Module 4: Equilibrium Chemistry PDF

Module 4: Equilibrium Chemistry At the end of the module, the students are expected to: 1. analyze reversible reactions, chemical equilibria, and thermodynamics; 2. compute worded problems related to equilibrium chemistry; and 3. appreciate the importance of equilibrium chemistry....

Module 4: Equilibrium Chemistry At the end of the module, the students are expected to: 1. analyze reversible reactions, chemical equilibria, and thermodynamics; 2. compute worded problems related to equilibrium chemistry; and 3. appreciate the importance of equilibrium chemistry. Regardless of the problem on which an analytical chemist is working, its solution requires a knowledge of chemistry and the ability to use that knowledge. For example, an analytical chemist who is studying the effect of pollution on spruce trees needs to know, or know where to find, the chemical differences between p‑hydroxybenzoic acid and p‑hydroxyacetophenone, two common phenols found in the needles of spruce trees. The ability to “think as a chemist” is a product of your experience in the classroom and in the laboratory. For the most part, the material in this text assumes you are familiar with topics covered in earlier courses; however, because of its importance to analytical chemistry, this chapter provides a review of equilibrium chemistry. Much of the material in this chapter should be familiar to you, although some topics—ladder diagrams and activity, for example—likely afford you with new ways to look at equilibrium chemistry. Reversible Reactions and Chemical Equilibria In 1798, the chemist Claude Berthollet accompanied Napoleon’s military expedition to Egypt. While visiting the Natron Lakes, a series of salt water lakes carved from limestone, Berthollet made an observation that led him to an important discovery. When exploring the lake’s shore, Berthollet found deposits of Na2CO3, a result he found surprising. Why did Berthollet find this result surprising and how did it contribute to an important discovery? Answering these questions provides us with an example of chemical reasoning and introduces us to the topic of this chapter. Napoleon’s expedition to Egypt was the first to include a significant scientific presence. The Commission of Sciences and Arts, which included Claude Berthollet, began with 151 members, and operated in Egypt for three years. In addition to Berthollet’s work, other results included a publication on mirages and a detailed catalogs of plant and animal life, mineralogy, and archeology. For a review of the Commission’s contributions, see Gillispie, C. G. “Scientific Aspects of the French Egyptian Expedition, 1798-1801,” Proc. Am. Phil. Soc. 1989, 133, 447–474. Napoleon’s expedition to Egypt was the first to include a significant scientific presence. The Commission of Sciences and Arts, which included Claude Berthollet, began with 151 members, and operated in Egypt for three years. In addition to Berthollet’s work, other results included a publication on mirages and a detailed catalogs of plant and animal life, mineralogy, and archeology. For a review of the Commission’s contributions, see Gillispie, C. G. “Scientific Aspects of the French Egyptian Expedition, 1798-1801,” Proc. Am. Phil. Soc. 1989, 133, 447–474. At the end of the 18th century, chemical reactivity was explained in terms of elective affinities. If, for example, substance A reacts with substance BC to form AB A+BC AB+C then A and B were said to have an elective affinity for each other. With elective affinity as the driving force for chemical reactivity, reactions were understood to proceed to completion and to proceed in one direction. Once formed, the compound AB could not revert to A and BC. From his experience in the laboratory, Berthollet knew that adding solid Na2CO3 to a solution of CaCl2 produces a precipitate of CaCO3. Natron is another name for the mineral sodium carbonate, Na2CO3 10H2O. In nature, it usually contains impurities of NaHCO3 and NaCl. In ancient Egypt, natron was mined and used for a variety of purposes, including as a cleaning agent and in mummification. Understanding this, Berthollet was surprised to find solid Na2CO3 forming on the edges of the lake, particularly since the deposits formed only when the lake’s salt water, NaCl(aq), was in contact with limestone, CaCO3. Where the lake was in contact with clay soils, there was little or no Na2CO3. Berthollet’s important insight was recognizing that the chemistry leading to the formation of Na2CO3 is the reverse of that seen in the laboratory. Using this insight Berthollet reasoned that the reaction is reversible, and that the relative amounts of NaCl, CaCO3, Na2CO3, and CaCl2 determine the direction in which the reaction occurs and the final composition of the reaction mixture. We recognize a reaction’s ability to move in both directions by using a double arrow when we write the reaction. Berthollet’s reasoning that reactions are reversible was an important step in understanding chemical reactivity. When we mix together solutions of Na2CO3 and CaCl2 they react to produce NaCl and CaCO3. As the reaction takes place, if we monitor the mass of Ca2+ that remains in solution and the mass of CaCO3 that precipitates, the result looks something like Figure 6.1. At the start of the reaction the mass of Ca 2+ decreases and the mass of CaCO3 increases. Eventually the reaction reaches a point after which there is no further change in the amounts of these species. Such a condition is called a state of equilibrium. Although a system at equilibrium appears static on a macroscopic level, it is important to remember that the forward and the reverse reactions continue to occur. A reaction at equilibrium exists in a steady-state, in which the rate at which a species forms equals the rate at which it is consumed. Thermodynamics and Equilibrium Chemistry Thermodynamics is the study of thermal, electrical, chemical, and mechanical forms of energy. The study of thermodynamics crosses many disciplines, including physics, engineering, and chemistry. Of the various branches of thermodynamics, the most important to chemistry is the study of how energy changes during a chemical reaction. Consider, for example, the general equilibrium reaction shown in equation 6.1, which involves the species A, B, C, and D, with stoichiometric coefficients of a, b, c, and d.  By convention, we identify the species on the left side of the equilibrium arrow as reactants and those on the right side of the equilibrium arrow as products. As Berthollet discovered, writing a reaction in this fashion does not guarantee that the reaction of A and B to produce C and D is favorable. Depending on initial conditions the reaction may move to the left, it may move to the right, or it may exist in a state of equilibrium. Understanding the factors that determine the reaction’s final equilibrium position is one of the goals of chemical thermodynamics. The direction of a reaction is that which lowers the overall free energy. At a constant temperature and pressure, which is typical of many benchtop chemical reactions, a reaction’s free energy is given by the Gibb’s free energy function where T is the temperature in kelvin, and G, H, and S are the differences in the Gibb's free energy, the enthalpy, and the entropy between the products and the reactants. Enthalpy is a measure of the flow of energy, as heat, during a chemical reaction. A reaction that releases heat has a negative ΔH and is called exothermic. An endothermic reaction absorbs heat from its surroundings and has a positive ΔH. Entropy is a measure of energy that is unavailable for useful, chemical work. The entropy of an individual species is always positive and generally is larger for gases than for solids, and for more complex molecules than for simpler molecules. Reactions that produce a large number of simple, gaseous products usually have a positive ΔS. The sign of ΔG indicates the direction in which a reaction moves to reach its equilibrium position. A reaction is thermodynamically favorable when its enthalpy, ΔH, decreases and its entropy, ΔS, increases. Substituting the inequalities ΔH < 0 and ΔS > 0 into equation 6.2 shows that a reaction is thermodynamically favorable when ΔG is negative. When ΔG is positive the reaction is unfavorable as written (although the reverse reaction is favorable). A reaction at equilibrium has a ΔG of zero. As a reaction moves from its initial, non-equilibrium condition to its equilibrium position, its value of ΔG approaches zero. At the same time, the chemical species in the reaction experience a change in their concentrations.  The Gibb's free energy, therefore, must be a function of the concentrations of reactants and products. As shown in equation 6.3, we can divide the Gibb’s free energy, ΔG, into two terms. The first term, ΔG°, is the change in the Gibb’s free energy when each species in the reaction is in its standard state, which we define as follows: gases with unit partial pressures, solutes with unit concentrations, and pure solids and pure liquids. The second term includes the reaction quotient, Q, which accounts for non-standard state pressures and concentrations. For reaction 6.1 the reaction quotient is where the terms in brackets are the concentrations of the reactants and products. Note that we define the reaction quotient with the products in the numerator and the reactants in the denominator. In addition, we raise the concentration of each species to a power equivalent to its stoichiometry in the balanced chemical reaction. For a gas, we use partial pressure in place of concentration. Pure solids and pure liquids do not appear in the reaction quotient. At equilibrium the Gibb’s free energy is zero, and equation 6.3 simplifies to where K is an equilibrium constant that defines the reaction’s equilibrium position. The equilibrium constant is just the reaction quotient’s numerical value when we substitute equilibrium concentrations into equation 6.4. Here we include the subscript “eq” to indicate a concentration at equilibrium. Although we will omit the “eq” when we write an equilibrium constant expressions, it is important to remember that the value of K is determined by equilibrium concentrations. Manipulating Equilibrium Constants We will take advantage of two useful relationships when we work with equilibrium constants. First, if we reverse a reaction’s direction, the equilibrium constant for the new reaction is the inverse of that for the original reaction. For example, the equilibrium constant for the reaction Second, if we add together two reactions to form a new reaction, the equilibrium constant for the new reaction is the product of the equilibrium constants for the original reactions. Example Equilibrium Constants for Chemical Reactions Several types of chemical reactions are important in analytical chemistry, either in preparing a sample for analysis or during the analysis. The most significant of these are precipitation reactions, acid–base reactions, complexation reactions, and oxidation– reduction reactions. In this section we review these reactions and their equilibrium constant expressions. Another common name for an oxidation– reduction reaction is a redox reaction, where “red” is short for reduction and “ox” is short for oxidation. Another common name for an oxidation– reduction reaction is a redox reaction, where “red” is short for reduction and “ox” is short for oxidation. Precipitation Reactions In a precipitation reaction, two or more soluble species combine to form an insoluble precipitate. The most common precipitation reaction is a metathesis reaction in which two soluble ionic compounds exchange parts. For example, if we add a solution of lead nitrate, Pb(NO3)2, to a solution of potassium chloride, KCl, a precipitate of lead chloride, PbCl2, forms. We usually write a precipitation reaction as a net ionic equation, which shows only the precipitate and those ions that form the precipitate; thus, the precipitation reaction for PbCl2 is When we write the equilibrium constant for a precipitation reaction, we focus on the precipitate’s solubility; thus, for PbCl2, the solubility reaction is and its equilibrium constant, or solubility product, Ksp, is Even though it does not appear in the Ksp expression, it is important to remember that equation 6.6 is valid only if PbCl2(s) is present and in equilibrium with Pb2+ and Cl–. You will find values for selected solubility products in Appendix 10. Acid–Base Reactions A useful definition of acids and bases is that independently introduced in 1923 by Johannes Brønsted and Thomas Lowry. In the Brønsted-Lowry definition, an acid is a proton donor and a base is a proton acceptor. Note the connection between these definitions—defining a base as a proton acceptor implies there is an acid available to donate the proton. For example, in reaction 6.7 acetic acid, CH3COOH, donates a proton to ammonia, NH3, which serves as the base. When an acid and a base react, the products are a new acid and a new base. For example, the acetate ion, CH3COO–, in reaction 6.7 is a base that can accept a proton from the acidic ammonium ion, NH4+, forming acetic acid and ammonia. We call the acetate ion the conjugate base of acetic acid, and we call the ammonium ion the conjugate acid of ammonia. Strong and Weak Acids The reaction of an acid with its solvent (typically water) is an acid dissociation reaction. We divide acids into two categories—strong and weak— based on their ability to donate a proton to the solvent. A strong acid, such as HCl, almost completely transfers its proton to the solvent, which acts as the base. We use a single arrow ( ) in place of the equilibrium arrow because we treat HCl as if it dissociates completely in an aqueous solution. In water, the common strong acids are hydrochloric acid (HCl), hydroiodic acid (HI), hydrobromic acid (HBr), nitric acid (HNO3), perchloric acid (HClO4), and the first proton of sulfuric acid (H2SO4). A weak acid, of which aqueous acetic acid is one example, does not completely donate its acidic proton to the solvent. Instead, most of the acid remains undissociated with only a small fraction present as the conjugate base. The equilibrium constant for this reaction is an acid dissociation constant, Ka, which we write as The magnitude of Ka provides information about a weak acid’s relative strength, with a smaller Ka corresponding to a weaker acid. The ammonium ion, NH4+, for example, has a Ka of 5.702 × 10–10 and is a weaker acid than acetic acid. A monoprotic weak acid, such as acetic acid, has only a single acidic proton and a single acid dissociation constant. Other acids, such as phosphoric acid, have multiple acidic protons, each characterized by an acid dissociation constant. We call such acids polyprotic. Phosphoric acid, for example, has three acid dissociation reactions and three acid dissociation constants. The decrease in the acid dissociation constants from Ka1 to Ka3 tells us that each successive proton is harder to remove. Consequently, H3PO4 is a stronger acid than H2PO4- , and H2PO4 - is a stronger acid than HPO4 2-. Strong and Weak Bases The most common example of a strong base is an alkali metal hydroxide, such as sodium hydroxide, NaOH, which completely dissociates to produce hydroxide ion. A weak base, such as the acetate ion, CH3COO–, only partially accepts a proton from the solvent, and is characterized by a base dissociation constant, Kb. For example, the base dissociation reaction and the base dissociation constant for the acetate ion are A polyprotic weak base, like a polyprotic acid, has more than one base dissociation reaction and more than one base dissociation constant. Amphiprotic Species Some species can behave as either a weak acid or as a weak base. For example, the following two reactions show the chemical reactivity of the bicarbonate ion, HCO3- , in water. A species that is both a proton donor and a proton acceptor is called amphiprotic. Whether an amphiprotic species behaves as an acid or as a base depends on the equilibrium constants for the competing reactions. For bicarbonate, the acid dissociation constant for reaction 6.8 Because bicarbonate is a stronger base than it is an acid, we expect that an aqueous solution of HCO3- is basic. Dissociation of Water Water is an amphiprotic solvent because it can serve as an acid or as a base. An interesting feature of an amphiprotic solvent is that it is capable of reacting with itself in an acid–base reaction. We identify the equilibrium constant for this reaction as water’s dissociation constant, Kw, Example Equation 6.11 allows us to develop a pH scale that indicates a solution’s acidity. When the concentrations of H3O+ and OH– are equal a solution is neither acidic nor basic; that is, the solution is neutral. Letting [H3O+] = [OH-] substituting into equation 6.11 and solving for [H3O+] gives A neutral solution has a hydronium ion concentration of 1.00 × 10-7 M and a pH of 7.00. In an acidic solution the concentration of H3O+ is greater than that for OH–, which means that The pH of an acidic solution, therefore, is less than 7.00. A basic solution, on the other hand, has a pH greater than 7.00. Figure 6.2 shows the pH scale and pH values for some representative solutions. Tabulating Values for Ka and Kb A useful observation about weak acids and weak bases is that the strength of a weak base is inversely proportional to the strength of its conjugate weak acid. Consider, for example, the dissociation reactions of acetic acid and acetate. Adding together these two reactions gives the reaction for which the equilibrium constant is Kw. Because adding together two reactions is equivalent to multiplying their respective equilibrium constants, we may express Kw as the product of Ka for CH3COOH and Kb for CH3COO–. For any weak acid, HA, and its conjugate weak base, A–, we can generalize this to the following equation where HA and A- are a conjugate acid–base pair. The relationship between Ka and Kb for a conjugate acid–base pair simplifies our tabulation of acid and base dissociation constants. Appendix 11 includes acid dissociation constants for a variety of weak acids. To find the value of Kb for a weak base, use equation 6.14 and the Ka value for its corresponding weak acid. A common mistake when using equation 6.14 is to forget that it applies to a conjugate acid–base pair only. Example Complexation Reactions A more general definition of acids and bases was proposed in 1923 by G. N. Lewis. The Brønsted-Lowry definition of acids and bases focuses on an acid’s proton-donating ability and a base’s proton-accepting ability. Lewis theory, on the other hand, uses the breaking and the forming of covalent bonds to describe acids and bases. In this treatment, an acid is an electron pair acceptor and a base in an electron pair donor. Although we can apply Lewis theory to the treatment of acid–base reactions, it is more useful for treating complexation reactions between metal ions and ligands. The following reaction between the metal ion Cd2+ and the ligand NH3 is typical of a complexation reaction. The product of this reaction is a metal–ligand complex. In writing this reaction we show ammonia as :NH3, using a pair of dots to emphasize the pair of electrons that it donates to Cd2+. In subsequent reactions we will omit this notation. Metal-Ligand Formation Constants We characterize the formation of a metal–ligand complex by a formation constant, Kf. The complexation reaction between Cd2+ and NH3, for example, has the following equilibrium constant. The reverse of reaction 6.15 is a dissociation reaction, which we characterize by a dissociation constant, Kd, that is the reciprocal of Kf. Many complexation reactions occur in a stepwise fashion. For example, the reaction between Cd2+ and NH3 involves four successive reactions. To avoid ambiguity, we divide formation constants into two categories. A stepwise formation constant, which we designate as Ki for the ith step, describes the successive addition of one ligand to the metal–ligand complex from the previous step. Thus, the equilibrium constants for reactions 6.17–6.20 are, respectively, K1, K2, K3, and K4. An overall, or cumulative formation constant, which we designate as bi, describes the addition of i ligands to the free metal ion. The equilibrium constant in equation 6.16 is correctly identified as β4, where In general Stepwise and overall formation constants for selected metal–ligand complexes are in Appendix 12. Metal-Ligand Complexation and Solubility A formation constant describes the addition of one or more ligands to a free metal ion. To find the equilibrium constant for a complexation reaction that includes a solid, we combine appropriate Ksp and Kf expressions. For example, the solubility of AgCl increases in the presence of excess chloride ions as the result of the following complexation reaction. We can write this reaction as the sum of three other equilibrium reactions with known equilibrium constants—the solubility of AgCl, which is described by its Ksp reaction and the stepwise formation of AgCl2 - , which is described by K1 and K2 reactions. The equilibrium constant for reaction 6.21, therefore, is Ksp × K1 × K2. Example Oxidation–Reduction (Redox) Reactions An oxidation–reduction reaction occurs when electrons move from one reactant to another reactant. As a result of this transfer of electrons, the reactants undergo a change in oxidation state. Those reactant that increases its oxidation state undergoes oxidation, and the reactant that decreases its oxidation state undergoes reduction. For example, in the following redox reaction between Fe3+ and oxalic acid, H2C2O4, iron is reduced because its oxidation state changes from +3 to +2. Oxalic acid, on the other hand, is oxidized because the oxidation state for carbon increases from +3 in H2C2O4 to +4 in CO2. We can divide a redox reaction, such as reaction 6.22, into separate half-reactions that show the oxidation and the reduction processes. It is important to remember, however, that an oxidation reaction and a reduction reaction always occur as a pair. We formalize this relationship by identifying as a reducing agent the reactant that is oxidized, because it provides the electrons for the reduction half-reaction. Conversely, the reactant that is reduced is an oxidizing agent. In reaction 6.22, Fe3+ is the oxidizing agent and H2C2O4 is the reducing agent. The products of a redox reaction also have redox properties. For example, the Fe2+ in reaction 6.22 is oxidized to Fe3+ when CO2 is reduced to H2C2O4. Borrowing some terminology from acid–base chemistry, Fe2+ is the conjugate reducing agent of the oxidizing agent Fe3+, and CO2 is the conjugate oxidizing agent of the reducing agent H2C2O4. Thermodynamics of Redox Reactions Unlike precipitation reactions, acid–base reactions, and complexation reactions, we rarely express the equilibrium position of a redox reaction with an equilibrium constant. Because a redox reaction involves a transfer of electrons from a reducing agent to an oxidizing agent, it is convenient to consider the reaction’s thermodynamics in terms of the electron. For a reaction in which one mole of a reactant undergoes oxidation or reduction, the net transfer of charge, Q, in coulombs is where n is the moles of electrons per mole of reactant, and F is Faraday’s constant (96 485 C/mol). The free energy, ΔG, to move this charge, Q, over a change in potential, E, is The change in free energy (in kJ/mole) for a redox reaction, therefore, is where ΔG has units of kJ/mol. The minus sign in equation 6.23 is the result of a different convention for assigning a reaction’s favorable direction. In thermodynamics, a reaction is favored when ΔG is negative, but an oxidation-reduction reaction is favored when E is positive. Substituting equation 6.23 into equation 6.3 where E° is the potential under standard-state conditions. Substituting appropriate values for R and F, assuming a temperature of 25°C (298 K), and switching from ln to log gives the potential in volts as Standard Potentials A redox reaction’s standard potential, E °, provides an alternative way of expressing its equilibrium constant and, therefore, its equilibrium position. Because a reaction at equilibrium has a ΔG of zero, the potential, E, also is zero at equilibrium. Substituting these values into equation 6.24 and rearranging provides a relationship between E° and K. A standard potential is the potential when all species are in their standard states. You may recall that we define standard state conditions as follows: all gases have unit partial pressures, all solutes have unit concentrations, and all solids and liquids are pure. We generally do not tabulate standard potentials for redox reactions. Instead, we calculate E° using the standard potentials for the corresponding oxidation half-reaction and reduction half-reaction. By convention, standard potentials are provided for reduction half-reactions. The standard potential for a redox reaction, E°, is where E°red and E°ox are the standard reduction potentials for the reduction half- reaction and the oxidation half-reaction. Because we cannot measure the potential for a single half-reaction, we arbitrarily assign a standard reduction potential of zero to a reference half-reaction and report all other reduction potentials relative to this reference. Appendix 13 contains a list of selected standard reduction potentials. The more positive the standard reduction potential, the more favorable the reduction reaction is under standard state conditions. For example, under standard state conditions the reduction of Cu2+ to Cu (E° = +0.3419 V) is more favorable than the reduction of Zn2+ to Zn (E°= –0.7618 V). Example When writing precipitation, acid–base, and metal–ligand complexation reactions, we represent acidity as H3O+. Redox reactions more commonly are written using H+ instead of H3O+. For the reaction in Practice Exercise 6.4, we could replace H+ with H3O+ and increase the stoichiometric coefficient for H2O from 4 to 12. Le Châtelier’s Principle At a temperature of 25°C, acetic acid’s dissociation reaction has an equilibrium constant of Because equation 6.26 has three variables—[CH3COOH], [CH3COO–], and [H3O+]—it does not have a unique mathematical solution. Nevertheless, although two solutions of acetic acid may have different values for [CH3COOH], [CH3COO–], and [H3O+], each solution has the same value of Ka. If we add sodium acetate to a solution of acetic acid, the concentration of CH3COO– increases, which suggests there is an increase in the value of Ka; however, because Ka must remain constant, the concentration of all three species in equation 6.26 must change to restore Ka to its original value. In this case, a partial reaction of CH3COO– and H3O+ decreases their concentrations, increases the concentration of CH3COOH, and reestablishes the equilibrium. The observation that a system at equilibrium responds to an external action by reequilibrating itself in a manner that diminishes that action, is formalized as Le Châtelier’s principle. One common action is to change the concentration of a reactant or product for a system at equilibrium. As noted above for a solution of acetic acid, if we add a product to a reaction at equilibrium the system responds by converting some of the products into reactants. Adding a reactant has the opposite effect, resulting in the conversion of reactants to products. When we add sodium acetate to a solution of acetic acid, we directly apply the action to the system. It is also possible to apply a change concentration indirectly. Consider, for example, the solubility of AgCl.  The effect on the solubility of AgCl of adding AgNO3 is obvious, but what is the effect of if we add a ligand that forms a stable, soluble complex with Ag+? Ammonia, for example, reacts with Ag+ as shown here Adding ammonia decreases the concentration of Ag+ as the Ag(NH3)2 complex forms. In turn, a decrease in the concentration of Ag+ increases the solubility of AgCl as reaction 6.27 reestablishes its equilibrium position. Adding together reaction 6.27 and reaction 6.28 clarifies the effect of ammonia on the solubility of AgCl, by showing ammonia as a reactant. So what is the effect on the solubility of AgCl of adding AgNO3? Adding AgNO3 increases the concentration of Ag+ in solution. To reestablish equilibrium, some of the Ag+ and Cl– react to form additional AgCl; thus, the solubility of AgCl decreases. The solubility product, Ksp, of course, remains unchanged. Example Increasing or decreasing the partial pressure of a gas is the same as increasing or decreasing its concentration. Because the concentration of a gas depends on its partial pressure, and not on the total pressure of the system, adding or removing an inert gas has no effect on a reaction’s equilibrium position. Most reactions involve reactants and products dispersed in a solvent. If we change the amount of solvent by diluting the solution, then the concentrations of all reactants and products must increase; conversely, if we allow the solvent to evaporate partially, then the concentration of the solutes must increase. The effect of simultaneously changing the concentrations of all reactants and products is not intuitively as obvious as when we change the concentration of a single reactant or product. As an example, let’s consider how diluting a solution affects the equilibrium position for the formation of the aqueous silver-amine complex (reaction 6.28). The equilibrium constant for this reaction is where we include the subscript “eq” for clarification. If we dilute a portion of this solution with an equal volume of water, each of the concentration terms in equation 6.30 is cut in half. The reaction quotient, Q, becomes Because Q is greater than b2, equilibrium is reestablished by shifting the reaction to the left, decreasing the concentration of Ag(NH3)2. Note that the new equilibrium position lies toward the side of the equilibrium reaction that has the greatest number of solute particles (one Ag+ ion and two molecules of NH3 versus a single metal-ligand complex). If we concentrate the solution of Ag(NH3)2 by evaporating some of the solvent, equilibrium is reestablished in the opposite direction. This is a general conclusion that we can apply to any reaction. Increasing volume always favors the direction that produces the greatest number of particles, and decreasing volume always favors the direction that produces the fewest particles. If the number of particles is the same on both sides of the reaction, then the equilibrium position is unaffected by a change in volume. Ladder Diagrams When we develop or evaluate an analytical method, we often need to understand how the chemistry that takes place affects our results. Suppose we wish to isolate Ag+ by precipitating it as AgCl. If we also need to control pH, then we must use a reagent that does not adversely affect the solubility of AgCl. It is a mistake to use NH3 to adjust the pH, for example, because it increases the solubility of AgCl (reaction 6.29). In this section we introduce the ladder diagram as a simple graphical tool for visualizing equilibrium chemistry. We will use ladder diagrams to determine what reactions occur when we combine several reagents, to estimate the approximate composition of a system at equilibrium, and to evaluate how a change to solution conditions might affect an analytical method. Ladder Diagrams for Acid–Base Equilibria Let’s use acetic acid, CH3COOH, to illustrate the process we will use to draw and to interpret an acid–base ladder diagram. Before we draw the diagram, however, let’s consider the equilibrium reaction in more detail. The equilibrium constant expression for acetic acid’s dissociation reaction Now, let’s replace -log[H3O+] with pH and rearrange the equation to obtain the result shown here. Equation 6.31 tells us a great deal about the relationship between pH and the relative amounts of acetic acid and acetate at equilibrium. If the concentrations of CH3COOH and CH3COO– are equal, then equation 6.31 reduces to If the concentration of CH3COO– is greater than that of CH3COOH, then the log term in equation 6.31 is positive and the pH is greater than 4.76. This is a reasonable result because we expect the concentration of the conjugate base, CH3COO–, to increase as the pH increases. Similar reasoning will convince you that the pH is less than 4.76 when the concentration of CH3COOH exceeds that of CH3COO–. Now we are ready to construct acetic acid’s ladder diagram (Figure 6.3). First, we draw a vertical arrow that represents the solution’s pH, with smaller (more acidic) pH levels at the bottom and larger (more basic) pH levels at the top. Second, we draw a horizontal line at a pH equal to acetic acid’s pKa value. This line, or step on the ladder, divides the pH axis into regions where either CH3COOH or CH3COO– is the predominate species. This completes the ladder diagram. Using the ladder diagram, it is easy to identify the predominate form of acetic acid at any pH. At a pH of 3.5, for example, acetic acid exists primarily as CH3COOH. If we add sufficient base to the solution such that the pH increases to 6.5, the predominate form of acetic acid is CH3COO–. Example  A ladder diagram is particularly useful for evaluating the reactivity between a weak acid and a weak base. Figure 6.5, for example, shows a single ladder diagram for acetic acid/acetate and for p-nitrophenol/p-nitrophenolate. An acid and a base can not co-exist if their respective areas of predominance do not overlap. If we mix together solutions of acetic acid and sodium p-nitrophenolate, the reaction occurs because the areas of predominance for acetic acid and p-nitrophenolate do not overlap. The solution’s final composition depends on which species is the limiting reagent. The following example shows how we can use the ladder diagram in Figure 6.5 to evaluate the result of mixing together solutions of acetic acid and p-nitrophenolate. Example If the areas of predominance for an acid and a base overlap, then we do not expect that much of a reaction will occur. For example, if we mix together solutions of CH3COO– and p-nitrophenol, we do not expect a significant change in the moles of either reagent. Furthermore, the pH of the mixture must be between 4.76 and 7.15, with the exact pH depending upon the relative amounts of CH3COO– and p-nitrophenol. We also can use an acid–base ladder diagram to evaluate the effect of pH on other equilibria. For example, the solubility of CaF2 is affected by pH because F– is a weak base. From Le Châtelier’s principle, we know that converting F– to HF will increase the solubility of CaF2. To minimize the solubility of CaF2 we need to maintain the solution’s pH so that F– is the predominate species. The ladder diagram for HF (Figure 6.6) shows us that maintaining a pH of more than 3.17 will minimize solubility losses. Ladder Diagrams for Complexation Equilibria We can apply the same principles for constructing and interpreting an acid– base ladder diagram to equilibria that involve metal–ligand complexes. For a complexation reaction we define the ladder diagram’s scale using the concentration of uncomplexed, or free ligand, pL. Using the formation of Cd(NH3)2+ as an example we can show that log K1 is the dividing line between the areas of predominance for Cd2+ and for Cd(NH3)2+. Thus, Cd2+ is the predominate species when pNH3 is greater than 2.55 (a concentration of NH3 smaller than 2.82 × 10–3 M) and for a pNH3 value less than 2.55, Cd(NH3)2+ is the predominate species. Figure 6.7 shows a complete metal–ligand ladder diagram for Cd2+ and NH3 that includes additional Cd–NH3 complexes. Example The metal–ligand ladder diagram in Figure 6.7 uses stepwise formation constants. We also can construct a ladder diagram using cumulative formation constants. For example, the first three stepwise formation constants for the reaction of Zn2+ with NH3 suggests that the formation of Zn(NH3) is more favorable than the formation of Zn(NH3) or Zn(NH3). For this reason, the equilibrium is best represented by the cumulative formation reaction shown here. To see how we incorporate this cumulative formation constant into a ladder diagram, we begin with the reaction’s equilibrium constant expression. Ladder Diagram for Oxidation/Reduction Equilibria We also can construct ladder diagrams to help us evaluate redox equilibria. Figure 6.10 shows a typical ladder diagram for two half-reactions in which the scale is the potential, E. The Nernst equation defines the areas of predominance. Using the Fe3+/Fe2+ half- reaction as an example, we write At a potential more positive than the standard state potential, the predominate species is Fe3+, whereas Fe2+ predominates at potentials more negative than E°. When coupled with the step for the Sn4+/Sn2+ half-reaction we see that Sn2+ is a useful reducing agent for Fe3+. If Sn2+ is in excess, the potential of the resulting solution is near +0.154 V. Because the steps on a redox ladder diagram are standard state potentials, a complication arises if solutes other than the oxidizing agent and reducing agent are present at non-standard state concentrations. For example, the potential for the half- reaction depends on the solution’s pH. To define areas of predominance in this case we begin with the Nernst equation Solving Equilibrium Problems 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(IO3)2 in deionized water, the solid dissolves until the concentrations of Pb2+ and IO3 - satisfy the solubility product for Pb(IO3)2. At equilibrium the solution is saturated with Pb(IO3)2, which means simply that no more solid can dissolve. How do we determine the equilibrium concentrations of Pb2+ and IO3- , and what is the molar solubility of Pb(IO3)2 in this saturated solution? We begin by writing the equilibrium reaction and the solubility product expression for Pb(IO3)2. As Pb(IO3)2 dissolves, two IO3 - ions form for each ion of Pb2+. If we assume that the change in the molar concentration of Pb2+ at equilibrium is x, then the change in the molar concentration of IO3 - is 2x. The following table helps us keep track of the initial concentrations, the change in concentrations, and the equilibrium concentrations of Pb2+ and IO3-. Because one mole of Pb(IO3)2 contains one mole of Pb2+, the molar solubility of Pb(IO3)2 is equal to the concentration of Pb2+, or 4.0 × 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. A More Complex Problem—The Common Ion Effect Calculating the solubility of Pb(IO3)2 in deionized water is a straightforward problem because the solid’s dissolution is the only source of Pb2+ and IO3-. But what if we add Pb(IO3)2 to a solution of 0.10 M Pb(NO3)2? Before we set-up and solve this problem algebraically, think about the system’s chemistry and decide whether the solubility of Pb(IO3)2 will increase, decrease, or remain the same. We begin by setting up a table to help us keep track of the concentrations of Pb2+ and IO3- as this system moves toward and reaches equilibrium. This is a more difficult equation to solve than that for the solubility of Pb(IO3)2 in deionized water, and its solution is not immediately obvious. We can find a rigorous solution to equation 6.34 using computational software packages and spreadsheets, some of which are described in Section 6.J. How might we solve equation 6.34 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 Pb2+ will decrease significantly the solubility of Pb(IO3)2. One reasonable assumption is that the initial concentration of Pb2+ is very close to its equilibrium concentration. If this assumption is correct, then the following approximation is reasonable Before we accept this answer, we must verify that our approximation is reasonable. The difference between the actual concentration of Pb2+, which is 0.10 + x M, and our assumption that the concentration of Pb2+ is 0.10 M is 7.9 × 10–7, or 7.9 × 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 Pb2+ and IO3 - are The molar solubility of Pb(IO3)2 is equal to the additional concentration of Pb2+ in solution, or 7.9 × 10–4 mol/L. As expected, we find that Pb(IO3)2 is less soluble in the presence of a solution that already contains one of its ions. This is known as the common ion effect. 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. Example A Systematic Approach to Solving Equilibrium Problems Calculating the solubility of Pb(IO3)2 in a solution of Pb(NO3)2 is more complicated than calculating its solubility in deionized water. The calculation, however, is still relatively easy to organize and the simplifying assumption 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 6.1, this approach involves four steps. 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 mass balance equation, 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, CH3COOH, and the conjugate weak base, CH3COO–, must equal acetic acid’s initial concentration, CCH3COOH. The second equation is a charge balance equation, which requires that the total positive charge from the cations equal the total negative charge from the anions. Mathematically, the charge balance equation is Example