CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA. Questions

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Questions 8. MX(s) M n+ (aq) + X n (aq) K sp = [M n+ [X n ; the K sp reaction always refers to a solid breaking up into its ions. The representations all show 1 : 1 salts, i.e., the formula of the solid contains 1 cation for every 1 anion (either +1 and 1, or + and, or + and ). The solution with the largest number of ions (largest [M n+ and [X n ) will have the largest K sp value. From the representations, the second beaker has the largest number of ions present, so this salt has the largest K sp value. Conversely, the third beaker, with the fewest number of hydrated ions, will have the smallest K sp value. 9. K sp values can only be directly compared to determine relative solubilities when the salts produce the same number of ions (have the same stoichiometry). Here, Ag S and CuS do not produce the same number of ions when they dissolve, so each has a different mathematical relationship between the K sp value and the molar solubility. To determine which salt has the larger molar solubility, you must do the actual calculations and compare the two molar solubility values. 10. The solubility product constant (K sp ) is an equilibrium constant that has only one value for a given solid at a given temperature. Solubility, on the other hand, can have many values for a given solid at a given temperature. In pure water, the solubility is some value, yet the solubility is another value if a common ion is present. And the actual solubility when a common ion is present varies according to the concentration of the common ion. However, in all cases the product of the ion concentrations must satisfy the K sp expression and give that one unique K sp value at that particular temperature. 11. i. This is the result when you have a salt that breaks up into two ions. Examples of these salts include AgCl, SrSO, BaCrO, and ZnCO. ii. This is the result when you have a salt that breaks up into three ions, either two cations and one anion or one cation and two anions. Some examples are SrF, Hg I, and Ag SO. iii. This is the result when you have a salt that breaks up into four ions, either three cations and one anion (Ag PO ) or one cation and three anions (ignoring the hydroxides, there are no examples of this type of salt in Table 16.1). iv. This is the result when you have a salt that breaks up into five ions, either three cations and two anions [Sr (PO ) or two cations and three anions (no examples of this type of salt are in Table 16.1). 61

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 1. The obvious choice is that the metal ion reacts with PO and forms an insoluble phosphate salt. The other possibility is due to the weak base properties of PO (PO is the conjugate base of the weak acid HPO, so it is a weak base). Because PO is a weak base in water, OH ions are present at a fairly large concentration. Hence the other potential precipitate is the metal ion reacting with OH to form an insoluble hydroxide salt. 1. For the K sp reaction of a salt dissolving into its respective ions, a common ion would be one of the ions in the salt added from an outside source. When a common ion (a product in the K sp reaction) is present, the K sp equilibrium shifts to the left, resulting in less of the salt dissolving into its ions (solubility decreases). 1. S is a very basic anion and reacts significantly with H + to form HS (S + H + HS ). Thus, the actual concentration of S in solution depends on the amount of H + present. In basic solutions, little H + is present, which shifts the above reaction to the left. In basic solutions, the S concentration is relatively high. So, in basic solutions, a wider range of sulfide salts will precipitate. However, in acidic solutions, added H + shifts the equilibrium to the right resulting in a lower S concentration. In acidic solutions, only the least soluble sulfide salts will precipitate out of solution. 15. Some people would automatically think that an increase in temperature would increase the solubility of a salt. This is not always the case as some salts show a decrease in solubility as temperature increases. The two major methods used to increase solubility of a salt both involve removing one of the ions in the salt by reaction. If the salt has an ion with basic properties, adding H + will increase the solubility of the salt because the added H + will react with the basic ion, thus removing it from solution. More salt dissolves in order to to make up for the lost ion. Some examples of salts with basic ions are AgF, CaCO, and Al(OH). The other way to remove an ion is to form a complex ion. For example, the Ag + ion in silver salts forms the complex ion Ag(NH ) + as ammonia is added. Silver salts increase their solubility as NH is added because the Ag + ion is removed through complex ion formation. 16. Because the formation constants are usually very large numbers, the stepwise reactions can be assumed to essentially go to completion. Thus an equilibrium mixture of a metal ion and a specific ligand will mostly contain the final complex ion in the stepwise formation reactions. 1. In.0 M NH, the soluble complex ion Ag(NH ) + forms, which increases the solubility of AgCl(s). The reaction is AgCl(s) + NH Ag(NH ) + + Cl. In.0 M NH NO, NH is only formed by the dissociation of the weak acid NH +. There is not enough NH produced by this reaction to dissolve AgCl(s) by the formation of the complex ion. 18. Unlike AgCl(s), PbCl (s) shows a significant increase in solubility with an increase in temperature. Hence add NaCl to the solution containing the metal ion to form the chloride salt precipitate, and then heat the solution. If the precipitate dissolves, then PbCl is present, and the metal ion is Pb +. If the precipitate does not dissolve with an increase in temperature, then AgCl is the precipitate, and Ag + is the metal ion present.

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 Exercises Solubility Equilibria 19. a. AgC H O (s) Ag + (aq) + C H O (aq) K sp = [Ag + [C H O b. Al(OH) (s) Al + (aq) + OH (aq) K sp = [Al + [OH c. Ca (PO ) (s) Ca + (aq) + PO (aq) K sp = [Ca + [PO 0. a. Ag CO (s) Ag + (aq) + CO (aq) K sp = [Ag + [CO b. Ce(IO ) (s) Ce + (aq) + IO (aq) K sp = [Ce + [IO c. BaF (s) Ba + (aq) + F - (aq) K sp = [Ba + [F 1. In our setup, s = solubility of the ionic solid in mol/l. This is defined as the maximum amount of a salt that can dissolve. Because solids do not appear in the K sp expression, we do not need to worry about their initial and equilibrium amounts. a. CaC O (s) Ca + (aq) + C O (aq) Initial 0 0 s mol/l of CaC O (s) dissolves to reach equilibrium Change s +s +s Equil. s s From the problem, s =.8 5 10 mol/l. K sp = [Ca + [C O = (s)(s) = s 5, K sp = (.8 10 ) =. b. BiI (s) Bi + (aq) + I - (aq) Initial 0 0 s mol/l of BiI (s) dissolves to reach equilibrium Change s +s +s Equil. s s 10 9 K sp = [Bi + [I = (s)(s) = s 5, K sp = (1. 10 ) = 8.0. a. Pb (PO ) (s) Pb + (aq) + PO (aq) 10 19 Initial 0 0 s mol/l of Pb (PO ) (s) dissolves to reach equilibrium = molar solubility Change s +s +s Equil. s s K sp = [Pb + [PO = (s) (s) = 108 s 5, K sp = 108(6. 10 1 ) 5 = 9.9 10 55

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA b. Li CO (s) Li + (aq) + CO - (aq) Equil. s s K sp = [Li + [CO = (s) (s) = s, K sp = (. 10 ) = 1.6 10 0.1g Ni(OH) 1mol Ni(OH). Solubility = s = 1.5 10 mol/ L L 9.1g Ni(OH) Ni(OH) (s) Ni + (aq) + OH (aq) Initial 0 1.0 10 M (from water) s mol/l of Ni(OH) dissolves to reach equailibrium Change s +s +s Equil. s 1.0 10 + s From the calculated molar solubility, 1.0 10 + s s. K sp = [Ni + [OH = s(s) = s, K sp = (1.5 10 ) = 1. 10 8. M X (s) M + (aq) + X (aq) K sp = [M + [X s mol/l of M X (s) dissolves to reach equilibrium Change s +s +s Equil. s s K sp = (s) (s) = 108s 5 ; s =.60 10 L g 1molM X 88g = 1.5 9 10 mol/l K sp = 108(1.5 9 10 ) 5 =.0 10 5. PbBr (s) Pb + (aq) + Br - (aq) Initial 0 0 s mol/l of PbBr (s) dissolves to reach equilibrium Change s +s +s Equil. s s From the problem, s = [Pb + =.1 10 M. So: K sp = [Pb + [Br = s(s) = s, K sp = (.1 10 ) =.9 10 5 6. Ag C O (s) Ag + (aq) + C O (aq) Equil. s s

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 65 From problem, [Ag + = s =. 10 M, s = 1.1 10 M K sp = [Ag + [C O = (s) (s) = s = (1.1 10 ) = 5. 10 1. In our setup, s = solubility in mol/l. Because solids do not appear in the K sp expression, we do not need to worry about their initial or equilibrium amounts. a. Ag PO (s) Ag + (aq) + PO (aq) Initial 0 0 s mol/l of Ag PO (s) dissolves to reach equilibrium Change s +s +s Equil. s s K sp = 1.8 18 10 = [Ag + [PO = (s) (s) = s s = 1.8 10 18, s = (6. 0 10 ) 1/ = 1.6 5 10 mol/l = molar solubility b. CaCO (s) Ca + (aq) + CO (aq) Equil. s s K sp = 8. 9 10 = [Ca + [CO = s 5, s = 9. 10 mol/l c. Hg Cl (s) Hg + (aq) + Cl (aq) Equil. s s K sp = 1.1 18 10 = [Hg + [Cl = (s)(s) = s, s = 6.5 10 mol/l 8. a. PbI (s) Pb + (aq) + I (aq) Equil. s s K sp = 1. s = (1. 8 10 = [Pb + [I = s(s) = s 8 10 /) 1/ = 1.5 10 mol/l = molar solubility b. CdCO (s) Cd + (aq) + CO (aq) Equil. s s K sp = 5. 1 10 = [Cd + [CO - = s 6, s =. 10 mol/l

66 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA c. Sr (PO ) (s) Sr + (aq) + PO - (aq) Equil. s s K sp = 1 1 10 = [Sr + [PO - = (s) (s) = 108s 5, s = 10 mol/l 9. KBT dissolves to form the potassium ion (K + ) and the bitartrate ion (abbreviated as BT ). KBT(s) K + (aq) + BT (aq) K sp =.8 Equil. s s.8 10 = [K + [ BT = s(s) = s, s = 1.9 10 mol/l 10 1.9 10 mol KBT 0.500L L 188. g KBT mol = 0.89 g KBT 0. BaSO (s) Ba + (aq) + SO (aq) K sp = 1.5 Equil. s s 1.5 9 10 = [Ba + [SO = s 5, s =.9 10 mol/l 10 9 5 0.9 10.1000L mol BaSO L. g BaSO mol = 9.1 10 g BaSO 1. Mg(OH) (s) Mg + (aq) + OH (aq) Initial s = solubility (mol/l) 0 1.0 10 M (from water) Equil. s 1.0 10 + s K sp = [Mg + [OH = s(1.0 10 + s) ; assume that 1.0 10 + s s, then: K sp = 8.9 10 1 = s(s) = s, s = 1. 10 mol/l Assumption is good (1.0 10 is 0.0% of s). Molar solubility = 1. 10 mol/l. Cd(OH) (s) Cd + (aq) + OH (aq) K sp = 5.9 Initial s = solubility (mol/l) 0 1.0 10 M Equil. s 1.0 10 + s K sp = [Cd + [OH = s(1.0 10 + s) ; assume that 1.0 10 + s s, then: 10 15

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 K sp = 5.9 Assumption is good (1.0 15 10 = s(s) = s 5, s = 1.1 10 mol/l 10 5 is 0.% of s). Molar solubility = 1.1 10 mol/l. Let s = solubility of Al(OH) in mol/l. Note: Because solids do not appear in the K sp expression, we do not need to worry about their initial or equilibrium amounts. Al(OH) (s) Al + (aq) + OH (aq) Initial 0 1.0 10 - M (from water) s mol/l of Al(OH) (s) dissolves to reach equilibrium = molar solubility Change s +s +s Equil. s 1.0 10 + s K sp = 10 = [Al + [OH = (s)(1.0 10 + s) s(1.0 10 ) s = 10 1.0 10 1 = 10 11 mol/l; assumption good (1.0 10 + s 1.0 10 ).. Let s = solubility of Co(OH) in mol/l. Co(OH) (s) Co + (aq) + OH (aq) Initial 0 1.0 10 M (from water) s mol/l of Co(OH) (s) dissolves to reach equilibrium = molar solubility Change s +s +s Equil. s 1.0 10 + s K sp =.5 10 = [Co + [OH = (s)(1.0.5 10 s = =.5 1 1.0 10 10 + s) s(1.0 10 mol/l; assumption good (1.0 10 ) 10 + s 1.0 10 ). 5. a. Because both solids dissolve to produce three ions in solution, we can compare values of K sp to determine relative solubility. Because the K sp for CaF is the smallest, CaF (s) has the smallest molar solubility. b. We must calculate molar solubilities because each salt yields a different number of ions when it dissolves. Ca (PO ) (s) Ca + (aq) + PO (aq) K sp = 1. 10 Equil. s s K sp = [Ca + [PO = (s) (s) = 108s 5, s = (1. 10 /108) 1/5 = 1.6 10 mol/l FePO (s) Fe + (aq) + PO (aq) K sp = 1.0 10 Equil. s s

68 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA K sp = [Fe + [PO = s, s = 1.0 10 = 1.0 10 11 mol/l FePO has the smallest molar solubility. 6. a. FeC O (s) Fe + (aq) + C O (aq) Equil. s = solubility (mol/l) s s K sp =.1 10 = [Fe + [C O = s, s =.6 10 mol/l Cu(IO ) (s) Cu + (aq) + IO (aq) Equil. s s K sp = 1. 10 = [Cu + [IO = s(s) = s, s = (1. 10 /) 1/ =. 10 mol/l By comparing calculated molar solubilities, FeC O (s) is less soluble (in mol/l). b. Each salt produces three ions in solution, so we can compare K sp values to determine relative molar solubilities. Therefore, Mn(OH) (s) will be less soluble (in mol/l) because it has a smaller K sp value.. a. Fe(OH) (s) Fe + (aq) + OH (aq) Initial 0 1 10 M (from water) s mol/l of Fe(OH) (s) dissolves to reach equilibrium = molar solubility Change s +s +s Equil. s 1 10 + s K sp = s = 8 10 = [Fe + [OH = (s)(1 10 + s) s(1 10 ) 1 10 mol/l; assumption good (s << 1 10 ) b. Fe(OH) (s) Fe + (aq) + OH (aq) ph = 5.0, [OH = 1 9 10 M Initial 0 1 10-9 M (buffered) s mol/l dissolves to reach equilibrium Change s +s (assume no ph change in buffer) Equil. s 1 10 9 8 K sp = 10 = [Fe + [OH 9 = (s)(1 10 ) 11, s = 10 mol/l = molar solubility c. Fe(OH) (s) Fe + (aq) + OH - (aq) ph = 11.0, [OH = 1 Initial 0 0.001 M (buffered) s mol/l dissolves to reach equilibrium Change s +s (assume no ph change) Equil. s 0.001 10 M

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 69 K sp = 8 10 = [Fe + [OH = (s)(0.001) 9, s = 10 mol/l = molar solubility Note: As [OH increases, solubility decreases. This is the common ion effect. 8. Co(OH) (s) Co + (aq) + OH (aq) ph = 11.00, [OH = 1.0 10 M Initial s = solubility (mol/l) 0 1.0 10 (buffered) Equil. s 1.0 10 (assume no ph change) K sp =.5 10 16 = [Co + [OH = s(1.0 10 ), s =.5 10 10 mol/l 9. a. Ag SO (s) Ag + (aq) + SO - (aq) Equil. s s K sp = 1. 5 10 = [Ag + [SO = (s) s = s, s = 1. 10 mol/l b. Ag SO (s) Ag + (aq) + SO (aq) Initial s = solubility (mol/l) 0.10 M 0 Equil. 0.10 + s s K sp = 1. 5 10 = (0.10 + s) (s) (0.10) (s), s = 1. 10 mol/l; assumption good. c. Ag SO (s) Ag + (aq) + SO (aq).0 M Equil s 0.0 + s 1. 5 10 = (s) (0.0 + s) s (0.0), s =.9 10 mol/l; assumption good. Note: Comparing the solubilities of parts b and c to that of part a illustrates that the solubility of a salt decreases when a common ion is present. 0. a. PbI (s) Pb + (aq) + I (aq) Equil. s s K sp = 1. 10 8 = [Pb + [I = s, s = 1.5 10 mol/l b. PbI (s) Pb + (aq) + I (aq) Initial s = solubility (mol/l) 0.10 M 0 Equil. 0.10 + s s 1. 10 8 = (0.10 + s)(s) (0.10)(s) = (0.0)s, s = 1.9 10 mol/l; assumption good.

60 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA c. PbI (s) Pb + (aq) + I (aq).010 M Equil. s 0.010 + s 1. 10 8 = (s)(0.010 + s) (s)(0.010), s = 1. 10 mol/l; assumption good. Note that in parts b and c, the presence of a common ion decreases the solubility as compared to the solubility of PbI (s) in water. 1. Ca (PO ) (s) Ca + (aq) + PO - (aq) Initial 0 0.0 M s mol/l of Ca (PO ) (s) dissolves to reach equilibrium Change s +s +s Equil. s 0.0 + s K sp = 1. 10 = [Ca + [PO = (s) (0.0 + s) Assuming 0.0 + s 0.0: 1. s = molar solubility =. 10 = (s) (0.0) = s (0.00) 11 10 mol/l; assumption good.. Pb (PO ) (s) Pb + (aq) + PO (aq) K sp = 1 10 5 Initial s = solubility (mol/l) 0.10 M 0 Equil. 0.10 + s s 1 10 5 = (0.10 + s) (s) (0.10) (s), s = 10 6 mol/l; assumption good.. Ce(IO ) (s) Ce + (aq) + IO (aq).0 M Equil. s 0.0 + s K sp = [Ce + [IO = s(0.0 + s) From the problem, s =. K sp = (. 8 10 mol/l; solving for K sp : 8 10 )[0.0 + (. 10 8 ) =.5 10 10. Pb(IO ) (s) Pb + (aq) + IO (aq).10 M Equil. s 0.10 + s K sp = [Pb + [IO = (s)(0.10 + s) From the problem, s =.6 10 11 mol/l; solving for K sp : K sp = (.6 10 11 )[0.10 + (.6 10 11 ) =.6 10 1

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 61 5. If the anion in the salt can act as a base in water, then the solubility of the salt will increase as the solution becomes more acidic. Added H + will react with the base, forming the conjugate acid. As the basic anion is removed, more of the salt will dissolve to replenish the basic anion. The salts with basic anions are Ag PO, CaCO, CdCO and Sr (PO ). Hg Cl and PbI do not have any ph dependence because Cl and I are terrible bases (the conjugate bases of a strong acids). Ag PO (s) + H + (aq) Ag + (aq) + HPO (aq) excess H + Ag + (aq) + H PO (aq) CaCO (s) + H + Ca + + HCO CdCO (s) + H + Cd + + HCO excess H + excess H + Ca + + H CO [H O(l) + CO (g) Cd + + H CO [H O(l) + CO (g) Sr (PO ) (s) + H + Sr + + HPO excess H + Sr + + H PO 6. a. AgF b. Pb(OH) c. Sr(NO ) d. Ni(CN) All these salts have anions that are bases. The anions of the other choices are conjugate bases of strong acids. They have no basic properties in water and, therefore, do not have solubilities that depend on ph. Precipitation Conditions. ZnS(s) Zn + (aq) + S (aq) K sp = [Zn + [S Initial s = solubility (mol/l) 0.050 M 0 Equil. 0.050 + s s K sp =.5 10 1 = (0.050 + s)(s) (0.050)s, s = 5.0 10 mol/l; assumption good. Mass ZnS that dissolves = 0.000 L 1 5.0 10 mol ZnS L 9.5g ZnS mol = 1.5 19 10 g 8. For 99% of the Mg + to be removed, we need, at equilibrium, [Mg + = 0.01(0.05 M). Using the K sp equilibrium constant, calculate the [OH required to reach this reduced [Mg +. Mg(OH) (s) Mg + (aq) + OH (aq) K sp = 8.9 8.9 10 1 1 10 = [Mg + [OH = [0.01(0.05 M) [OH, [OH = 1. 10 M (extra sig. fig.) poh = log(1. 10 ) =.89; ph = 10.11; at a ph = 10.1, 99% of the Mg + in seawater will be removed as Mg(OH) (s). 9. The formation of Mg(OH) (s) is the only possible precipitate. Mg(OH) (s) will form if Q > K sp.

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Mg(OH) (s) Mg + (aq) + OH (aq) K sp = [Mg + [OH = 8.9 10 1 [Mg + 0 = [OH 0 = 100.0 ml.0 10 mmolmg 100.0 ml 100.0 ml /ml 100.0 ml.0 10 mmoloh /ml 00.0 ml =.0 = 1.0 10 M 10 M Q = [Mg + 0 [OH 0 = (.0 10 M)(1.0 10 ) =.0 10 1 Because Q < K sp, Mg(OH) (s) will not precipitate, so no precipitate forms. 50. AgCN(s) Ag + (aq) + CN (aq) K sp =. 10 1 Q = [Ag + 0 [CN 0 = (1.0 10 5 )(.0 10 6 ) =.0 10 11 Because Q > K sp, AgCN(s) will form as a precipitate. 51. PbF (s) Pb + (aq) + F (aq) K sp = 10 8 [Pb + 0 = mmolpb (aq) totalmlsolution 1.0 10 mmolpb 100.0 ml ml 100.0 ml 100.0 ml 5.0 10 M 1.0 10 mmolf 100.0 ml [F mmolf 0 = ml 5.0 10 M totalmlsolution 00.0 ml Q = [Pb [F 0 0 = (5.0 10 )(5.0 10 ) = 1. 10 9 Because Q < K sp, PbF (s) will not form as a precipitate. 5. Ce(IO ) (s) Ce + (aq) + IO (aq) K sp =. 10 10 Q = [Ce + 0 [IO 0 = (.0 10 )(1.0 10 ) =.0 10 9 Because Q > K sp, Ce(IO ) (s) will form as a precipitate. 5. The concentrations of ions are large, so Q will be greater than K sp, and BaC O (s) will form. To solve this problem, we will assume that the precipitation reaction goes to completion; then we will solve an equilibrium problem to get the actual ion concentrations. This makes the math reasonable. 100. ml 150. ml 0.00mmolK CO = 0.0 mmol K C O ml 0.50mmolBaBr =.5 mmol BaBr ml

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 Ba + (aq) + C O (aq) BaC O (s) K = 1/K sp >> 1 Before.5 mmol 0.0 mmol 0 Change 0.0 0.0 +0.0 Reacts completely (K is large). After 1.5 0 0.0 New initial concentrations (after complete precipitation) are: [Ba + = [K + = 1.5 mmol =.00 10 M; [C O = 0 M 50. ml (0.0 mmol) 50. ml = 0.160 M; [Br = (.5 mmol) 50. ml = 0.00 M For K + and Br -, these are also the final concentrations. We can t have 0 M C O. For Ba + and C O, we need to perform an equilibrium calculation. BaC O (s) Ba + (aq) + C O (aq) K sp =. Initial 0.000 M 0 s mol/l of BaC O (s) dissolves to reach equilibrium Equil. 0.000 + s s K sp =. s = [C O =. 5. [Ba + 0 = [SO 0 = 8 10 = [Ba + [C O = (0.000 + s)(s) (0.000)s 5.0 ml 15mL 10 8 10 mol/l; [Ba + = 0.000 M; assumption good (s << 0.000). 0.00mmol ml 00. ml 0.00mmol ml 00. ml =.5 10 M =.5 10 M Q = [Ba + 0 [SO 0 = (.5 10 )(.5 10 ) = 1.9 10 > K sp (1.5 10 9 ) A precipitate of BaSO (s) will form. BaSO (s) Ba + + SO Before 0.005 M 0.05 M Let 0.005 mol/l Ba + react with SO to completion because K sp << 1. Change 0.005 0.005 Reacts completely After 0 0.015 New initial (carry extra sig. fig.) s mol/l BaSO dissolves to reach equilibrium Change s +s +s Equil. s 0.015 + s K sp = 1.5 10 9 = [Ba + [SO = (s)(0.015 + s) s(0.015) s = 8.6 10 8 mol/l; [Ba + = 8.6 10 8 M; [SO = 0.018 M; assumption good.

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 55. 50.0 ml 0.0000 M = 0.100 mmol Ag + ; 50.0 ml 0.0100 M = 0.500 mmol IO From the small K sp value, assume AgIO (s) precipitates completely. After reaction, 0.00 mmol IO is remaining. Now, let some AgIO (s) dissolve in solution with excess IO to reach equilibrium. AgIO (s) Ag + (aq) + IO (aq) 0.00mmol Initial 0 =.00 10 M 100.0 ml s mol/l AgIO (s) dissolves to reach equilibrium Equil. s.00 10 + s K sp = [Ag + [IO =.0 10 8 = s(.00 10 + s) s(.00 10 ) s =.5 10 6 mol/l = [Ag + ; assumption good. 56. 50.0 ml 0.10 M = 5.0 mmol Pb + ; 50.0 ml 1.0 M = 50. mmol Cl. For this solution, Q > K sp, so PbCl precipitates. Assume precipitation of PbCl (s) is complete. 5.0 mmol Pb + requires 10. mmol of Cl for complete precipitation, which leaves 0. mmol Cl in excess. Now, let some of the PbCl (s) redissolve to establish equilibrium PbCl (s) Pb + (aq) + Cl (aq) 0. mmol Initial 0 = 0.0 M 100.0 ml s mol/l of PbCl (s) dissolves to reach equilibrium Equil. s 0.0 + s K sp =[Pb + [Cl, 1.6 s = 1.0 5 10 = s(0.0 + s) s(0.0) 10 mol/l; assumption good. At equilibrium: [Pb + = s = 1.0 10 mol/l; [Cl = 0.0 + s, 0.0 + (1.0 10 ) = 0.0 M 5. Ag PO (s) Ag + (aq) + PO (aq); when Q is greater than K sp, precipitation will occur. We will calculate the [Ag + 0 necessary for Q = K sp. Any [Ag + 0 greater than this calculated number will cause precipitation of Ag PO (s). In this problem, [PO - 0 = [Na PO 0 = 1.0 10 5 M. 18 K sp = 1.8 10 18 ; Q = 1.8 10 = [Ag 0 [PO 0 = [Ag 0 (1.0 5 10 M) [Ag + 0 = 1.8 10 1.0 10 18 5 1/, [Ag + 0 = 5.6 5 10 M When [Ag + 0 = [AgNO 0 is greater than 5.6 5 10 M, precipitation of Ag PO (s) will occur.

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 65 58. Al(OH) (s) Al + (aq) + OH (aq) K sp = 10 Q = 10 = [Al 0 [OH 0 = (0.)[OH 0, [OH 0 =.6 11 10 (carrying extra sig. fig.) poh = log(.6 10 11 ) = 10.; when the poh of the solution equals 10., K sp = Q. For precipitation, we want Q > K sp. This will occur when [OH 0 >.6 10 11 or when poh < 10.. Because ph + poh = 1.00, precipitation of Al(OH) (s) will begin when ph >. because this corresponds to a solution with poh < 10.. 59. For each lead salt, we will calculate the [Pb + 0 necessary for Q = K sp. Any [Pb + 0 greater than this value will cause precipitation of the salt (Q > K sp ). PbF (s) Pb + (aq) + F (aq) K sp = 10 8 ; Q = 10 8 = [Pb 0 [F 0 8 [Pb + 10 0 = (1 10 ) = M PbS(s) Pb + (aq) + S (aq) K sp = 10 9 ; Q = 10 9 = [Pb + 0 [S 0 [Pb + 0 = 10 1 10 9 = 10 5 M Pb (PO ) (s) Pb + (aq) + PO (aq) K sp = 1 10 5 Q = 1 10 5 = [Pb 0 [PO 0 [Pb + 0 = 1 10 (1 10 5 ) 1/ = 5 10 16 M From the calculated [Pb + 0, the least soluble salt is PbS(s), and it will form first. Pb (PO ) (s) will form second, and PbF (s) will form last because it requires the largest [Pb + 0 in order for precipitation to occur. 60. From Table 16.1, K sp for NiCO = 1. 10 10 and K sp for CuCO =.5 10. From the K sp values, CuCO will precipitate first since it has the smaller K sp value and will be the least soluble. For CuCO (s), precipitation begins when: [CO = K 10 sp, CuCO [Cu.5 10 = 1.0 0.5M 9 10 M CO For NiCO (s) to precipitate: [CO = K sp, NiCO [Ni 1. 10 = 5.6 0.5M 10 M CO

66 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Determining the [Cu + when CuCO (s) begins to precipitate: [Cu + = K sp, CuCO [CO 10.5 10 =.5 5.6 10 M 10 M Cu + For successful separation, 1% Cu + or less of the initial amount of Cu + (0.5 M) must be present before NiCO (s) begins to precipitate. The percent of Cu + present when NiCO (s) begins to precipitate is:.5 10 M 0.5M 100 = 0.18% Cu + Because less than 1% of the initial amount of Cu + remains, the metals can be separated through slow addition of Na CO (aq). Complex Ion Equilibria 61. a. Ni + + CN NiCN + K 1 NiCN + + CN Ni(CN) K Ni(CN) + CN Ni(CN) K Ni(CN) + CN Ni(CN) K Ni + + CN Ni(CN) K f = K 1 K K K Note: The various K constants are included for your information. Each NH adds with a corresponding K value associated with that reaction. The overall formation constant K f for the overall reaction is equal to the product of all the stepwise K values. b. V + + + C O VC O VC O + + C O V(C O ) V(C O ) + C O V(C O ) V + + C O V(C O ) K 1 K K K f = K 1 K K 6. a. Co + + F CoF + K 1 CoF + + F CoF + K CoF + + F CoF K CoF + F CoF K CoF + F CoF 5 K 5 CoF 5 + F CoF 6 K 6 Co + + 6 F CoF 6 K f = K 1 K K K K 5 K 6

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 b. Zn + + + NH ZnNH ZnNH + + + NH Zn(NH ) Zn(NH ) + + NH + Zn(NH ) Zn(NH ) + + NH + Zn(NH ) Zn + + + NH Zn(NH ) K 1 K K K K f = K 1 K K K 6. Fe + (aq) + 6 CN (aq) Fe(CN) 6 1.5 10 K = 0 6 (8.5 10 )(0.11) = 1.0 10 [Fe(CN) 6 K = 6 [Fe [CN 6. Cu + (aq) + NH (aq) Cu(NH ) + (aq) K = K = 1.0 10 (1.8 10 1 )(1.5) 1.1 10 1 [Cu(NH [Cu ) [NH 65. Hg + (aq) + I (aq) HgI (s); HgI (s) + I (aq) HgI (aq) orange ppt soluble complex ion 66. Ag + (aq) + Cl (aq) AgCl(s), white ppt.; AgCl(s) + NH (aq) Ag(NH ) + (aq) + Cl (aq) Ag(NH ) + (aq) + Br - (aq) AgBr(s) + NH (aq), pale yellow ppt. AgBr(s) + S O (aq) Ag(S O ) (aq) + Br - (aq) Ag(S O ) (aq) + I (aq) AgI(s) + S O (aq), yellow ppt. The least soluble salt (smallest K sp value) must be AgI because it forms in the presence of Cl and Br. The most soluble salt (largest K sp value) must be AgCl since it forms initially but never re-forms. The order of K sp values is K sp (AgCl) > K sp (AgBr) > K sp (AgI). 6. The formation constant for HgI is an extremely large number. Because of this, we will let the Hg + and I ions present initially react to completion and then solve an equilibrium problem to determine the Hg + concentration. Hg + (aq) + I (aq) HgI (aq) K = 1.0 10 0 Before 0.010 M 0.8 M 0 Change 0.010 0.00 +0.010 Reacts completely (K is large). After 0 0. 0.010 New initial x mol/l HgI dissociates to reach equilibrium Change +x +x x Equil. x 0. + x 0.010 x

68 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA K = 1.0 10 0 [HgI (0.010 x) = [Hg [I ( x) (0. x) ; making normal assumptions: 1.0 10 0 (0.010) =, x = [Hg + =. 10 M ; assumptions good. ( x) (0.) Note:. 10 mol/l corresponds to one Hg + ion per 5 10 L. It is very reasonable to approach this problem in two steps. The reaction does essentially go to completion. 68. Ni + (aq) + 6 NH (aq) Ni(NH ) 6 + (aq) K = 5.5 10 8 Initial 0.0 M 0.10 mol/0.50 L = 0.0 M x mol/l Ni(NH ) 6 + dissociates to reach equilibrium Change +x +6x x Equil. x.0 + 6x 0.0 x K = 5.5 10 8 = ) 6 6 [Ni(NH [Ni [NH (0.0 x) ( x)(.0 6x) 6, 8 5.5 10 (0.0) 6 ( x)(.0) x = [Ni + = 5.0 10 1 M; [Ni(NH ) 6 + = 0.0 M x = 0.0 M; assumptions good. 69. [X 0 = 5.00 M and [Cu + 0 = 1.0 10 M because equal volumes of each reagent are mixed. Because the K values are much greater than 1, assume the reaction goes completely to CuX, and then solve an equilibrium problem. Cu + (aq) + X (aq) CuX (aq) K = K 1 K K = 1.0 10 9 Before 1.0 10 M 5.00 M 0 After 0 5.00 ( 10 ) 5.00 1.0 10 Reacts completely Equil. x 5.00 + x 1.0 10 x K = (1.0 10 x) ( x)(5.00 x) = 1.0 10 9 1.0 10, ( x)(5.00) x = [Cu + = 8.0 15 10 M Assumptions good. [CuX = 1.0 10 8.0 15 10 = 1.0 10 M K = [CuX [CuX Summarizing: [X = 1.0 10 = (1.0 10 ), [CuX =.0 [CuX (5.00) 10 M [CuX - = 1.0 [CuX - =.0 [Cu + = 8.0 10 M (answer a) 10 M (answer b) 15 10 M (answer c)

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 69 0. [Be + 5 0 = 5.0 10 M and [F 0 =.0 M because equal volumes of each reagent are mixed, so all concentrations given in the problem are diluted by a factor of one-half. Because the K values are large, assume all reactions go to completion, and then solve an equilibrium problem. Be + (aq) + F (aq) BeF (aq) K = K 1 K K K =.5 10 1 Before 5 5.0 10 M.0 M 0 After 0.0 M 5 5.0 10 M Equil. x.0 + x 5 5.0 10 x K =.5 1 10 = [Be [BeF [F = 5.0 10 5 x x (.0 x) 5.0 10 x(.0) 5 x = [Be + =.6 0 10 M; assumptions good. [F =.0 M; [BeF = 5.0 10 5 M Now use the stepwise K values to determine the other concentrtations. K 1 =.9 K = 5.8 K = 6.1 10 = 10 = 10 = [Be [BeF [BeF [F [F [ BeF [BeF [BeF [F = = = [BeF (.6 10 (8. 10 0 [BeF [BeF (1.9 10 15 10 )(.0) )(.0) )(.0), [BeF + = 8. 10 15 M, [BeF = 1.9 10 10 M, [BeF =.6 10 M 1. a. AgI(s) Ag + (aq) + I (aq) K sp = [Ag + [I = 1.5 Equil. s s K sp = 1.5 16 10 = s 8, s = 1. 10 mol/l b. AgI(s) Ag + + I K sp = 1.5 10 Ag + + + NH Ag(NH ) K f = 1. 10 AgI(s) + NH (aq) Ag(NH ) + (aq) + I (aq) K = K sp K f =.6 AgI(s) + NH Ag(NH ) + + I Initial.0 M 0 0 s mol/l of AgBr(s) dissolves to reach equilibrium = molar solubility Equil..0 s s s 16 10 9 16 10 [Ag(NH) [I K = [NH s = (.0 s) 9 =.6 10 s (.0), s = 1.5 10 mol/l Assumption good.

60 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA c. The presence of NH increases the solubility of AgI. Added NH removes Ag + from solution by forming the complex ion, Ag(NH ) +. As Ag + is removed, more AgI(s) will dissolve to replenish the Ag + concentration.. AgBr(s) Ag + + Br K sp = 5.0 Ag + + S O Ag(S O ) K f =.9 10 1 10 1 AgBr(s) + S O Ag(S O ) + Br - K = K sp K f = 1.5 (Carry extra sig. figs.) AgBr(s) + S O (aq) Ag(S O ) (aq) + Br aq) Initial 0.500 M 0 0 s mol/l AgBr(s) dissolves to reach equilibrium Change s s +s +s Equil. 0.500 s s s s K = ( 0.500 s) = 1.5; taking the square root of both sides: s 0.500 s =.81, s = 1.91 (.6)s, s = 0. mol/l 1.00 L 0.mol AgBr L 18.8 g AgBr = 1. g AgBr = g AgBr mol AgBr. AgCl(s) Ag + + Cl K sp = 1.6 Ag + + + NH Ag(NH ) K f = 1. 10 10 10 AgCl(s) + NH (aq) Ag(NH ) + (aq) + Cl (aq) K = K sp K f =. 10 AgCl(s) + NH Ag(NH ) + + Cl Initial 1.0 M 0 0 s mol/l of AgCl(s) dissolves to reach equilibrium = molar solubility Equil. 1.0 s s s K =. 10 [Ag(NH) [Cl = [NH s = (. 1.0 s 10 ) 1/ = 5. s = ( 1.0 s) 10, s =. ; taking the square root: 10 mol/l 10 In pure water, the solubility of AgCl(s) is (1.6 10 ) 1/ 5 = 1. 10 mol/l. Notice how the presence of NH increases the solubility of AgCl(s) by over a factor of 500.. a. CuCl(s) Cu + (aq) + Cl (aq) Equil. s s

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 61 K sp = 1. 6 10 = [Cu + [Cl = s, s = 1.1 10 mol/l b. Cu + forms the complex ion CuCl - in the presence of Cl -. We will consider both the K sp reaction and the complex ion reaction at the same time. CuCl(s) Cu + (aq) + Cl (aq) K sp = 1. Cu + (aq) + Cl (aq) CuCl (aq) K f = 8. 10 10 6 CuCl(s) + Cl (aq) CuCl (aq) K = K sp K f = 0.10 CuCl(s) + Cl CuCl Initial 0.10 M 0 Equil. 0.10 s s where s = solubility of CuCl(s) in mol/l K = 0.10 = [ CuCl [Cl = s, 1.0 0.10 s 10 0.10 s = s, s = 9.1 10 mol/l 5. Test tube 1: Added Cl reacts with Ag + to form a silver chloride precipitate. The net ionic equation is Ag + (aq) + Cl (aq) AgCl(s). Test tube : Added NH reacts with Ag + ions to form a soluble complex ion, Ag(NH ) +. As this complex ion forms, Ag + is removed from the solution, which causes the AgCl(s) to dissolve. When enough NH is added, all the silver chloride precipitate will dissolve. The equation is AgCl(s) + NH (aq) Ag(NH ) + (aq) + Cl (aq). Test tube : Added H + reacts with the weak base, NH, to form NH +. As NH is removed from the Ag(NH ) + complex ion, Ag + ions are released to solution and can then react with Cl to re-form AgCl(s). The equations are Ag(NH ) + (aq) + H + (aq) Ag + (aq) + NH + (aq) and Ag + (aq) + Cl (aq) AgCl(s). 6. In NH, Cu + forms the soluble complex ion Cu(NH ) +. This increases the solubility of Cu(OH) (s) because added NH removes Cu + from the equilibrium causing more Cu(OH) (s) to dissolve. In HNO, H + removes OH from the K sp equilibrium causing more Cu(OH) (s) to dissolve. Any salt with basic anions will be more soluble in an acid solution. AgC H O (s) will be more soluble in either NH or HNO. This is because Ag + forms the complex ion Ag(NH ) +, and C H O is a weak base, so it will react with added H +. AgCl(s) will be more soluble only in NH due to Ag(NH ) + formation. In acid, Cl - is a horrible base, so it doesn t react with added H +. AgCl(s) will not be more soluble in HNO. Additional Exercises. Mol Ag + 0. mol AgNO 1molAg added = 0.00 L = 0.08 mol Ag + L mol AgNO The added Ag + will react with the halogen ions to form a precipitate. Because the K sp values are small, we can assume these precipitation reactions go to completion. The order of precipitation will be AgI(s) first (the least soluble compound since K sp is the smallest), followed by AgBr(s), with AgCl(s) forming last [AgCl(s) is the most soluble compound listed since it has the largest K sp.

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Let the Ag + react with I to completion. Ag + (aq) + I (aq) AgI(s) K = 1/K sp >> 1 Before 0.08 mol 0.018 mol 0 Change 0.018 0.018 +0.018 I is limiting. After 0.00 mol 0 0.018 mol Let the Ag + remaining react next with Br to completion. Ag + (aq) + Br (aq) AgBr(s) K = 1/K sp >> 1 Before 0.00 mol 0.018 mol 0 Change 0.018 0.018 +0.018 Br is limiting. After 0.01 mol 0 0.018 mol Finally, let the remaining Ag + react with Cl to completion. Ag + (aq) + Cl (aq) AgCl(s) K = 1/K sp >> 1 Before 0.01 mol 0.018 mol 0 Change 0.01 0.01 +0.01 Ag + is limiting. After 0 0.006 mol 0.01 mol Some of the AgCl will redissolve to produce some Ag + ions; we can t have [Ag + = 0 M. Calculating how much AgCl(s) redissolves: AgCl(s) Ag + (aq) + Cl (aq) K sp = 1.6.006 mol/0.00 L = 0.0 M s mol/l of AgCl dissolves to reach equilibrium Change s +s +s Equil. s 0.0 + s K sp = 1.6 s = 5 10 10 = [Ag + [Cl = s(0.0 + s) (0.0)s 9 10 mol/l; the assumption that 0.0 + s 0.0 is good. Mol AgCl present = 0.01 mol 5 Mass AgCl present = 0.01 mol AgCl [Ag + = s = 5 9 10 mol/l 9 10 mol = 0.01 mol 8. AgX(s) Ag + (aq) + X (aq) K sp = [Ag + [X AgY(s) Ag + (aq) + Y (aq) K sp = [Ag + [Y 1. g = 1. g AgCl mol AgCl 10 10 For conjugate acid-base pairs, the weaker the acid, the stronger is the conjugate base. Because HX is a stronger acid (has a larger K a value) than HY, Y will be a stronger base

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 than X. In acidic solution, Y will have a greater affinity for the H + ions. Therefore, AgY(s) will be more soluble in acidic solution because more Y will be removed through reaction with H +, which will cause more AgY(s) to dissolve. 9. Ca 5 (PO ) OH(s) 5 Ca + + PO + OH 1.0 10 from water Equil. 5s s s + 1.0 10 s K sp = 6.8 6.8 10 = [Ca + 5 [PO [OH = (5s) 5 (s) (s) 10 = (15)()s 9 5, s =. 10 mol/l; assumption is good. The solubility of hydroxyapatite will increase as the solution gets more acidic because both phosphate and hydroxide can react with H +. Ca 5 (PO ) F(s) 5 Ca + (aq) + PO (aq) + F (aq) 0 Equil. 5s s s K sp = 1 60 10 = (5s) 5 (s) (s) = (15)()s 9 8, s = 6 10 mol/l The hydroxyapatite in tooth enamel is converted to the less soluble fluorapatite by fluoridetreated water. The less soluble fluorapatite is more difficult to remove, making teeth less susceptible to decay. 1mg F 1g 1molF 80. L 1000mg 19.00g F = 5. 5 10 M F 5 = 5 10 M F CaF (s) Ca + (aq) + F (aq) K sp = [Ca + [ F 11 =.0 10 ; precipitation will occur when Q > K sp. Let s calculate [Ca + so that Q = K sp. 11 Q =.0 10 = [Ca + 0 [ F 0 = [Ca + 0 (5 5 10 ), [Ca + 0 = 10 M CaF (s) will precipitate when [Ca + 0 > 10 M. Therefore, hard water should have a calcium ion concentration of less than 10 M in order to avoid CaF (s) formation. 81. CaF (s) Ca + (aq) + F (aq) K sp = [Ca + [F We need to determine the F concentration present in a 1.0 M HF solution. Solving the weak acid equilibrium problem: HF(aq) H + (aq) + F (aq) [H [F K a = [HF Initial 1.0 M ~0 0 Equil. 1.0 x x x

6 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA K a =. 10 x(x) x =, 1.0 x 1.0 x = [F =. 10 M; assumption good. Next, calculate the Ca + concentration necessary for Q = K sp, CaF. Q = [Ca 11 8 0[F 0,.0 10 [Ca 0(. 10 ), [Ca 0 5.5 10 mol/l Mass Ca(NO ) 8 ) 1.0 L 5.5 10 molca L 1molCa(NO molca 16.10g Ca(NO) mol Mass Ca(NO ) = 9.0 10 g Ca(NO) 6 For precipitation of CaF (s) to occur, we need Q > K sp. When 9.0 10 g Ca(NO) has been added to 1.0 L of solution, Q = K sp. So precipitation of CaF (s) will begin to occur 6 when just over 9.0 10 g Ca(NO) has been added. 6 8. Mn(OH) (s) Mn + (aq) + OH (aq) K sp = [Mn + [OH Initial s = solubility (mol/l) 0 1.0 10 M (from water) Equil. s 1.0 10 + s K sp =.0 10 1 = s(1.0 10 + s) s(s) = s, s =. 10 5 mol/l; assumption good. 1. L 5. 10 mol Mn(OH) 88.96g Mn(OH) L mol. 10 g Mn(OH) 1molPbI 0.g PbI 61.0 g 8. s = solubility =.6 10 M 0.000L PbI (s) Pb + (aq) + I (aq) K sp = [Pb + [I Equil. s s K sp = s(s) = s, K sp = (.6 10 ) =.0 10 8 8. C H 5 BiO (s) + H O(l) C H O (aq) + Bi + (aq) + OH (aq) 1.0 10 M (from water) Equil. s s 1.0 10 + s K = [C H O [Bi + [OH = s(s)(1.0 10 + s); from the problem, s =. 10 19 mol/l: K = (. 10 19 ) (1.0 10 +. 10 19 ) = 1.0 10

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 65 85. Mn + + C O MnC O K 1 =.9 10 - MnC O + C O Mn(C O ) K =.9 10 1 Mn + (aq) + C O (aq) Mn(C O ) (aq) K f = K 1 K = 6. 10 5 86. Cr + + H EDTA CrEDTA + H + Before 0.0010 M 0.050 M 0 1.0 10 6 M (Buffer) Change 0.0010 0.0010 +0.0010 No change Reacts completely After 0 0.09 0.0010 1.0 10 6 New initial x mol/l CrEDTA dissociates to reach equilibrium Change +x +x x Equil. x 0.09 + x 0.0010 x 1.0 10 6 (Buffer) K f = 1.0 10 = [CrEDT A [Cr [H [H EDTA = (0.0010 x)(1.0 10 ( x)(0.09 x) 6 ) 1.0 10 (0.0010)(1.0 10 x(0.09) 1 ), x = [Cr + =.0 10 M; assumptions good. 8. a. Pb(OH) (s) Pb + + OH Initial s = solubility (mol/l) 0 1.0 10 M from water Equil. s 1.0 10 + s K sp = 1. 10 15 = [Pb + [OH = s(1.0 10 + s) s(s ) = s s = [Pb + = 6. 10 6 M; assumption is good by the 5% rule. b. Pb(OH) (s) Pb + + OH Initial 0 0.10 M ph = 1.00, [OH = 0.10 M s mol/l Pb(OH) (s) dissolves to reach equilibrium Equil. s 0.10 (Buffered solution) 1. 10 15 = (s)(0.10), s = [Pb + = 1. 10 1 M c. We need to calculate the Pb + concentration in equilibrium with EDTA -. Since K is large for the formation of PbEDTA, let the reaction go to completion and then solve an equilibrium problem to get the Pb + concentration. Pb + + EDTA PbEDTA K = 1.1 10 18 Before 0.010 M 0.050 M 0 0.010 mol/l Pb + reacts completely (large K) Change 0.010 0.010 +0.010 Reacts completely After 0 0.00 0.010 New initial x mol/l PbEDTA dissociates to reach equilibrium Equil. x 0.00 + x 0.010 x

66 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 1.1 10 18 = (0.010 x) ( x)(0.00 x) 0.010 x(0.00), x = [Pb + =. 10 19 M; assumptions good. 88. 1.0 ml Now calculate the solubility quotient for Pb(OH) to see if precipitation occurs. The concentration of OH is 0.10 M since we have a solution buffered at ph = 1.00. Q = [Pb + 0 [OH 0 = (. 10 19 )(0.10) =. 10 1 < K sp (1. 10 15 ) Pb(OH) (s) will not form since Q is less than K sp. 1.0 mmol = 1.0 mmol Cd + added to the ammonia solution ml Thus [Cd + 0 = 1.0 10 mol/l. We will first calculate the equilibrium Cd + concentration using the complex ion equilibrium and then determine if this Cd + concentration is large enough to cause precipitation of Cd(OH) (s). Cd + + NH Cd(NH ) + K f = 1.0 10 Before 1.0 10 M 5.0 M 0 Change 1.0 10.0 10 +1.0 10 Reacts completely After 0.996 5.0 1.0 10 New initial x mol/l Cd(NH ) + dissociates to reach equilibrium Change +x +x x Equil. x 5.0 + x 0.0010 x K f = 1.0 10 (0.010 x) (0.010) = ( x)(5.0 x) ( x)(5.0) x = [Cd + = 1.6 10 1 M; assumptions good. This is the maximum [Cd + possible. Now we will determine if Cd(OH) (s) forms at this concentration of Cd +. In 5.0 M NH we can calculate the ph: NH + H O NH + Initial 5.0 M 0 ~0 Equil. 5.0 y y y K b = 1.8 10 5 [NH [OH y = = [NH 5.0 y + OH K b = 1.8 10 5 y 5.0 We now calculate the value of the solubility quotient, Q: Q = [Cd + [OH = (1.6 10 1 )(9.5 10 ), y = [OH = 9.5 10 M; assumptions good. Q = 1. 10 1 < K sp (5.9 10 15 ); therefore, no precipitate forms. 89. a. Cu(OH) Cu + + OH K sp = 1.6 Cu + + + NH Cu(NH ) K f = 1.0 10 1 10 19 Cu(OH) (s) + NH (aq) Cu(NH ) + (aq) + OH (aq) K = K sp K f = 1.6 10 6

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 6 b. Cu(OH) (s) + NH Cu(NH ) + + OH K = 1.6 Initial 5.0 M 0 0.0095 M s mol/l Cu(OH) dissolves to reach equilibrium Equil. 5.0 s s 0.0095 + s 6 K = 1.6 10 = [Cu(NH) [OH s(0.0095 s) [NH (5.0 s) 10 6 6 If s is small: 1.6 10 s(0.0095) =, (5.0) s = 11. mol/l Assumptions are horrible. We will solve the problem by successive approximations. 6 1.6 10 (5.0 s s calc = (0.0095 s ) guess guess) ; the results from six trials are: s guess : 0.10, 0.050, 0.060, 0.055, 0.056 s calc : 1.6 10 -, 0.01, 0.09, 0.058, 0.056 Thus the solubility of Cu(OH) is 0.056 mol/l in 5.0 M NH. 90. a.

68 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA b. c. 91. Ba(OH) (s) Ba + (aq) + OH (aq) K sp = [Ba + [OH = 5.0 Initial s = solubility (mol/l) 0 ~0 Equil. s s K sp = 5.0 10 = s(s) = s, s = 0.11 mol/l; assumption good. [OH = s = (0.11) = 0. mol/l; poh = 0.66, ph = 1. Sr(OH) (s) Sr + (aq) + OH (aq) K sp = [Sr + [OH =. Equil. s s K sp =. 10 = s, s = 0.0 mol/l; asssumption good. [OH = (0.0) = 0.086 M; poh = 1.0, ph = 1.9 10 10

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 69 Ca(OH) (s) Ca + (aq) + OH (aq) K sp = [Ca + [OH = 1. Equil. s s 10 6 K sp = 1. 9. K sp = 6. [OH = (6.9 6 10 = s, s = 6.9 10 mol/l; assumption good. 10 ) = 1. 10 mol/l; poh = 1.85, ph = 1.15 9 10 = [Mg + [F 9, 6. 10 = (0.005 y)(0.065 y) This is a cubic equation. No simplifying assumptions can be made since y is relatively large. Solving cubic equations is difficult unless you have a graphing calculator. However, if you don t have a graphing calculator, one way to solve this problem is to make the simplifying assumption to run the precipitation reaction to completion. This assumption is made because of the very small value for K, indicating that the ion concentrations are very small. Once this assumption is made, the problem becomes much easier to solve. ChemWork Problems The answers to the problems 9-98 (or a variation to these problems) are found in OWL. These problems are also assignable in OWL. Challenge Problems 99. a. CuBr(s) Cu + + Br K sp = 1.0 10 5 Cu+ + CN Cu(CN) K f = 1.0 10 11 CuBr(s) + CN Cu(CN) + Br K = 1.0 10 6 Because K is large, assume that enough CuBr(s) dissolves to completely use up the 1.0 M CN ; then solve the back equilibrium problem to determine the equilibrium concentrations. CuBr(s) + CN Cu(CN) + Br Before x 1.0 M 0 0 x mol/l of CuBr(s) dissolves to react completely with 1.0 M CN Change x x +x +x After 0 1.0 x x x For reaction to go to completion, 1.0 x = 0 and x = 0. mol/l. Now solve the backequilibrium problem.

650 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA CuBr(s) + CN Cu(CN) + Br Initial 0 0. M 0. M Let y mol/l of Cu(CN) react to reach equilibrium. Change +y y y Equil. y 0. y 0. y K = 1.0 10 6 = (0. y) (y) (0.) y, y = 1.6 10 M; assumption good. Of the initial 1.0 M CN, only (1.6 10 ) =.8 10 M is present at equilibrium. Indeed, enough CuBr(s) did dissolve to essentially remove the initial 1.0 M CN. This amount, 0. mol/l, is the solubility of CuBr(s) in 1.0 M NaCN. b. [Br = 0. y = 0. 1.6 10 = 0. M c. [CN = y = (1.6 10 ) =.8 10 M 100. Ag + + NH AgNH + AgNH + + NH Ag(NH ) + Ag + + NH Ag(NH ) + K 1 =.1 10 K = 8. 10 K = K 1 K = 1. 10 The initial concentrations are halved because equal volumes of the two solutions are mixed. Let the reaction go to completion since K is large; then solve an equilibrium problem. Ag + (aq) + NH (aq) Ag(NH ) + (aq) Before 0.0 M.0 M 0 After 0 1.6 0.0 Equil. x 1.6 + x 0.0 x K = 1. 10 [Ag(NH) = [Ag [NH [Ag + = x =.6 0.0 x = x(1.6 x) 0.0 x(1.6), x =.6 9 10 M; [NH = 1.6 M; [Ag(NH ) + = 0.0 M Use either the K 1 or K equilibrium expression to calculate [AgNH +. 9 10 M; assumptions good. AgNH + + NH Ag(NH ) + K = 8. 10 8. 10 = [Ag(NH [AgNH ) [NH 0.0, [AgNH + 5 = 1.5 10 M [AgNH (1.6) 101. a. AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + [Br - = 5.0 Equil. s s K sp = 5.0 1 10 = s, s =.1 10 mol/l 1 10

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 651 b. AgBr(s) Ag + + Br - K sp = 5.0 Ag + + + NH Ag(NH ) K f = 1. 10 10 1 AgBr(s) + NH (aq) Ag(NH ) + (aq) + Br - (aq) K = K sp K f = 8.5 AgBr(s) + NH Ag(NH ) + + Br - Initial.0 M 0 0 s mol/l of AgBr(s) dissolves to reach equilibrium = molar solubility Equil..0 s s s 10 6 [Ag(NH) [Br s K = [NH (.0 s) Assumption good. 6 = 8.5 10 s, s = 8. 10 mol/l (.0) c. The presence of NH increases the solubility of AgBr. Added NH removes Ag + from solution by forming the complex ion, Ag(NH ) +. As Ag + is removed, more AgBr(s) will dissolve to replenish the Ag + concentration. d. Mass AgBr = 0.500 L 8. 10 mol AgBr L 18.8 g AgBr mol AgBr = 0.1 g AgBr e. Added HNO will have no effect on the AgBr(s) solubility in pure water. Neither H + nor NO react with Ag + or Br ions. Br is the conjugate base of the strong acid HBr, so it is a terrible base. However, added HNO will reduce the solubility of AgBr(s) in the 5 ammonia solution. NH is a weak base (K b = 1.8 10 ). Added H + will react with NH to form NH + +. As NH is removed, a smaller amount of the Ag(NH ) complex ion will form, resulting in a smaller amount of AgBr(s) that will dissolve. 10. [NH 0 =.00 M = 1.50 M; [Cu + 0 =.00 10 M = 1.00 10 M Because [NH 0 >> [Cu + 0, and because K 1, K, K and K are all large, Cu(NH ) + will be the dominant copper-containing species. The net reaction will be Cu + + NH Cu(NH ) +. Here, 1.00 10 M Cu + plus (1.00 10 M) NH will produce 1.00 10 M Cu(NH ) +. At equilibrium: [Cu(NH ) + 1.00 10 M [NH = [NH 0 [NH reacted = 1.50 M (1.00 Calculate [Cu(NH ) + from the K reaction: 10 M) = 1.50 M 1.55 10 = [Cu(NH [Cu(NH ) ) [NH 1.00 10, [Cu(NH ) [Cu(NH) (1.50).0 10 6 M

65 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA Calculate [Cu(NH ) + from K reaction: 1.00 10 = [Cu(NH).0 10, [Cu(NH ) [Cu(NH) [NH [Cu(NH) (1.50).8 10 Calculate [Cu(NH ) + from the K reaction:.88 10 = [Cu(NH).8 10, [CuNH [CuNH [NH [CuNH (1.50).9 10 Calculate [Cu + from the K 1 reaction: [CuNH.9 10 1 1.86 10 =, [Cu 1. 10 M [Cu [NH [Cu (1.50) The assumptions are valid. Cu(NH ) + is clearly the dominant copper-containing component. 10. AgCN(s) Ag + (aq) + CN (aq) K sp =. H + (aq) + CN (aq) HCN(aq) K = 1/K 1.6 1 9 6 10 1 9 a, HCN 10 1 M 9 M AgCN(s) + H + (aq) Ag + (aq) + HCN(aq) K =. 10 1 9 (1.6 10 ).5 10 AgCN(s) + H + (aq) Ag + (aq) + HCN(aq) Initial 1.0 M 0 0 s mol/l AgCN(s) dissolves to reach equilibrium Equil. 1.0 s s s.5 10 [Ag [HCN [H s(s) s, s 5.9 10 1.0 s 1.0 Assumption fails the 5% rule (s is 5.9% of 1.0 M). Using the method of successive approximations:.5 10 1.0 s, 0.059 s 5. 10.5 10 1.0 s, 0.05 s 5. 10 (consistent answer) The molar solubility of AgCN(s) in 1.0 M H + is 5. 10 10. Solubility in pure water: mol/l. CaC O (s) Ca + (aq) + C O (aq) K sp = 10 9 Equil. s s

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 65 K sp = s = 10 9, s = solubility =. 10 5 = 10 5 mol/l Solubility in 0.10 M H + : CaC O (s) Ca + + C O K sp = 10 9 C O + H + HC O K = 1/ K a = 1.6 10 HC O + H + H C O K = 1/ K = 15 CaC O (s) + H + Ca + + H C O K overall = 5 10 CaC O (s) + H + Ca + + H C O Initial 0.10 M 0 0 s mol/l of CaC O (s) dissolves to reach equilibrium Equil. 0.10 s s s 5 10 s s 1/ =, (5 10 ), s 10 mol/ L (0.10 s) 0.10 s a 1 Solubility in 0.10M H Solubility in pure water 10 10 5 mol/ L mol/ L = 50 CaC O (s) is 50 times more soluble in 0.10 M H + than in pure water. This increase in solubility is due to the weak base properties of C O. 105. H S(aq) H + (aq) + HS (aq) HS (aq) H + (aq) + S (aq) H S(aq) H + (aq) + S (aq) K = K a 1 = 1.0 10 - K = 1 a 10 19 6 K K = 1 10 = a1 a [H [S [H S Because K is very small, only a tiny fraction of the H S will react. At equilibrium, [H S = 0.10 M and [H + = 1 10. 6 [S K[H = S (1 10 )(0.10) = 1 [H (1 10 ) 1 10 M NiS(s) Ni + (aq) + S (aq) K sp = [Ni + [S = 10 1 Precipitation of NiS will occur when Q > K sp. We will calculate [Ni + for Q = K sp. Q = K sp = [Ni + [S 1 =.0 10, [Ni +.0 10 = 1 1 10 1 = M = maximum concentration 106. We need to determine [S 0 that will cause precipitation of CuS(s) but not MnS(s). For CuS(s): CuS(s) Cu + (aq) + S (aq) K sp = [Cu + [S = 8.5 10 5

65 CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA K 5 [Cu + 0 = 1.0 10 sp 8.5 10 M, 8.5 10 M [S [Cu 1.0 10 0 This [S represents the concentration that we must exceed to cause precipitation of CuS because if [S 0 > 8.5 10 M, Q > K sp. For MnS(s): MnS(s) Mn + (aq) + S (aq) K sp = [Mn + [ S =. 10 1 K 1 [Mn + sp. 10 10 0 = 1.0 10 M,. 10 M [S [Mn 1.0 10 This value of [S represents the largest concentration of sulfide that can be present without causing precipitation of MnS. That is, for this value of [S, Q = K sp, and no precipitatation of MnS occurs. However, for any [S 10 0 >. 10 M, MnS(s) will form. We must have [S 0 > 8.5 10 M to precipitate CuS, but [S 10 0 <. 10 M to prevent precipitation of MnS. The question asks for a ph that will precipitate CuS(s) but not MnS(s). We need to first choose an initial concentration of S that will do this. Let s choose [S 10 0 = 1.0 10 M because this will clearly cause CuS(s) to precipitate but is still less than the [S 0 required for MnS(s) to precipitate. The problem now is to determine the ph necessary for a 0.1 M H S solution to have [S 10 = 1.0 10 M. Let s combine the K a 1 and K a equations for H S to determine the required [H +. H S(aq) H + (aq) + HS (aq) HS (aq) H + (aq) + S (aq) K 1 a 1.0 10 a 1 10 K H S(aq) H + (aq) + S (aq) K = 10 a 19 Ka K 1.0 10 6 [H [S [H (1 10 ) 1 10, [H 10 [H S 0.10 ph = log( 10 9 10 ) = 8.5. So, if ph = 8.5, [S 1 10 M, which will cause precipitation of CuS(s) but not MnS(s). 1 9 M 6 Note: Any ph less than 8. would be a correct answer to this problem. 10. Mg + + P O 10 5 MgP O 10 K =.0 10 8 [Mg + 0 = 50. 10 L g 1mol.1g =.1 10 M

CHAPTER 16 SOLUBILITY AND COMPLEX ION EQUILIBRIA 655 [P O 10 5 0 = 0. g Na5PO 10 1mol = 0.11 M L 6.86g Assume the reaction goes to completion because K is large. Then solve the back-equilibrium problem to determine the small amount of Mg + present. Mg + + P O 10 5 MgP O 10 Before.1 10 M 0.11 M 0 Change.1 10.1 10 +.1 10 React completely After 0 0.11.1 10 New initial condition x mol/l MgP O 10 dissociates to reach equilibrium Change +x +x x Equil. x 0.11 + x.1 10 x K =.0 10 8 = [MgP O [Mg.0 10 8.1 10, x(0.11) 10 [P O 10.1 10 x (assume x <<.1 10 ) 5 x(0.11 x) x = [Mg + =.8 11 10 M; assumptions good. 108. MX(s) M n+ (aq) + X n ΔT 0.08 C (aq); T =K f m, m = = 0.015 mol/kg o Kf 1.86 C/ molal 0.015mol 1kg 50 g = 0.005 mol total solute particles (carrying extra sig. fig.) kg 1000g 0.05 mol = mol M n+ + mol X n, mol M n+ = mol X n = 0.05/ Because the density of the solution is 1.0 g/ml, 50 g = 50 ml of solution. [M n+ = (0. 005/ ) molm 0.5L K sp = [M n+ [X n = (.5 n =.5 10 ) = 5.6 10 5 10 M, [X n = 109. a. SrF (s) Sr + (aq) + F (aq) Initial 0 0 s mol/l SrF dissolves to reach equilibrium Equil. s s o n (0.005/ ) molx 0.5L [Sr + [F = K sp =.9 10 10 = s, s = 5.8 10 mol/l in pure water b. Greater, because some of the F would react with water: =.5 10 M F + H O HF + OH K b = K K w a, HF = 1. 10 11 This lowers the concentration of F, forcing more SrF to dissolve.