}{c} \) (rate of formation of C) = \( \dfrac{1}{d} \) (rate of formation of D) Even though the concentrations of A, B, C and D may all change at different rates, there is only one average rate of reaction. To get this unique rate, choose any one rate and divide it by the stoichiometric coefficient. When the reaction has the formula: \[ C_{R1}R_1 + \dots + C_{Rn}R_n \rightarrow C_{P1}P_1 + \dots + C_{Pn}P_n \] The general case of the unique average rate of reaction has the form: rate of reaction = \( - \dfrac{1}{C_{R1}}\dfrac{\Delta [R_1]}{\Delta t} = \dots = - \dfrac{1}{C_{Rn}}\dfrac{\Delta [R_n]}{\Delta t} = \dfrac{1}{C_{P1}}\dfrac{\Delta [P_1]}{\Delta t} = \dots = \dfrac{1}{C_{Pn}}\dfrac{\Delta [P_n]}{\Delta t} \) Average Reaction Rates: Rather than performing a whole set of initial rate experiments, one can gather information about orders of reaction by following a particular reaction from start to finish. There are two different ways this can be accomplished. These approaches must be considered separately. Consider that bromoethane reacts with sodium hydroxide solution as follows: \[ CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^-\] During the course of the reaction, both bromoethane and sodium hydroxide are consumed. However, it is relatively easy to measure the concentration of sodium hydroxide at any one time by performing a titration with a standard acid: for example, with hydrochloric acid of a known concentration. The process starts with known concentrations of sodium hydroxide and bromoethane, and it is often convenient for them to be equal. Because the reaction is 1:1, if the concentrations are equal at the start, they remain equal throughout the reaction. Samples are taken with a pipette at regular intervals during the reaction, and titrated with standard hydrochloric acid in the presence of a suitable indicator. The problem with this approach is that the reaction is still proceeding in the time required for the titration. In addition, only one titration attempt is possible, because by the time another sample is taken, the concentrations have changed. There are two ways around this problem: At this point the resulting solution is titrated with standard sodium hydroxide solution to determine how much hydrochloric acid is left over in the mixture. This allows one to calculate how much acid was used, and thus how much sodium hydroxide must have been present in the original reaction mixture. This technique is known as a . This process generates a set of values for concentration of (in this example) sodium hydroxide over time. The concentrations of bromoethane are, of course, the same as those obtained if the same concentrations of each reagent were used. These values are plotted to give a concentration-time graph, such as that below: The rates of reaction at a number of points on the graph must be calculated; this is done by drawing tangents to the graph and measuring their slopes. These values are then tabulated. The quickest way to proceed from here is to plot a log graph as described further up the page. All rates are converted to log(rate), and all the concentrations to log(concentration). Then, log(rate) is plotted against log(concentration). The slope of the graph is equal to the order of reaction. In the example of the reaction between bromoethane and sodium hydroxide solution, the order is calculated to be 2. Notice that this is the overall order of the reaction, not just the order with respect to the reagent whose concentration was measured. The rate of reaction decreases because the concentrations of both of the reactants decrease. A familiar example is the catalytic decomposition of hydrogen peroxide (used above as an example of an initial rate experiment). This time, measure the oxygen given off using a gas syringe, recording the volume of oxygen collected at regular intervals. The practical side of this experiment is straightforward, but the calculation is not. The problem is that the volume of the product is measured, whereas the concentration of the reactants is used to find the reaction order. This means that the concentration of hydrogen peroxide remaining in the solution must be determined for each volume of oxygen recorded. This requires ideal gas law and stoichiometric calculations. The table of concentrations and times is processed as described above. This is an example of measuring the initial rate of a reaction producing a gas. A simple set-up for this process is given below: The reason for the weighing bottle containing the catalyst is to avoid introducing errors at the beginning of the experiment. The catalyst must be added to the hydrogen peroxide solution without changing the volume of gas collected. If it is added to the flask using a spatula before replacing the bung, some gas might leak out before the bung is replaced. Alternatively, air might be forced into the measuring cylinder. Either would render results meaningless. To start the reaction, the flask is shaken until the weighing bottle falls over, and then shaken further to make sure the catalyst mixes evenly with the solution. Alternatively, a special flask with a divided bottom could be used, with the catalyst in one side and the hydrogen peroxide solution in the other. The two are easily mixed by tipping the flask. Using a 10 cm measuring cylinder, initially full of water, the time taken to collect a small fixed volume of gas can be accurately recorded. A small gas syringe could also be used. To study the effect of the concentration of hydrogen peroxide on the rate, the concentration of hydrogen peroxide must be changed and everything else held constant—the temperature, the total volume of the solution, and the mass of manganese(IV) oxide. The manganese(IV) oxide must also always come from the same bottle so that its state of division is always the same. The same apparatus can be used to determine the effects of varying the temperature, catalyst mass, or state of division due to the catalyst Mixing dilute hydrochloric acid with sodium thiosulphate solution causes the slow formation of a pale yellow precipitate of sulfur. \[ Na_2S_2O_{2(aq)} + 2HCl_{(aq)} \rightarrow 2NaCl_{(aq)} + H_2O_{(l)} + S_{(s)} + SO_{2(g)}\] A very simple, but very effective, way of measuring the time taken for a small fixed amount of precipitate to form is to stand the flask on a piece of paper with a cross drawn on it, and then look down through the solution until the cross disappears. A known volume of sodium thiosulphate solution is placed in a flask. Then a small known volume of dilute hydrochloric acid is added, a timer is started, the flask is swirled to mix the reagents, and the flask is placed on the paper with the cross. The timer is used to determine the time for the cross to disappear. The process is repeated using a smaller volume of sodium thiosulphate, but topped up to the same original volume with water. Everything else is exactly as before. The actual concentration of the sodium thiosulphate does not need to be known. In each case the relative concentration could be recorded. The solution with 40 cm of sodium thiosulphate solution plus 10 cm of water has a concentration which is 80% of the original, for example. The one with 10 cm of sodium thiosulphate solution plus 40 cm of water has a concentration 20% of the original. The quantity 1/t can again be plotted as a measure of the rate, and the volume of sodium thiosulphate solution as a measure of concentration. Alternatively, relative concentrations could be plotted. In either case, the shape of the graph is the same. The effect of temperature on this reaction can be measured by warming the sodium thiosulphate solution before adding the acid. The temperature must be measured after adding the acid, because the cold acid cools the solution slightly.This time, the temperature is changed between experiments, keeping everything else constant. To get reasonable times, a diluted version of the sodium thiosulphate solution must be used. Using the full strength, hot solution produces enough precipitate to hide the cross almost instantly. There are several reactions bearing the name "iodine clock." Each produces iodine as one of the products. This is the simplest of them, because it involves the most familiar reagents. The reaction below is the oxidation of iodide ions by hydrogen peroxide under acidic conditions: \[ H_2O_{2(aq)} + 2I_{(aq)}^- + 2H^+ \rightarrow I_{2(aq)} + 2H_2O_{(l)}\] The iodine is formed first as a pale yellow solution, darkening to orange and then dark red before dark gray solid iodine is precipitated. Iodine reacts with starch solution to give a deep blue solution. If starch solution is added to the reaction above, as soon as the first trace of iodine is formed, the solution turns blue. This gives no useful information. thiosulphate \[ 2S_2O^{2-}_{3(aq)} + I_{2(aq)} \rightarrow S_2O_{6(aq)}^{2-} + 2I^-_{(aq)}\] If a very small amount of sodium thiosulphate solution is added to the reaction mixture (including the starch solution), it reacts with the iodine that is initially produced, so the iodine does not affect the starch, and there is no blue color. However, when that small amount of sodium thiosulphate is consumed, nothing inhibits further iodine produced from reacting with the starch. The mixture turns blue. Reaction rates have the general form of (change of concentration / change of time). There are two types of reaction rates. One is called the average rate of reaction, often denoted by (Δ[conc.] / Δt), while the other is referred to as the instantaneous rate of reaction, denoted as either: \[ \lim_{\Delta t \rightarrow 0} \dfrac{\Delta [concentration]}{\Delta t} \] or \[ \dfrac{d [concentration]}{dt} \] The average rate of reaction, as the name suggests, is an average rate, obtained by taking the change in concentration over a time period, for example: -0.3 M / 15 minutes. This is an approximation of the reaction rate in the interval; it does not necessarily mean that the reaction has this specific rate throughout the time interval or even at any instant during that time. The instantaneous rate of reaction, on the other hand, depicts a more accurate value. The instantaneous rate of reaction is defined as the change in concentration of an infinitely small time interval, expressed as the limit or derivative expression above. Instantaneous rate can be obtained from the experimental data by first graphing the concentration of a system as function of time, and then finding the slope of the tangent line at a specific point which corresponds to a time of interest. Alternatively, experimenters can measure the change in concentration over a very small time period two or more times to get an average rate close to that of the instantaneous rate. The reaction rate for that time is determined from the slope of the tangent lines. The initial rate of reaction is the rate at which the reagents are first brought together. Like the instantaneous rate mentioned above, the initial rate can be obtained either experimentally or graphically. To experimentally determine the initial rate, an experimenter must bring the reagents together and measure the reaction rate as quickly as possible. If this is not possible, the experimenter can find the initial rate graphically. To do this, he must simply find the slope of the line tangent to the reaction curve when t=0. The simplest initial rate experiments involve measuring the time taken for some recognizable event to happen early in a reaction. This could be the time required for 5 cm of gas to be produced, for a small, measurable amount of precipitate to form, or for a dramatic color change to occur. Examples of these three indicators are discussed below. The concentration of one of the components of the reaction could be changed, holding everything else constant: the concentrations of other reactants, the total volume of the solution and the temperature. The time required for the event to occur is then measured. This process is repeated for a range of concentrations of the substance of interest. A reasonably wide range of concentrations must be measured.This process could be repeated by altering a different property. Consider a simple example of an initial rate experiment in which a gas is produced. This might be a reaction between a metal and an acid, for example, or the catalytic decomposition of hydrogen peroxide. If volume of gas evolved is plotted against time, the first graph below results. A measure of the rate of the reaction at any point is found by measuring the slope of the graph. The steeper the slope, the faster the rate. Because the initial rate is important, the slope at the beginning is used. In the second graph, an enlarged image of the very beginning of the first curve, the curve is approximately straight. This is only a reasonable approximation when considering an early stage in the reaction. As the reaction progresses, the curvature of the graph increases. Suppose the experiment is repeated with a different (lower) concentration of the reagent. Again, the time it takes for the same volume of gas to evolve is measured, and the initial stage of the reaction is studied. Instantaneous Rates: Determine the initial rate of the reaction using the table below. initial rate of reaction = \( \dfrac{-(0-2.5) M}{(195-0) sec} \) = 0.0125 M per sec Use the points [A]=2.43 M, t= 0 and [A]=1.55, t=100 initial rate of reaction = \( - \dfrac{\Delta [A]}{\Delta t} = \dfrac{-(1.55-2.43) M }{\ (100-0) sec} \) = 0.0088 M per sec Jim Clark ( ) | 16,073 | 3,677 |
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With the , we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. For gases the density varies with the number of gas molecules in a constant volume. The ideal-gas equation can be manipulated to solve a variety of different types of problems. To determine the density, \(\rho\), of a gas, we rearrange the equation to \[\rho=\dfrac{n}{V}=\dfrac{P}{RT}\label{6.4.1}\] Density of a gas is generally expressed in g/L. Multiplication of the left and right sides of the equation by the molar mass (\(M\)) of the gas gives \[ \dfrac{g}{L}=\dfrac{PM}{RT} \label{6.4.2}\] This allows us to determine the density of a gas when we know the molar mass, or vice versa. What is the density of nitrogen gas (\(N_2\)) at 248.0 Torr and 18º C? \[(248 \; \rm{Torr}) \times \dfrac{1 \; \rm{atm}}{760 \; \rm{Torr}} = 0.3263 \; \rm{atm}\] \[18ºC + 273 = 291 K\] *Write down all known equations: \[PV = nRT\] \[\rho=\dfrac{m}{V}\] \[m=M \times n\] *Now take the density equation. \[\rho=\dfrac{m}{V}\] *Keeping in mind \(m=M \times n\)...replace \((M \times n)\) for \(mass\) within the density formula. \[\rho=\dfrac{M \times n}{V}\] \[\dfrac{\rho}{M} = \dfrac{n}{V}\] *Now manipulate the Ideal Gas Equation \(PV = nRT\) \[\dfrac{n}{V} = \dfrac{P}{RT}\] *\((n/V)\) is in both equations. \[\dfrac{n}{V} = \dfrac{\rho}{M}\] \[\dfrac{n}{V} = \dfrac{P}{RT}\] *Now combine them please. \[\dfrac{\rho}{M} = \dfrac{P}{RT}\] *Isolate density. \[\rho = \dfrac{PM}{RT}\] \[\rho = \dfrac{PM}{RT}\] \[\rho = \dfrac{(0.3263\; \rm{atm})(2*14.01 \; \rm{g/mol})}{(0.08206 L atm/K mol)(291 \; \rm{K})}\] \[\rho = 0.3828 \; g/L\] An example of varying density for a useful purpose is the hot air balloon, which consists of a bag (called the envelope) that is capable of containing heated air. As the air in the envelope is heated, it becomes less dense than the surrounding cooler air (Equation 6.4.1), which is has enough lifting power (due to buoyancy) to cause the balloon to float and rise into the air. Constant heating of the air is required to keep the balloon aloft. As the air in the balloon cools, it contracts, allowing outside cool air to enter, and the density increases. When this is carefully controlled by the pilot, the balloon can land as gently as it rose. Density and the Molar Mass of Gases: The ideal gas law can be used to calculate volume of gases consumed or produced. The ideal-gas equation frequently is used to interconvert between volumes and molar amounts in chemical equations. What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g \(CaCO_3\) via the equation: \[ CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}\] Begin by converting the mass of calcium carbonate to moles. \[ \dfrac{0.150\;g}{100.1\;g/mol} = 0.0015\; mol\] The stoichiometry of the reaction dictates that the number of moles \(CaCO_3\) decomposed equals the number of moles \(CO_2\) produced. Use the ideal-gas equation to convert moles of \(CO_2\) to a volume. \[V = \dfrac{nRT}{R} = \dfrac{(0.0015\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} = 0.0336\;L \; or \; 33.6\;mL\] A 3.00 L container is filled with \(Ne_{(g)}\) at 770 mmHg at 27 C. A \(0.633\;\rm{g}\) sample of \(CO_2\) vapor is then added. What is the partial pressure of \(CO_2\) and \(Ne\) in atm? What is the total pressure in the container in atm? , Before: Other Unknowns: \(n_{CO_2}\)= ? \[n_{CO_2} = 0.633\; \rm{g} \;CO_2 \times \dfrac{1 \; \rm{mol}}{44\; \rm{g}} = 0.0144\; \rm{mol} \; CO_2\] \[n_{Ne} = \dfrac{PV}{RT}\] \[n_{Ne} = \dfrac{(1.01\; \rm{atm})(3.00\; \rm{L})}{(0.08206\;atm\;L/mol\;K)(300\; \rm{K})}\] \[n_{Ne} = 0.123 \; \rm{mol}\] Because the pressure of the container before the \(CO_2\) was added contained only \(Ne\), that is your partial pressure of \(Ne\). After converting it to atm, you have already answered part of the question! \[P_{Ne} = 1.01\; \rm{atm}\] Step 3: Now that have pressure for Ne, you must find the partial pressure for \(CO_2\). Use the ideal gas equation. \[ \dfrac{P_{Ne}V}{n_{Ne}RT} = \dfrac{P_{CO_2}V}{n_{CO_2}RT}\] but because both gases share the same Volume (\(V\)) and Temperature (\(T\)) and since the Gas Constant (\(R\)) is constants, all three terms cancel and can be removed them from the equation. \[\dfrac{P}{n_{Ne}} = \dfrac{P}{n_{CO_2}}\] \[\dfrac{1.01 \; \rm{atm}}{0.123\; \rm{mol} \;Ne} = \dfrac{P_{CO_2}}{0.0144\; \rm{mol} \;CO_2} \] \[P_{CO_2} = 0.118 \; \rm{atm}\] \(CO_2\). \[P_{total}= P_{Ne} + P_{CO_2}\] \[P_{total}= 1.01 \; \rm{atm} + 0.118\; \rm{atm}\] \[P_{total}= 1.128\; \rm{atm} \approx 1.13\; \rm{atm} \; \text{(with appropriate significant figures)} \] Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO , followed by the reaction of SO with O in the presence of a catalyst to give SO , which reacts with water to give H SO . The overall chemical equation is as follows: \[\rm 2S_{(s)}+3O_{2(g)}+2H_2O_{(l)}\rightarrow 2H_2SO_{4(aq)}\] What volume of O (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H SO ? reaction, temperature, pressure, and mass of one product volume of gaseous reactant Calculate the number of moles of H SO in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O required. Use the ideal gas law to determine the volume of O required under the given conditions. Be sure that all quantities are expressed in the appropriate units. We begin by calculating the number of moles of H SO in 1.00 ton: \[\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4\] We next calculate the number of moles of O required: \[\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2\] After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O : \[V=\dfrac{nRT}{P}=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}}=3.43\times10^5\;L\] The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry. Charles used a balloon containing approximately 31,150 L of H for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation: \[Fe_{(s)} + 2 HCl_{(aq)} \rightarrow H_{2(g)} + FeCl_{2(aq)}\] How much iron (in kilograms) was needed to produce this volume of H if the temperature was 30°C and the atmospheric pressure was 745 mmHg? 68.6 kg of Fe (approximately 150 lb) Sodium azide (\(NaN_3\)) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation: \[ 2NaN_3 \rightarrow 2Na_{(s)} + 3N_{2\; (g)}\] This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the \(N_2\) gas that results from the decomposition of a 5.00 g sample of \(NaN_3\) could be collected by displacing water from an inverted flask, as in Figure \(\Page {4}\), what volume of gas would be produced at 21°C and 762 mmHg? reaction, mass of compound, temperature, and pressure volume of nitrogen gas produced Calculate the number of moles of N gas produced. From the data in Table \(\Page {4}\), determine the partial pressure of N gas in the flask. Use the ideal gas law to find the volume of N gas produced. Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N gas produced: \[\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2\] The pressure given (762 mmHg) is the pressure in the flask, which is the sum of the pressures due to the N gas and the water vapor present. Table \(\Page {4}\) tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the N gas in the flask is only \[\rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;atm}= 743.4\; mmHg \times\dfrac{1\;atm}{760\;atm}= 0.978\; atm.\] Solving the ideal gas law for and substituting the other quantities (in the appropriate units), we get \[V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L\] A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H gas according to the equation Zn(s) + 2 HCl(aq) → H (g) + ZnCl (aq). The resulting H gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy? 0.397 L Ideal Gas law Equation and Reaction Stoichiometry: The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. | 10,585 | 3,679 |
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Many oxyacids are rather unstable and cannot be isolated in pure form. An example is carbonic acid, H CO , which decomposes to water and carbon dioxide: \[ \text{H}_2 \text{CO}_3(aq) \rightleftharpoons \text{H}_2\text{O}(l) + \text{CO}_2(g) \nonumber \] This decomposition process is familiar to us, as it is responsible for the fizzy nature of soda. When the CO is dissolved in the soda, it becomes H CO , but when the soda is released from high pressure, the decomposition process occurrs rapidly, forming bubbles of CO and water. Since it can be made by removing H O from H CO , CO is called the of H CO . (The term anhydride is derived from anhydrous, meaning “not containing water.”) Acid anhydrides are usually oxides of nonmetallic elements. Some common examples and their corresponding oxyacids are SO —H SO ; SO —H SO ; P O —H PO ; N O —HNO . Any of these anhydrides increases the hydronium-ion concentration when dissolved in water; for example, \[\text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4 \text{H}_3\text{PO}_4(aq) \nonumber \] \[\text{H}_3\text{PO}_4(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{H}_2\text{PO}_4^-(aq) \nonumber \] In the Arrhenius sense, then, acid anhydrides are acids, but according to the Brönsted-Lowry definition, they are not acids because they contain no hydrogen. In 1923, at the same time that the Brönsted-Lowry definition was proposed, suggested another definition which includes the acid anhydrides and a number of other substances as acids. According to the , an acid is any species which can accept a lone pair of electrons, and a base is any species which can donate a lone pair of electrons. An acid-base reaction in the Lewis sense involves formation of a (where one atom provides both shared electrons). The Lewis definition has little effect on the types of molecules we expect to be basic. All the Brönsted-Lowry bases, for example, NH , O , H , contain at least one lone pair. Lewis’ idea does expand the number of acids, though. The proton is not the only species which can form a coordinate covalent bond with a lone pair. Cations of the transition metals, which are strongly hydrated, do the same thing: \[\begin{align} & \underset{\text{Lewis acid}}{\text{Cr}^{\text{3+}}}\text{ + }\underset{\text{Lewis base}}{\text{6H}_{\text{2}}\text{O}}\text{ }\to \text{ Cr(H}_{\text{2}}\text{O)}_{\text{6}}^{\text{3+}} \\ & \underset{\text{Lewis acid}}{\text{Cu}^{\text{2+}}}\text{ + }\underset{\text{Lewis base}}{\text{4NH}_{\text{3}}}\text{ }\to \text{ Cu(NH}_{\text{3}}\text{)}_{\text{4}}^{\text{2+}} \\ \end{align} \nonumber \] So can electron deficient compounds such as boron trifluoride:
Many Lewis acid-base reactions occur in media other than aqueous solution. The Brönsted-Lowry theory accounts for almost all aqueous acid-base chemistry. Therefore the Brönsted-Lowry concept is most often intended when the words acid or base are used. The Lewis definition is useful when discussing transition-metal ions, however, and is discussed again in the sections on . Identify the Lewis acids and bases in the following list. Write an equation for the combination of each acid with the Lewis base H O.(a) BeCl ( ); (b) CH OH; (c) SO ; (d) CF . The Lewis diagram The S atom in SO can accept an extra pair of electrons, and so SO is a Lewis acid. The O atoms have lone pairs but are only weakly basic for the same reason as the Cl atoms in part (a).
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In a gravimetric analysis, a measurement of mass or a change in mass provides quantitative information about the analyte. The most common form of gravimetry uses a precipitation reaction to generate a product whose mass is proportional to the amount of analyte. In many cases the precipitate includes the analyte; however, an indirect analysis in which the analyte causes the precipitation of another compound also is possible. Precipitation gravimetric procedures must be carefully controlled to produce precipitates that are easy to filter, free from impurities, and of known stoichiometry. In volatilization gravimetry, thermal or chemical energy decomposes the sample containing the analyte. The mass of residue that remains after decomposition, the mass of volatile products collected using a suitable trap, or a change in mass due to the loss of volatile material are all gravimetric measurements. When the analyte is already present in a particulate form that is easy to separate from its matrix, then a particulate gravimetric analysis is feasible. Examples include the determination of dissolved solids and the determination of fat in foods. coagulation definitive technique electrogravimetry ignition occlusion precipitant
relative supersaturation surface adsorbate volatilization gravimetry conservation of mass digestion gravimetry inclusion particulate gravimetry precipitation gravimetry reprecipitation thermogram coprecipitate direct analysis homogeneous precipitation indirect analysis peptization quartz crystal microbalance supernatant thermogravimetry | 1,596 | 3,681 |
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A water well is an excavation or structure created in the ground by digging, driving, boring, or drilling to access groundwater in underground aquifers. The well water is often drawn by a pump (Figure \(\Page {1}\)). Unfortunately, it impossible to pump water from very deep in the ground with just a surface pump. The key to understanding why is realizing that suction generated by the pump is not a force, but simply removing an opposing force to the force of air pressure which is already there. When you stick a pipe down a deep hole into a pool of water at the bottom of a well, air inside the pipe is pushing down on the water in the pipe, and air outside the pipe is pushing down on the water outside the pipe, which in turn pushes up on water inside the pipe - all is in a balance. Let's say you suck out the air inside the pipe. The water is pushed up the the same as it was before, but there is no counter acting force pushing the water down, so it begins to rise inside the pipe (Figure \(\Page {2}\)). So far so good, but the water stops rising at some height since the water is pulled down by gravity (i.e., the more water in the pipe, the more it weighs). Since the force of the air outside the pipe is not changing, eventually the weight of the water is equal to the air pressure outside the pipe. When this happens, the system is in balance again and water stops flowing. Suction is not a force, the atmospheric pressure is. Water is pumped from a well by creating a partial vacuum above the water by the pump. The amount of vacuum is equal to the weight of the column of water from the water table to the surface. Atmospheric pressure at sea level is 760 mm of mercury (\(1.01 \times 10^5 \,Pascals\)), which is equivalent to a 10.3-meter column of water. This is how deep water can be pumped from (with a surface pump; other pressurized pumps can go deeper). | 1,900 | 3,682 |
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Careful measurements on the metal–ligand complex Fe(SCN) suggest its stability decreases in the presence of inert ions [Lister, M. W.; Rivington, D. E. , , 1572–1590]. We can demonstrate this by adding an inert salt to an equilibrium mixture of Fe and SCN . Figure 6.9.1
a shows the result of mixing together equal volumes of 1.0 mM FeCl and 1.5 mM KSCN, both of which are colorless. The solution’s reddish–orange color is due to the formation of Fe(SCN) . \[\mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{SCN})^{2+}(a q) \label{6.1}\] Adding 10 g of KNO to the solution and stirring to dissolve the solid, produces the result shown in Figure 6.9.1
b. The solution’s lighter color suggests that adding KNO shifts reaction \ref{6.1} to the left, decreasing the concentration of Fe(SCN) and increasing the concentrations of Fe and SCN . The result is a decrease in the complex’s formation constant, . \[K_{1}=\frac{\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]} \label{6.2}\] Why should adding an inert electrolyte affect a reaction’s equilibrium position? We can explain the effect of KNO on the formation of Fe(SCN) if we consider the reaction on a microscopic scale. The solution in Figure 6.9.1
b contains a variety of cations and anions: Fe , SCN , K , \(\text{NO}_3^-\), H O , and OH . Although the solution is homogeneous, on average, there are slightly more anions in regions near the Fe ions, and slightly more cations in regions near the SCN ions. As shown in Figure 6.9.2
, each Fe ion and each SCN ion is surrounded by an ionic atmosphere of opposite charge (\(\delta^–\) and \(\delta^+\)) that partially screen the ions from each other. Because each ion’s apparent charge at the edge of its ionic atmosphere is less than its actual charge, the force of attraction between the two ions is smaller. As a result, the formation of Fe(SCN) is slightly less favorable and the formation constant in Equation \ref{6.2} is slightly smaller. Higher concentrations of KNO increase \(\delta^–\) and \(\delta^+\), resulting in even smaller values for the formation constant. To factor the concentration of ions into the formation constant for Fe(SCN) , we need a way to express that concentration in a meaningful way. Because both an ion’s concentration and its charge are important, we define the solution’s , \(\mu\) as \[\mu=\frac{1}{2} \sum_{i=1}^{n} c_{i} z_{i}^{2} \nonumber\] where and are the concentration and charge of the ion. Calculate the ionic strength of a solution of 0.10 M NaCl. Repeat the calculation for a solution of 0.10 M Na SO . The ionic strength for 0.10 M NaCl is \[\begin{array}{c}{\mu=\frac{1}{2}\left\{\left[\mathrm{Na}^{+}\right] \times(+1)^{2}+\left[\mathrm{Cl}^{-}\right] \times(-1)^{2}\right\}} \\ {\mu=\frac{1}{2}\left\{(0.10) \times(+1)^{2}+(0.10) \times(-1)^{2}\right\}=0.10 \ \mathrm{M}}\end{array} \nonumber\] For 0.10 M Na SO the ionic strength is \[\begin{array}{c}{\mu=\frac{1}{2}\left\{\left[\mathrm{Na}^{+}\right] \times(+1)^{2}+\left[\mathrm{SO}_{4}^{2-}\right] \times(-2)^{2}\right\}} \\ {\mu=\frac{1}{2}\left\{(0.20) \times(+1)^{2}+(0.10) \times(-2)^{2}\right\}=0.30 \ \mathrm{M}}\end{array} \nonumber\] In calculating the ionic strengths of these solutions we are ignoring the presence of H O and OH , and, in the case of Na SO , the presence of \(\text{HSO}_4^-\) from the base dissociation reaction of \(\text{SO}_4^{2-}\). In the case of 0.10 M NaCl, the concentrations of H O and OH are \(1.0 \times 10^{-7}\), which is significantly smaller than the concentrations of Na and Cl . Because \(\text{SO}_4^{2-}\) is a very weak base ( = \(1.0 \times 10^{-12}\)), the solution is only slightly basic (pH = 7.5), and the concentrations of H O , OH , and \(\text{HSO}_4^-\) are negligible. Although we can ignore the presence of H O , OH , and \(\text{HSO}_4^-\) when we calculate the ionic strength of these two solutions, be aware that an equilibrium reaction can generate ions that might affect the solution’s ionic strength. Note that the unit for ionic strength is molarity, but that a salt’s ionic strength need not match its molar concentration. For a 1:1 salt, such as NaCl, ionic strength and molar concentration are identical. The ionic strength of a 2:1 electrolyte, such as Na SO , is three times larger than the electrolyte’s molar concentration. shows that adding KNO to a mixture of Fe and SCN decreases the formation constant for Fe(SCN) . This creates a contradiction. Earlier in this chapter we showed that there is a relationship between a reaction’s standard‐state free energy, ∆ , and its equilibrium constant, . \[\triangle G^{\circ}=-R T \ln K \nonumber\] Because a reaction has only one standard‐state, its equilibrium constant must be independent of solution conditions. Although ionic strength affects the formation constant for Fe(SCN) , reaction \ref{6.1} must have an underlying formation constant that is independent of ionic strength. The apparent formation constant for Fe(SCN) , as shown in Equation \ref{6.2}, is a function of concentrations. In place of concentrations, we define the true thermodynamic equilibrium constant using activities. The activity of species , , is the product of its concentration, [ ], and a solution‐dependent activity coefficient, \(\gamma_A\) \[a_{A}=[A] \gamma_{A} \nonumber\] The true thermodynamic formation constant for Fe(SCN) , therefore, is \[K_{1}=\frac{a_{\mathrm{Fe}(S \mathrm{CN})^{2+}}}{a_{\mathrm{Fe}^{3+}} \times a_{\mathrm{SCN}^-}}=\frac{\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right] \gamma_{\mathrm{Fe}(\mathrm{SCN})^{2+}}}{\left[\mathrm{Fe}^{3+}\right] \gamma_{\mathrm{Fe}^{3+}}\left[\mathrm{SCN}^{-}\right] \gamma_{\mathrm{SCN}^{-}}} \nonumber\] Unless otherwise specified, the equilibrium constants in the appendices are thermodynamic equilibrium constants. A species’ corrects for any deviation between its physical concentration and its ideal value. For a gas, a pure solid, a pure liquid, or a non‐ionic solute, the activity coefficient is approximately one under most reasonable experimental conditions. For a gas the proper terms are fugacity and fugacity coefficient, instead of activity and activity coefficient. For a reaction that involves only these species, the difference between activity and concentration is negligible. The activity coefficient for an ion, however, depends on the solution’s ionic strength, the ion’s charge, and the ion’s size. It is possible to estimate activity coefficients using the extended \[\log \gamma_{A}=\frac{-0.51 \times z_{A}^{2} \times \sqrt{\mu}}{1+3.3 \times \alpha_{A} \times \sqrt{\mu}} \label{6.3}\] where is the ion’s charge, \(\alpha_A\) is the hydrated ion’s effective diameter in nanometers (Table 6.2), \(\mu\) is the solution’s ionic strength, and 0.51 and 3.3 are constants appropriate for an aqueous solution at 25 C. A hydrated ion’s effective radius is the radius of the ion plus those water molecules closely bound to the ion. The effective radius is greater for smaller, more highly charged ions than it is for larger, less highly charged ions. H O Li Na , \(\text{IO}_3^-\), \(\text{HSO}_3^-\), \(\text{HCO}_3^-\), \(\text{H}_2\text{PO}_4^-\) OH , F , SCN , HS , \(\text{ClO}_3^-\), \(\text{ClO}_4^-\), \(\text{MnO}_4^-\) K , Cl , Br , I , CN , \(\text{NO}_2^-\), \(\text{NO}_3^-\) Cs , Tl , Ag , \(\text{NH}_4^+\) Mg , Be Ca , Cu , Zn , Sn , Mn , Fe , Ni , Co Sr , Ba , Cd , Hg , S Pb , \(\text{SO}_4^{2-}\), \(\text{SO}_3^{2-}\) \(\text{Hg}_2^{2+}\), \(\text{SO}_4^{2-}\), \(\text{S}_22\text{O}_3^{2-}\), \(\text{CrO}_4^{2-}\), \(\text{HPO}_4^{2-}\) Al , Fe , Cr \(\text{PO}_4^{3-}\), \(\text{Fe(CN)}_6^{3-}\) Zr , Ce , Sn \(\text{Fe(CN)}_6^{4-}\) Kielland, J. , , 1675–1678. Several features of Equation \ref{6.3} deserve our attention. First, as the ionic strength approaches zero an ion’s activity coefficient approaches a value of one. In a solution where \(\mu = 0\), an ion’s activity and its concentration are identical. We can take advantage of this fact to determine a reaction’s thermodynamic equilibrium constant by measuring the apparent equilibrium constant for several increasingly smaller ionic strengths and extrapolating back to an ionic strength of zero. Second, an activity coefficient is smaller, and the effect of activity is more important, for an ion with a higher charge and a smaller effective radius. Finally, the extended Debye‐Hückel equation provides a reasonable estimate of an ion’s activity coefficient when the ionic strength is less than 0.1. Modifications to Equation \ref{6.3} extend the calculation of activity coefficients to higher ionic strengths [Davies, C. W. , Butterworth: London, 1962]. Earlier in this chapter we calculated the solubility of Pb(IO ) in deionized water, obtaining a result of \(4.0 \times 10^{-5}\) mol/L. Because the only significant source of ions is from the solubility reaction, the ionic strength is very low and we can assume that \(\gamma \approx 1\) for both Pb and \(\text{IO}_3^-\). In calculating the solubility of Pb(IO ) in deionized water, we do not need to account for ionic strength. But what if we need to know the solubility of Pb(IO ) in a solution that contains other, inert ions? In this case we need to include activity coefficients in our calculation. Calculate the solubility of Pb(IO ) in a matrix of 0.020 M Mg(NO ) . We begin by calculating the solution’s ionic strength. Since Pb(IO ) is only sparingly soluble, we will assume we can ignore its contribution to the ionic strength; thus \[\mu=\frac{1}{2}\left\{(0.020)(+2)^{2}+(0.040)(-1)^{2}\right\}=0.060 \ \mathrm{M} \nonumber\] Next, we use Equation \ref{6.3} to calculate the activity coefficients for Pb and \(\text{IO}_3^-\). \[\log \gamma_{\mathrm{Pb}^{2+}}=\frac{-0.51 \times(+2)^{2} \times \sqrt{0.060}}{1+3.3 \times 0.45 \times \sqrt{0.060}}=-0.366 \nonumber\] \[\gamma_{\mathrm{Pb}^{2+}}=0.431 \nonumber\] \[\log \gamma_{\mathrm{IO}_{3}^{-}}=\frac{-0.51 \times(-1)^{2} \times \sqrt{0.060}}{1+3.3 \times 0.45 \times \sqrt{0.060}}=-0.0916 \nonumber\] \[\gamma_{\mathrm{IO}_{3}^-}=0.810 \nonumber\] Defining the equilibrium concentrations of Pb and \(\text{IO}_3^-\) in terms of the variable Concentrations and substituting into the thermodynamic solubility product for Pb(IO ) leaves us with \[K_{\mathrm{sp}}=a_{\mathrm{Pb}^{2+}} \times a_{\mathrm{IO}_{3}^-}^{2}=\gamma_{\mathrm{Pb}^{2+}}\left[\mathrm{Pb}^{2+}\right] \times \gamma_{\mathrm{IO}_3^-}^{2}\left[\mathrm{IO}_{3}^{-}\right]^{2}=2.5 \times 10^{-13} \nonumber\] \[K_{\mathrm{sp}}=(0.431)(x)(0.810)^{2}(2 x)^{2}=2.5 \times 10^{-13} \nonumber\] \[K_{\mathrm{sp}}=1.131 x^{3}=2.5 \times 10^{-13} \nonumber\] Solving for gives \(6.0 \times 10^{-5}\) and a molar solubility of \(6.0 \times 10^{-5}\) mol/L for Pb(IO ) . If we ignore activity, as we did in our earlier calculation, we report the molar solubility as \(4.0 \times 10^{-5}\) mol/L. Failing to account for activity in this case underestimates the molar solubility of Pb(IO ) by 33%. The solution’s equilibrium composition is \[\begin{array}{c}{\left[\mathrm{Pb}^{2+}\right]=6.0 \times 10^{-5} \ \mathrm{M}} \\ {\left[\mathrm{IO}_{3}^{-}\right]=1.2 \times 10^{-4} \ \mathrm{M}} \\ {\left[\mathrm{Mg}^{2+}\right]=0.020 \ \mathrm{M}} \\ {\left[\mathrm{NO}_{3}^{-}\right]=0.040 \ \mathrm{M}}\end{array} \nonumber\] Because the concentrations of both Pb and \(\text{IO}_3^-\) are much smaller than the concentrations of Mg and \(\text{NO}_3^-\) our decision to ignore the contribution of Pb and \(\text{IO}_3^-\) to the ionic strength is reasonable. How do we handle the calculation if we can not ignore the concentrations of Pb and \(\text{IO}_3^-\) when calculating the ionic strength. One approach is to use the method of successive approximations. First, we recalculate the ionic strength using the concentrations of all ions, including Pb and \(\text{IO}_3^-\). Next, we recalculate the activity coefficients for Pb and \(\text{IO}_3^-\) using this new ionic strength and then recalculate the molar solubility. We continue this cycle until two successive calculations yield the same molar solubility within an acceptable margin of error. Calculate the molar solubility of Hg Cl in 0.10 M NaCl, taking into account the effect of ionic strength. Compare your answer to that from in which you ignored the effect of ionic strength. We begin by calculating the solution’s ionic strength. Because NaCl is a 1:1 ionic salt, the ionic strength is the same as the concentration of NaCl; thus \(\mu\) = 0.10 M. This assumes, of course, that we can ignore the contributions of \(\text{Hg}_2^{2+}\) and Cl from the solubility of Hg Cl . Next we use Equation \ref{6.3} to calculate the activity coefficients for \(\text{Hg}_2^{2+}\) and Cl . \[\log \gamma_{\mathrm{Hg}_{2}^{2+}}=\frac{-0.51 \times(+2)^{2} \times \sqrt{0.10}}{1+3.3 \times 0.40 \times \sqrt{0.10}}=-0.455 \nonumber\] \[\gamma_{\mathrm{H} \mathrm{g}_{2}^{2+}}=0.351 \nonumber\] \[\log \gamma_{\mathrm{Cl}^{-}}=\frac{-0.51 \times(-1)^{2} \times \sqrt{0.10}}{1+3.3 \times 0.3 \times \sqrt{0.10}}=-0.12 \nonumber\] \[\gamma_{\mathrm{Cl}^-}=0.75 \nonumber\] Defining the equilibrium concentrations of \(\text{Hg}_2^{2+}\) and Cl in terms of the variable and substituting into the thermodynamic solubility product for Hg Cl , leave us with \[K_{\mathrm{sp}}=a_{\mathrm{Hg}_{2}^{2+}}\left(a_{\mathrm{Cl}^-}\right)^{2} = \gamma_{\mathrm{Hg}_{2}^{2+}}\left[\mathrm{Hg}_{2}^{2+}\right]\left(\gamma_{\mathrm{Cl}^{-}}\right)^{2}\left[\mathrm{Cl}^{-}\right]^{2}=1.2 \times 10^{-18} \nonumber\] Because the value of likely is small, let’s simplify this equation to \[(0.351)(x)(0.75)^{2}(0.1)^{2}=1.2 \times 10^{-18} \nonumber\] Solving for gives its value as \(6.1 \times 10^{-16}\). Because is the concentration of \(\text{Hg}_2^{2+}\) and 2 is the concentration of Cl , our decision to ignore their contributions to the ionic strength is reasonable. The molar solubility of Hg Cl in 0.10 M NaCl is \(6.1 \times 10^{-16}\) mol/L. In , where we ignored ionic strength, we determined that the molar solubility of Hg Cl is \(1.2 \times 10^{-16}\) mol/L, a result that is \(5 \times\) smaller than the its actual value. As Example 6.9.2
and Exercise 6.9.1
show, failing to correct for the effect of ionic strength can lead to a significant error in an equilibrium calculation. Nevertheless, it is not unusual to ignore activities and to assume that the equilibrium constant is expressed in terms of concentrations. There is a practical reason for this—in an analysis we rarely know the exact composition, much less the ionic strength of aqueous samples or of solid samples brought into solution. Equilibrium calculations are a useful guide when we develop an analytical method; however, it only is when we complete an analysis and evaluate the results that can we judge whether our theory matches reality. In the end, work in the laboratory is the most critical step in developing a reliable analytical method. This is a good place to revisit the meaning of pH. In we defined pH as \[\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber\] Now we see that the correct definition is \[\begin{array}{c}{\mathrm{pH}=-\log a_{\mathrm{H}_{3} \mathrm{O}^{+}}} \\ {\mathrm{pH}=-\log \gamma_{\mathrm{H}_{3} \mathrm{O}^{+}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\end{array} \nonumber\] Failing to account for the effect of ionic strength can lead to a significant error in the reported concentration of H O . For example, if the pH of a solution is 7.00 and the activity coefficient for H O is 0.90, then the concentration of H O is \(1.11 \times 10^{-7}\) M, not \(1.00 \times 10^{-7}\) M, an error of +11%. Fortunately, when we develop and carry out an analytical method, we are more interested in controlling pH than in calculating [H O ]. As a result, the difference between the two definitions of pH rarely is of significant concern. | 16,034 | 3,683 |
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We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. As you have learned, the molecules of a gas are stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, , is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses Helium ( = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon \(\sqrt{\dfrac{29}{4.00}}=2.7\) times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days. At a given temperature, heavier molecules move more slowly than lighter molecules. During World War , scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of \(\ce{^{235}U}\). Naturally occurring uranium is only 0.720% \(\ce{^{235}U}\), whereas most of the rest (99.275%) is \(\ce{^{238}U}\), which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound \(UF_6\) (boiling point = 56°C). Given: isotopic content of naturally occurring uranium and atomic masses of U and U Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF Strategy: A The first step is to calculate the molar mass of UF containing U and U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of UF is The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation \(\ref{10.8.1}\): \[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043 \nonumber \] To obtain 99.0% pure UF requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give \[1.0043^n=\dfrac{0.990}{0.00720}=137.50 \nonumber \] Taking the logarithm of both sides gives \[\begin{align} n\ln(1.0043)&=\ln(137.50) \\[4pt] n &=\dfrac{\ln(137.50)}{\ln(1.0043)} \\[4pt]&=1148 \end{align} \nonumber \] Thus at least a thousand effusion steps are necessary to obtain highly enriched U. Below is a small part of a system that is used to prepare enriched uranium on a large scale. Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. ratio of effusion rates = 1.15200; one step gives 0.000154% He 96 steps Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{10.8.2} \] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{10.8.3} \] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{10.8.4} \] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation \(\ref{10.8.4}\) shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure \(\Page {3}\) for some common gases. Because all gases have the same average kinetic energy, according to the , molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure \(\Page {4}\). The average distance traveled by a molecule between collisions is the . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The the gas, the the mean free path. Calculate the rms speed of a sample -butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.8.5 to calculate the rms speed of the gas. To use Equation 10.8.4, we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[\begin{align} u_{\rm rms} &= \sqrt{\dfrac{3RT}{M}} \\[4pt] &=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}\\[4pt] &=361\;m/s \end{align} \nonumber \] or approximately 810 mi/h. Calculate the rms speed of a sample of radon gas at 23°C. \(1.82 \times 10^2\; m/s\) (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collisions. | 9,655 | 3,684 |
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Once scientists concluded that all matter contains negatively charged electrons, it became clear that atoms, which are electrically neutral, must also contain positive charges to balance the negative ones. Thomson proposed that the electrons were embedded in a uniform sphere that contained both the positive charge and most of the mass of the atom, much like raisins in plum pudding or chocolate chips in a cookie (Figure \(\Page {1}\)). In a single famous experiment, however, Rutherford showed unambiguously that Thomson’s model of the atom was incorrect. Rutherford aimed a stream of α particles at a very thin gold foil target (part (a) in Figure \(\Page {2}\)) and examined how the α particles were scattered by the foil. Gold was chosen because it could be easily hammered into extremely thin sheets, minimizing the number of atoms in the target. If Thomson’s model of the atom were correct, the positively-charged α particles should crash through the uniformly distributed mass of the gold target like cannonballs through the side of a wooden house. They might be moving a little slower when they emerged, but they should pass essentially straight through the target (part (b) in Figure \(\Page {2}\)). To Rutherford’s amazement, a small fraction of the α particles were deflected at large angles, and some were reflected directly back at the source (part (c) in Figure \(\Page {2}\)). According to Rutherford, “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Rutherford’s results were not consistent with a model in which the mass and positive charge are distributed uniformly throughout the volume of an atom. Instead, they strongly suggested that both the mass and positive charge are concentrated in a tiny fraction of the volume of an atom, which Rutherford called the nucleus. It made sense that a small fraction of the α particles collided with the dense, positively charged nuclei in either a glancing fashion, resulting in large deflections, or almost head-on, causing them to be reflected straight back at the source. Although Rutherford could not explain why repulsions between the positive charges in nuclei that contained more than one positive charge did not cause the nucleus to disintegrate, he reasoned that repulsions between negatively charged electrons would cause the electrons to be uniformly distributed throughout the atom’s volume.Today it is known that strong nuclear forces, which are much stronger than electrostatic interactions, hold the protons and the neutrons together in the nucleus. For this and other insights, Rutherford was awarded the Nobel Prize in Chemistry in 1908. Unfortunately, Rutherford would have preferred to receive the Nobel Prize in Physics because he considered physics superior to chemistry. In his opinion, “All science is either physics or stamp collecting.” The historical development of the different models of the atom’s structure is summarized in Figure \(\Page {3}\). Rutherford established that the nucleus of the hydrogen atom was a positively charged particle, for which he coined the name proton in 1920. He also suggested that the nuclei of elements other than hydrogen must contain electrically neutral particles with approximately the same mass as the proton. The neutron, however, was not discovered until 1932, when James Chadwick (1891–1974, a student of Rutherford; Nobel Prize in Physics, 1935) discovered it. As a result of Rutherford’s work, it became clear that an α particle contains two protons and neutrons, and is therefore the nucleus of a helium atom. Rutherford’s model of the atom is essentially the same as the modern model, except that it is now known that electrons are not uniformly distributed throughout an atom’s volume. Instead, they are distributed according to a set of principles described by Quantum Mechanics. Figure \(\Page {4}\) shows how the model of the atom has evolved over time from the indivisible unit of Dalton to the modern view taught today. The precise physical nature of atoms finally emerged from a series of elegant experiments carried out between 1895 and 1915. The most notable of these achievements was Ernest Rutherford's famous 1911 alpha-ray scattering experiment, which established that The nucleus is itself composed of two kinds of particles. are the carriers of positive electric charge in the nucleus; the proton charge is exactly the same as the electron charge, but of opposite sign. This means that in any [electrically neutral] atom, the number of protons in the nucleus (often referred to as the ) is balanced by the number of electrons outside the nucleus. The other nuclear particle is the . As its name implies, this particle carries no electrical charge. Its mass is almost the same as that of the proton. Most nuclei contain roughly equal numbers of neutrons and protons, so we can say that these two particles together account for almost all the mass of the atom. Because the electrons of an atom are in contact with the outside world, it is possible for one or more electrons to be lost, or some new ones to be added. The resulting electrically-charged atom is called an . Atoms consist of electrons, protons, and neutrons. This is an oversimplification that ignores the other subatomic particles that have been discovered, but it is sufficient for discussion of chemical principles. Some properties of these subatomic particles are summarized in Table \(\Page {1}\) which illustrates three important points: The discovery of the electron and the proton was crucial to the development of the modern model of the atom and provides an excellent case study in the application of the scientific method. In fact, the elucidation of the atom’s structure is one of the greatest detective stories in the history of science. The Nuclear Atom: Atoms, the smallest particles of an element that exhibit the properties of that element, consist of negatively charged electrons around a central nucleus composed of more massive positively charged protons and electrically neutral neutrons. Radioactivity is the emission of energetic particles and rays (radiation) by some substances. Three important kinds of radiation are α particles (helium nuclei), β particles (electrons traveling at high speed), and γ rays (similar to x-rays but higher in energy). Please be sure you are familiar with the topics discussed in Essential Skills 1 ( ) before proceeding to the Numerical Problems. a. electrons. b. protons. a. electrons. b. protons. | 6,558 | 3,685 |
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Calibrating a balance does not eliminate all sources of determinate error that might affect the signal. Because of the buoyancy of air, an object always weighs less in air than it does in a vacuum. If there is a difference between the object’s density and the density of the weights used to calibrate the balance, then we can make a correction for buoyancy [Battino, R.; Williamson, A. G. , , 51–52]. An object’s true weight in vacuo, , is related to its weight in air, , by the equation \[W_v = W_a \times \left[ 1 + \left( \frac {1} {D_o} - \frac {1} {D_w} \right) \times 0.0012 \right] \label{16.1}\] where is the object’s density, is the density of the calibration weight, and 0.0012 is the density of air under normal laboratory conditions (all densities are in units of g/cm ). The greater the difference between and the more serious the error in the object’s measured weight. The buoyancy correction for a solid is small and frequently ignored. The correction may be significant, however, for low density liquids and gases. This is particularly important when calibrating glassware. For example, we can calibrate a volumetric pipet by carefully filling the pipet with water to its calibration mark, dispensing the water into a tared beaker, and determining the water’s mass. After correcting for the buoyancy of air, we use the water’s density to calculate the volume dispensed by the pipet. A 10-mL volumetric pipet is calibrated following the procedure outlined above, using a balance calibrated with brass weights with a density of 8.40 g/cm . At 25 C the pipet dispenses 9.9736 g of water. What is the actual volume dispensed by the pipet and what is the determinate error in this volume if we ignore the buoyancy correction? At 25 C the density of water is 0.997 05 g/cm . Using Equation \ref{16.1} the water’s true weight is \[W_v = 9.9736 \text{ g} \times \left[ 1 + \left( \frac {1} {0.99705} - \frac {1} {8.40} \right) \times 0.0012 \right] = 9.9842 \text{ g} \nonumber\] and the actual volume of water dispensed by the pipet is \[\frac {9.9842 \text{ g}} {0.99705 \text{ g/cm}^{3}} = 10.014 \text{ cm}^{3} \nonumber\] If we ignore the buoyancy correction, then we report the pipet’s volume as \[\frac {9.9736 \text{ g}} {0.99705 \text{ g/cm}^{3}} = 10.003 \text{ cm}^{3} \nonumber\] introducing a negative determinate error of –0.11%. To calibrate a 10-mL pipet a measured volume of water is transferred to a tared flask and weighed, yielding a mass of 9.9814 grams. (a) Calculate, with and without correcting for buoyancy, the volume of water delivered by the pipet. Assume the density of water is 0.99707 g/cm and that the density of the weights is 8.40 g/cm . (b) What is the absolute error and the relative error introduced if we fail to account for the effect of buoyancy? Is this a significant source of determinate error for the calibration of a pipet? Explain. For (a), without accounting for buoyancy, the volume of water is \[\frac {9.9814 \text{ g}} {0.99707 \text{ g/cm}^3} = 10.011 \text{ cm}^3 = 10.011 \text{ mL} \nonumber\] When we correct for buoyancy, however, the volume is \[W_v = 9.9814 \text{ g} \times \left[ 1 + \left( \frac {1} {0.99707 \text{ g/cm}^3} - \frac {1} {8.40 \text{ g/cm}^3} \right) \times 0.0012 \text{ g/cm}^3 \right] = 9.920 \text{ g} \nonumber\] For (b), the absolute and relative errors in the mass are \[10.011 \text{ mL} - 10.021 \text{ mL} = -0.010 \text{ mL} \nonumber\] \[\frac {- 0.010 \text{ mL}} {10.021 \text{ mL}} \times 100 = -0.10\% \nonumber\] shows us that the standard deviation for the calibration of a 10-mL pipet is on the order of ±0.006 mL. Failing to correct for the effect of buoyancy gives a determinate error of –0.010 mL that is slightly larger than ±0.006 mL, suggesting that it introduces a small, but significant determinate error. Repeat the questions in Exercise 16.9.1
for the case where a mass of 0.2500 g is measured for a solid that has a density of 2.50 g/cm . The sample’s true weight is \[W_v = 0.2500 \text{ g} \times \left[ 1 + \left( \frac {1} {2.50 \text{ g/cm}^3} - \frac {1} {8.40 \text{ g/cm}^3} \right) \times 0.0012 \text{ g/cm}^3 \right] = 0.2501 \text{ g} \nonumber\] In this case the absolute and relative errors in mass are –0.0001 g and –0.040%. Is the failure to correct for buoyancy a constant or proportional source of determinate error? The true weight is the product of the weight measured in air and the buoyancy correction factor, which makes this a proportional error. The percentage error introduced when we ignore the buoyancy correction is independent of mass and a function only of the difference between the density of the object being weighed and the density of the calibration weights. What is the minimum density of a substance necessary to keep the buoyancy correction to less than 0.01% when using brass calibration weights with a density of 8.40 g/cm ? To determine the minimum density, we note that the buoyancy correction factor equals 1.00 if the density of the calibration weights and the density of the sample are the same. The correction factor is greater than 1.00 if is smaller than ; thus, the following inequality applies \[\left( \frac {1} {D_o} - \frac {1} {8.40} \right) \times 0.0012 \le (1.00)(0.0001) \nonumber\] Solving for shows that the sample’s density must be greater than 4.94 g/cm to ensure an error of less than 0.01%. | 5,398 | 3,686 |
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In we examined Nessler’s original method for matching the color of a sample to the color of a standard. Matching colors is a labor intensive process for the analyst and, not surprisingly, spectroscopic methods of analysis were slow to find favor. The 1930s and 1940s saw the introduction of photoelectric transducers for ultraviolet and visible radiation, and thermocouples for infrared radiation. As a result, modern instrumentation for absorption spectroscopy routinely became available in the 1940s—further progress has been rapid ever since. Frequently an analyst must select from among several instruments of different design, the one instrument best suited for a particular analysis. In this section we examine several different instruments for molecular absorption spectroscopy, with an emphasis on their advantages and limitations. Methods of sample introduction also are covered in this section. The simplest instrument for molecular UV/Vis absorption is a (Figure 10.3.1
), which uses an absorption or interference filter to isolate a band of radiation. The filter is placed between the source and the sample to prevent the sample from decomposing when exposed to higher energy radiation. A filter photometer has a single optical path between the source and detector, and is called a instrument. The instrument is calibrated to 0% T while using a shutter to block the source radiation from the detector. After opening the shutter, the instrument is calibrated to 100% T using an appropriate blank. The blank is then replaced with the sample and its transmittance measured. Because the source’s incident power and the sensitivity of the detector vary with wavelength, the photometer is recalibrated whenever the filter is changed. Photometers have the advantage of being relatively inexpensive, rugged, and easy to maintain. Another advantage of a photometer is its portability, making it easy to take into the field. Disadvantages of a photometer include the inability to record an absorption spectrum and the source’s relatively large effective bandwidth, which limits the calibration curve’s linearity. The percent transmittance varies between 0% and 100%. As we learned from , we use a blank to determine , which corresponds to 100%T. Even in the absence of light the detector records a signal. Closing the shutter allows us to assign 0%T to this signal. Together, setting 0% T and 100%T calibrates the instrument. The amount of light that passes through a sample produces a signal that is greater than or equal to 0%T and smaller than or equal to 100%T.
. Schematic diagram of a filter photometer. The analyst either inserts a removable filter or the filters are placed in a carousel, an example of which is shown in the photographic inset. The analyst selects a filter by rotating it into place. An instrument that uses a monochromator for wavelength selection is called a . The simplest spectrophotometer is a single-beam instrument equipped with a fixed-wavelength monochromator (Figure 10.3.2
). Single-beam spectrophotometers are calibrated and used in the same manner as a photometer. One example of a single-beam spectrophotometer is Thermo Scientific’s Spectronic 20D+, which is shown in the photographic insert to Figure 10.3.2
. The Spectronic 20D+ has a wavelength range of 340–625 nm (950 nm when using a red-sensitive detector), and a fixed effective bandwidth of 20 nm. Battery-operated, hand-held single-beam spectrophotometers are available, which are easy to transport into the field. Other single-beam spectrophotometers also are available with effective bandwidths of 2–8 nm. Fixed wavelength single-beam spectrophotometers are not practical for recording spectra because manually adjusting the wavelength and recalibrating the spectrophotometer is awkward and time-consuming. The accuracy of a single-beam spectrophotometer is limited by the stability of its source and detector over time. The limitations of a fixed-wavelength, single-beam spectrophotometer is minimized by using a spectrophotometer (Figure 10.3.3
). A chopper controls the radiation’s path, alternating it between the sample, the blank, and a shutter. The signal processor uses the chopper’s speed of rotation to resolve the signal that reaches the detector into the transmission of the blank, , and the sample, . By including an opaque surface as a shutter, it also is possible to continuously adjust 0%T. The effective bandwidth of a double-beam spectrophotometer is controlled by adjusting the monochromator’s entrance and exit slits. Effective bandwidths of 0.2–3.0 nm are common. A scanning monochromator allows for the automated recording of spectra. Double-beam instruments are more versatile than single-beam instruments, being useful for both quantitative and qualitative analyses, but also are more expensive and not particularly portable. An instrument with a single detector can monitor only one wavelength at a time. If we replace a single photomultiplier with an array of photodiodes, we can use the resulting detector to record a full spectrum in as little as 0.1 s. In a diode array spectrometer the source radiation passes through the sample and is dispersed by a grating (Figure 10.3.4
). The photodiode array detector is situated at the grating’s focal plane, with each diode recording the radiant power over a narrow range of wavelengths. Because we replace a full monochromator with just a grating, a diode array spectrometer is small and compact. One advantage of a diode array spectrometer is the speed of data acquisition, which allows us to collect multiple spectra for a single sample. Individual spectra are added and averaged to obtain the final spectrum. This improves a spectrum’s signal-to-noise ratio. If we add together spectra, the sum of the signal at any point, , increases as , where is the signal. The noise at any point, , is a random event, which increases as \(\sqrt{n} N_x\) when we add together spectra. The after scans, (S/N) is \[\left(\frac{S}{N}\right)_{n}=\frac{n S_{x}}{\sqrt{n} N_{x}}=\sqrt{n} \frac{S_{x}}{N_{x}} \nonumber\] where / is the signal-to-noise ratio for a single scan. The impact of signal averaging is shown in Figure 10.3.5
. The first spectrum shows the signal after one scan, which consists of a single, noisy peak. Signal averaging using 4 scans and 16 scans decreases the noise and improves the signal-to-noise ratio. One disadvantage of a photodiode array is that the effective bandwidth per diode is roughly an order of magnitude larger than that for a high quality monochromator. For more details on signals and noise, see Introduction to Signals and Noise by Steven Petrovic, an on-line resource that is part of the Analytical Sciences Digital Library. The sample compartment provides a light-tight environment that limits stray radiation. Samples normally are in a liquid or solution state, and are placed in cells constructed with UV/Vis transparent materials, such as quartz, glass, and plastic (Figure 10.3.6
). A quartz or fused-silica cell is required when working at a wavelength <300 nm where other materials show a significant absorption. The most common pathlength is 1 cm (10 mm), although cells with shorter (as little as 0.1 cm) and longer pathlengths (up to 10 cm) are available. Longer pathlength cells are useful when analyzing a very dilute solution or for gas samples. The highest quality cells allow the radiation to strike a flat surface at a 90 angle, minimizing the loss of radiation to reflection. A test tube often is used as a sample cell with simple, single-beam instruments, although differences in the cell’s pathlength and optical properties add an additional source of error to the analysis. If we need to monitor an analyte’s concentration over time, it may not be possible to remove samples for analysis. This often is the case, for example, when monitoring an industrial production line or waste line, when monitoring a patient’s blood, or when monitoring an environmental system, such as stream. With a we can analyze samples . An example of a remote sensing fiber-optic probe is shown in Figure 10.3.7
. The probe consists of two bundles of fiber-optic cable. One bundle transmits radiation from the source to the probe’s tip, which is designed to allow the sample to flow through the sample cell. Radiation from the source passes through the solution and is reflected back by a mirror. The second bundle of fiber-optic cable transmits the nonabsorbed radiation to the wavelength selector. Another design replaces the flow cell shown in Figure 10.3.7
with a membrane that contains a reagent that reacts with the analyte. When the analyte diffuses into the membrane it reacts with the reagent, producing a product that absorbs UV or visible radiation. The nonabsorbed radiation from the source is reflected or scattered back to the detector. Fiber optic probes that show chemical selectivity are called optrodes [(a) Seitz, W. R. , , 16A–34A; (b) Angel, S. M. , , 38–48]. The simplest instrument for IR absorption spectroscopy is a filter photometer similar to that shown in for UV/Vis absorption. These instruments have the advantage of portability and typically are used as dedicated analyzers for gases such as HCN and CO. Infrared instruments using a monochromator for wavelength selection use double-beam optics similar to that shown in . Double-beam optics are preferred over single-beam optics because the sources and detectors for infrared radiation are less stable than those for UV/Vis radiation. In addition, it is easier to correct for the absorption of infrared radiation by atmospheric CO and H O vapor when using double-beam optics. Resolutions of 1–3 cm are typical for most instruments. In a Fourier transform infrared spectrometer, or FT–IR, the monochromator is replaced with an interferometer ( ). Because an FT-IR includes only a single optical path, it is necessary to collect a separate spectrum to compensate for the absorbance of atmospheric CO and H O vapor. This is done by collecting a background spectrum without the sample and storing the result in the instrument’s computer memory. The background spectrum is removed from the sample’s spectrum by taking the ratio the two signals. In comparison to other instrument designs, an FT–IR provides for rapid data acquisition, which allows for an enhancement in signal-to-noise ratio through signal-averaging. Infrared spectroscopy routinely is used to analyze gas, liquid, and solid samples. Sample cells are made from materials, such as NaCl and KBr, that are transparent to infrared radiation. Gases are analyzed using a cell with a pathlength of approximately 10 cm. Longer pathlengths are obtained by using mirrors to pass the beam of radiation through the sample several times. A liquid sample may be analyzed using a variety of different sample cells (Figure 10.3.8
). For non-volatile liquids a suitable sample is prepared by placing a drop of the liquid between two NaCl plates, forming a thin film that typically is less than 0.01 mm thick. Volatile liquids are placed in a sealed cell to prevent their evaporation. The analysis of solution samples is limited by the solvent’s IR absorbing properties, with CCl , CS , and CHCl being the most common solvents. Solutions are placed in cells that contain two NaCl windows separated by a Teflon spacer. By changing the Teflon spacer, pathlengths from 0.015–1.0 mm are obtained. Transparent solid samples are analyzed by placing them directly in the IR beam. Most solid samples, however, are opaque, and are first dispersed in a more transparent medium before recording the IR spectrum. If a suitable solvent is available, then the solid is analyzed by preparing a solution and analyzing as described above. When a suitable solvent is not available, solid samples are analyzed by preparing a mull of the finely powdered sample with a suitable oil. Alternatively, the powdered sample is mixed with KBr and pressed into an optically transparent pellet. The analysis of an aqueous sample is complicated by the solubility of the NaCl cell window in water. One approach to obtaining an infrared spectrum of an aqueous solution is to use instead of transmission. Figure 10.3.9
shows a diagram of a typical attenuated total reflectance (ATR) FT–IR instrument. The ATR cell consists of a high refractive index material, such as ZnSe or diamond, sandwiched between a low refractive index substrate and a lower refractive index sample. Radiation from the source enters the ATR crystal where it undergoes a series of internal reflections before exiting the crystal. During each reflection the radiation penetrates into the sample to a depth of a few microns, which results in a selective attenuation of the radiation at those wavelengths where the sample absorbs. ATR spectra are similar, but not identical, to those obtained by measuring the transmission of radiation. Solid samples also can be analyzed using an ATR sample cell. After placing the solid in the sample slot, a compression tip ensures that it is in contact with the ATR crystal. Examples of solids analyzed by ATR include polymers, fibers, fabrics, powders, and biological tissue samples. Another reflectance method is diffuse reflectance, in which radiation is reflected from a rough surface, such as a powder. Powdered samples are mixed with a non-absorbing material, such as powdered KBr, and the reflected light is collected and analyzed. As with ATR, the resulting spectrum is similar to that obtained by conventional transmission methods. Further details about these, and other methods for preparing solids for infrared analysis can be found in this chapter’s . The determination of an analyte’s concentration based on its absorption of ultraviolet or visible radiation is one of the most frequently encountered quantitative analytical methods. One reason for its popularity is that many organic and inorganic compounds have strong absorption bands in the UV/Vis region of the electromagnetic spectrum. In addition, if an analyte does not absorb UV/Vis radiation—or if its absorbance is too weak—we often can react it with another species that is strongly absorbing. For example, a dilute solution of Fe does not absorb visible light. Reacting Fe with -phenanthroline, however, forms an orange–red complex of \(\text{Fe(phen)}_3^{2+}\) that has a strong, broad absorbance band near 500 nm. An additional advantage to UV/Vis absorption is that in most cases it is relatively easy to adjust experimental and instrumental conditions so that Beer’s law is obeyed. A quantitative analysis based on the absorption of infrared radiation, although important, is encountered less frequently than with UV/Vis absorption. One reason is the greater tendency for instrumental deviations from Beer’s law when using infrared radiation. Because an infrared absorption band is relatively narrow, any deviation due to the lack of monochromatic radiation is more pronounced. In addition, infrared sources are less intense than UV/Vis sources, which makes stray radiation more of a problem. Differences between the pathlengths for samples and for standards when using thin liquid films or KBr pellets are a problem, although an internal standard can correct for any difference in pathlength. Finally, establishing a 100%T ( = 0) baseline often is difficult because the optical properties of NaCl sample cells may change significantly with wavelength due to contamination and degradation. We can minimize this problem by measuring absorbance relative to a baseline established for the absorption band. Figure 10.3.10
shows how this is accomplished. Another approach is to use a cell with a fixed pathlength, such as that shown in . The analysis of waters and wastewaters often relies on the absorption of ultraviolet and visible radiation. Many of these methods are outlined in Table 10.3.1
. Several of these methods are described here in more detail. react with Eriochrome cyanide R dye at pH 6; forms red to pink complex reduce to AsH using Zn and react with silver diethyldithiocarbamate; forms red complex extract into CHCl containing dithizone from a sample made basic with NaOH; forms pink to red complex oxidize to Cr(VI) and react with diphenylcarbazide; forms 540 red-violet product react with neocuprine in neutral to slightly acid solution and extract into CHCl /CH OH; forms yellow complex reduce to Fe and react with -phenanthroline; forms orange-red complex extract into CHCl containing dithizone from sample made basic with NH / NH buffer; forms cherry red complex extract into CHCl containing dithizone from acidic sample; forms orange complex reaction with hypochlorite and phenol using a manganous 630 salt catalyst; forms blue indophenol as product react with chloroamine-T to form CNCl and then with a pyridine-barbituric acid; forms a red-blue dye react with red Zr-SPADNS lake; formation of ZrF decreases color of the red lake react with leuco crystal violet; forms blue product react with Cd to form NO and then react with sulfanilamide and -(1-napthyl)-ethylenediamine; forms red azo 543 dye react with ammonium molybdate and then reduce with SnCl ; forms molybdenum blue react with 4-aminoantipyrine and K Fe(CN) ; forms yellow antipyrine dye react with cationic methylene blue dye and extract into CHCl ; forms blue ion pair Although the quantitative analysis of metals in waters and wastewaters is accomplished primarily by atomic absorption or atomic emission spectroscopy, many metals also can be analyzed following the formation of a colored metal–ligand complex. One advantage to these spectroscopic methods is that they easily are adapted to the analysis of samples in the field using a filter photometer. One ligand used for the analysis of several metals is diphenylthiocarbazone, also known as dithizone. Dithizone is not soluble in water, but when a solution of dithizone in CHCl is shaken with an aqueous solution that contains an appropriate metal ion, a colored metal–dithizonate complex forms that is soluble in CHCl . The selectivity of dithizone is controlled by adjusting the sample’s pH. For example, Cd is extracted from solutions made strongly basic with NaOH, Pb from solutions made basic with an NH / NH buffer, and Hg from solutions that are slightly acidic. The structure of dithizone is shown below. See for a discussion of extracting metal ions using dithizone. When chlorine is added to water the portion available for disinfection is called the chlorine residual. There are two forms of chlorine residual. The free chlorine residual includes Cl , HOCl, and OCl . The combined chlorine residual, which forms from the reaction of NH with HOCl, consists of monochloramine, NH Cl, dichloramine, NHCl , and trichloramine, NCl . Because the free chlorine residual is more efficient as a disinfectant, there is an interest in methods that can distinguish between the total chlorine residual’s different forms. One such method is the leuco crystal violet method. The free residual chlorine is determined by adding leuco crystal violet to the sample, which instantaneously oxidizes to give a blue-colored compound that is monitored at 592 nm. Completing the analysis in less than five minutes prevents a possible interference from the combined chlorine residual. The total chlorine residual (free + combined) is determined by reacting a separate sample with iodide, which reacts with both chlorine residuals to form HOI. When the reaction is complete, leuco crystal violet is added and oxidized by HOI, giving the same blue-colored product. The combined chlorine residual is determined by difference. In Chapter 9 we explored how the total chlorine residual can be determined by a redox titration; see for further details. The method described here allows us to divide the total chlorine residual into its component parts. The concentration of fluoride in drinking water is determined indirectly by its ability to form a complex with zirconium. In the presence of the dye SPADNS, a solution of zirconium forms a red colored compound, called a lake, that absorbs at 570 nm. When fluoride is added, the formation of the stable \(\text{ZrF}_6^{2-}\) complex causes a portion of the lake to dissociate, decreasing the absorbance. A plot of absorbance versus the concentration of fluoride, therefore, has a negative slope. SPADNS, the structure of which is shown below, is an abbreviation for the sodium salt of 2-(4-sulfophenylazo)-1,8-dihydroxy-3,6-napthalenedisulfonic acid, which is a mouthful to say. Spectroscopic methods also are used to determine organic constituents in water. For example, the combined concentrations of phenol and ortho- and meta-substituted phenols are determined by using steam distillation to separate the phenols from nonvolatile impurities. The distillate reacts with 4-aminoantipyrine at pH 7.9 ± 0.1 in the presence of K Fe(CN) to a yellow colored antipyrine dye. After extracting the dye into CHCl , its absorbance is monitored at 460 nm. A calibration curve is prepared using only the unsubstituted phenol, C H OH. Because the molar absorptivity of substituted phenols generally are less than that for phenol, the reported concentration represents the minimum concentration of phenolic compounds. 4-aminoantipyrene Molecular absorption also is used for the analysis of environmentally significant airborne pollutants. In many cases the analysis is carried out by collecting the sample in water, converting the analyte to an aqueous form that can be analyzed by methods such as those described in . For example, the concentration of NO is determined by oxidizing NO to \(\text{NO}_3^-\). The concentration of \(\text{NO}_3^-\) is then determined by first reducing it to \(\text{NO}_2^-\) with Cd, and then reacting \(\text{NO}_2^-\) with sulfanilamide and -(1-naphthyl)-ethylenediamine to form a red azo dye. Another important application is the analysis for SO , which is determined by collecting the sample in an aqueous solution of \(\text{HgCl}_4^{2-}\) where it reacts to form \(\text{Hg(SO}_3)_2^{2-}\). Addition of -rosaniline and formaldehyde produces a purple complex that is monitored at 569 nm. Infrared absorption is useful for the analysis of organic vapors, including HCN, SO , nitrobenzene, methyl mercaptan, and vinyl chloride. Frequently, these analyses are accomplished using portable, dedicated infrared photometers. The analysis of clinical samples often is complicated by the complexity of the sample’s matrix, which may contribute a significant background absorption at the desired wavelength. The determination of serum barbiturates provides one example of how this problem is overcome. The barbiturates are first extracted from a sample of serum with CHCl and then extracted from the CHCl into 0.45 M NaOH (pH ≈ 13). The absorbance of the aqueous extract is measured at 260 nm, and includes contributions from the barbiturates as well as other components extracted from the serum sample. The pH of the sample is then lowered to approximately 10 by adding NH Cl and the absorbance remeasured. Because the barbiturates do not absorb at this pH, we can use the absorbance at pH 10, , to correct the absor-ance at pH 13, \[A_\text{barb} = A_\text{pH 13} - \frac {V_\text{samp} + V_{\text{NH}_4\text{Cl}}} {V_\text{samp}} \times A_\text{pH 10} \nonumber\] where is the absorbance due to the serum barbiturates and and \(V_{\text{NH}_4\text{Cl}}\) are the volumes of sample and NH Cl, respectively. Table 10.3.2
provides a summary of several other methods for analyzing clinical samples. react with NaOH and Cu ; forms blue-violet complex react with phosphotungstic acid; forms tungsten blue react with -toludine at 100 C; forms blue-green complex decompose protein to release iodide, which catalyzes redox reaction between Ce and As ; forms yellow colored Ce UV/Vis molecular absorption is used for the analysis of a diverse array of industrial samples including pharmaceuticals, food, paint, glass, and metals. In many cases the methods are similar to those described in and in Table 10.3.2
. For example, the amount of iron in food is determined by bringing the iron into solution and analyzing using the -phenanthroline method listed in . Many pharmaceutical compounds contain chromophores that make them suitable for analysis by UV/Vis absorption. Products analyzed in this fashion include antibiotics, hormones, vitamins, and analgesics. One example of the use of UV absorption is in determining the purity of aspirin tablets, for which the active ingredient is acetylsalicylic acid. Salicylic acid, which is produced by the hydrolysis of acetylsalicylic acid, is an undesirable impurity in aspirin tablets, and should not be present at more than 0.01% w/w. Samples are screened for unacceptable levels of salicylic acid by monitoring the absorbance at a wavelength of 312 nm. Acetylsalicylic acid absorbs at 280 nm, but absorbs poorly at 312 nm. Conditions for preparing the sample are chosen such that an absorbance of greater than 0.02 signifies an unacceptable level of salicylic acid. UV/Vis molecular absorption routinely is used for the analysis of narcotics and for drug testing. One interesting forensic application is the determination of blood alcohol using the Breathalyzer test. In this test a 52.5-mL breath sample is bubbled through an acidified solution of K Cr O , which oxidizes ethanol to acetic acid. The concentration of ethanol in the breath sample is determined by a decrease in the absorbance at 440 nm where the dichromate ion absorbs. A blood alcohol content of 0.10%, which is above the legal limit, corresponds to 0.025 mg of ethanol in the breath sample. To develop a quantitative analytical method, the conditions under which Beer’s law is obeyed must be established. First, the most appropriate wavelength for the analysis is determined from an absorption spectrum. In most cases the best wavelength corresponds to an absorption maximum because it provides greater sensitivity and is less susceptible to instrumental limitations. Second, if the instrument has adjustable slits, then an appropriate slit width is chosen. The absorption spectrum also aids in selecting a slit width by choosing a width that is narrow enough to avoid instrumental limita- tions to Beer’s law, but wide enough to increase the throughput of source radiation. Finally, a calibration curve is constructed to determine the range of concentrations for which Beer’s law is valid. Additional considerations that are important in any quantitative method are the effect of potential interferents and establishing an appropriate blank. 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 iron in water and waste- water provides an instructive example of a typical procedure. The description here is based on Method 3500-Fe B as published in , 20th Ed., American Public Health Association: Washington, D. C., 1998. Iron in the +2 oxidation state reacts with -phenanthroline to form the orange-red \(\text{Fe(phen)}_3^{2+}\) complex. The intensity of the complex’s color is independent of the solution’s acidity between a pH of 3 and 9. Because the complex forms more rapidly at lower pH levels, the reaction usually is carried out within a pH range of 3.0–3.5. Any iron present in the +3 oxidation state is reduced with hydroxylamine before adding -phenanthroline. The most important interferents are strong oxidizing agents, polyphosphates, and metal ions such as Cu , Zn , Ni , and Cd . An interference from oxidizing agents is minimized by adding an excess of hydroxylamine, and an interference from polyphosphate is minimized by boiling the sample in the presence of acid. The absorbance of samples and standards are measured at a wavelength of 510 nm using a 1-cm cell (longer pathlength cells also may be used). Beer’s law is obeyed for concentrations of within the range of 0.2–4.0 mg Fe/L. For a sample that contains less than 2 mg Fe/L, directly transfer a 50-mL portion to a 125-mL Erlenmeyer flask. Samples that contain more than 2 mg Fe/L are diluted before acquiring the 50-mL portion. Add 2 mL of concentrated HCl and 1 mL of hydroxylamine to the sample. Bring the solution to a boil and continue boiling until the solution’s volume is reduced to between 15 and 20 mL. After cooling to room temperature, transfer the solution to a 50-mL volumetric flask, add 10 mL of an ammonium acetate buffer, 2 mL of a 1000 ppm solution of -phenanthroline, and dilute to volume. Allow 10–15 minutes for color development before measuring the absorbance, using distilled water to set 100% T. Calibration standards, including a blank, are prepared by the same procedure using a stock solution that contains a known concentration of Fe . 1. Explain why strong oxidizing agents are interferents and why an excess of hydroxylamine prevents the interference. A strong oxidizing agent will oxidize some Fe to Fe . Because \(\text{Fe(phen)}_3^{3+}\) does not absorb as strongly as \(\text{Fe(phen)}_3^{2+}\), the absorbance is smaller than expected, which produces a negative determinate error. The excess hydroxylamine reacts with the oxidizing agents, removing them from the solution. 2. The color of the complex is stable between pH levels of 3 and 9. What are some possible complications at more acidic or at more basic pH’s? Because -phenanthroline is a weak base, its conditional formation constant for \(\text{Fe(phen)}_3^{2+}\) becomes smaller at more acidic pH levels, where -phenanthroline is present in its protonated form. The result is a decrease in absorbance and a less sensitive analytical method. When the pH is greater than 9, competition between OH and -phenanthroline for Fe also decreases the absorbance. In addition, if the pH is sufficiently basic there is a risk that the iron will precipitate as Fe(OH) . 3. Cadmium is an interferent because it forms a precipitate with -phenanthroline. What effect does the formation of precipitate have on the determination of iron? Because -phenanthroline is present in large excess (2000 μg of -phenanthroline for 100 μg of Fe2 ), it is not likely that the interference is due to an insufficient amount of -phenanthroline being available to react with the Fe . The presence of a precipitate in the sample cell results in the scattering of radiation, which causes an apparent increase in absorbance. Because the measured absorbance increases, the reported concentration is too high. Although scattering is a problem here, it can serve as the basis of a useful analytical method. See for further details. 4. Even high quality ammonium acetate contains a significant amount of iron. Why is this source of iron not a problem? Because all samples and standards are prepared using the same volume of ammonium acetate buffer, the contribution of this source of iron is accounted for by the calibration curve’s reagent blank. To determine the concentration of an analyte we measure its absorbance and apply Beer’s law using any of the standardization methods described in . The most common methods are a normal calibration curve using external standards and the method of standard additions. A single point standardization also is possible, although we must first verify that Beer’s law holds for the concentration of analyte in the samples and the standard. The determination of iron in an industrial waste stream is carried out by the -phenanthroline described in . Using the data in the following table, determine the mg Fe/L in the waste stream. Linear regression of absorbance versus the concentration of Fe in the standards gives the calibration curve and calibration equation shown here \[A=0.0006+\left(0.1817 \ \mathrm{mg}^{-1} \mathrm{L}\right) \times(\mathrm{mg} \mathrm{Fe} / \mathrm{L}) \nonumber\] Substituting the sample’s absorbance into the calibration equation gives the concentration of Fe in the waste stream as 1.48 mg Fe/L The concentration of Cu in a sample is determined by reacting it with the ligand cuprizone and measuring its absorbance at 606 nm in a 1.00-cm cell. When a 5.00-mL sample is treated with cuprizone and diluted to 10.00 mL, the resulting solution has an absorbance of 0.118. A second 5.00-mL sample is mixed with 1.00 mL of a 20.00 mg/L standard of Cu , treated with cuprizone and diluted to 10.00 mL, giving an absorbance of 0.162. Report the mg Cu /L in the sample. For this standard addition we write equations that relate absorbance to the concentration of Cu in the sample before the standard addition \[0.118=\varepsilon b \left[ C_{\mathrm{Cu}} \times \frac{5.00 \text{ mL}}{10.00 \text{ mL}}\right] \nonumber\] and after the standard addition \[0.162=\varepsilon b\left(C_{\mathrm{Cu}} \times \frac{5.00 \text{ mL}}{10.00 \text{ mL}}+\frac{20.00 \ \mathrm{mg} \ \mathrm{Cu}}{\mathrm{L}} \times \frac{1.00 \ \mathrm{mL}}{10.00 \ \mathrm{mL}}\right) \nonumber\] in each case accounting for the dilution of the original sample and for the standard. The value of \(\varepsilon b\) is the same in both equation. Solving each equation for \(\varepsilon b\) and equating \[\frac{0.162}{C_{\mathrm{Cu}} \times \frac{5.00 \text{ mL}}{10.00 \text{ mL}}+\frac{20.00 \ \mathrm{mg} \ \mathrm{Cu}}{\mathrm{L}} \times \frac{1.00 \ \mathrm{mL}}{10.00 \ \mathrm{mL}}}=\frac{0.118}{C_{\mathrm{Cu}} \times \frac{5.00 \text{ mL}}{10.00 \text{ mL}}} \nonumber\] leaves us with an equation in which is the only variable. Solving for gives its value as \[\frac{0.162}{0.500 \times C_{\mathrm{Cu}}+2.00 \ \mathrm{mg} \ \mathrm{Cu} / \mathrm{L}}=\frac{0.118}{0.500 \times C_{\mathrm{Cu}}} \nonumber\] \[0.0810 \times C_{\mathrm{Cu}}=0.0590 \times C_{\mathrm{Ca}}+0.236 \ \mathrm{mg} \ \mathrm{Cu} / \mathrm{L} \nonumber\] \[0.0220 \times C_{\mathrm{Cu}}=0.236 \ \mathrm{mg} \ \mathrm{Cu} / \mathrm{L} \nonumber\] \[C_{\mathrm{Cu}}=10.7 \ \mathrm{mg} \ \mathrm{Cu} / \mathrm{L} \nonumber\] Suppose we need to determine the concentration of two analytes, and , in a sample. If each analyte has a wavelength where the other analyte does not absorb, then we can proceed using the approach in Example 10.3.5
. Unfortunately, UV/Vis absorption bands are so broad that frequently it is not possible to find suitable wavelengths. Because Beer’s law is additive the mixture’s absorbance, , is \[\left(A_{m i x}\right)_{\lambda_{1}}=\left(\varepsilon_{x}\right)_{\lambda_{1}} b C_{X}+\left(\varepsilon_{Y}\right)_{\lambda_{1}} b C_{Y} \label{10.1}\] where \(\lambda_1\) is the wavelength at which we measure the absorbance. Because Equation \ref{10.1} includes terms for the concentration of both and , the absorbance at one wavelength does not provide enough information to determine either or . If we measure the absorbance at a second wavelength \[\left(A_{m i x}\right)_{\lambda_{2}}=\left(\varepsilon_{x}\right)_{\lambda_{2}} b C_{X}+\left(\varepsilon_{Y}\right)_{\lambda_{2}} b C_{Y} \label{10.2}\] then we can determine and by solving simultaneously Equation \ref{10.1} and Equation \ref{10.2}. Of course, we also must determine the value for \(\varepsilon_X\) and \(\varepsilon_Y\) at each wavelength. For a mixture of components, we must measure the absorbance at different wavelengths. The concentrations of Fe and Cu in a mixture are determined following their reaction with hexacyanoruthenate (II), \(\text{Ru(CN)}_6^{4-}\), which forms a purple-blue complex with Fe (\(\lambda_\text{max}\) = 550 nm) and a pale-green complex with Cu (\(\lambda_\text{max}\) = 396 nm) [DiTusa, M. R.; Schlit, A. A. , , 541–542]. The molar absorptivities (M cm ) for the metal complexes at the two wavelengths are summarized in the following table. When a sample that contains Fe and Cu is analyzed in a cell with a pathlength of 1.00 cm, the absorbance at 550 nm is 0.183 and the absorbance at 396 nm is 0.109. What are the molar concentrations of Fe and Cu in the sample? Substituting known values into Equation \ref{10.1} and Equation \ref{10.2} gives \[\begin{aligned} A_{550} &=0.183=9970 C_{\mathrm{Fe}}+34 C_{\mathrm{Cu}} \\ A_{396} &=0.109=84 C_{\mathrm{Fe}}+856 C_{\mathrm{Cu}} \end{aligned} \nonumber\] To determine and we solve the first equation for \[C_{\mathrm{Cu}}=\frac{0.183-9970 C_{\mathrm{Fe}}}{34} \nonumber\] and substitute the result into the second equation. \[\begin{aligned} 0.109 &=84 C_{\mathrm{Fe}}+856 \times \frac{0.183-9970 C_{\mathrm{Fe}}}{34} \\ &=4.607-\left(2.51 \times 10^{5}\right) C_{\mathrm{Fe}} \end{aligned} \nonumber\] Solving for gives the concentration of Fe as \(1.8 \times 10^{-5}\) M. Substituting this concentration back into the equation for the mixture’s absorbance at 396 nm gives the concentration of Cu as \(1.3 \times 10^{-4}\) M. Another approach to solving Example 10.3.2
is to multiply the first equation by 856/34 giving \[4.607=251009 C_{\mathrm{Fe}}+856 C_\mathrm{Cu} \nonumber\] Subtracting the second equation from this equation \[\begin{aligned} 4.607 &=251009 C_{\mathrm{Fe}}+856 C_{\mathrm{Cu}} \\-0.109 &=84 C_{\mathrm{Fe}}+856 C_{\mathrm{Cu}} \end{aligned} \nonumber\] gives \[4.498=250925 C_{\mathrm{Fe}} \nonumber\] and we find that is \(1.8 \times 10^{-5}\). Having determined we can substitute back into one of the other equations to solve for , which is \(1.3 \times 10^{-5}\). The absorbance spectra for Cr and Co overlap significantly. To determine the concentration of these analytes in a mixture, its absorbance is measured at 400 nm and at 505 nm, yielding values of 0.336 and 0.187, respectively. The individual molar absorptivities (M cm ) for Cr3 are 15.2 at 400 nm and 0.533 at 505 nm; the values for Co are 5.60 at 400 nm and 5.07 at 505 nm. Substituting into Equation \ref{10.1} and Equation \ref{10.2} gives \[A_{400} = 0.336 = 15.2C_\text{Cr} + 5.60C_\text{Co} \nonumber\] \[A_{400} = 0187 = 0.533C_\text{Cr} + 5.07C_\text{Co} \nonumber\] To determine and we solve the first equation for \[C_{\mathrm{Co}}=\frac{0.336-15.2 \mathrm{C}_{\mathrm{Co}}}{5.60} \nonumber\] and substitute the result into the second equation. \[0.187=0.533 C_{\mathrm{Cr}}+5.07 \times \frac{0.336-15.2 C_{\mathrm{Co}}}{5.60} \nonumber\] \[0.187=0.3042-13.23 C_{\mathrm{Cr}} \nonumber\] Solving for gives the concentration of Cr as \(8.86 \times 10^{-3}\) M. Substituting this concentration back into the equation for the mixture’s absorbance at 400 nm gives the concentration of Co as \(3.60 \times 10^{-2}\) M. To obtain results with good accuracy and precision the two wavelengths should be selected so that \(\varepsilon_X > \varepsilon_Y\) at one wavelength and \(\varepsilon_X < \varepsilon_Y\) at the other wavelength. It is easy to appreciate why this is true. Because the absorbance at each wavelength is dominated by one analyte, any uncertainty in the concentration of the other analyte has less of an impact. Figure 10.3.11
shows that the choice of wavelengths for Practice Exercise 10.3.2
are reasonable. When the choice of wavelengths is not obvious, one method for locating the optimum wavelengths is to plot \(\varepsilon_X / \varepsilon_y\) as function of wavelength, and determine the wavelengths where \(\varepsilon_X / \varepsilon_y\) reaches maximum and minimum values [Mehra, M. C.; Rioux, J. , , 688–689]. When the analyte’s spectra overlap severely, such that \(\varepsilon_X \approx \varepsilon_Y\) at all wavelengths, other computational methods may provide better accuracy and precision. In a multiwavelength linear regression analysis, for example, a mixture’s absorbance is compared to that for a set of standard solutions at several wavelengths [Blanco, M.; Iturriaga, H.; Maspoch, S.; Tarin, P. , , 178–180]. If and are the absorbance values for standard solutions of components and at any wavelength, then \[A_{SX}=\varepsilon_{X} b C_{SX} \label{10.3}\] \[A_{SY}=\varepsilon_{Y} b C_{SY} \label{10.4}\] where and are the known concentrations of and in the standard solutions. Solving Equation \ref{10.3} and Equation \ref{10.4} for \(\varepsilon_X\) and for \(\varepsilon_Y\), substituting into Equation \ref{10.1}, and rearranging, gives \[\frac{A_{\operatorname{mix}}}{A_{S X}}=\frac{C_{X}}{C_{S X}}+\frac{C_{Y}}{C_{S Y}} \times \frac{A_{S Y}}{A_{S X}} \nonumber\] To determine and the mixture’s absorbance and the absorbances of the standard solutions are measured at several wavelengths. Graphing / versus / gives a straight line with a slope of / and a -intercept of / . This approach is particularly helpful when it is not possible to find wavelengths where \(\varepsilon_X > \varepsilon_Y\) and \(\varepsilon_X < \varepsilon_Y\). The approach outlined here for a multiwavelength linear regression uses a single standard solution for each analyte. A more rigorous approach uses multiple standards for each analyte. The math behind the analysis of such data—which we call a multiple linear regression—is beyond the level of this text. For more details about multiple linear regression see Brereton, R. G. , Wiley: Chichester, England, 2003. Figure \(Page {10.11}\) shows visible absorbance spectra for a standard solution of 0.0250 M Cr , a standard solution of 0.0750 M Co , and a mixture that contains unknown concentrations of each ion. The data for these spectra are shown here. Use a multiwavelength regression analysis to determine the composition of the unknown. First we need to calculate values for and for / . Let’s define as Co and as Cr . For example, at a wavelength of 375 nm is 0.53/0.01, or 53 and / is 0.26/0.01, or 26. Completing the calculation for all wavelengths and graphing / versus / gives the calibration curve shown in Figure 10.3.12
. Fitting a straight-line to the data gives a regression model of \[\frac{A_{\operatorname{mix}}}{A_{S X}}=0.636+2.01 \times \frac{A_{S Y}}{A_{S X}} \nonumber\] Using the -intercept, the concentration of Co is \[\frac{C_{X}}{C_{S X}}=\frac{\left[\mathrm{Co}^{2+}\right]}{0.0750 \mathrm{M}}=0.636 \nonumber\] or [Co ] = 0.048 M; using the slope the concentration of Cr is \[\frac{C_{Y}}{C_{S Y}}=\frac{\left[\mathrm{Cr}^{3+}\right]}{0.0250 \mathrm{M}}=2.01 \nonumber\] or [Cr ] = 0.050 M. A mixture of \(\text{MnO}_4^{-}\) and \(\text{Cr}_2\text{O}_7^{2-}\), and standards of 0.10 mM KMnO and of 0.10 mM K Cr O give the results shown in the following table. Determine the composition of the mixture. The data for this problem is from Blanco, M. C.; Iturriaga, H.; Maspoch, S.; Tarin, P. , , 178–180. Letting represent \(\text{MnO}_4^{-}\) and letting represent \(\text{Cr}_2\text{O}_7^{2-}\), we plot the equation \[\frac{A_{\operatorname{mix}}}{A_{SX}}=\frac{C_{X}}{C_{SX}}+\frac{C_{Y}}{C_{S Y}} \times \frac{A_{S Y}}{A_{SX}} \nonumber\] placing / on the -axis and / on the -axis. For example, at a wavelength of 266 nm the value / of is 0.766/0.042, or 18.2, and the value of / is 0.410/0.042, or 9.76. Completing the calculations for all wavelengths and plotting the data gives the result shown here Fitting a straight-line to the data gives a regression model of \[\frac{A_{\text { mix }}}{A_{\text { SX }}}=0.8147+1.7839 \times \frac{A_{SY}}{A_{SX}} \nonumber\] Using the -intercept, the concentration of \(\text{MnO}_4^{-}\) is \[\frac{C_{X}}{C_{S X}}=0.8147=\frac{\left[\mathrm{MnO}_{4}^{-}\right]}{1.0 \times 10^{-4} \ \mathrm{M} \ \mathrm{MnO}_{4}^{-}} \nonumber\] or \(8.15 \times 10^{-5}\) M \(\text{MnO}_4^{-}\), and using the slope, the concentration of \(\text{Cr}_2\text{O}_7^{2-}\) is \[\frac{C_{Y}}{C_{S Y}}=1.7839=\frac{\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]}{1.00 \times 10^{-4} \ \mathrm{M} \ \text{Cr}_{2} \mathrm{O}_{7}^{2-}} \nonumber\] or \(1.78 \times 10^{-4}\) M \(\text{Cr}_2\text{O}_7^{2-}\). As discussed in , ultraviolet, visible, and infrared absorption bands result from the absorption of electromagnetic radiation by specific valence electrons or bonds. The energy at which the absorption occurs, and the intensity of that absorption, is determined by the chemical environment of the absorbing moiety. For example, benzene has several ultraviolet absorption bands due to \(\pi \rightarrow \pi^*\) transitions. The position and intensity of two of these bands, 203.5 nm (\(\varepsilon\) = 7400 M cm ) and 254 nm (\(\varepsilon\) = 204 M cm ), are sensitive to substitution. For benzoic acid, in which a carboxylic acid group replaces one of the aromatic hydrogens, the two bands shift to 230 nm (\(\varepsilon\) = 11600 M cm ) and 273 nm (\(\varepsilon\) = 970 M cm ). A variety of rules have been developed to aid in correlating UV/Vis absorption bands to chemical structure. Similar correlations are available for infrared absorption bands. For example a carbonyl’s C=O stretch is sensitive to adjacent functional groups, appearing at 1650 cm for acids, 1700 cm for ketones, and 1800 cm for acid chlorides. The interpretation of UV/ Vis and IR spectra receives adequate coverage elsewhere in the chemistry curriculum, notably in organic chemistry, and is not considered further in this text. With the availability of computerized data acquisition and storage it is possible to build digital libraries of standard reference spectra. The identity of an a unknown compound often can be determined by comparing its spectrum against a library of reference spectra, a process known as . Comparisons are made using an algorithm that calculates the cumulative difference between the sample’s spectrum and a reference spectrum. For example, one simple algorithm uses the following equation \[D = \sum_{i = 1}^n | (A_{sample})_i - (A_{reference})_i | \nonumber\] where is the cumulative difference, is the sample’s absorbance at wavelength or wavenumber , is the absorbance of the reference compound at the same wavelength or wavenumber, and is the number of digitized points in the spectra. The cumulative difference is calculated for each reference spectrum. The reference compound with the smallest value of is the closest match to the unknown compound. The accuracy of spectral searching is limited by the number and type of compounds included in the library, and by the effect of the sample’s matrix on the spectrum. Another advantage of computerized data acquisition is the ability to subtract one spectrum from another. When coupled with spectral searching it is possible to determine the identity of several components in a sample without the need of a prior separation step by repeatedly searching and sub- tracting reference spectra. An example is shown in Figure 10.3.13
in which the composition of a two-component mixture is determined by successive searching and subtraction. Figure 10.3.13
a shows the spectrum of the mixture. A search of the spectral library selects cocaine•HCl (Figure 10.3.13
b) as a likely component of the mixture. Subtracting the reference spectrum for cocaine•HCl from the mixture’s spectrum leaves a result (Figure 10.3.13
c) that closely matches mannitol’s reference spectrum (Figure 10.3.13
d). Subtracting the reference spectrum for mannitol leaves a small residual signal (Figure 10.3.13
e). Molecular absorption, particularly in the UV/Vis range, has been used for a variety of different characterization studies, including determining the stoichiometry of metal–ligand complexes and determining equilibrium constants. Both of these examples are examined in this section. We can determine the stoichiometry of the metal–ligand complexation reaction \[\mathrm{M}+y \mathrm{L} \rightleftharpoons \mathrm{ML}_{y} \nonumber\] using one of three methods: the method of continuous variations, the mole-ratio method, and the slope-ratio method. Of these approaches, the , also called Job’s method, is the most popular. In this method a series of solutions is prepared such that the total moles of metal and of ligand, , in each solution is the same. If ( ) and ( ) are, respectively, the moles of metal and ligand in solution , then \[n_{\text { total }}=\ \left(n_{\mathrm{M}}\right)_{i} \ + \ \left(n_{\mathrm{L}}\right)_{i} \nonumber\] The relative amount of ligand and metal in each solution is expressed as the mole fraction of ligand, ( ) and the mole fraction of metal, ( ) , \[\left(X_{\mathrm{L}}\right)_{i}=\frac{\left(n_{\mathrm{L}}\right)_{i}}{n_{\mathrm{total}}} \nonumber\] \[\left(X_{M}\right)_{i}=1-\frac{\left(n_\text{L}\right)_{i}}{n_{\text { total }}}=\frac{\left(n_\text{M}\right)_{i}}{n_{\text { total }}} \nonumber\] The concentration of the metal–ligand complex in any solution is determined by the limiting reagent, with the greatest concentration occurring when the metal and the ligand are mixed stoichiometrically. If we monitor the complexation reaction at a wavelength where only the metal–ligand complex absorbs, a graph of absorbance versus the mole fraction of ligand has two linear branches—one when the ligand is the limiting reagent and a second when the metal is the limiting reagent. The intersection of the two branches represents a stoichiometric mixing of the metal and the ligand. We use the mole fraction of ligand at the intersection to determine the value of for the metal–ligand complex ML . \[y=\frac{n_{\mathrm{L}}}{n_{\mathrm{M}}}=\frac{X_{\mathrm{L}}}{X_{\mathrm{M}}}=\frac{X_{\mathrm{L}}}{1-X_{\mathrm{L}}} \nonumber\] You also can plot the data as absorbance versus the mole fraction of metal. In this case, is equal to (1 – )/ . To determine the formula for the complex between Fe and -phenanthroline, a series of solutions is prepared in which the total concentration of metal and ligand is held constant at \(3.15 \times 10^{-4}\) M. The absorbance of each solution is measured at a wavelength of 510 nm. Using the following data, determine the formula for the complex. A plot of absorbance versus the mole fraction of ligand is shown in Figure 10.3.14
. To find the maximum absorbance, we extrapolate the two linear portions of the plot. The two lines intersect at a mole fraction of ligand of 0.75. Solving for gives \[y=\frac{X_{L}}{1-X_{L}}=\frac{0.75}{1-0.75}=3 \nonumber\] The formula for the metal–ligand complex is \(\text{Fe(phen)}_3^{2+}\). Use the continuous variations data in the following table to determine the formula for the complex between Fe and SCN . The data for this problem is adapted from Meloun, M.; Havel, J.; Högfeldt, E. , Ellis Horwood: Chichester, England, 1988, p. 236. The figure below shows a continuous variations plot for the data in this exercise. Although the individual data points show substantial curvature—enough curvature that there is little point in trying to draw linear branches for excess metal and excess ligand—the maximum absorbance clearly occurs at ≈ 0.5. The complex’s stoichiometry, therefore, is Fe(SCN) . Several precautions are necessary when using the method of continuous variations. First, the metal and the ligand must form only one metal–ligand complex. To determine if this condition is true, plots of absorbance versus are constructed at several different wavelengths and for several different values of . If the maximum absorbance does not occur at the same value of for each set of conditions, then more than one metal–ligand complex is present. A second precaution is that the metal–ligand complex’s absorbance must obey Beer’s law. Third, if the metal–ligand complex’s formation constant is relatively small, a plot of absorbance versus may show significant curvature. In this case it often is difficult to determine the stoichiometry by extrapolation. Finally, because the stability of a metal–ligand complex may be influenced by solution conditions, it is necessary to control carefully the composition of the solutions. When the ligand is a weak base, for example, each solutions must be buffered to the same pH. In the the moles of one reactant, usually the metal, is held constant, while the moles of the other reactant is varied. The absorbance is monitored at a wavelength where the metal–ligand complex absorbs. A plot of absorbance as a function of the ligand-to-metal mole ratio, / , has two linear branches that intersect at a mole–ratio corresponding to the complex’s formula. Figure 10.3.15
a shows a mole-ratio plot for the formation of a 1:1 complex in which the absorbance is monitored at a wavelength where only the complex absorbs. Figure 10.3.15
b shows a mole-ratio plot for a 1:2 complex in which all three species—the metal, the ligand, and the complex—absorb at the selected wavelength. Unlike the method of continuous variations, the mole-ratio method can be used for complexation reactions that occur in a stepwise fashion if there is a difference in the molar absorptivities of the metal–ligand complexes, and if the formation constants are sufficiently different. A typical mole-ratio plot for the step-wise formation of ML and ML is shown in Figure 10.3.15
c. For both the method of continuous variations and the mole-ratio method, we determine the complex’s stoichiometry by extrapolating absorbance data from conditions in which there is a linear relationship between absorbance and the relative amounts of metal and ligand. If a metal–ligand complex is very weak, a plot of absorbance versus or / becomes so curved that it is impossible to determine the stoichiometry by extrapolation. In this case the slope-ratio is used. In the two sets of solutions are prepared. The first set of solutions contains a constant amount of metal and a variable amount of ligand, chosen such that the total concentration of metal, , is much larger than the total concentration of ligand, . Under these conditions we may assume that essentially all the ligand reacts to form the metal–ligand complex. The concentration of the complex, which has the general form M L , is \[\left[\mathrm{M}_{x} \mathrm{L_y}\right]=\frac{C_{\mathrm{L}}}{y} \nonumber\] If we monitor the absorbance at a wavelength where only M L absorbs, then \[A=\varepsilon b\left[\mathrm{M}_{x} \mathrm{L}_{y}\right]=\frac{\varepsilon b C_{\mathrm{L}}}{y} \nonumber\] and a plot of absorbance versus is linear with a slope, , of \[s_{\mathrm{L}}=\frac{\varepsilon b}{y} \nonumber\] A second set of solutions is prepared with a fixed concentration of ligand that is much greater than a variable concentration of metal; thus \[\left[\mathrm{M}_{x} \mathrm{L}_{y}\right]=\frac{C_{\mathrm{M}}}{x} \nonumber\] \[A=\varepsilon b\left[\mathrm{M}_{x} \mathrm{L}_{y}\right]=\frac{\varepsilon b C_{\mathrm{M}}}{x} \nonumber\] \[s_{M}=\frac{\varepsilon b}{x} \nonumber\] A ratio of the slopes provides the relative values of and . \[\frac{s_{\text{M}}}{s_{\text{L}}}=\frac{\varepsilon b / x}{\varepsilon b / y}=\frac{y}{x} \nonumber\] An important assumption in the slope-ratio method is that the complexation reaction continues to completion in the presence of a sufficiently large excess of metal or ligand. The slope-ratio method also is limited to systems in which only a single complex forms and for which Beer’s law is obeyed. Another important application of molecular absorption spectroscopy is the determination of equilibrium constants. Let’s consider, as a simple example, an acid–base reaction of the general form \[\operatorname{HIn}(a q)+ \ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \ \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{In}^{-}(a q) \nonumber\] where HIn and In are the conjugate weak acid and weak base forms of an acid–base indicator. The equilibrium constant for this reaction is \[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right,\mathrm{A^-}]}{[\mathrm{HA}]} \nonumber\] To determine the equilibrium constant’s value, we prepare a solution in which the reaction is in a state of equilibrium and determine the equilibrium concentration for H O , HIn, and In . The concentration of H O is easy to determine by measuring the solution’s pH. To determine the concentration of HIn and In we can measure the solution’s absorbance. If both HIn and In absorb at the selected wavelength, then, from Beer's law, we know that \[A=\varepsilon_{\mathrm{Hln}} b[\mathrm{HIn}]+\varepsilon_{\mathrm{ln}} b[\mathrm{In}^-] \label{10.5}\] where \(\varepsilon_\text{HIn}\) and \(\varepsilon_{\text{In}}\) are the molar absorptivities for HIn and In . The indicator’s total concentration, , is given by a mass balance equation \[C=[\mathrm{HIn}]+ [\text{In}^-] \label{10.6}\] Solving Equation \ref{10.6} for [HIn] and substituting into Equation \ref{10.5} gives \[A=\varepsilon_{\mathrm{Hln}} b\left(C-\left[\mathrm{In}^{-}\right]\right)+\varepsilon_{\mathrm{ln}} b\left[\mathrm{In}^{-}\right] \nonumber\] which we simplify to \[A=\varepsilon_{\mathrm{Hln}} bC- \varepsilon_{\mathrm{Hln}}b\left[\mathrm{In}^{-}\right]+\varepsilon_{\mathrm{ln}} b\left[\mathrm{In}^{-}\right] \nonumber\] \[A=A_{\mathrm{HIn}}+b\left[\operatorname{In}^{-}\right]\left(\varepsilon_{\mathrm{ln}}-\varepsilon_{\mathrm{HIn}}\right) \label{10.7}\] where , which is equal to \(\varepsilon_\text{HIn}bC\), is the absorbance when the pH is acidic enough that essentially all the indicator is present as HIn. Solving Equation \ref{10.7} for the concentration of In gives \[\left[\operatorname{In}^{-}\right]=\frac{A-A_{\mathrm{Hln}}}{b\left(\varepsilon_{\mathrm{ln}}-\varepsilon_{\mathrm{HIn}}\right)} \label{10.8}\] Proceeding in the same fashion, we derive a similar equation for the concentration of HIn \[[\mathrm{HIn}]=\frac{A_{\mathrm{In}}-A}{b\left(\varepsilon_{\mathrm{ln}}-\varepsilon_{\mathrm{Hln}}\right)} \label{10.9}\] where , which is equal to \(\varepsilon_{\text{In}}bC\), is the absorbance when the pH is basic enough that only In contributes to the absorbance. Substituting Equation \ref{10.8} and Equation \ref{10.9} into the equilibrium constant expression for HIn gives \[K_a = \frac {[\text{H}_3\text{O}^+,\text{In}^-]} {[\text{HIn}]} = [\text{H}_3\text{O}^+] \times \frac {A - A_\text{HIn}} {A_{\text{In}} - A} \label{10.10}\] We can use Equation \ref{10.10} to determine in one of two ways. The simplest approach is to prepare three solutions, each of which contains the same amount, , of indicator. The pH of one solution is made sufficiently acidic such that [HIn] >> [In ]. The absorbance of this solution gives . The value of is determined by adjusting the pH of the second solution such that [In ] >> [HIn]. Finally, the pH of the third solution is adjusted to an intermediate value, and the pH and absorbance, , recorded. The value of is calculated using Equation \ref{10.10}. The acidity constant for an acid–base indicator is determined by preparing three solutions, each of which has a total concentration of indicator equal to \(5.00 \times 10^{-5}\) M. The first solution is made strongly acidic with HCl and has an absorbance of 0.250. The second solution is made strongly basic and has an absorbance of 1.40. The pH of the third solution is 2.91 and has an absorbance of 0.662. What is the value of for the indicator? The value of is determined by making appropriate substitutions into 10.20 where [H O ] is \(1.23 \times 10^{-3}\); thus \[K_{\mathrm{a}}=\left(1.23 \times 10^{-3}\right) \times \frac{0.662-0.250}{1.40-0.662}=6.87 \times 10^{-4} \nonumber\] To determine the of a merocyanine dye, the absorbance of a solution of \(3.5 \times 10^{-4}\) M dye was measured at a pH of 2.00, a pH of 6.00, and a pH of 12.00, yielding absorbances of 0.000, 0.225, and 0.680, respectively. What is the value of for this dye? The data for this problem is adapted from Lu, H.; Rutan, S. C. , , , 1381–1386. The value of is \[K_{\mathrm{a}}=\left(1.00 \times 10^{-6}\right) \times \frac{0.225-0.000}{0.680-0.225}=4.95 \times 10^{-7} \nonumber\] A second approach for determining is to prepare a series of solutions, each of which contains the same amount of indicator. Two solutions are used to determine values for and . Taking the log of both sides of Equation \ref{10.10} and rearranging leave us with the following equation. \[\log \frac{A-A_{\mathrm{Hin}}}{A_{\mathrm{ln}^{-}}-A}=\mathrm{pH}-\mathrm{p} K_{\mathrm{a}} \label{10.11}\] A plot of log[( – )/( – )] versus pH is a straight-line with a slope of +1 and a -intercept of –p . To determine the for the indicator bromothymol blue, the absorbance of each a series of solutions that contain the same concentration of bromothymol blue is measured at pH levels of 3.35, 3.65, 3.94, 4.30, and 4.64, yielding absorbance values of 0.170, 0.287, 0.411, 0.562, and 0.670, respectively. Acidifying the first solution to a pH of 2 changes its absorbance to 0.006, and adjusting the pH of the last solution to 12 changes its absorbance to 0.818. What is the value of for bromothymol blue? The data for this problem is from Patterson, G. S. , , , 395–398. To determine we use Equation \ref{10.11}, plotting log[( – )/( – )] versus pH, as shown below. Fitting a straight-line to the data gives a regression model of \[\log \frac{A-A_{\mathrm{HIn}}}{A_{\mathrm{ln}}-A}=-3.80+0.962 \mathrm{pH} \nonumber\] The -intercept is –p ; thus, the p is 3.80 and the is \(1.58 \times 10^{-4}\). In developing these approaches for determining we considered a relatively simple system in which the absorbance of HIn and In are easy to measure and for which it is easy to determine the concentration of H O . In addition to acid–base reactions, we can adapt these approaches to any reaction of the general form \[X(a q)+Y(a q)\rightleftharpoons Z(a q) \nonumber\] including metal–ligand complexation reactions and redox reactions, provided we can determine spectrophotometrically the concentration of the product, , and one of the reactants, either or , and that we can determine the concentration of the other reactant by some other method. With appropriate modifications, a more complicated system in which we cannot determine the concentration of one or more of the reactants or products also is possible [Ramette, R. W. , Addison-Wesley: Reading, MA, 1981, Chapter 13]. Molecular UV/Vis absorption routinely is used for the analysis of trace analytes in macro and meso samples. Major and minor analytes are determined by diluting the sample before analysis, and concentrating a sample may allow for the analysis of ultratrace analytes. The scale of operations for infrared absorption is generally poorer than that for UV/Vis absorption. Under normal conditions a relative error of 1–5% is easy to obtained with UV/Vis absorption. Accuracy usually is limited by the quality of the blank. Examples of the type of problems that are encountered include the pres- ence of particulates in the sample that scatter radiation, and the presence of interferents that react with analytical reagents. In the latter case the interferent may react to form an absorbing species, which leads to a positive determinate error. Interferents also may prevent the analyte from reacting, which leads to a negative determinate error. With care, it is possible to improve the accuracy of an analysis by as much as an order of magnitude. In absorption spectroscopy, precision is limited by indeterminate errors—primarily instrumental noise—which are introduced when we measure absorbance. Precision generally is worse for low absorbances where ≈ , and for high absorbances where approaches 0. We might expect, therefore, that precision will vary with transmittance. We can derive an expression between precision and transmittance by applying the propagation of uncertainty as described in . To do so we rewrite Beer’s law as \[C=-\frac{1}{\varepsilon b} \log T \label{10.12}\] in Chapter 4 helps us complete the propagation of uncertainty for Equation \ref{10.12}; thus, the absolute uncertainty in the concentration, , is \[s_{c}=-\frac{0.4343}{\varepsilon b} \times \frac{s_{T}}{T} \label{10.13}\] where is the absolute uncertainty in the transmittance. Dividing Equation \ref{10.13} by Equation \ref{10.12} gives the relative uncertainty in concentration, / , as \[\frac{s_c}{C}=\frac{0.4343 s_{T}}{T \log T} \nonumber\] If we know the transmittance’s absolute uncertainty, then we can determine the relative uncertainty in concentration for any measured transmittance. Determining the relative uncertainty in concentration is complicated because is a function of the transmittance. As shown in Table 10.3.3
, three categories of indeterminate instrumental error are observed [Rothman, L. D.; Crouch, S. R.; Ingle, J. D. Jr. , , 1226–1233]. A constant is observed for the uncertainty associated with reading % on a meter’s analog or digital scale. Typical values are ±0.2–0.3% (a of ±0.002–0.003) for an analog scale and ±0.001% a ( of ±0.00001) for a digital scale. %T readout resolution noise in thermal detectors positioning of sample cell fluctuations in source intensity A constant also is observed for the thermal transducers used in infrared spectrophotometers. The effect of a constant on the relative uncertainty in concentration is shown by curve A in Figure 10.3.16
. Note that the relative uncertainty is very large for both high absorbances and low absorbances, reaching a minimum when the absorbance is 0.4343. This source of indeterminate error is important for infrared spectrophotometers and for inexpensive UV/Vis spectrophotometers. To obtain a relative uncertainty in concentration of ±1–2%, the absorbance is kept within the range 0.1–1. Values of are a complex function of transmittance when indeterminate errors are dominated by the noise associated with photon detectors. Curve B in Figure 10.3.16
shows that the relative uncertainty in concentration is very large for low absorbances, but is smaller at higher absorbances. Although the relative uncertainty reaches a minimum when the absorbance is 0.963, there is little change in the relative uncertainty for absorbances between 0.5 and 2. This source of indeterminate error generally limits the precision of high quality UV/Vis spectrophotometers for mid-to-high absorbances. Finally, the value of is directly proportional to transmittance for indeterminate errors that result from fluctuations in the source’s intensity and from uncertainty in positioning the sample within the spectrometer. The latter is particularly important because the optical properties of a sample cell are not uniform. As a result, repositioning the sample cell may lead to a change in the intensity of transmitted radiation. As shown by curve C in Figure 10.3.16
, the effect is important only at low absorbances. This source of indeterminate errors usually is the limiting factor for high quality UV/Vis spectrophotometers when the absorbance is relatively small. When the relative uncertainty in concentration is limited by the % readout resolution, it is possible to improve the precision of the analysis by redefining 100% T and 0% T. Normally 100% T is established using a blank and 0% T is established while preventing the source’s radiation from reaching the detector. If the absorbance is too high, precision is improved by resetting 100% T using a standard solution of analyte whose concentration is less than that of the sample (Figure 10.3.17
a). For a sample whose absorbance is too low, precision is improved by redefining 0% T using a standard solution of the analyte whose concentration is greater than that of the analyte (Figure 10.3.17
b). In this case a calibration curve is required because a linear relationship between absorbance and concentration no longer exists. Precision is further increased by combining these two methods (Figure 10.3.17
c). Again, a calibration curve is necessary since the relation- ship between absorbance and concentration is no longer linear. The sensitivity of a molecular absorption method, which is the slope of a Beer’s law calibration curve, is the product of the analyte’s absorptivity and the pathlength of the sample cell (\(\varepsilon b\)). You can improve a method’s sensitivity by selecting a wavelength where absorbance is at a maximum or by increasing pathlength. See for an example of how the choice of wavelength affects a calibration curve’s sensitivity. Selectivity rarely is a problem in molecular absorption spectrophotometry. In many cases it is possible to find a wavelength where only the analyte absorbs. When two or more species do contribute to the measured absorbance, a multicomponent analysis is still possible, as shown in and . The analysis of a sample by molecular absorption spectroscopy is relatively rapid, although additional time is required if we need to convert a nonabsorbing analyte into an absorbing form. The cost of UV/Vis instrumentation ranges from several hundred dollars for a simple filter photometer, to more than $50,000 for a computer-controlled, high-resolution double-beam instrument equipped with variable slit widths, and operating over an extended range of wavelengths. Fourier transform infrared spectrometers can be obtained for as little as $15,000–$20,000, although more expensive models are available. | 69,110 | 3,687 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/13%3A_Air/13.03%3A_Chemistry_of_the_Atmosphere |
All life requires nitrogen-compounds, e.g., proteins and nucleic acids. Air, which is 79% nitrogen gas (N ), is the major reservoir of nitrogen. But most organisms cannot use nitrogen in this form. Figure \(\Page {1}\) illustrates the entire nitrogen cycle. Plants must secure their nitrogen in "fixed" form, i.e., incorporated in compounds such as: nitrate ions (NO ), ammonium ions (NH ) and urea (NH ) CO. Animals secure their nitrogen (and all other) compounds from plants (or animals that have fed on plants). Four processes participate in the cycling of nitrogen through the biosphere: (1) nitrogen fixation, (2) decay, (3) nitrification, and (4) denitrification. Microorganisms play major roles in all four of these. The nitrogen molecule (N ) is quite inert. To break it apart so that its atoms can combine with other atoms requires the input of substantial amounts of energy. Three processes are responsible for most of the nitrogen fixation in the biosphere: The enormous energy of lightning breaks nitrogen molecules and enables their atoms to combine with oxygen in the air forming nitrogen oxides. These dissolve in rain, forming nitrates, that are carried to the earth. Atmospheric nitrogen fixation probably contributes some 5– 8% of the total nitrogen fixed. Under great pressure, at a temperature of 600°C, and with the use of a catalyst, atmospheric nitrogen and hydrogen (usually derived from natural gas or petroleum) can be combined to form ammonia (NH ). Ammonia can be used directly as fertilizer, but most of its is further processed to urea and ammonium nitrate (NH NO ). The ability to fix nitrogen in the soil (Figure \(\Page {2}\) ) only in certain bacteria and archaea. Biological nitrogen fixation requires a complex set of enzymes and a huge expenditure of ATP. Although the first stable product of the process is ammonia, this is quickly incorporated into protein and other organic nitrogen compounds. The proteins made by plants enter and pass through food webs just as carbohydrates do. At each trophic level, their metabolism produces organic nitrogen compounds that return to the environment, chiefly in excretions. The final beneficiaries of these materials are microorganisms of decay. They break down the molecules in excretions and dead organisms into . Ammonia can be taken up directly by plants — usually through their roots. However, most of the ammonia produced by decay is converted into . Until recently this was thought always to be accomplished in two steps: These two groups of autotrophic bacteria are called . Through their activities (which supply them with all their energy needs), nitrogen is made available to the roots of plants. However, in 2015, two groups reported finding that bacteria in the genus were able to carry out both steps: ammonia to nitrite and nitrite to nitrate. This ability is called "comammox" (for complete ammonia oxidation). In addition, both soil and the ocean contain microbes, assigned to the Crenarchaeota, that convert ammonia to nitrites. They are more abundant than the nitrifying bacteria and may turn out to play an important role in the nitrogen cycle. Many legumes, in addition to fixing atmospheric nitrogen, also perform nitrification - converting some of their organic nitrogen to nitrites and nitrates. These reach the soil when they shed their leaves. The three processes above remove nitrogen from the atmosphere and pass it through ecosystems. Denitrification reduces nitrates and nitrites to nitrogen gas, thus replenishing the atmosphere. In the process several intermediates are formed: Once again, bacteria are the agents. They live deep in soil and in aquatic sediments where conditions are anaerobic. They use nitrates as an alternative to oxygen for the final electron acceptor in their respiration. Under anaerobic conditions in marine and freshwater sediments, other species of bacteria are able to oxidize ammonia (with \(\ce{NO2^{−}}\)) forming nitrogen gas. \[\ce{NH4^{+} + NO2^{−} → N2 + 2H2O} \nonumber \] The anammox reaction may account for as much as 50% of the denitrification occurring in the oceans. All of these processes participate in closing the nitrogen cycle. Agriculture may now be responsible for one-half of the nitrogen fixation on earth through the use of fertilizers produced by industrial fixation and the the growing of legumes like soybeans and alfalfa. This is a remarkable influence on a natural cycle. Are the denitrifiers keeping up the nitrogen cycle in balance? Probably not. Certainly, there are examples of nitrogen enrichment in ecosystems. One troubling example: the "blooms" of algae in lakes and rivers as nitrogen fertilizers leach from the soil of adjacent farms (and lawns). The accumulation of dissolved nutrients in a body of water is called . Oxygen is the most abundant element on the earth’s crust. The earth’s surface is composed of the crust, atmosphere, and hydrosphere. About 50% of the mass of the earth’s crust consists of oxygen (combined with other elements, principally silicon). Oxygen occurs as O molecules and, to a limited extent, as O (ozone) molecules in air. It forms about 20% of the mass of the air. About 89% of water by mass consists of combined oxygen. In combination with carbon, hydrogen, and nitrogen, oxygen is a large part of plants and animals. Oxygen is a colorless, odorless, and tasteless gas at ordinary temperatures. It is slightly denser than air. Although it is only slightly soluble in water (49 mL of gas dissolves in 1 L at STP), oxygen’s solubility is very important to aquatic life. Oxygen is essential in combustion processes such as the burning of fuels. Plants and animals use the oxygen from the air in respiration ( . The main way free oxygen is lost from the atmosphere is via and , mechanisms in which life and consume oxygen and release carbon dioxide. The respiration process is represented as: \[\ce{6O2+C6H12O6→6CO2+6H2O} \nonumber \] Green plants continually replenish the oxygen in the atmosphere by a process called ( . The products of photosynthesis may vary, but, in general, the process converts carbon dioxide and water into glucose (a sugar) and oxygen using the energy of light: Overview of and photosynthesis (green) and respiration (red). Water (at right), together with carbon dioxide (CO ), form oxygen and organic compounds (at left), which can be respired to water and (CO ). Thus, the oxygen that became carbon dioxide and water by the metabolic processes in plants and animals returns to the atmosphere by photosynthesis. Photosynthesizing organisms include the plant life of the land areas as well as the of the oceans. The tiny marine was discovered in 1986 and accounts for more than half of the photosynthesis of the open ocean. Oxygen is a key reactant in various oxidation reactions mentioned in section 8.5. Atmospheric free oxygen is also consumed by chemical weathering and surface reactions. An example of surface weathering is formation of rust: forms naturally in the upper atmosphere by the action of ultraviolet light from the sun on the oxygen there. Most atmospheric ozone occurs in the stratosphere, a layer of the atmosphere extending from about 10 to 50 kilometers above the earth’s surface. This ozone acts as a barrier to harmful ultraviolet light from the sun by absorbing it via a chemical decomposition reaction: \[\ce{O3}(g)\xrightarrow{\ce{ultraviolet\: light}}\ce{O}(g)+\ce{O2}(g) \nonumber \] In meteorology, an , also known as a , is a deviation from the normal change of an amospheric property with altitude. It almost always refers to an inversion of the thermal lapse rate. Normally, air temperature decreases with an increase in altitude. During an inversion, warmer air is held above cooler air; the normal temperature profile with altitude is inverted. An inversion traps air pollution, such as smog, close to the ground. An inversion can also suppress convection by acting as a "cap". If this cap is broken for any of several reasons, convection of any moisture present can then erupt into violent thunderstorms Temperature inversion can notoriously result in freezing rain in cold climates. ). ( ) | 8,201 | 3,688 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/03%3A__The_Vocabulary_of_Analytical_Chemistry/3.06%3A_Protocols |
Earlier we defined a protocol as a set of stringent written guidelines that specify an exact procedure that we must follow if an agency is to accept the results of our analysis. In addition to the considerations that went into the procedure’s design, a protocol also contains explicit instructions regarding internal and external quality assurance and quality control ( ) procedures [Amore, F. , , 1105A–1110A; Taylor, J. K. , , 1588A–1593A]. The goal of internal QA/QC is to ensure that a laboratory’s work is both accurate and precise. External QA/QC is a process in which an external agency certifies a laboratory. As an example, let’s outline a portion of the Environmental Protection Agency’s protocol for determining trace metals in water by graphite furnace atomic absorption spectroscopy as part of its Contract Laboratory Program (CLP). The CLP protocol (see Figure 3.6.1
) calls for an initial calibration using a method blank and three standards, one of which is at the detection limit. The resulting calibration curve is verified by analyzing initial calibration verification (ICV) and initial calibration blank (ICB) samples. The lab’s result for the ICV sample must fall within ±10% of its expected concentration. If the result is outside this limit the analysis is stopped and the problem identified and corrected before continuing. After a successful analysis of the ICV and ICB samples, the lab reverifies the calibration by analyzing a continuing calibration verification (CCV) sample and a continuing calibration blank (CCB). Results for the CCV also must be within ±10% of its expected concentration. Again, if the lab’s result for the CCV is outside the established limits, the analysis is stopped, the problem identified and corrected, and the system recalibrated as described above. Additional CCV and the CCB samples are analyzed before the first sample and after the last sample, and between every set of ten samples. If the result for any CCV or CCB sample is unacceptable, the results for the last set of samples are discarded, the system is recalibrated, and the samples reanalyzed. By following this protocol, each result is bound by successful checks on the calibration. Although not shown in Figure 3.6.1
, the protocol also contains instructions for analyzing duplicate or split samples, and for using spike tests to verify accuracy. | 2,381 | 3,689 |
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