1. Chapter 15 Applications of Aqueous Equilibria Addition of base: Normal human blood pH is 7.4 and has a narrow range of about +/- 0.2

2. When an acid or base is added to water, the pH changes drastically. A buffer solution resists a change in pH when an acid or base is added. Buffer Solutions Buffers: Mixture of a weak acid and its salt Mixture of a weak base and its salt pH = 4 pH = 10.5 pH = 6.9 pH = 7.1

3. The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion Understanding Buffer Solutions

4. Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid] pH of a Buffer:

5. Using Henderson-Hasselbalch equation pH = pK a + log [HCOO - ] [HCOOH] HCOOH : K a = 1.67 X 10 -4, pK a = 3.78 pH = 3.77 + log [0.52] [0.30] = 4.02 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (K a = 1.67 X 10 -4 ).

6. A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Add strong acid H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

7. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution

8. Blood Plasma pH = 7.4

9. Buffer Solutions02

10. = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? (K a = 5.62 X 10 -10 ). NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) Initial (moles) End (moles) 0.36 X0.0 800 =0.0290.050 x 0.0200 = 0.001 0.30 X.0800 = 0.024 0.029-0.001 = 0.0280.001-0.001 =0.0 0.024 + 0.001 = 0.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL, 0.1 L [NH 3 ] = 0.025 0.10 -0.001 +0.001 Change

11. Titrations Slowly add base to unknown acid UNTIL The indicator changes color (pink)

12. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

13. Acid–Base Titration Curves01

14. Strong Acid-Strong Base Titrations Equivalence Point: The point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

16. Strong Acid-Strong Base Titrations 1.Before Addition of Any NaOH 0.100 M HCl

17. 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+

18. 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

19. 4.Beyond the Equivalence Point Strong Acid-Strong Base Titrations Excess OH 1-

20. Strong Acid-Strong Base Titrations

21. Weak Acid-Strong Base Titrations

22. 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)

23. 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

24. 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

25. 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-

27. Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

28. Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): H + (aq) + NH 3 (aq) NH 4 + (aq)

31. Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

32. Polyprotic Acids: Titration of Diprotic Acid + Strong Base

33. Titration of a Triprotic Acid: H 3 PO 4 + Strong Base

35. Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

36. Solubility Equilibria AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 2 - (aq) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO 4 3 - (aq) K sp = [Ca 2+ ] 3 [PO 4 3 - ] 2 Dissolution of an ionic solid in aqueous solution: Q (IP) = K sp Saturated solution Q (IP) < K sp Unsaturated solution No precipitate Q (IP) > K sp Supersaturated solution Precipitate will form K sp = 1.8 x 10 -10 K sp : Solubility Product (IP) : Ion product

37. Measuring K sp and Calculating Solubility from K sp

38. Molar solubility (mol/L): Number of moles of a solute that dissolve to produce a litre of saturated solution Solubility (g/L) is the number of grams of solute that dissolve to produce a litre of saturated solution

39. Measuring 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)

40. What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) 0.00 +s+s +s+s ss K sp = s 2 s = K sp  s = 1.3 x 10 -5 [Ag + ] = 1.3 x 10 -5 M [Cl - ] = 1.3 x 10 -5 M Solubility of AgCl = 1.3 x 10 -5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10 -3 g/L K sp = 1.8 x 10 -10 Calculating Solubility from K sp

42. Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

43. If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl 2, will a precipitate form? The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = 0.100 M [OH - ] 0 = 4.0 x 10 -4 M K sp = [Ca 2+ ][OH - ] 2 = 4.7 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.100 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form If this solution also contained 0.100 M AlCl3, Aluminum could be precipitated, with Calcium remaining in the solution.

44. 1) Use the difference between Ksp’s to separate one precipitate at a time. 2) Use Common Ion effect to Precipitate one Ion at a time Fractional Precipitation01

45. The Common-Ion Effect and Solubility01

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

47. 1. Use the difference between Ksp’s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 3. Use pH to Increase or decrease solubility and separate one ion at a time Fractional Precipitation01

48. 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)

49. 1. Use the difference between Ksp’s to separate one precipitate at a time. 2. Use Common Ion effect to Precipitate one Ion at a time 3. Use pH to Increase or decrease solubility and separate one ion at a time. 4. Use complex formation to prevent certain Ions to precipitate Fractional Precipitation01

50. 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)

51. 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 =

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

53. 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.

54. Factors That Affect Solubility Solubility and Amphoterism

55. The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? AgBr (s) Ag + (aq) + Br - (aq) K sp = 5.4 x 10 -13 s 2 = K sp s = 7.3 x 10 -7 NaBr (s) Na + (aq) + Br - (aq) [Br - ] = 0.0010 M AgBr (s) Ag + (aq) + Br - (aq) [Ag + ] = s [Br - ] = 0.0010 + s  0.0010 K sp = 0.0010 x s s = 5.4 x 10 -10 AgBr (s) Ag + (aq) + Br - (aq) M M

56. The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solution. pH and Solubility

57. Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 5.6 x 10 -12 K sp = (s)(2s) 2 = 4s 3 4s 3 = 5.6 x 10 -12 s = 1.1 x 10 -4 M [OH - ] = 2s = 2.2 x 10 -4 M pOH = 3.65 pH = 10.35 At pH less than 10.35 Lower [OH - ], Increase solubility of Mg(OH) 2 OH - (aq) + H + (aq) H 2 O (l) remove At pH greater than 10.35 Raise [OH - ] add Decrease solubility of Mg(OH) 2 What is the pH of a solution containing Mg(OH) 2 ? what pH Would reduce the solubility of this precipitate?

58. Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- K f = [CoCl 4 ] [Co 2+ ][Cl - ] 4 2- The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. Co(H 2 O) 6 2+ CoCl 4 2- KfKf stability of complex

59. --

61. Qualitative Analysis

62. Qualitative Analysis of Cations

63. Fractional precipitation is a method of removing one ion type while leaving others in solution. Ions are added that will form an insoluble product with one ion and a soluble one with others. When both products are insoluble, their relative K sp values can be used for separation. Fractional Precipitation01

64. Flame Test for Cations lithium sodium potassiumcopper Nessler's is K 2 [HgI 4 ] in a basic solution, [HgI 4 ] 2- reacts with the NH4 + to form a yellow brown precipitate.

65. Fractional Precipitation01

66. Separation of Ions by Selective Precipitation MS(s) + 2 H 3 O + (aq) ae M 2+ (aq) + H 2 S(aq) + 2 H 2 O(l) (see page 631) : K spa = [Cd +2 ] [H 2 S ] / [H 3 O + ] 2 K spa = For ZnS, K spa = 3 x 10 ­-2 ; for CdS, K spa = 8 x 10 ­-7 [Cd +2 ] = [Zn +2 ] = 0.005 M Because the two cation concentrations are equal, Q spa is :[Cd +2 ] [H 2 S ] / [H 3 O + ] 2 =(0.005).(0.1)/ (0.3) 2 Q spa = 6 x 10 ­-3 For CdS, Q spa > K spa ; CdS will precipitate. For ZnS, Q spa < K spa ; Zn +2 will remain in solution. Determine whether Cd +2 can be separated from Zn +2 by bubbling H 2 S through ([H 2 S ] = 0.1 ) a 0.3 M HCl solutions that contains 0.005 M Cd +2 and 0.005 M Zn +2 ?

67. What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (K a = 1.67 X 10 -4 ). HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 Mixture of weak acid and conjugate base! X (0.52 + X) 0.30 - X = 1.67 X 10 -4 X = 9.77 X 10 -5 = [ H 3 O + ] pH = - log ( 9.77 X 10 -5 ) = 4.01

68. End