1. 15 - 1 Ionic Equilibrium When a slightly soluble or insoluble salt is mixed with water, a saturated solution quickly results and a dynamic equilibrium results. This is a dynamic equilibrium because the rate of the forward reaction (dissociation of the salt) equals the rate of the reverse reaction (salt precipitating).  Reactants are converted to products at the same rate as products are converted to reactants.

2. 15 - 2 For example, when solutions of AgNO 3 and K 2 CrO 4 are mixed, the insoluble salt, Ag 2 CrO 4, is immediately formed. This is shown in the following equations: Ag 2 CrO 4 (s) → 2Ag + (aq) + CrO 4 2- (aq)(1) 2Ag + (aq) + CrO 4 2- (aq) → Ag 2 CrO 4 (s)(2) Equation (1) represents Ag 2 CrO 4 initially put into water and dissolving. As Ag 2 CrO 4 continues to dissolve, some of the ions start to precipitate shown by Equation (2).

3. 15 - 3  At this point the salt is dissolving at a faster rate than the precipitate is forming. Eventually the rate of dissolving will equal the rate of precipitating and the system will be in state of dynamic equilibrium shown in the following equation: Ag 2 CrO 4 (s) 2Ag + (aq) + CrO 4 2- (aq) The solubility product constant and the solubility product equilibrium expression is given by:

4. 15 - 4 K sp (Ag 2 CrO 4 ) = [Ag + ] 2 [CrO 4 2- ] = 1.2 × 10 -12  K sp is used for a saturated solution of a slightly soluble salt at a given temperature.  The smaller the K sp, the less soluble the salt.  Remember that solids are never included in the equilibrium expression.

5. 15 - 5  The concentration units are moles per liter (M) and the concentration of the precipitate is included in the K sp value. It is important not to confuse the terms solubility product and solubility.  Solubility product is an equilibrium constant related to the equilibrium between a solid salt and its ions in solution.  Solubility is the amount of solute needed to form a saturated solution.

6. 15 - 6 Solubility and Solubility Product The solubility of lead(II) sulfate is 0.038 g/L. Determine the solubility product for lead(II) sulfate. s(PbSO 4 ) = 0.038 g/LK sp = ? [PbSO 4 ] = n V 0.038 g PbSO 4 L × 1 mol PbSO 4 303.27 g PbSO 4 = 1.2 × 10 -4 M

7. 15 - 7 PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) [ ] i 0 0 [ ] c +1.2 × 10 -4 +1.2 × 10 -4 [ ] e 1.2 × 10 -4 1.2 × 10 -4 K sp = [Pb 2+ ][SO 4 2- ] K sp = 1.2 × 10 -4 × 1.2 × 10 -4 = 1.4 x 10 -8

8. 15 - 8 The K sp (Pb(IO 3 ) 2 ) = 2.5 x 10 -13. (a)Determine its solubility in pure water. Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3 - (aq) [ ] i 0 0 [ ] c +x +2x [ ] e x 2x K sp = [Pb 2+ ][IO 3 - ] 2 2.5 x 10 -13 = x (2x) 2 × = 4x 3

9. 15 - 9 x = 4.0 x 10 -5 M s = 4.0 x 10 -5 mol Pb(IO 3 ) 2 L (b) Determine the solubility in g/L in pure water. s = 4.0 x 10 -5 mol Pb(IO 3 ) 2 L × 557.00 g Pb(IO 3 ) 2 1 mol Pb(IO 3 ) 2 = 2.2 x 10 -2 g Pb(IO 3 ) 2 L

10. 15 - 10 Common-Ion Effect The common-ion effect makes it possible to shift the equilibrium to favor the reactants.  The solubility of a solid is lowered if the solution contains an ion that is common to the salt.  The saturated solution will contain more undissolved solid and a lower concentration of the other ions.

11. 15 - 11 Using silver chloride as an example, the equilibrium existing in pure water would be given by: AgCl(s) Ag + (aq) + Cl - (aq) In deionized water, the [Ag + ] and [Cl - ] would attain the maximum concentration dictated by its K sp. But what if you tried using an aqueous solution of sodium chloride?

12. 15 - 12 The sodium chloride dissolves given by: NaCl(aq) → Na + (aq) + Cl - (aq) Cl - being the common ion between the insoluble salt and the aqueous solution, would suppress the amount of Cl - coming from the AgCl. The common-ion effect is Le Chatelier’s principle applied to ionic equilibria.

13. 15 - 13 (a)Calculate the solubility in g/L of Mg 3 (PO 4 ) 2 in pure water if its K sp = 1 x 10 -24. Mg 3 (PO 4 ) 2 (s) 3Mg 2+ (aq) + 2PO 4 3- (aq) [ ] i 0 0 [ ] c +3x +2x [ ] e 3x 2x K sp = [Mg 2+ ] 3 [PO 4 3- ] 2 1 x 10 -24 = (3x) 3 × (2x) 2 = 108x 5 x = 6 × 10 -6 M

14. 15 - 14. s = 6 × 10 -6 mol Mg 3 (PO 4 ) 2 L × 262.84 g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 2 × 10 -3 g Mg 3 (PO 4 ) 2 s =

15. 15 - 15 (b) Calculate the solubility in g/L of Mg 3 (PO 4 ) 2 in 0.010 M Mg(NO 3 ) 2. Mg 3 (PO 4 ) 2 (s) 3Mg 2+ (aq) + 2PO 4 3- (aq) [ ] i 0.010 0 [ ] c +3x + 0.010 +2x [ ] e 3x + 0.010 2x K sp = [Mg 2+ ] 3 [PO 4 3- ] 2 1 x 10 -24 = (3x + 0.010) 3 × (2x) 2

16. 15 - 16 Chemists must bond together and continue to look for legitimate shortcuts! The expression (3x + 0.010) 3 would be pretty nasty to expand and even more nasty to calculate! We use the approximation 3x + 0.010 ≈ 0.010 to greatly simplify matters. The task becomes much more manageable now that 1 x 10 -24 = (0.010) 3 × (2x) 2

17. 15 - 17 x = 5 x 10 -10 M s = 5 x 10 -10 mol Mg 3 (PO 4 ) 2 L × 262.84 g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 s = 1 x 10 -7 g Mg 3 (PO 4 ) 2 /L This drastic drop in solubility from 2 x 10 -3 to 1 x 10 -7 g/L is exactly what is predicted from Le Chatelier’s principle.

18. 15 - 18 (c) Calculate the solubility in g/L of Mg 3 (PO 4 ) 2 in 0.020 M Na 3 PO 4. Mg 3 (PO 4 ) 2 (s) 3Mg 2+ (aq) + 2PO 4 3- (aq) [ ] i 0 0.020 [ ] c +3x +0.020 + 2x [ ] e 3x 0.020 + 2x K sp = [Mg 2+ ] 3 [PO 4 3- ] 2 1 x 10 -24 = (3x) 3 × (0.020 + 2x) 2 (0.020 + 2x) 2 ≈ 0.020 2

19. 15 - 19 1 x 10 -24 = (3x) 3 × (0.020) 2 x = 5 x 10 -8 M s = 5 x 10 -8 mol Mg 3 (PO 4 ) 2 L × 262.84 g Mg 3 (PO 4 ) 2 1 mol Mg 3 (PO 4 ) 2 s = 1 x 10 -5 g Mg 3 (PO 4 ) 2 /L

20. 15 - 20 Selective Precipitation The difference in the solubility of two salts containing a common anion can be used to separate a pair of cations. The difference in the solubility of two salts containing a common cation can be used to separate a pair of anions. To illustrate this method, assume you have a solution of sodium chloride and potassium chromate.

21. 15 - 21  Your task is to remove the chloride from the solution.  What could you do? One thing you could do is to boil the solution in a beaker covered with a watch glass to prevent splattering.  The solid remaining on the bottom of the beaker is likely to be a mixture of Na 2 CrO 4, KCl, NaCl, and K 2 CrO 4.

22. 15 - 22 An alternative method is using selective precipitation. A salt whose cation will form a precipitate with Cl - and CrO 4 2- is AgNO 3. For a given problem, one would determine which anion would required the smallest concentration of Cl - to form the precipitate.

23. 15 - 23 Selective Precipitation Problem A solution is prepared by using 0.10 M NaCl and 0.010 M K 2 CrO 4. AgNO 3 is slowly added to the solution. K sp (AgCl) = 1.6 x 10 -10 and K sp (Ag 2 CrO 4 ) = 9.0 x 10 -12. (a)Which precipitates first, the AgCl or the Ag 2 CrO 4 ? AgCl(s) Ag + (aq) + Cl - (aq) [ ] i 0 0.10 [ ] c +x +x [ ] e x0.10+x

24. 15 - 24 K sp = [Ag + ][Cl - ] 1.6 x 10 -10 = x (0.10+x) ≈ 0.10x [Ag + ] = 1.6 x 10 -9 M Ag 2 CrO 4 (s)2Ag + (aq) + CrO 4 2- (aq) [ ] i 0 0.010 [ ] c +2x 0.010 + x [ ] e 2x 0.010 + x ×

25. 15 - 25 K sp = [Ag + ] 2 [CrO 4 2- ] 9.0 x 10 -12 = 4x 2 (0.010 + x) ≈ 0.040x 2 [Ag + ] = 1.5 x 10 -5 M × 2 = 3.0 x 10 -5 M The AgCl will precipitate first because it requires the smaller [Ag + ] = 1.6 x 10 -9 M. (b) What is the [Ag + ] when precipitation begins? [Ag + ] = 1.6 x 10 -9 M ×

26. 15 - 26 (c) What is [Cl-] when the Ag 2 CrO 4 starts to precipitate? K sp = [Ag + ] 2 [CrO 4 2- ] When Ag 2 CrO 4 starts to precipitate, the [Ag + ] = 3.0 x 10 -5 M (see Part (a)). When [Ag + ] = 3.0 x 10 -5 M, K sp = [Ag + ][Cl - ] [Cl - ] = 1.6 x 10 -10 3.0 x 10 -5 = 5.3 x 10 -6 M

27. 15 - 27 Ion Product and Precipitation When mixing two solutions, how does one determine if a precipitate forms? The procedure is the same as determining the reaction quotient, Q, in a gaseous equilibrium. When aqueous solutions are involved, you determine the ion product, P.

28. 15 - 28 The ion product of a silver chloride solution is given by P = [Ag + ][Cl - ] where P is compared to the K sp of AgCl. If P is:  > K sp, the ion concentrations are greater than the maximum for that temperature. Precipitation occurs until P = K sp.

29. 15 - 29  = K sp, the ion concentrations are a maximum for the given temperature and equilibrium exists.  < K sp, the ion concentrations are less than the maximum and no precipitate forms.

30. 15 - 30 Ion Product Problem 0.100 L of 0.0015 M K 2 CrO 4 is mixed with 0.200 L of 0.0060 M Sr(NO 3 ) 2. Does a precipitate form? [K 2 CrO 4 ] = 0.015 M [Sr(NO 3 ) 2 ] = 0.0060 M V 1 = 0.100 L V 2 = 0.200 L [K 2 CrO 4 ] = n V n = 0.015 mol K 2 CrO 4 L × 0.100 L

31. 15 - 31 n = 1.5 x 10 -3 mol K 2 CrO 4 n = 1.2 x 10 -3 mol Sr(NO 3 ) 2 n = × 0.200 L 0.0060 mol Sr(NO 3 ) 2 L [Sr(NO 3 ) 2 ] = n V = 1.2 x 10 -3 mol Sr(NO 3 ) 2 0.300 L

32. 15 - 32. [Sr(NO 3 ) 2 ] = 4.0 x 10 -3 M [K 2 CrO 4 ] = 1.5 x 10 -3 mol K 2 CrO 4 0.300 L [K 2 CrO 4 ] = 5.0 x 10 -3 M

33. 15 - 33. SrCrO 4 (s)Sr 2+ (aq) + CrO 4 2- (aq) [ ] i 0 0 [ ] c +4.0 x 10 -3 +5.0 x 10 -3 [ ] e 4.0 x 10 -3 5.0 x 10 -3 P = [Sr 2+ ][CrO 4 2- ] P = 4.0 x 10 -3 × 5.0 x 10 -3 = 2.0 x 10 -5 P = 2.0 x 10 -5 < K sp (SrCrO 4 ) = 3.6 x 10 -5 No precipitate will form.

34. 15 - 34 Life After A Precipitate 35.0 mL of 0.150 M Pb(NO 3 ) 2 is mixed with 15.0 mL of 0.800 M KIO 3. [Pb(NO 3 ) 2 ] = 0.150 M[KIO 3 ] = 0.800 M V 1 = 35.0 mLV 2 = 15.0 mL K sp (Pb(IO 3 ) 2 ) = 2.6 x 10 -13 (a)Does a precipitate form? [Pb(NO 3 ) 2 ] c × V c = [Pb(NO 3 ) 2 ] d × V d

35. 15 - 35. [Pb(NO 3 ) 2 ] d = 0.150 M × 35.0 mL 50.0 mL = 0.105 M [KIO 3 ] d = 0.800 M × 15.0 mL 50.0 mL = 0.240 M Pb(IO 3 ) 2 (s)Pb 2+ (aq) + 2IO 3 - (aq) P = [Pb 2+ ] [ IO 3 - ] 2 = 0.105 × 0.240 2 P = 6.0 x 10 -3

36. 15 - 36 P = 6.0 x 10 -3 > K sp = 2.6 x 10 -13, therefore a precipitate forms. (b) If a precipitate forms, what are the concentrations of Pb 2+ and IO 3 - in the final solution? This part is a little tricky because your first thought might be to think the initial concentrations of Pb 2+ and IO 3 2- will be zero, use the K sp, and determine the concentrations from that calculation.

37. 15 - 37 However, P > K sp, meaning that the concentrations of Pb 2+ (0.105 M) and IO 3 2- (0.240 M) are too large. Before equilibrium can be reestablished, there is a small window in which a stoichiometry (limiting reactant) scenario exists. A precipitate forms, meaning n b 0.105 0.240 Pb 2+ (aq) + 2IO 3 - (aq) → Pb(IO 3 ) 2 (s) n a 0 0.030

38. 15 - 38 This stoichiometry is very important because it says that when the shift occurs to reestablish the equilibrium, [Pb 2+ ] = 0 and [IO 3 - ] = 0.030 M. PbIO 3 (s)Pb 2+ (aq) + IO 3 2- (aq) [ ] i 0 0.030 [ ] c +x +2x [ ] e x 0.030 + 2x K sp = [Pb 2+ ][IO 3 2- ]

39. 15 - 39 2.6 x 10 -13 = x 0.030 + 2x ≈ 0.030 [Pb 2+ ] = 2.9 x 10 -10 M [IO 3 2- ] = 0.030 M + 2 2.9 x 10 -10 M × (0.030 + 2x) 2 × ≈ 0.030 M

40. 15 - 40 pH and Solubility The pH or pOH of a solution can have a drastic affect on the solubility of a salt. If HCl is added to a solution of the slightly soluble salt, CaCO 3, the solubility of the salt is increased.  This is an example of Le Chatelier’s principle.  Before adding HCl, the following equilibrium exists:

41. 15 - 41 CaCO 3 (s) Ca 2+ (aq) + CO 3 2- (aq)  Adding HCl results in the ionization of HCl. HCl(aq) → H + (aq) + Cl - (aq)  CO 3 2- is the conjugate base of carbonic acid which decreases the concentration of CO 3 2- as a result of the following reaction: CaCO 3 (s) + 2H + (aq) → Ba 2+ (aq) + CO 2 (g) + H 2 O(l)

42. 15 - 42  The system will respond to the stress brought about by the lowering of the [CO 3 2- ] by dissolving more of the slightly soluble salt. A similar situation arises when HCl is added to a solution of the slightly soluble salt, Fe(OH) 3.  In this case, the [OH - ] is decreased because of the formation of H 2 O as shown below: H + (aq) + OH - (aq) → H 2 O(l)

43. 15 - 43 A similar situation arises when KOH is added to a solution of the slightly soluble salt, Fe(OH) 3.  In this case, the [OH - ] is increased by the addition of KOH which suppresses the dissolving of Fe(OH) 3.

44. 15 - 44 pH and Solubility Problem (a)Determine the solubility of iron(III) hydroxide in pure water. K sp = 4 x 10 -38 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH - (aq) [ ] i 0 1.0 x 10 -7 [ ] c +x +3x [ ] e x 1.0 x 10 -7 + 3x K sp = [Fe 3+ ][OH - ] 3 4 x 10 -38 = x 27x 3 x = s = 4 x 10 -17 M ×

45. 15 - 45 Some observations from the previous problem:  Considering the K sp is minuscule, it is absolutely necessary to consider the [OH - ] found in pure water.  The approximation (1.0 x 10 -7 + 3x) ≈ 1.0 x 10 -7 is an excellent approximation when considering the K sp value.

46. 15 - 46 (b) Determine the solubility of iron(III) hydroxide in an aqueous solution whose pH = 5.0. pH + pOH = 14.00 pOH = 9.0 [OH-] = 10 -9 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH - (aq) [ ] i 0 10 -9 [ ] c +x +3x [ ] e x 10 -9 + 3x

47. 15 - 47 K sp = [Fe 3+ ][OH - ] 3 4 x 10 -38 = x (10 -9 + 3x) 3 ≈ 10 -27 x x = s = 4 x 10 -11 M Unless told otherwise, assume that there is no change in the pH. ×

48. 15 - 48 (c) Determine the solubility of iron(III) hydroxide in an aqueous solution whose pH = 11.0. pH + pOH = 14.00 pOH = 3.0 [OH-] = 10 -3 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH - (aq) [ ] i 0 10 -3 [ ] c +x +3x [ ] e x 10 -3 + 3x

49. 15 - 49 K sp = [Fe 3+ ][OH - ] 3 4 x 10 -38 = x (10 -3 + 3x) 3 ≈ 10 -9 x x = s = 4 x 10 -29 M Unless told otherwise, assume that there is no change in the pH. ×

50. 15 - 50 In conclusion: In pure water, s = 4 x 10 -17 M.  This very small solubility is not surprising when considering the K sp. In an acidic solution, s = 4 x 10 -11 M.  This drastic increase is not surprising when considering Le Chatelier’s principle.

51. 15 - 51 The stress on the system results from the formation of water which decreases the [OH - ]. The equilibrium shifts to the right resulting in an increased solubility. In a basic solution, s = 4 x 10 -29 M.  Because of the large [OH - ] resulting from the pH of the solution, the dissolving of the salt will be suppressed.