1. Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE Chapter 15 Applications of Aqueous Equilibria John E. McMurry Robert C. Fay CHEMISTRY Fifth Edition Copyright © 2008 Pearson Prentice Hall, Inc.

2. Chapter 15/2 Neutralization Reactions H 2 O(l)H 1+ (aq) + OH 1- (aq) Assuming complete dissociation: Strong Acid-Strong Base After neutralization: pH = 7 (net ionic equation) H 2 O(l) +NaOH(aq)HCl(aq)+NaCl(aq) 2H 2 O(l)H 3 O 1+ (aq) + OH 1- (aq) or

3. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/3 Neutralization Reactions H 2 O(l) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + OH 1- (aq) Assuming complete dissociation: Weak Acid-Strong Base After neutralization: pH > 7 (net ionic equation) H 2 O(l) +CH 3 CO 2 H(aq) +NaOH(aq)NaCH 3 CO 2 (aq)

4. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/4 NH 4 1+ (aq)H 1+ (aq) + NH 3 (aq) Neutralization Reactions Assuming complete dissociation: Strong Acid-Weak Base After neutralization: pH < 7 NH 4 Cl(aq)HCl(aq)+ NH 3 (aq) or H 2 O(l) +H 3 O 1+ (aq) + OH 1- (aq)NH 4 1+ (aq) (net ionic equation)

5. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/5 NH 4 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + NH 3 (aq) Neutralization Reactions Weak Acid-Weak Base After neutralization: pH = ? NH 4 CH 3 CO 2 (aq)CH 3 CO 2 H(aq) + NH 3 (aq) (net ionic equation)

6. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/6 The Common-Ion Effect H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) Common-Ion Effect: The shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium.

7. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/7 The Common-Ion Effect The pH of 0.10 M acetic acid is 2.89. Calculate the pH of a solution that is prepared by dissolved 0.10 mol of acetic acid and 0.10 mol sodium acetate in enough water to make 1.00 L of solution. H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) I 0.10≈00.10 C-x+x E0.10 - xx0.10 + x [H 3 O 1+ ][CH 3 CO 2 1- ] [CH 3 CO 2 H] K a =

8. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/8 The Common-Ion Effect x(0.10) 0.10 ≈1.8 x 10 -5 = (x)(0.10 + x) (0.10 - x) pH =-log([H 3 O 1+ ]) = -log(1.8 x 10 -5 ) = 4.74 x = [H 3 O 1+ ] = 1.8 x 10 -5 M Why is there a difference in pH?

9. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/9 The Common-Ion Effect H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) The addition of acetate ion to a solution of acetic acid suppresses the dissociation of the acid. The equilibrium shifts to the left. Le Châtelier’s Principle

11. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/11 Buffer Solutions Buffer Solution: A solution which contains a weak acid and its conjugate base and resists drastic changes in pH. CH 3 CO 2 H + CH 3 CO 2 1- HF + F 1- NH 4 1+ + NH 3 H 2 PO 4 2- + HPO 4 2- Weak acid + Conjugate base For Example:

12. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/12 Buffer Solutions Conjugate base (NaCH 3 CO 2 ) Weak acid H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l) H 2 O(l) + CH 3 CO 2 H(aq)CH 3 CO 2 1- (aq) + H 3 O 1+ (aq) 100% Addition of H 3 O 1+ to a buffer: Addition of OH 1- to a buffer: H 2 O(l) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + OH 1- (aq) 100%

13. Buffer Solutions

14. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/14 The Henderson-Hasselbalch Equation Conjugate baseWeak acid H 3 O 1+ (aq) + Base(aq)Acid(aq) + H 2 O(l) [Acid] [Base] [H 3 O 1+ ] = K a K a = [H 3 O 1+ ][Base] [Acid] H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l)

15. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/15 The Henderson-Hasselbalch Equation [Base] [Acid] pH = pK a + log [Acid] [Base] [H 3 O 1+ ] = K a -log([H 3 O 1+ ]) = -log(K a ) - log [Acid] [Base]

16. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/16 pH Titration Curves Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. Erlenmeyer flask buret unknown concentration solution standard solution (known concentration) An indicator is added which changes color once the reaction is complete

17. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/17 Strong Acid-Strong Base Titrations Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

19. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/19 Strong Acid-Strong Base Titrations 1.Before Addition of Any NaOH 0.100 M HCl

20. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/20 2.Before the Equivalence Point Strong Acid-Strong Base Titrations 2H 2 O(l)H 3 O 1+ (aq) + OH 1- (aq) 100% Excess H 3 O 1+

21. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/21 3.At the Equivalence Point Strong Acid-Strong Base Titrations The quantity of H 3 O 1+ is equal to the quantity of added OH 1-. pH = 7

22. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/22 4.Beyond the Equivalence Point Strong Acid-Strong Base Titrations Excess OH 1-

23. Strong Acid-Strong Base Titrations

24. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/24 Weak Acid-Strong Base Titrations

25. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/25 Weak Acid-Strong Base Titrations 1.Before Addition of Any NaOH 0.100 M CH 3 CO 2 H H 3 O 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + H 2 O(l)

26. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/26 Weak Acid-Strong Base Titrations 2.Before the Equivalence Point H 2 O(l) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) + OH 1- (aq) 100% Excess CH 3 CO 2 H Buffer region

27. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/27 Weak Acid-Strong Base Titrations 3.At the Equivalence Point OH 1- (aq) + CH 3 CO 2 H(aq)CH 3 CO 2 1- (aq) + H 2 O(l) Notice how the pH at the equivalence point is shifted up versus the strong acid titration. pH > 7

28. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/28 Notice that the strong and weak acid titration curves correspond after the equivalence point. Weak Acid-Strong Base Titrations 4.Beyond the Equivalence Point Excess OH 1-

29. Weak Acid-Strong Base Titrations

30. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/30 Weak Base-Strong Acid Titrations 1.Before Addition of Any HCl 0.100 M NH 3

31. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/31 Weak Base-Strong Acid Titrations 2.Before the Equivalence Point Excess NH 3 NH 4 1+ (aq) + H 2 O(l)NH 3 (aq) + H 3 O 1+ (aq) 100% Buffer region

32. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/32 Weak Base-Strong Acid Titrations 3.At the Equivalence Point H 3 O 1+ (aq) + NH 3 (aq)NH 4 1+ (aq) + H 2 O(l) pH < 7

33. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/33 Weak Base-Strong Acid Titrations 4.Beyond the Equivalence Point Excess H 3 O 1+

34. Polyprotic Acid-Strong Base Titrations

35. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/35 A saturated solution of calcium fluoride in contact with solid CaF 2 contains constant equilibrium concentrations of Ca 2+ (aq) and F 1- (aq) because at equilibrium the ions crystallize at the same rate as the solid dissolves. Solubility Equilibria Ca 2+ (aq) + 2F 1- (aq)CaF 2 (s) K sp = [Ca 2+ ][F 1- ] 2

36. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/36 Solubility Equilibria mM n+ (aq) + xX y- (aq)MmXx(s)MmXx(s) K sp = [M n+ ] m [X y- ] x

37. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/37 Measuring K sp and Calculating Solubility from K sp If the concentrations of Ca 2+ (aq) and F 1- (aq) in a saturated solution of calcium fluoride are known, K sp may be calculated. Ca 2+ (aq) + 2F 1- (aq)CaF 2 (s) K sp = [Ca 2+ ][F 1- ] 2 = (3.3 x 10 -4 )(6.7 x 10 -4 ) 2 = 1.5 x 10 -10 [F 1- ] = 6.7 x 10 -4 M[Ca 2+ ] = 3.3 x 10 -4 M (at 25 °C)

38. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/38 Measuring K sp and Calculating Solubility from K sp

39. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/39 Measuring K sp and Calculating Solubility from K sp Calculate the molar solubility of MgF 2 in water at 25 °C. x = [Mg 2+ ] = Molar solubility = 2.6 x 10 -4 M K sp = 7.4 x 10 -11 = [Mg 2+ ][F 1- ] 2 = (x)(2x) 2 4x 3 = 7.4 x 10 -11 Mg 2+ (aq) + 2F 1- (aq)MgF 2 (s) E x2x Solubility and the Common-Ion Effect

40. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/40 Factors That Affect Solubility Calculate the molar solubility of MgF 2 in 0.10 M NaF at 25 °C. Solubility and the Common-Ion Effect Mg 2+ (aq) + 2F 1- (aq)MgF 2 (s) E x0.10 + 2x 7.4 x 10 -11 = (x)(0.10 + 2x) 2 ≈ (x)(0.10) 2 K sp = 7.4 x 10 -11 = [Mg 2+ ][F 1- ] 2 = (x)(0.10 + 2x) 2 = 7.4 x 10 -9 M (0.10) 2 7.4 x 10 -9 x = [Mg 2+ ] = Molar solubility =

41. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/41 Factors That Affect Solubility Mg 2+ (aq) + 2F 1- (aq)MgF 2 (s) Solubility and the Common-Ion Effect Molar solubility in 0.10 M NaF: Molar solubility: 7.4 x 10 -9 M 2.6 x 10 -4 M Why does the solubility decrease in the presence of a common ion? Le Châtelier’s Principle

42. Factors That Affect Solubility Mg 2+ (aq) + 2F 1- (aq)MgF 2 (s) Solubility and the Common-Ion Effect

43. Factors That Affect Solubility Solubility and the pH of the Solution Ca 2+ (aq) + HCO 3 1- (aq) + H 2 O(l)CaCO 3 (s) + H 3 O 1+ (aq)

44. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/44 Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag(NH 3 ) 2 1+ (aq)Ag(NH 3 ) 1+ (aq) + NH 3 (aq) Ag(NH 3 ) 2 1+ (aq)Ag 1+ (aq) + 2NH 3 (aq)K f = 1.7 x 10 7 K 1 = 2.1 x 10 3 Ag(NH 3 ) 1+ (aq)Ag 1+ (aq) + NH 3 (aq) K 2 = 8.1 x 10 3 K f is the formation constant. How is it calculated? (at 25 °C)

45. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/45 Factors That Affect Solubility Solubility and the Formation of Complex Ions = (2.1 x 10 3 )(8.1 x 10 3 ) = 1.7 x 10 7 [Ag(NH 3 ) 2 1+ ] [Ag 1+ ][NH 3 ] 2 K 1 = [Ag(NH 3 ) 1+ ] [Ag 1+ ][NH 3 ] K 2 = [Ag(NH 3 ) 2 1+ ] [Ag(NH 3 ) 1+ ][NH 3 ] K f = K 1 K 2 =

46. Factors That Affect Solubility Solubility and the Formation of Complex Ions Ag(NH 3 ) 2 1+ (aq)Ag 1+ (aq) + 2NH 3 (aq)

47. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/47 Factors That Affect Solubility Solubility and Amphoterism Al(OH) 4 1- (aq)Al(OH) 3 (s) + OH 1- (aq) Al 3+ (aq) + 6H 2 O(l)Al(OH) 3 (s) + 3H 3 O 1+ (aq) In base: In acid: Aluminum hydroxide is soluble both in strongly acidic and in strongly basic solutions.

48. Factors That Affect Solubility Solubility and Amphoterism

49. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/49 Precipitation of Ionic Compounds Ca 2+ (aq) + 2F 1- (aq)CaF 2 (s) Q c = [Ca 2+ ] t [F 1- ] t 2 K sp = [Ca 2+ ][F 1- ] 2 Q c is called the ion-product (IP).

50. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/50 Precipitation of Ionic Compounds The solution is supersaturated and precipitation will occur. The solution is saturated and equilibrium exists already. The solution is unsaturated and precipitation will not occur. If IP > K sp : If IP = K sp : If IP < K sp : Compare the IP to K sp

51. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/51 Separation of Ions by Selective Precipitation M 2+ (aq) + H 2 S(aq) + 2H 2 O(l)MS(s) + 2H 3 O 1+ (aq) K spa = [M 2+ ][H 2 S] [H 3 O 1+ ] 2

52. Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 15/52 Separation of Ions by Selective Precipitation M 2+ (aq) + H 2 S(aq) + 2H 2 O(l)MS(aq) + 2H 3 O 1+ (aq) Q c = [M 2+ ] t [H 2 S] t [H 3 O 1+ ] t 2 K spa = [M 2+ ][H 2 S] [H 3 O 1+ ] 2 Adjusting the pH changes Q c (ion-product).

53. Qualitative Analysis