Chapter 16

1. Chapter 16 Precipitation Equilibrium Solubility Product Principle

2. 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.

3. Solubility Product Constants • The equilibrium constant expression for this dissolution is called a solubility product constant. • Ksp=solubility product constant • Molar concentration of ions raised to their stoichiometric powers at equilibrium

4. Solubility Product Constants • Solubility product constant 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.

5. Solubility Product Constants • Its solubility product expression is

6. Solubility Product Constants • The dissolution of solid calcium phosphate in water is represented as

7. Solubility Product Constants • Its solubility product constant expression is

8. Solubility Product Constants • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as • Ksp has a fixed value for a given system at a given temperature

9. Solubility Product Constants • The same rules apply for compounds that have more than two kinds of ions. • An example is calcium ammonium phosphate.

10. Determination of Solubility Product Constants • Example: One liter of saturated silver chloride solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl. • Molar solubility can be calculated from the data:

11. Determination of Solubility Product Constants • The equation for the dissociation of silver chloride and its solubility product expression are

12. Determination of Solubility Product Constants • Substitution into the solubility product expression gives

13. Uses of Solubility Product Constants • We can use the solubility product constant to calculate the solubility of a compound at 25oC. • Example: 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. Ksp= 1.1 x 10-10.

14. Uses of Solubility Product Constants • Example: 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. Ksp= 1.1 x 10-10.

15. Uses of Solubility Product Constants • Substitute into solubility product expression and solve for x, giving the ion concentrations.

16. Uses of Solubility Product Constants • Now we can calculate the mass of BaSO4 in 1.00 L of saturated solution.

17. The Reaction Quotient in Precipitation Reactions • Use solubility product constants to calculate the concentration of ions in a solution and whether or not a precipitate will form. • Example: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

18. The Reaction Quotient in Precipitation Reactions • Example: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

19. The Reaction Quotient in Precipitation Reactions • Calculate the Qsp for PbSO4. • Solution volumes are additive. • Concentrations of the important ions are:

20. The Reaction Quotient in Precipitation Reactions • Finally, we calculate Qsp for PbSO4.

21. Fractional Precipitation • Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. • Look at a solution that contains Cu+, Ag+, and Au+ • We could precipitate them as chlorides

22. Fractional Precipitation • Fractional precipitation is a method of precipitating some ions from solution while leaving others in solution. • Look at a solution that contains Cu+, Ag+, and Au+ • We could precipitate them as chlorides

23. Fractional Precipitation • Example: 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 I chlorides.

24. Fractional Precipitation • Example: 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 I chlorides.

25. Fractional Precipitation • Example: 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 I chlorides.

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

27. Fractional Precipitation • For copper (I) chloride to precipitate.

28. Fractional Precipitation • We have calculated the [Cl-] required to precipitate AuCl, [Cl-] >2.0 x 10-11 M to precipitate AgCl, [Cl-] >1.8 x 10-8 M to precipitate CuCl, [Cl-] >1.9 x 10-5 M • We can calculate the amount of Au+ precipitated before Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before Cu+ begins to precipitate.

29. Fractional Precipitation • Example: Calculate the percent of Au+ ions that precipitate before AgCl begins to precipitate. • Use the [Cl-] from before to determine the [Au+] remaining in solution just before AgCl begins to precipitate.

30. Fractional Precipitation • Example: Calculate the percent of Au+ ions that precipitate before AgCl begins to precipitate. • Use the [Cl-] from before to determine the [Au+] remaining in solution just before AgCl begins to precipitate.

31. Fractional Precipitation • The percent of Au+ ions unprecipitated just before AgCl precipitates is

32. Fractional Precipitation • The percent of Au+ ions unprecipitated just before AgCl precipitates is • Therefore, 99.99989% of the Au+ ions precipitates before AgCl begins to precipitate.

33. Fractional Precipitation • Similar calculations for the concentration of Ag+ ions unprecipitated before CuCl begins to precipitate gives

34. Fractional Precipitation • The percent of Au+ ions unprecipitated just before AgCl precipitates is

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

36. Factors that Affect Solubility CaF2 (s)  Ca+2 (aq)+ 2F- (aq) • Addition of a Common ion (F- from NaF) • Solubility decreases • Equilibrium shifts to left • Changes in pH (H+ reacts with F-) • Solubility increases (with increasing pH) • Equilibrium shifts to right

37. Factors that Affect Solubility Ag+(aq) + 2NH3(aq)  Ag(NH3)2+ (aq) complex ion • Formation of a complex ion • Lewis Acid base chemistry • Calculate Kf Formation Constant

38. Factors that Affect Solubility Ag+(aq) + 2NH3(aq)  Ag(NH3)2+ (aq) complex ion AgCl(s)  Ag+(aq) + Cl-(aq) • In formation of complex ion • Removed Ag+ from the equilibrium • Equilibrium shifts to right • Favors dissolving AgCl

39. 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?

40. Synthesis Question

41. 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?