More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions

1. Chapter Sixteen More Equilibria in Aqueous Solutions:Slightly Soluble Salts and Complex Ions

2. BaSO4(s) Ba2+(aq) + SO42–(aq) The Solubility Product Constant, Ksp • Solubility product constant, Ksp: the equilibrium constant expression for the dissolving of a slightly soluble solid. • Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble” – Chapter 4). • An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution: Ksp= [ Ba2+ ][ SO42–]

4. Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4, and (b) chromium(III) hydroxide, Cr(OH)3.

5. Example 16.1 Write a solubility product constant expression for equilibrium in a saturated aqueous solution of the slightly soluble salts (a) iron(III) phosphate, FePO4 and (b) chromium(III) hydroxide, Cr(OH)3. Strategy We first write chemical equations for the solubility equilibrium, basing each equation on one mole of solid on the left side. The coefficients on the right side are then the numbers of moles of cations and anions per mole of the compound. As with other equilibrium constant expressions, coefficients in the balanced equation appear as exponents in the Ksp expression. Solution Exercise 16.1A Exercise 16.1B Write a Ksp expression for equilibrium in a saturated aqueous solution of (a) MgF2, (b) Li2CO3, and (c) Cu3(AsO4)2. Write a Ksp expression for equilibrium in a saturated solution of (a) magnesium hydroxide (commonly known as Milk of magnesia), (b) scandium fluoride, ScF3 (used in the preparation of scandium metal), and (c) zinc phosphate (used in dental cements).

6. Ksp and Molar Solubility • Ksp is an equilibrium constant • Molar solubility is the number of moles of compound that will dissolve per liter of solution. • Molar solubility is related to the value of Ksp, but molar solubility and Ksp are not the same thing. • In fact, “smaller Ksp” doesn’t always mean “lower molar solubility.” • Solubility depends on both Ksp and the form of the equilibrium constant expression.

7. Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Kspfor Ag2CO3 at 20 °C. The balanced equation is Ag2CO3(s) 2 Ag+(aq) + CO32–(aq) Ksp= ? Example 16.3 From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C. Ag2SO4(s) 2 Ag+(aq) + SO42–(aq) Ksp= 1.4 x 10–5 at 25 °C

8. Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag2CO3 per liter of solution. Calculate Ksp for Ag2CO3 at 20 °C. The balanced equation is Strategy To evaluate Ksp, we need the molarities of the individual ions in the saturated solution, but we are given only the solubility of the solute. Thus, the heart of the calculation is to obtain ion molarities from a concentration in milligrams of solute per liter. Once we have these ion molarities, we can get a value for Ksp by substituting them into the solubility constant expression. Solution From the equation for the solubility equilibrium, we see that 2 mol Ag+ and 1 mol CO32– appear in solution for every 1 mol Ag2CO3 that dissolves. We can represent this fact by conversion factors (shown in blue).

9. Example 16.2 continued Now we can write the Ksp expression for Ag2CO3 and substitute in the equilibrium concentrations of the ions. Ksp = [Ag+]2[CO32–] = (2.3 x 10–4)2(1.2 x 10–4) = 6.3 x 10–12 Exercise 16.2A Exercise 16.2B In Example 15.4, we determined the molar solubility of magnesium hydroxide from the measured pH of its saturated solution. Use data from that example to determine Ksp for Mg(OH)2. A saturated aqueous solution of silver(I) chromate contains 14 ppm of Ag+ by mass. Determine Ksp for silver(I) chromate. Assume the solution has a density of 1.00 g/mL.

10. Example 16.3 From the Ksp value for silver sulfate, calculate its molar solubility at 25 °C. Strategy The equation shows that when 1 mol Ag2SO4(s) dissolves, 1 mol SO42– and 2 mol Ag+ appear in solution. If we let s represent the number of moles of Ag2SO4 that dissolve per liter of solution, the two ion concentrations are [SO42–] = s and [Ag+] = 2s We then substitute these concentrations into the Ksp expression and solve for s, which, because 1 mol of Ag2SO4 produces 1 mol of SO42–, is the molar solubility we are seeking. Solution First, we write the Ksp expression for Ag2SO4 based on the solubility equilibrium equation. Ksp = [Ag+]2[SO42–] = 1.4 x 10–5 Then, we substitute the values [Ag+] = 2s and [SO42–] = s to obtain the equation that must be solved.

11. Example 16.3 continued Exercise 16.3A Now, by recalling that s = [SO42–] and that [SO42–] is the same as the molar solubility of Ag2SO4, we obtain our final result. Calculate the molar solubility of silver arsenate, given that Exercise 16.3B Using the Ksp value in Table 16.1, determine the quantity of I–, in parts per million, in a saturated aqueous solution of PbI2. Assume the solution has a density of 1.00 g/mL. Assessment In substituting ion concentrations into the Ksp expression, it is important to note that (i) the factor 2 appears in 2s because the concentration of Ag+ is twice that of SO42–, and (ii) the exponent 2 appears in the term (2s)2 because the concentration of Ag+ (that is, 2s) must be raised to the second power in the Ksp expression. Be alert for similar situations that might arise in other problems.

12. Example 16.4 A Conceptual Example Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI.

13. Example 16.4 Without doing detailed calculations, but using data from Table 16.1, establish the order of increasing solubility of these silver halides in water: AgCl, AgBr, AgI. Analysis and Conclusions The key to solving this problem is to recognize that all three solutes are of the same type; that is, the ratio of number of cations to anions in each is 1:1. As suggested by the following equation for the solubility equilibrium of AgX, the ion concentrations are equal to each other and to the molar solubility, s. The order of increasing molar solubility is the same as the order of increasing Ksp values: Increasing molar solubilities: AgI < AgBr < AgCl Increasing Ksp: 8.5 x 10–17 < 5.0 x 10–13 < 1.8 x 10–10 This method works only if we are comparing solutes that are all of the same general formula, that is, all MX, all MX2, and so on.

14. The Common Ion Effectin Solubility Equilibria • The common ion effect affects solubility equilibria as it does other aqueous equilibria. • The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.

15. Common Ion Effect Illustrated The added sulfate ion reduces the solubility of Ag2SO4. Na2SO4(aq) Saturated Ag2SO4(aq) Ag2SO4 precipitates

16. Common Ion Effect Illustrated When Na2SO4(aq) is added to the saturated solution of Ag2SO4 … … [Ag+] attains a new, lower equilibrium concentration as Ag+ reacts with SO42–to produce Ag2SO4.

17. Example 16.5 Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq).

18. Example 16.5 Calculate the molar solubility of Ag2SO4 in 1.00 M Na2SO4(aq). Strategy The molar solubility of Ag2SO4 will be influenced by the concentration of SO42– in solution coming from the 1.00 M Na2SO4(aq). We will assume that the Na2SO4 is completely dissociated and that the presence of Ag2SO4 has no effect on this dissociation. As in Example 16.3, we will use s to represent the number of moles of Ag2SO4 dissolved per liter of saturated solution. We will then represent [Ag+] and [SO42–] in terms of s and solve the Ksp expressions for s. Solution We can tabulate the relevant data, including the already present 1.00 mol SO42–/L, in the ICE format introduced in Chapter 14.

19. Example 16.5 continued The equilibrium concentrations from the ICE table must satisfy the Ksp expression. Ksp = [Ag+]2[SO42–] 1.4 x 10–5 = (2s)2(1.00 + s) To simplify the equation, let us assume that s is much smaller than 1.00 M, so that (1.00 + s)  1.00. (2s)2(1.00) = 1.4 x 10–5 4s2 = 1.4 x 10–5 We obtain our final result by solving for s. s2 = 3.5 x 10–6 s = molar solubility = (3.5 x 10–6)1/2 = 1.9 x 10–3 mol Ag2SO4/L Assessment When we assumed that (1.00 + s)  1.00, we were saying that essentially all the common ion comes from the added source (1.00 M) and essentially none comes from the slightly soluble solute (s). The assumption is valid here: (1.00 + 1.9 x 10–3) = 1.00 (to two decimal places). This type of assumption will often be valid in calculations of this type. Notice that the solubility of Ag2SO4 found here is only about one-eighth of that found for Ag2SO4 in pure water in Example 16.3, illustrating, as we had intended, the common ion effect. Exercise 16.5A Exercise 16.5B Calculate the molar solubility of Ag2SO4 in 1.00 M AgNO3(aq). How many grams of silver nitrate must be added to 250.0 mL of a saturated solution of Ag2SO4(aq) to reduce the solubility of the Ag2SO4 to 1.0 x 10–3 M (Ksp of Ag2SO4 = 1.4 x 10–5)?

20. Solubility and Activities • Ions that are not common to the precipitate can also affect solubility. • CaF2 is more soluble in 0.010 M Na2SO4 than it is in water. • Increased solubility occurs because of interionic attractions. • Each Ca2+ and F– is surrounded by ions of opposite charge, which impede the reaction of Ca2+ with F–. • The effective concentrations, or activities, of Ca2+ and F– are lower than their actual concentrations.

21. Will Precipitation Occur? Is It Complete? • Qip can then be compared to Ksp. • Precipitation should occur if Qip > Ksp. • Precipitation cannot occur if Qip < Ksp. • A solution is just saturated if Qip = Ksp. • In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. • Qip is the ion product reaction quotient and is based on initial conditions of the reaction. Qip andQc: new look, same great taste!

22. Example 16.6 If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3, will a precipitate form? Ag2CrO4(s) 2 Ag+(aq) + CrO42–(aq) Ksp= 1.1 x 10–12

23. Example 16.6 If 1.00 mg of Na2CrO4 is added to 225 mL of 0.00015 M AgNO3(aq), will a precipitate form? Strategy We can apply the three-step strategy just outlined. Solution Step 1:Initial concentrations of ions: The initial concentration of Ag+ is simply 1.5 x 10–4 M. To establish CrO42–, we first need to determine the number of moles of CrO42– placed in solution. We then determine [CrO42–] in the 225 mL (0.225 L) of solution that contains 6.17 x 10–6 mol CrO42–. Step 2:Evaluation of Qip: Qip = [Ag+]initial2[CrO42–]initial = (1.5 x 10–4)2(2.74 x 10–5) = 6.2 x 10–13

24. Example 16.6 continued Step 3:Comparison of Qip andKsp: Qip = 6.2 x 10–13 is less than Ksp = 1.1 x 10–12 Because Qip < Ksp, we conclude that no precipitation occurs. Exercise 16.6A Exercise 16.6B If 1.00 g Pb(NO3)2 and 1.00 g MgI2 are both added to 1.50 L of H2O, should a precipitate form (Ksp for PbI2 = 7.1 x 10–9)? Will precipitation occur from a solution that has 2.5 ppm MgCl2 and a pH of 10.35? Assume the solution density is 1.00 g/mL [Ksp for Mg(OH)2 = 1.8 x 10–11].

25. Example 16.7 A Conceptual Example Pictured here is the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2. What is the solid that first appears? Explain why it then disappears. Example 16.8 If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form? MgF2(s) Mg2+(aq) + 2 F–(aq) Ksp= 3.7 x 10–8

26. Example 16.7 Figure 16.4 shows the result of adding a few drops of concentrated KI(aq) to a dilute solution of Pb(NO3)2(aq). What is the yellow solid that appears at first? Explain why it then disappears. Analysis and Conclusions Where the drops of concentrated KI(aq) first enter the Pb(NO3) solution, there is a localized excess of iodide ion. In this region of the solution, [I–] is large enough so that Qip = [Pb2+][I–]2 > Ksp Because Qip > Ksp, yellow PbI2(s) precipitates. When the PbI2(s) that initially formed settles below the point of entry of the KI(aq), however, [I–] is no longer large enough to maintain a saturated solution for the concentration of Pb2+ present, and therefore the solid redissolves. Based on a uniform distribution of I– ion throughout the solution, the following holds true: Qip = [Pb2+][I–]2 < Ksp Exercise 16.7A Exercise 16.7B Describe how you might use observations of the type suggested by the photographs of Figure 16.4 to estimate the value of Ksp for PbI2. Consider the following situation similar to Example 16.7: Initially, the 100.0 mL of dilute solution in the beaker is 1 x 10–4 M and the more concentrated solution added dropwise is 0.10 M. (a) If the dilute solution is Pb(NO3)2(aq) and the more concentrated one is KI(aq), will the observations made resemble Figure 16.4? Explain. (b) What should be observed if the dilute solution is KI(aq) and the more concentrated one is Pb(NO3)2? Explain.

27. Example 16.8 If 0.100 L of 0.0015 M MgCl2 and 0.200 L of 0.025 M NaF are mixed, should a precipitate of MgF2 form? Strategy We can apply the same three-step strategy as in Example 16.6, but now we will incorporate into Step 1 the assumption that the volume of the mixture of solutions is 0.100 L + 0.200 L = 0.300 L. Solution Step 1:Initial concentrations of ions: First, we need to determine [Mg2+] and [F–] as they initially exist in this mixed solution.

28. Example 16.8 continued Step 2:Evaluation of Qip: Qip = [Mg2+][F–]2 = (5.0 x 10–4)(1.7 x 10–2)2 = 1.4 x 10–7 Step 3:Comparison of Qip and Ksp: Because Qip > Ksp, we conclude that precipitation should occur. Qip = 1.4 x 10–7 is greater than Ksp = 3.7 x 10–8. Exercise 16.8A Exercise 16.8B Should a precipitate of MgF2(s) form when equal volumes of 0.0010 M MgCl2(aq) and 0.020 M NaF(aq) are mixed? How many grams of KI(s) should be dissolved in 10.5 L of 0.00200 M Pb(NO3)2(aq) so that precipitation of PbI2(s) will just begin?

29. To Determine Whether Precipitation Is Complete • A slightly soluble solid does not precipitate totally from solution … • … but we generally consider precipitation to be “complete” if about 99.9% of the target ion is precipitated (0.1% or less left in solution). • Three conditions generally favor completeness of precipitation: • A very small value of Ksp. • A high initial concentration of the target ion. • A concentration of common ion that greatly exceeds that of the target ion.

30. Example 16.9 To a solution with [Ca2+] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), to make the initial [C2O42–] = 0.0051 M. Will precipitation of Ca2+ as CaC2O4(s) be complete? CaC2O4(s) Ca2+(aq) + C2O42–(aq) Ksp= 2.7 x 10–9

31. Example 16.9 To a solution with [Ca2+] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), to make the initial [C2O42–] = 0.0051 M. Will the precipitation of Ca2+as CaC2O4(s) be complete? Strategy We can begin with the three-step approach to determine whether precipitation occurs. If no precipitation occurs, we need not proceed further—precipitation obviously will not be complete. If precipitation does occur, we will need to make further calculations, which we will do in two parts: First, we will treat the situation as if complete precipitation occurred. This will enable us to determine the excess [C2O42–] present. Then, we will solve for the solubility of a slightly soluble solute in the presence of a common ion in a calculation similar to that of Example 16.5. Solution First, we use our usual three-step approach to determine if precipitation occurs. Step 1:Initial concentrations of ions: These are given. [Ca2+] = 0.0050 M, and [C2O42–] = 0.0051 M. Step 2:Evaluation of Qip: Qip = [Ca2+][C2O42–] = (5.0 x 10–3)(5.1 x 10–3) = 2.6 x 10–5

32. Example 16.9 continued Step 3:Comparison of Qip and Ksp: Qip = 2.6 x 10–5 exceeds Ksp = 2.7 x 10–9 Because Qip > Ksp, precipitation should occur. Next, we follow the two-part approach outlined in our basic strategy. Assume complete precipitation. We describe the stoichiometry of the precipitation in a format resembling the ICE format for equilibrium calculations (see Example 15.17b). The reaction: Ca2+(aq) + C2O42–(aq)  CaC2O4(s) Initial concentrations, M: 0.0050 0.0051 Changes, M: –0.0050 –0.0050 After precipitation, M: 0 0.0001 The solubility equilibrium. Here we enter the appropriate data into the ICE format for solubility equilibrium in the presence of a common ion.

33. Example 16.9 continued As usual, we assume that s << 0.0001 and that (0.0001 + s)  0.0001. (We test the validity of this assumption in the Assessment.) Ksp = [Ca2+][C2O42–] = s(0.0001 + s)  s x 0.0001 = 2.7 x 10–9 s = [Ca2+] = 3 x 10–5 M We calculate the percentage of Ca2+ remaining in solution as the ratio of the remaining Ca2+ concentration to the initial Ca2+ concentration (5 x 10–3 M). Our rule for complete precipitation requires that less than 0.1% of an ion should remain in solution. We therefore conclude that precipitation is incomplete. Assessment In this problem, we made the assumption that the contribution of s to the term 0.0001 + s was negligible. Such assumptions are usually valid in common ion situations because the concentration of the common ion is generally relatively large, but 0.0001 M is not a high concentration. If we had not made the simplifying assumption and solved a quadratic equation, we would have obtained s = 2 x 10–5 M. From this result, the percent Ca2+ remaining in solution is 0.4%, but our conclusion that precipitation is incomplete is still valid. If our task had been to calculate [Ca2+] remaining in solution, we would not have made the simplifying assumption.

34. Example 16.9 continued Exercise 16.9A Consider a solution in which [Ca2+] = 0.0050 M. If we add sufficient solid ammonium oxalate, (NH4)2C2O4(s), so that the solution is also [C2O42–] = 0.0100 M, will the precipitation of Ca2+ as CaC2O4(s) be complete? Exercise 16.9B The two members of each pair of ions are brought together in solution. Without doing detailed calculations, arrange the pairs according to how close to completeness the cation precipitation is—farthest from completeness first and closest to completeness last. (a) [Ca2+] = 0.110 M and [SO42–] = 0.090 M (b) [Pb2+] = 0.12 M and [Cl–] = 0.25 M (c) [Mg2+] = 0.12 M and [SO42–] = 0.15 M (d) [Ag+] = 0.050 M and [Cl–] = 0.055 M

35. Selective Precipitation AgNO3 added to a mixture containing Cl– and I–

36. Example 16.10 An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is 0.0100 M in Cl– and also 0.0100 M in I–. • Which ion, Cl– or I–, is the first to precipitate from solution? • When the second ion begins to precipitate, what is the remaining concentration of the first ion? • Is separation of the two ions by selective precipitation feasible? AgCl(s) Ag+(aq) + Cl–(aq) Ksp= 1.8 x 10–10 AgI(s) Ag+(aq) + I–(aq) Ksp= 8.5 x 10–17

37. Example 16.10 An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is 0.0100 M in Cl– and 0.0100 M in I– (Figure 16.5). (a) Which ion, Cl– or I–, is the first to precipitate from solution? (b) When the second ion begins to precipitate, what is the remaining concentration of the first ion? (c) Is separation of the two ions by selective precipitation feasible?

38. Example 16.10 continued Strategy The molarity of the silver nitrate solution is 200 times greater than [Cl–] or [I–], which means that only a small volume of AgNO3(aq) is required in the precipitation. Because of this, we can neglect throughout the calculation the slight dilution that occurs as the AgNO3(aq) is added to the beaker. Solution (a) We must find the [Ag+] necessary to precipitate each anion. Precipitation will begin when the solution has just been saturated, a point at which Qip = Ksp. To precipitate Cl–. When Qip is equal to Ksp for silver chloride, [Cl–] = 0.0100 M. We then solve the Ksp expression for [Ag+]. To precipitate I–. When Qip is equal to Ksp for silver chloride, [I–] = 0.0100 M. We then solve the Ksp expression for [Ag+].

39. Example 16.10 continued Precipitation of AgI(s) begins as soon as [Ag+] = 1.8 x 10–15 M, a concentration much too low for AgCl(s) to precipitate. Therefore I– is the first ion to precipitate. (b) Next, we determine [I–] remaining in solution when [Ag+] = 1.8 x 10–8 M, the [Ag+] at which AgCl(s) begins to precipitate. To determine this quantity, we rearrange the Ksp expression for silver iodide and solve for [I–]. Assessment This example illustrates a general rule: The greater the difference in magnitude of the relevant Ksp values, the more likely it is that ion separations by selective precipitation can be achieved. The Ksp values of AgI and AgCl differ by a factor of about 2 x 106. (c) Now, let us see what percent of the I– is still in solution when AgCl(s) begins to precipitate. The amount of I– remaining is far below the 0.1% we require for completeness of precipitation. Thus, complete precipitation of I– occurs before Cl– precipitation begins. Separation of the two anions is feasible.

40. CaF2(s) Ca2+(aq) + 2 F–(aq) AgCl(s) Ag+(aq) + Cl–(aq) Effect of pH on Solubility • If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered: Added H+ reacts with, and removes, F–; LeChâtelier’s principle says more F– forms. • If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate. H+ does not consume Cl– ; acid does not affect the equilibrium.

41. Example 16.11 What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x 10–5 M, that is, pH = 9.00? Mg(OH)2(s) Mg2+(aq) + 2 OH–(aq) Ksp= 1.8 x 10–11 Example 16.12 A Conceptual Example Without doing detailed calculations, determine in which of the following solutions Mg(OH)2(s) is most soluble: (a) 1.00 M NH3 (b) 1.00 M NH3 /1.00 M NH4+ (c) 1.00 M NH4Cl.

42. Example 16.11 What is the molar solubility of Mg(OH)2(s) in a buffer solution having [OH–] = 1.0 x10–5 M, that is, pH = 9.00? Strategy First, note that because the solution is a buffer, any OH– coming from the dissolution of Mg(OH)2(s) is neutralized by the acidic component of the buffer. Therefore, the pH will remain at 9.00 and the [OH–] at 1.0 x 10–5 M. We will use the Ksp expression and [OH–] = 1.0 x 10–5 M to calculate [Mg2+] in the solution. Strategy For every mole of Mg2+(aq) appearing in solution, 1 mol of Mg(OH)2(s) must have dissolved. Thus the molar solubility of Mg(OH)2 and the equilibrium concentration of Mg2+ are the same, meaning that the molar solubility of Mg(OH)2 at pH 9.00 is 0.18 mol Mg(OH)2/L.

43. Equilibria Involving Complex Ions Silver chloride becomes more soluble, not less soluble, in high concentrations of chloride ion.

44. Ag+(aq) + 2 Cl–(aq) [AgCl2]–(aq) Complex Ion Formation • A complex ion consists of a central metal atom or ion, with other groups called ligands bonded to it. • The metal ion acts as a Lewis acid (accepts electron pairs). • Ligands act as Lewis bases (donate electron pairs). • The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a formation constant, Kf: [AgCl2]– Kf = –––––––––– = 1.2 x 108 [Ag+][Cl–]2

45. Complex Ion Formation Concentrated NH3 added to a solution of pale-blue Cu2+ … … forms deep-blue Cu(NH3)42+.

47. Complex Ion Formationand Solubilities But if the concentration of NH3 is made high enough … … the AgCl forms the soluble [Ag(NH3)2]+ ion. AgCl is insoluble in water.

48. Ag+(aq) + 2 NH3(aq) [Ag(NH3)2]+(aq) Kf= 1.6 x 107 AgBr(s) Ag+(aq) + Br–(aq) Example 16.13 Calculate the concentration of free silver ion, [Ag+], in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3. Example 16.14 If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution? Ksp= 5.0 x 10–13

49. Example 16.13 Calculate the concentration of free silver ion, [Ag+], in an aqueous solution prepared as 0.10 M AgNO3 and 3.0 M NH3. Strategy First, because Kf is such a large number, we can assume that the formation of [Ag(NH3)2]+ goes to completion. We can then establish the concentrations of [Ag(NH3)2]+ and free NH3 through a straightforward stoichiometric calculation. Next, we can return to the formation reaction for the complex ion and consider the extent to which it proceeds in the reverse direction. For this calculation, we will use the Kf expression and solve for x = [Ag+]. Solution We complete the bottom row of our ICE tabulation to establish the concentrations of complex ion and free ammonia, assuming that the formation reaction goes to completion.

50. Example 16.13 continued Now, we restate the question in this way: What is [Ag+] in a solution that is 0.10 M [Ag(NH3)2]+ and 2.8 M NH3? To answer this question, we need to focus on the reverse direction in the reaction for [Ag(NH3)2]+ formation. When we substitute the equilibrium concentrations into the Kf expression, we obtain an algebraic equation that we must solve for x = [Ag+]. Because the formation constant is so large, practically all the silver will be present as [Ag(NH3)2]+ and very little as Ag+. We can therefore assume that x is very much smaller than 0.10 and that 2x is very much smaller than 2.8.