1. Chapter 17 sections 4-7 Solubility Equilibria © 2012 Pearson Education, Inc.

2. Aqueous Equilibria Solubility Product Constants Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) © 2012 Pearson Education, Inc. The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2  ] where the equilibrium constant, K sp, is called the solubility product constant.

3. Aqueous Equilibria Sample Exercise 17.9 Writing Solubility-Product (K sp ) Expressions Practice Exercise Give the solubility-product-constant expressions and K sp values (from Appendix D) for (a) barium carbonate, (b) silver sulfate. Write the expression for the solubility-product constant for CaF 2, and look up the corresponding K sp value in Appendix D.

4. Aqueous Equilibria Solubility Product Constants K sp is not the same as solubility. Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M). © 2012 Pearson Education, Inc.

5. Aqueous Equilibria A.AgI is the most soluble of the three. B.AgBr is the most soluble of the three. C.AgCl is the most soluble of the three. D. All of the compounds are totally insoluble based on their small K sp values.

6. Aqueous Equilibria Solubility Product Constants For solids dissolving to form aqueous solutions. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2  (aq) K sp = solubility product constant and K sp = [Bi 3+ ] 2 [S 2  ] 3 “molar Solubility” = s = concentration of Bi 2 S 3 that dissolves, which equals ½ [Bi 3+ ] and 1 / 3 [S 2  ]. Note: K sp is constant (at a given temperature) s is variable (especially with a common ion present)

7. Aqueous Equilibria Sample Exercise 17.10 Calculating K sp from Solubility Solid silver chromate is added to pure water at 25  C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag 2 CrO 4 (s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3  10  4 M. Assuming that Ag 2 CrO 4 dissociates completely in water and that there are no other important equilibria involving Ag + or CrO 4 2– ions in the solution, calculate K sp for this compound. Practice Exercise A saturated solution of Mg(OH) 2 in contact with undissolved Mg(OH) 2 (s) is prepared at 25  C. The pH of the solution is found to be 10.17. Assuming that Mg(OH) 2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg 2+ or OH – ions, calculate K sp for this compound.

8. Aqueous Equilibria Sample Exercise 17.11 Calculating Solubility from K sp The K sp for CaF 2 is 3.9  10  11 at 25  C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. Practice Exercise The K sp for LaF 3 is 2  10  19. What is the solubility of LaF 3 in water in moles per liter?

9. Aqueous Equilibria Factors Affecting Solubility The Common-Ion Effect –If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO 4 (s)Ba 2+ (aq) + SO 4 2  (aq) © 2012 Pearson Education, Inc.

10. Aqueous Equilibria The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. CH 3 COOH(aq) + H 2 O(l)H 3 O + (aq) + CH 3 COO  (aq) © 2012 Pearson Education, Inc.

11. Aqueous Equilibria The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.” © 2012 Pearson Education, Inc.

12. Aqueous Equilibria The Common-Ion Effect Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is 6.8  10  4. [H + ] [F  ] [HF] K a = = 6.8  10  4 © 2012 Pearson Education, Inc.

13. Aqueous Equilibria The Common-Ion Effect Because HCl, a strong acid, is also present, the initial [H + ] is not 0, but rather 0.10 M. [HF], M[H + ], M[F  ], M Initially0.200.100 Change At equilibrium HF(aq)H + (aq) + F  (aq) © 2012 Pearson Education, Inc. +x+x x +x+x-x-x 0.10 + x  0.100.20  x  0.20

14. Aqueous Equilibria The Common-Ion Effect = x 1.4  10  3 = x (0.10) (x) (0.20) 6.8  10  4 = (0.20) (6.8  10  4 ) (0.10) © 2012 Pearson Education, Inc.

15. Aqueous Equilibria The Common-Ion Effect Therefore, [F  ] = x = 1.4  10  3 [H 3 O + ] = 0.10 + x = 0.10 + 1.4  10  3 = 0.10 M So, pH =  log (0.10) pH = 1.00 © 2012 Pearson Education, Inc.

16. Aqueous Equilibria Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF 2 at 25  C in a solution that is (a) 0.010 M in Ca(NO 3 ) 2, (b) 0.010 M in NaF. Practice Exercise For manganese(II) hydroxide,Mn(OH) 2, K sp = 1.6  10  13. Calculate the molar solubility of Mn(OH) 2 in a solution that contains 0.020 M NaOH.

17. Aqueous Equilibria Factors Affecting Solubility pH –If a substance has a basic anion, it will be more soluble in an acidic solution. –Substances with acidic cations are more soluble in basic solutions. © 2012 Pearson Education, Inc.

18. Aqueous Equilibria Sample Exercise 17.13 Predicting the Effect of Acid on Solubility Which of these substances are more soluble in acidic solution than in basic solution: (a) Ni(OH) 2 (s), (b) CaCO 3 (s), (c) BaF 2 (s), (d) AgCl(s)? Practice Exercise Write the net ionic equation for the reaction between an acid and (a) CuS, (b) Cu(N 3 ) 2.

19. Aqueous Equilibria Factors Affecting Solubility Complex Ions –Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. © 2012 Pearson Education, Inc.

20. Aqueous Equilibria Factors Affecting Solubility Complex Ions –The formation of these complex ions increases the solubility of these salts. © 2012 Pearson Education, Inc.

21. Aqueous Equilibria Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag + present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO 3 to give an equilibrium concentration of [NH 3 ] = 0.20 M. Neglect the small volume change that occurs when NH 3 is added. Practice Exercise Calculate [Cr 3+ ] in equilibrium with Cr(OH) 4  when 0.010 mol of Cr(NO 3 ) 3 is dissolved in 1 L of solution buffered at pH 10.0.

22. Aqueous Equilibria Factors Affecting Solubility Amphoterism –Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. –Examples of such cations are Al 3+, Zn 2+, and Sn 2+. © 2012 Pearson Education, Inc.

23. Aqueous Equilibria A.Amphoteric substances are acids whereas amphiprotic substances are acids. B.Amphoteric substances are bases whereas amphiprotic substances are acids. C.Amphoteric substances can act both as an acid and a base whereas amphiprotic substances are acids. D.Amphoteric substances can act both as an acid and a base whereas amphiprotic substances can accept or donate a proton.

24. Aqueous Equilibria Will a Precipitate Form? In a solution, –If Q = K sp, the system is at equilibrium and the solution is saturated. –If Q < K sp, more solid can dissolve until Q = K sp. –If Q > K sp, the salt will precipitate until Q = K sp. © 2012 Pearson Education, Inc.

25. Aqueous Equilibria Sample Exercise 17.15 Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of 8.0  10  3 M Pb(NO 3 ) 2 is added to 0.40 L of 5.0  10  3 M Na 2 SO 4 ? Practice Exercise Does a precipitate form when 0.050 L of 2.0  10  2 M NaF is mixed with 0.010 L of 1.0  10  2 M Ca(NO 3 ) 2 ?

26. Aqueous Equilibria Precipitation After precipitation occurs, the equilibrium concentration of the remaining solution can be determined. First calculate concentration if all solid precipitates. Then determine how much solid dissolves.

27. Aqueous Equilibria Example Consider a solution formed by mixing 100.0 mL of 0.0500 M Pb(NO 3 ) 2 and 200.0 mL of 0.100 M NaI. The K sp =1.4 x 10 -8 for PbI 2

28. Aqueous Equilibria Solution Calculate the initial concentration:

29. Aqueous Equilibria Solution Pb 2+ I-I- Preliminary1.67 x 10 -2 M6.67 x 10 -2 M Reacted Initial 0 - 1.67 x 10 -2 M- 3.34 x 10 -2 M 3.33 x 10 -2 M To calculate the concentration of ions in solution, first assume that the precipitation reaction goes to completion

30. Aqueous Equilibria Solution Pb 2+ I-I- Initial0 mmol0.0333M Change Equilibrium +x+ 2x +x0.0333 + 2x Then determine the ‘bounce back’ or the amount of precipitate that re-dissolves. because the K sp is small, x<< 0.0333 0.0333 M + 1.3x10 -5 M+ 2(1.3 x 10 -5 M) 1.3 x10 -5 M+ 0.0333 M

31. Aqueous Equilibria Example A solution is prepared by mixing 150.0 mL of 1.00 x 10 -2 M Mg(NO 3 ) 2 and 250.0 mL of 1.00 x 10 -1 M NaF. Calaculate the concentrations of Mg 2+ and F - at equilibrium with solid MgF 2 (K sp = 6.4 x 10 -9 )

32. Aqueous Equilibria Solution Calculate the initial concentration:

33. Aqueous Equilibria Solution Mg 2+ F-F- Preliminary3.75 x 10 -3 M6.25 x 10 -2 M Reacted Initial 0 - 3.75 x 10 -3 M- 7.50 x 10 -3 M 5.50 x 10 -2 M Assume that the precipitation reaction goes to completion

34. Aqueous Equilibria Solution Mg 2+ F-F- Initial0 M0.0550 M Change Equilibrium +x+ 2x +x0.0550 + 2x because the K sp is small, x<< 0.0550 0.0550 M + 2.1x10 -6 M+ 2(2.1 x 10 -6 M) 2.1 x10 -6 M+ 0.0550 M Then determine the ‘bounce back’ or the amount of precipitate that re-dissolves.

35. Aqueous Equilibria Selective Precipitation Used to separate mixtures of metals in solution. Use anion that forms precipitate with one metal, but not another

36. Aqueous Equilibria Example A solution contains 1.0 x 10 -4 M Cu + and 2.0 x 10 -3 M Pb 2+. If a source of I - is added gradually to this solution, which compound will precipitate first? Specify the concentration of I - needed to begin precipitation. PbI 2 (K sp = 1.4 x 10 -8 ) CuI (K sp = 5.3 x 10 -12 )

37. Aqueous Equilibria Solution For PbI 2 : K sp = [Pb 2+ ][I - ] 2 = 1.4 x 10 -8 [Pb 2+ ] 0 = 2.0 x 10 -3 M, therefore, [Pb 2+ ][I - ] 2 = (2.0 x 10 -3 M)[I - ] 2 = 1.4 x 10 -8 [I - ]= 2.6 x 10 -3

38. Aqueous Equilibria Solution For CuI: K sp = [Cu + ][I - ] = 3.5 x 10 -12 [Cu + ] 0 = 1.0 x 10 -4 M, therefore, [Cu + ][I - ] = (1.0 x 10 -4 M)[I - ]= 3.5 x 10 -12 [I - ]= 5.3 x 10 -8

39. Aqueous Equilibria Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution contains 1.0  10  2 M and 2.0  10  2 M Pb 2+. When Cl  is added, both AgCl (K sp = 1.8  10  10 ) and PbCl 2 (K sp = 1.7  10  5 M) can precipitate. What concentration of Cl  is necessary to begin the precipitation of each salt? Which salt precipitates first? Practice Exercise A solution consists of 0.050 M Mg 2+ and Cu 2+. Which ion precipitates first as OH  is added? What concentration of OH  is necessary to begin the precipitation of each cation? [K sp = 1.8  10  11 for Mg(OH) 2, and K sp = 4.8  10  20 for Cu(OH) 2.]

40. Aqueous Equilibria A.ZnS and CuS both precipitate and separating the two ions is no longer feasible using given procedures. B.ZnS precipitates and separating the two ions is still feasible using given procedures. C.CuS precipitates and separating the two ions is still feasible using given procedures. D.No change in observations and ability to separate the two ions.

41. Aqueous Equilibria Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. © 2012 Pearson Education, Inc.

42. Aqueous Equilibria A.No, both would precipitate in step 1 and subsequently are not easily separated. B.No, both would precipitate in step 2 and subsequently are not easily separated. C.Yes, ZnS precipitates in step 1 and CuS in step 4. D.Yes, CuS precipitates in step 2 and Zn 2+ remains in solution.

43. Aqueous Equilibria A.The solution definitely contains either Pb 2+ or Hg 2 2+ cation. B.The solution definitely contains the Ag + cation. C.The solution must contain one or more of the following cations: Cu 2+, Bi 3+, or Cd 2+. D.The solution must contain one or more of the following cations: Ag +, Pb 2+ or Hg 2 2+.