1. 1 20 Ionic Equilibria III: The Solubility Product Principle

2. 2 Chapter Goals 1.Solubility Product Constants 2.Determination of Solubility Product Constants 3.Uses of Solubility Product Constants 4.Fractional Precipitation 5.Simultaneous Equilibria Involving Slightly Soluble Compounds 6.Dissolving Precipitates

3. 3 Solubility Product Constants Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

4. 4 Solubility Product Constants Silver chloride, AgCl,is rather insoluble in water. Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

5. 5 Solubility Product Constants The equilibrium constant expression for this dissolution is called a solubility product constant. –K sp = solubility product constant

6. 6 Solubility Product Constants The solubility product constant, K sp, for a compound is the product of the concentrations of the constituent ions, each raised to the power that corresponds to the number of ions in one formula unit of the compound. Consider the dissolution of silver sulfide in water.

7. 7 Solubility Product Constants The solubility product expression for Ag 2 S is:

8. 8 Solubility Product Constants The dissolution of solid calcium phosphate in water is represented as: The solubility product constant expression is: You do it!

9. 9 Solubility Product Constants In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

10. 10 Solubility Product Constants The same rules apply for compounds that have more than two kinds of ions. One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

11. 11 Determination of Solubility Product Constants Example 20-1: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25 o C. Calculate the molar solubility of, and K sp for, AgCl. The molar solubility can be easily calculated from the data:

12. 12 Determination of Solubility Product Constants The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are:

13. 13 Determination of Solubility Product Constants Substitution of the molar concentrations into the solubility product expression gives:

14. 14 Determination of Solubility Product Constants Example 20-2: One liter of saturated calcium fluoride solution contains 0.0167 gram of CaF 2 at 25 o C. Calculate the molar solubility of, and K sp for, CaF 2. 1.Calculate the molar solubility of CaF 2.

15. 15 Determination of Solubility Product Constants From the molar solubility, we can find the ion concentrations in saturated CaF 2. Then use those values to calculate the K sp. –Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!

16. 16 Uses of Solubility Product Constants The solubility product constant can be used to calculate the solubility of a compound at 25 o C. Example 20-3: Calculate the molar solubility of barium sulfate, BaSO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25 o C. For barium sulfate, K sp = 1.1 x 10 -10.

17. 17 Uses of Solubility Product Constants

18. 18 Uses of Solubility Product Constants Make the algebraic substitution of x’s into solubility product expression and solve for x, giving the ion concentrations.

19. 19 Uses of Solubility Product Constants Finally, to calculate the mass of BaSO 4 in 1.00 L of saturated solution, use the definition of molarity.

20. 20 Uses of Solubility Product Constants Example 20-4: The solubility product constant for magnesium hydroxide, Mg(OH) 2, is 1.5 x 10 -11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25 o C. You do it!

21. 21 Uses of Solubility Product Constants Be careful, do not forget the stoichiometric coefficient of 2!

22. 22 Uses of Solubility Product Constants Substitute the algebraic expressions into the solubility product expression.

23. 23 Uses of Solubility Product Constants Solve for the pOH and pH.

24. 24 The Common Ion Effect in Solubility Calculations Example 20-5: Calculate the molar solubility of barium sulfate, BaSO 4, in 0.010 M sodium sulfate, Na 2 SO 4, solution at 25 o C. Compare this to the solubility of BaSO 4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)

25. 25 The Common Ion Effect in Solubility Calculations 1.Write equations to represent the equilibria.

26. 26 The Common Ion Effect in Solubility Calculations 2.Substitute the algebraic representations of the concentrations into the K sp expression and solve for x.

27. 27 The Common Ion Effect in Solubility Calculations The molar solubility of BaSO 4 in 0.010 M Na 2 SO 4 solution is 1.1 x 10 -8 M. The molar solubility of BaSO 4 in pure water is 1.0 x 10 -5 M. –BaSO 4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate! –Adding sodium sulfate to a solution is a fantastic method to remove Ba 2+ ions from solution! If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

28. 28 The Reaction Quotient in Precipitation Reactions The reaction quotient, Q, and the K sp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K 2 SO 4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO 3 ) 2 solutions. Will a precipitate form?

29. 29 The Reaction Quotient in Precipitation Reactions 1.Write out the solubility expressions.

30. 30 The Reaction Quotient in Precipitation Reactions Calculate the Q sp for PbSO 4. –Assume that the solution volumes are additive. –Concentrations of the important ions are:

31. 31 The Reaction Quotient in Precipitation Reactions Finally, calculate Q sp for PbSO 4 and compare it to the K sp.

32. 32 The Reaction Quotient in Precipitation Reactions Example 20-7: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO 3 ) 2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na 2 S, is required to reduce the Hg 2+ concentration to 1.0 x 10 -8 M? For HgS, K sp =3.0 x 10 -53. You do it!

33. 33 The Reaction Quotient in Precipitation Reactions

34. 34 The Reaction Quotient in Precipitation Reactions Example 20-8: Refer to example 20-7. What volume of the solution (1.0 x 10 -8 M Hg 2+ ) contains 1.0 g of mercury?

35. 35 Fractional Precipitation The method of precipitating some ions from solution while leaving others in solution is called fractional precipitation. –If a solution contains Cu +, Ag +, and Au +, each ion can be precipitated as chlorides.

36. 36 Fractional Precipitation Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu +, Ag +, and Au + ions, which compound precipitates first? Calculate the concentration of Cl - required to initiate precipitation of each of these metal chlorides.

37. 37 Fractional Precipitation

38. 38 Fractional Precipitation Repeat the calculation for silver chloride.

39. 39 Fractional Precipitation Finally, for copper (I) chloride to precipitate.

40. 40 Fractional Precipitation These three calculations give the [Cl - ] required to precipitate AuCl ([Cl - ] >2.0 x 10 -11 M), to precipitate AgCl ([Cl - ] >1.8 x 10 -8 M), and to precipitate CuCl ([Cl - ] >1.9 x 10 -5 M). It is also possible to calculate the amount of Au + precipitated before the Ag + begins to precipitate, as well as the amounts of Au + and Ag + precipitated before the Cu + begins to precipitate.

41. 41 Fractional Precipitation Example 20-10: Calculate the percentage of Au + ions that precipitate before AgCl begins to precipitate. –Use the [Cl - ] from Example 20-9 to determine the [Au + ] remaining in solution just before AgCl begins to precipitate.

42. 42 Fractional Precipitation The percent of Au + ions unprecipitated just before AgCl precipitates is: Therefore, 99.9% of the Au + ions precipitates before AgCl begins to precipitate.

43. 43 Fractional Precipitation A similar calculation for the concentration of Ag + ions unprecipitated before CuCl begins to precipitate is:

44. 44 Fractional Precipitation The percent of Ag + ions unprecipitated just before AgCl precipitates is: Thus, 99.905% of the Ag + ions precipitates before CuCl begins to precipitate.

45. 45 Simultaneous Equilibria Involving Slightly Soluble Compounds Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

46. 46 Simultaneous Equilibria Involving Slightly Soluble Compounds Example 20-12: If 0.10 mole of ammonia and 0.010 mole of magnesium nitrate, Mg(NO 3 ) 2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? –For Mg(OH) 2, K sp = 1.5 x 10 -11. K b for NH 3 = 1.8 x 10 -5. 1.Calculate Q sp for Mg(OH) 2 and compare it to K sp. –Mg(NO 3 ) 2 is a soluble ionic compound so [Mg 2+ ] = 0.010 M. –Aqueous ammonia is a weak base that we can calculate [OH - ].

47. 47 Simultaneous Equilibria Involving Slightly Soluble Compounds

48. 48 Simultaneous Equilibria Involving Slightly Soluble Compounds Once the concentrations of both the magnesium and hydroxide ions are determined, the Q sp can be calculated and compared to the K sp.

49. 49 Simultaneous Equilibria Involving Slightly Soluble Compounds Example 20-13: How many moles of solid ammonium chloride, NH 4 Cl, must be used to prevent precipitation of Mg(OH) 2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO 3 ) 2 ? (Note the similarity between this problem and Example 20-12.) You do it! Calculate the maximum [OH - ] that can exist in a solution that is 0.010 M in Mg 2+.

50. 50 Simultaneous Equilibria Involving Slightly Soluble Compounds

51. 51 Simultaneous Equilibria Involving Slightly Soluble Compounds Using the maximum [OH - ] that can exist in solution, determine the number of moles of NH 4 Cl required to buffer 0.10 M aqueous ammonia so that the [OH - ] does not exceed 3.9 x 10 -5 M.

52. 52 Simultaneous Equilibria Involving Slightly Soluble Compounds

53. 53 Simultaneous Equilibria Involving Slightly Soluble Compounds

54. 54 Simultaneous Equilibria Involving Slightly Soluble Compounds Check these values by calculating Q sp for Mg(OH) 2.

55. 55 Simultaneous Equilibria Involving Slightly Soluble Compounds Use the ion product for water to calculate the [H + ] and the pH of the solution.

56. 56 Dissolving Precipitates For an insoluble solid, if the ion concentrations (of either the cation or anion) are decreased, a solid precipitate can be dissolved. –The trick is to make Q sp < K sp. One method is to convert the ions into weak electrolytes. –Make these ions more water soluble. If insoluble metal hydroxides are dissolved in strong acids, they form soluble salts and water.

57. 57 Dissolving Precipitates For example, look at the dissolution of Mg(OH) 2 in HCl.

58. 58 Dissolving Precipitates A second method is to dissolve insoluble metal carbonates in strong acids. –The carbonates will form soluble salts, carbon dioxide, and water.

59. 59 Dissolving Precipitates A third method is to convert an ion to another species by an oxidation-reduction reaction. For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur.

60. 60 Dissolving Precipitates A fourth method is complex ion formation. The cations in many slightly soluble compounds will form complex ions. –This is the method used to dissolve unreacted AgBr and AgCl on photographic film. Photographic “hypo” is Na 2 S 2 O 3.

61. 61 Dissolving Precipitates Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH 3 ) 4 ] 2+.

62. 62 Complex Ion Equilibria A metal ion coordinated to several neutral molecules or anions forms compounds called complex ions. Familiar examples of complex ions include:

63. 63 Complex Ion Equilibria The dissociation of complex ions can be represented similarly to equilibria. For example:

64. 64 Complex Ion Equilibria Complex ion equilibrium constants are called dissociation constants. Example 20-14: Calculate the concentration of silver ions in a solution that is 0.010 M in [Ag(NH 3 ) 2 ] +. –K d = 6.3 x 10 -8 1.Write the dissociation reaction and equilibrium concentrations.

65. 65 Complex Ion Equilibria 2.Substitute the algebraic quantities into the dissociation expression.

66. 66 Complex Ion Equilibria

67. 67 Complex Ion Equilibria

68. 68 Complex Ion Equilibria Example 20-15: How many moles of ammonia must be added to 2.00 L of water so that it will just dissolve 0.010 mole of silver chloride, AgCl? The reaction of interest is:

69. 69 Complex Ion Equilibria Two equilibria are involved when silver chloride dissolves in aqueous ammonia.

70. 70 Complex Ion Equilibria The [Ag + ] in the solution must satisfy both equilibrium constant expressions. Because the [Cl - ] is known, the equilibrium concentration of Ag + can be calculated from K sp for AgCl.

71. 71 Complex Ion Equilibria

72. 72 Complex Ion Equilibria Substitute the maximum [Ag + ] into the dissociation constant expression for [Ag(NH 3 ) 2 ] + and solve for the equilibrium concentration of NH 3.

73. 73 Complex Ion Equilibria The amount just calculated is the equilibrium concentration of NH 3 in the solution. But the total concentration of NH 3 is the equilibrium amount plus the amount used in the complex formation.

74. 74 Complex Ion Equilibria Finally, calculate the total number of moles of ammonia necessary.

75. 75 20 Ionic Equilibria III: The Solubility Product Principle