1. 1 16 Ionic Equilibria III: The Solubility Product Principle

2. 2 Determination of Solubility Product Constants Example 16-7: 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:

3. 3 Determination of Solubility Product Constants Example 16-8: 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.

4. 4 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!

5. 5 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!

6. 6 Uses of Solubility Product Constants Example 16-9: 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.

7. 7 Uses of Solubility Product Constants

8. 8 Make the algebraic substitution of x’s into solubility product expression and solve for x, giving the ion concentrations.

9. 9 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.

10. 10 Uses of Solubility Product Constants Example 16-10: 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.

11. 11 Uses of Solubility Product Constants Be careful, do not forget the stoichiometric coefficient of 2!

12. 12 Uses of Solubility Product Constants Substitute the algebraic expressions into the solubility product expression.

13. 13 Uses of Solubility Product Constants Solve for the pOH and pH.

14. 14 The Common Ion Effect in Solubility Calculations Example 16-11: 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?)

15. 15 The Common Ion Effect in Solubility Calculations 1.Write equations to represent the equilibria.

16. 16 The Common Ion Effect in Solubility Calculations 2.Substitute the algebraic representations of the concentrations into the K sp expression and solve for x.

17. 17 The Reaction Quotient in Precipitation Reactions Example 16-12: 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?

18. 18 The Reaction Quotient in Precipitation Reactions 1.Write out the solubility expressions.

19. 19 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:

20. 20 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:

21. 21 The Reaction Quotient in Precipitation Reactions Finally, calculate Q sp for PbSO 4 and compare it to the K sp.

22. 22 The Reaction Quotient in Precipitation Reactions Example 16-13: 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.

23. 23 The Reaction Quotient in Precipitation Reactions Example 16-14: Refer to example 16-13. What volume of the solution (1.0 x 10 -8 M Hg 2+ ) contains 1.0 g of mercury?

24. 24 Fractional Precipitation Example 16-15: 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.

25. 25 Fractional Precipitation

26. 26 Fractional Precipitation Repeat the calculation for silver chloride.

27. 27 Fractional Precipitation Finally, for copper (I) chloride to precipitate.

28. 28 Fractional Precipitation Example 16-16: Calculate the percentage of Au + ions that precipitate before AgCl begins to precipitate. –Use the [Cl - ] from Example 16-15 to determine the [Au + ] remaining in solution just before AgCl begins to precipitate.

29. 29 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.

30. 30 Fractional Precipitation A similar calculation for the concentration of Ag + ions unprecipitated before CuCl begins to precipitate is:

31. 31 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.

32. 32 Simultaneous Equilibria Involving Slightly Soluble Compounds Example 16-17: 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 - ].

33. 33 Simultaneous Equilibria Involving Slightly Soluble Compounds

34. 34 Simultaneous Equilibria Involving Slightly Soluble Compounds

35. 35 Simultaneous Equilibria Involving Slightly Soluble Compounds

36. 36 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.

37. 37 Simultaneous Equilibria Involving Slightly Soluble Compounds Example 16-18: 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 16-17.) Calculate the maximum [OH - ] that can exist in a solution that is 0.010 M in Mg 2+.

38. 38 Simultaneous Equilibria Involving Slightly Soluble Compounds

39. 39 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.

40. 40 Simultaneous Equilibria Involving Slightly Soluble Compounds

41. 41 Simultaneous Equilibria Involving Slightly Soluble Compounds

42. 42 Simultaneous Equilibria Involving Slightly Soluble Compounds Check these values by calculating Q sp for Mg(OH) 2.

43. 43 Simultaneous Equilibria Involving Slightly Soluble Compounds Use the ion product for water to calculate the [H + ] and the pH of the solution.

44. 44 Complex Ion Equilibria Example 16-19: 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:

45. 45 Complex Ion Equilibria Two equilibria are involved when silver chloride dissolves in aqueous ammonia.

46. 46 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.

47. 47 Complex Ion Equilibria

48. 48 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.

49. 49 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.

50. 50 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.

51. 51 Complex Ion Equilibria Finally, calculate the total number of moles of ammonia necessary.

52. 52 Synthesis Question Most kidney stones are made of calcium oxalate, Ca(O 2 CCO 2 ). Patients who have their first kidney stones are given an extremely simple solution to stop further stone formation. They are told to drink six to eight glasses of water a day. How does this stop kidney stone formation?

53. 53 Synthesis Question