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

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

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. 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. Solubility Product Constants

6. Solubility Product Constants

7. Solubility Product Constants
The equilibrium constant expression for this dissolution is called a solubility product constant.
Ksp = solubility product constant

8. Solubility Product Constants

9. Solubility Product Constants
The solubility product constant, Ksp, 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.

10. Solubility Product Constants
The solubility product expression for Ag2S is:

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

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

13. 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.

14. Determination of Solubility Product Constants
Example 20-1: One liter of saturated silver chloride solution contains g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.

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

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

17. Determination of Solubility Product Constants
Example 20-2: One liter of saturated calcium fluoride solution contains gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2.

18. Uses of Solubility Product Constants
The solubility product constant can be used to calculate the solubility of a compound at 25oC.

19. Uses of Solubility Product Constants
Example 20-3: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x

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

21. The Common Ion Effect in Solubility Calculations
Example 20-5: Calculate the molar solubility of barium sulfate, BaSO4, in M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)

22. The Common Ion Effect in Solubility Calculations
The molar solubility of BaSO4 in M Na2SO4 solution is 1.1 x 10-8 M.
The molar solubility of BaSO4 in pure water is 1.0 x 10-5 M.
BaSO4 is 900 times more soluble in pure water than in M sodium sulfate!
Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution!
If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

23. The Reaction Quotient in Precipitation Reactions
The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form.

24. The Reaction Quotient in Precipitation Reactions
Example 20-6: We mix 100 mL of M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

25. 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(NO3)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 Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0 x
You do it!

26. The Reaction Quotient in Precipitation Reactions
Example 20-7: What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0 x

27. The Reaction Quotient in Precipitation Reactions
Example 20-8: Refer to example What volume of the solution (1.0 x 10-8 M Hg2+ ) contains 1.0 g of mercury?

28. Fractional Precipitation
The method of precipitating some ions from a 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.

29. Fractional Precipitation
Example 20-9: If solid sodium chloride is slowly added to a solution that is 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.

30. Fractional Precipitation
These three calculations give the [Cl-] required to precipitate AuCl ([Cl-] >2.0 x M), to precipitate AgCl ([Cl-] >1.8 x M), and to precipitate CuCl ([Cl-] >1.9 x 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.

31. 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.

32. Fractional Precipitation
The percent of Au+ ions unprecipitated just before AgCl precipitates is:

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

34. Fractional Precipitation
The percent of Ag+ ions unprecipitated just before AgCl precipitates is:

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

36. Simultaneous Equilibria Involving Slightly Soluble Compounds
Example 20-12: If 0.10 mole of ammonia and mole of magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution?
For Mg(OH)2, Ksp = 1.5 x Kb for NH3 = 1.8 x 10-5.

37. Simultaneous Equilibria Involving Slightly Soluble Compounds
Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example )
You do it!

38. 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 Qsp < Ksp.
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.

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

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

41. 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.

42. 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 Na2S2O3.

43. Dissolving Precipitates
Copper(II) hydroxide, which is light blue colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+.

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

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

46. 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 M in [Ag(NH3)2]+.
Kd = 6.3 x 10-8
Write the dissociation reaction and equilibrium concentrations.

47. Complex Ion Equilibria
Substitute the algebraic quantities into the dissociation expression.

48. Complex Ion Equilibria

49. Complex Ion Equilibria

50. 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 mole of silver chloride, AgCl?
The reaction of interest is:

51. Synthesis Question
Most kidney stones are made of calcium oxalate, Ca(O2CCO2). 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?

52. Group Question
The cavities that we get in our teeth are a result of the dissolving of the material our teeth are made of, calcium hydroxy apatite. How does using a fluoride based toothpaste decrease the occurrence of cavities?