1. Acid-Base Equilibria and Solubility Equilibria Chapter 16

2. Acid-Base Equilibria and Solubility Equilibria We will continue to study acid-base reactions We will continue to study acid-base reactions Buffers Buffers Titrations Titrations We will also look at properties of slightly soluble compunds and their ions in solution We will also look at properties of slightly soluble compunds and their ions in solution

3. Homogeneous vs. Heterogeneous Homogeneous Solution Equilibria- Homogeneous Solution Equilibria- a solution that has the same composition throughout after equilibrium has been reached. a solution that has the same composition throughout after equilibrium has been reached. Heterogeneous Solution Equilibria- Heterogeneous Solution Equilibria- a solution that after equilibrium has been reached, results in components in more than one phase. a solution that after equilibrium has been reached, results in components in more than one phase.

4. The Common Ion Effect Acid-Base Solutions Acid-Base Solutions Common Ion Common Ion The Common Ion Effect- the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The Common Ion Effect- the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. pH pH

5. The Common Ion Effect 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

6. Henderson-Hasselbach Equation Relationship between pK a and K a. Relationship between pK a and K a. pK a = -log K a Henderson-Hasselbalch equation pH = pK a + log [conjugate base] [acid]

7. K a = [H + ] [Ac - ] [HAc] take the -log on both sides The Henderson-Hasselbalch Equation -log K a = -log [H + ] -log [Ac - ] [HAc] pH =pK a + log [Ac - ] [HAc] = pK a + log [Proton acceptor] [Proton donor] HAc H + + Ac - pK a = pH -log [Ac - ] [HAc] apply p(x) = -log(x) and finally solve for pH…

8. Acetic acid has a pK a of 4.8. How many ml of 0.1 M acetic acid and 0.1 M sodium acetate are required to prepare 1 liter of 0.1 M buffer with a pH of 5.8? Substitute the values for the pK a and pH into the Henderson-Hasselbalch equation: 5.8 = 4.8 + log [Acetate] [Acetic acid] 1.0 = log [Acetate] 10 x then *[Acetic acid] [Acetic acid] 10 [Acetic acid] = [Acetate] For each volume of acetic acid, 10 volumes of acetate must be added (total of 11 volumes). Acetic acid needed: 1/11 x 1,000 ml = 91 ml Acetate needed: 10/11 x 1,000 ml = 909 ml on both sides

9. Henderson-Hasselbach Equation pH = pK a + log [A - ] [HA] pOH = pK b + log [BH + ] [B] [HA] is the molar concentration of the undissociated weak acid, [A ⁻ ] is the molar concentration of this acid's conjugate base, K a and is the acid dissociation constantacid dissociation constant [B] is the molar concentration of the undissociated weak base, [BH + ] is the molar concentration of this base's conjugate acid, K b and is the base dissociation constantbase dissociation constant

10. At the pK a, [HAc] = [Ac - ] so the system is able to absorb the addition of HO - or H +. If we add HO - near the pH where [Hac] = [Ac - ] (ie pH ~= pK a ) then HAc can release H + to offset the HO - added but the ratio of HAc to Ac - does not change much. If we add H + then Ac - can absorb H + to form HAc. Hence, the pH does not change much. How does a buffer work? -

11. pH Calculations What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.30 -x-x 0.30 - x 0.000.52 +x+x+x+x x0.52 + x HCOOH pK a = 3.77

12. pH Calculations pH = pK a + log [HCOO - ] [HCOOH] pH = 3.77 + log [0.52] [0.30] Common ion effect 0.30 – x  0.30 0.52 + x  0.52 = 4.01

13. 13 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 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

14. Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na 2 CO 3 /NaHCO 3 (a) HF is a weak acid and F - is its conjugate base buffer solution (b) HCl 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

15. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 COOH and 0.300 mol CH 3 COONa to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

16. Calculating pH Changes in Buffers Before the reaction, since mol CH 3 COOH = mol CH 3 COO − pH = pK a = −log (1.8  10 −5 ) = 4.74

17. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: CH 3 COOH (aq) + OH − (aq)  CH 3 COO − (aq) + H 2 O (l) CH 3 COOHCH 3 COO − OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol New concentrations: 0.280 mol/ 1 L = 0.280 M CH 3 COOH 0.320 mol/ 1 L = 0.320 M CH 3 COO -

18. Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 280) pH = 4.74 + 0.06 pH = 4.80

19. To make a pH 7 buffer. You already have 0.1 M KH 2 PO 4. What concentration of K 2 HPO 4 do you need? pH = 7 = pK a + log 7 = 6.86 + log(x / 0.1 M) 0.14 = log(x / 0.1 M) 10 0.14 = x / 0.1 M x = 0.138 M = [K 2 HPO 4 ] [HPO 4 2- ] [H 2 PO 4 - ] H 2 PO 4 - HPO 4 2- + H + pK a = 6.86 KH 2 PO 4 H 2 PO 4 - + K + and K 2 HPO 4 HPO 4 2- + 2 K +

20. Make 200 ml of 0.1 M Na acetate (CH 3 COONa) buffer pH 5.1, starting with 5.0 M acetic acid (CH 3 COOH) and 1.0 M NaOH. Strategy 1.Calculate the total amount of acetic acid needed. 2. Calculate the ratio of the two forms of acetate (A - and HA) that will exist when the pH is 5.1. 3.Use this ratio to calculate the % of acetate that will be in the A - form. 4.Assume that each NaOH will convert one HAc to an Ac -. Use this plus the % A - to calculate the amount of NaOH needed to convert the correct amount of HAc to Ac -. Another HH calculation

21. (1) How much acetic acid is needed? 200 ml x 0.1 mol/l = 200 ml x 0.1 mmol/ml = 20 mmol 5.0 mol/l x x ml = 5.0 mmol/ml x x ml = 20 mmol x = 4.0 ml of 5.0 M acetic acid are 20 mmol (2) What is the ratio of Ac - to HAc at pH 5.1? 5.10 - 4.76 = log[A - ]/[HAc], thus [Ac - ]/[HAc] = 2.19 / 1 (3) What fraction of total acetate is Ac - at pH 5.1? [Ac - ] [Ac - ] 2.19 –––– = 2.19; –––––––––– = ––––––– = 0.687 or 68.7% [HAc] [HAc] + [Ac - ] 1 + 2.19 pH =pK a + log HH equation [Ac - ] [HAc]

22. (4) How much OH - is needed to obtain 68.7% Ac - ? Na + + OH - + HAc  Na + + Ac - + H 2 O mmol NaOH = 0.687 x 20 mmol = 13.7 mmol 1.0 mol/l x x ml = 1.0 mmol/ml x x ml = 13.7 mmol x = 13.7 ml of 1.0 M NaOH (5) Final answer (Jeopardy…) 4.0 ml of 5.0 M acetic acid 13.7 ml of 1.0 M NaOH Bring to final volume of 200 ml with water (ie add about 182.3 ml of H 2 O).

23. = 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? 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) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10

24. The Common Ion Effect and Buffer Solutions Example 16-1: Calculate the concentration of H + and the pH of a solution that is 0.15 M in CH3COOH and 0.15 M in CH3COONa. –This is another equilibrium problem with a starting concentration for both the acid and anion.

25. The Common Ion Effect and Buffer Solutions Substitute the quantities determined in the previous relationship into the ionization expression.

26. The Common Ion Effect and Buffer Solutions Apply the simplifying assumption to both the numerator and denominator.

27. Weak Bases plus Salts of Weak Bases Example 16-2: Calculate the concentration of OH- and the pH of the solution that is 0.15 M in aqueous ammonia, NH 3, and 0.30 M in ammonium nitrate, NH 4 NO 3.

28. Weak Bases plus Salts of Weak Bases Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

29. HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl -

30. Equilibria of Acid-Base Buffer Systems Calculating the Effect of Added H 3 O + or OH - on Buffer pH PROBLEM:Calculate the pH: (a) of a buffer solution consisting of 0.50M CH 3 COOH and 0.50M CH 3 COONa (b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a) (c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a) K a of CH 3 COOH = 1.8x10 -5. (Assume the additions cause negligible volume changes. PLAN:We know K a and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system. Initial Change Equilibrium 0.50 + x 0.50-x - - - 0.500 + x 0.50 +xx - x SOLUTION: CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) (a)

31. Equilibria of Acid-Base Buffer Systems Calculating the Effect of Added H 3 O + and OH - on Buffer pH continued (2 of 4) [CH 3 COOH] equil ≈ 0.50M[CH 3 COO - ] initial ≈ 0.50M[H 3 O + ] = x K a = [H 3 O + ][CH 3 COO - ] [CH 3 COOH] [H 3 O + ] = x = K a [CH 3 COO - ] [CH 3 COOH] = 1.8x10 -5 M Check the assumption:1.8x10 -5 /0.50 X 100 = 3.6x10 -3 % CH 3 COOH( aq ) + OH - ( aq ) CH 3 COO - ( aq ) + H 2 O ( l )Concentration (M) Before addition Addition After addition (b) [OH - ] added = 0.020 mol 1.0L soln = 0.020M NaOH 0.50- - - - -0.020- 0.4800.52

32. Equilibria of Acid-Base Buffer Systems continued (3 of 4) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) Initial Change Equilibrium 0.48- - x 0.48 -x - - 0.520 x + x 0.52 +x [H 3 O + ] = 1.8x10 -5 0.48 0.52 = 1.7x10 -5 pH = 4.77 Before addition Addition After addition (c)[H 3 O + ] added = 0.020 mol 1.0L soln = 0.020M H 3 O + 0.50- - - - -0.020- 0.4800.52 Calculating the Effect of Added H 3 O + and OH - on Buffer pH

33. Equilibria of Acid-Base Buffer Systems continued (4 of 4) Set up a reaction table with the new values. CH 3 COOH( aq ) + H 2 O( l ) CH 3 COO - ( aq ) + H 3 O + ( aq )Concentration (M) Initial Change Equilibrium 0.52- - x 0.52 -x - - 0.480 x + x 0.48 +x [H 3 O + ] = 1.8x10 -5 0.48 0.52 = 2.0x10 -5 pH = 4.70 Calculating the Effect of Added H 3 O + and OH - on Buffer pH

34. 19.1 - Equilibria of Acid-Base Buffer Systems The Henderson-Hasselbalch Equation HA + H 2 O H 3 O + + A - K a = [H 3 O + ] [A - ] [HA] [H 3 O + ] = K a [HA] [A - ] - log[H 3 O + ] = - log K a + log [A - ] [HA] pH = pK a + log [base] [acid]

35. Equilibria of Acid-Base Buffer Systems Buffer capacity is the ability to resist pH change. Buffer range is the pH range over which the buffer acts effectively. The more concentrated the components of a buffer, the greater the buffer capacity. The pH of a buffer is distinct from its buffer capacity. A buffer has the highest capacity when the component concentrations are equal. Buffers have a usable range within ± 1 pH unit of the pK a of its acid component.

36. Buffer Capacity Amount of H + or OH - it can absorb without a significant change in pH. For HA/A - system, buffer capacity depends on: — [HA] and [A - ](higher = higher) — [HA] ratio(closer to 1 = higher) [A - ] [A - ]

37. Equilibria of Acid-Base Buffer Systems The relation between buffer capacity and pH change.

38. Preparing a buffer 1.Choose and acid-conjugate base pair 2.Calculate the ratio of the buffer component pairs 3.Determine the buffer concentration How would you prepare a benzoic acid/benzoate buffer with pH = 4.25, starting with 5.0 L of 0.050 M sodium benzoate (C 6 H 5 COONa) solution and adding the acidic component? K a of benzoic acid (C 6 H 5 COOH) = 6.3 x 10 -5

39. Equilibria of Acid-Base Buffer Systems Preparing a Buffer SOLUTION: PROBLEM:An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-rich soils. How many grams of Na 2 CO 3 must she add to 1.5L of freshly prepared 0.20M NaHCO 3 to make the buffer? K a of HCO 3 - is 4.7x10 -11. PLAN:We know the K a and the conjugate acid-base pair. Convert pH to [H 3 O + ], find the number of moles of carbonate and convert to mass. HCO 3 - ( aq ) + H 2 O( l ) CO 3 2- ( aq ) + H 3 O + ( aq ) K a = [CO 3 2- ][H 3 O + ] [HCO 3 - ] pH = 10.00; [H 3 O + ] = 1.0x10 -10 4.7x10 -11 = [CO 3 2- ](0.20) 1.0x10 -10 [CO 3 2- ] = 0.094M moles of Na 2 CO 3 = (1.5L)(0.094mols/L)= 0.14 = 15 g Na 2 CO 3 0.14 moles 105.99g mol

40. Titration (pH) Curves Plot pH of solution vs. amount of titrant added Equivalence point: Enough titrant added to react exactly with solution being analyzed.

41. Titration of Strong Acid with Strong Base

42. Weak Acid-Strong Base Titration 1.Stoichiometry Reaction assumed to run to completion 2.Equilibrium Use weak acid equilibrium to find pH

43. Weak Acid-Strong Base Titration

44. Differences

45. Differences and K a

46. Titration Calculations 1.Solution of HA 2.Solution of HA and added base 3.Equivalent amounts of HA and added base 4.Excess base A chemist titrates 20.00 mL of 0.2000 M HBrO (K a = 2.3 x 10 -9 ) with 0.1000 M NaOH. Find the pH: (a)before any base is added (b)when 30.00 mL of NaOH is added (c)at the equivalence point (d)when the moles of OH - added are twice the moles of HBrO originally present?

47. Titration of Strong Base with Strong Acid

48. Weak Base-Strong Acid Titration

49. Acid-Base Indicator Indicates endpoint of a titration Endpoint is not necessarily the equivalence point

51. Curve for a strong acid-strong base titration

52. Acid-base Titration Curves Curve for a weak acid-strong base titration Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH [HPr] = [Pr - ] pH = 8.80 at equivalence point pK a of HPr = 4.89 methyl red

53. Curve for a weak base- strong acid titration Titration of 40.00mL of 0.1000M NH 3 with 0.1000M HCl pH = 5.27 at equivalence point pK a of NH 4 + = 9.25

54. pK a = 7.19 pK a = 1.85 Curve for the titration of a weak polyprotic acid (H 2 SO 3 ). Titration of 40.00mL of 0.1000M H 2 SO 3 with 0.1000M NaOH

55. Acid-base Titration Curves Calculating the pH During a Weak Acid-Strong Base Titration PROBLEM:Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; K a = 1.3x10 -5 ) after adding the following volumes of 0.1000M NaOH: (a) 0.00mL(b) 30.00mL(c) 40.00mL(d) 50.00mL PLAN:The amounts of HPr and Pr - will be changing during the titration. Remember to adjust the total volume of solution after each addition. SOLUTION:(a) Find the starting pH using the methods of Chapter 18. K a = [Pr - ][H 3 O + ]/[HPr][Pr - ] = x = [H 3 O + ] x = 1.1x10 -3 ; pH = 2.96 (b) Before addition Addition After addition 0.04000 0.03000 0.01000 0- - -0 - -- HPr( aq ) + OH - ( aq ) Pr - ( aq ) + H 2 O ( l )Amount (mol)

56. Acid-base Titration Curves Calculating the pH During a Weak Acid-Strong Base Titration continued [H 3 O + ] = 1.3x10 -5 0.001000 mol 0.003000 mol = 4.3x10 -6 MpH = 5.37 (c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr - ] will be (0.004000 mol) (0.004000L) + (0.004000L) = 0.05000M K a x K b = K w K b = K w /K a = 1.0x10 -14 /1.3x10 -5 = 7.7x10 -10 [H 3 O + ] = K w / = 1.6x10 -9 MpH = 8.80 (d) 50.00mL of NaOH will produce an excess of OH -. mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol M = 0.01111[H 3 O + ] = 1.0x10 -14 /0.01111 = 9.0x10 -11 M pH = 12.05

57. K sp Solubility product constant Solubility product constant –Shows the solubility of a given solid at a specific temperature

58. Solubility Constant, K sp For solids dissolving to form aqueous solutions. For slightly soluble salts: equilibrium between solid and component ions

59. K sp Write the expression for K sp for (a)CaSO 4 (b)Cr 2 CO 3 (c)Mg(OH) 2 (d)As 2 S 3

60. Solubility Equilibria 17.6 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 = K sp Saturated solution Q < K sp Unsaturated solution Q > K sp Supersaturated solution

61. 17.6

62. 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 = 1.6 x 10 -10 17.6

63. Determining K sp from Solubility PROBLEM:(a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION:PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq )(a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8

64. Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:1s2s2 s = (K sp ) 1/2 2PbSO 4 1:1s2s2 s = (K sp ) 1/2 2BaCrO 4 1:1s2s2 s = (K sp ) 1/2 3Ca(OH) 2 1:24s 3 s = (K sp /4) 1/3 3BaF 2 1:24s 3 s = (K sp /4) 1/3 4Al(OH) 3 1:327s 4 s = (K sp /27) 1/4 3Ca 3 (PO 4 ) 2 3:2108s 5 s = (K sp /108) 1/5 K sp = [s][ns] n

65. Table 16.3 Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:13.5 x 10 -8 1.9 x 10 -4 2PbSO 4 1:11.6 x 10 -8 1.3 x 10 -4 2BaCrO 4 1:12.1 x 10 -10 ? 3Ca(OH) 2 1:25.5 x 10 -6 1.2 x 10 -2 3BaF 2 1:21.5 x 10 -6 ? 3Al(OH) 3 1:31.8 x 10 -13 ? 3Ca 3 (PO4) 2 3:21.2 x 10 -26 ? K sp = [s][ns] n

66. 17.6

68. 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? 17.6 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 = 8.0 x 10 -6 Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x 10 -4 ) 2 = 1.6 x 10 -8 Q < K sp No precipitate will form

69. What concentration of Ag is required to precipitate ONLY AgBr in a solution that contains both Br - and Cl - at a concentration of 0.02 M? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] K sp = 1.6 x 10 -10 17.7 AgBr (s) Ag + (aq) + Br - (aq) K sp = 7.7 x 10 -13 K sp = [Ag + ][Br - ] [Ag + ] = K sp [Br - ] 7.7 x 10 -13 0.020 = = 3.9 x 10 -11 M [Ag + ] = K sp [Cl - ] 1.6 x 10 -10 0.020 = = 8.0 x 10 -9 M 3.9 x 10 -11 M < [Ag + ] < 8.0 x 10 -9 M

70. 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 = 7.7 x 10 -13 s 2 = K sp s = 8.8 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 = 7.7 x 10 -10 17.8

71. pH and Solubility The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) K sp = [Mg 2+ ][OH - ] 2 = 1.2 x 10 -11 K sp = (s)(2s) 2 = 4s 3 4s 3 = 1.2 x 10 -11 s = 1.4 x 10 -4 M [OH - ] = 2s = 2.8 x 10 -4 M pOH = 3.55 pH = 10.45 At pH less than 10.45 Lower [OH - ] OH - (aq) + H + (aq) H 2 O (l) Increase solubility of Mg(OH) 2 At pH greater than 10.45 Raise [OH - ] Decrease solubility of Mg(OH) 2 17.9

72. Equilibria of Slightly Soluble Ionic Compounds Writing Ion-Product Expressions for Slightly Soluble Ionic Compounds SOLUTION: PROBLEM:Write the ion-product expression for each of the following: (a) Magnesium carbonate(b) Iron ( II ) hydroxide (c) Calcium phosphate(d) Silver sulfide PLAN:Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d). K sp = [Mg 2+ ][CO 3 2- ](a) MgCO 3 ( s ) Mg 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Fe 2+ ][OH - ] 2 (b) Fe(OH) 2 ( s ) Fe 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 (c) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3- ( aq ) (d) Ag 2 S( s ) 2Ag + ( aq ) + S 2- ( aq ) S 2- ( aq ) + H 2 O( l ) HS - ( aq ) + OH - ( aq ) Ag 2 S( s ) + H 2 O( l ) 2Ag + ( aq ) + HS - ( aq ) + OH - ( aq ) K sp = [Ag + ] 2 [HS - ][OH - ]

73. Equilibria of Slightly Soluble Ionic Compounds Determining K sp from Solubility PROBLEM:(a) Lead ( II ) sulfate is a key component in lead-acid car batteries. Its solubility in water at 25 0 C is 4.25x10 -3 g/100mL solution. What is the K sp of PbSO 4 ? (b) When lead ( II ) fluoride (PbF 2 ) is shaken with pure water at 25 0 C, the solubility is found to be 0.64g/L. Calculate the K sp of PbF 2. PLAN:Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression. K sp = [Pb 2+ ][SO 4 2- ] = 1.40x10 -4 M PbSO 4 K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 = SOLUTION:PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq )(a) 1000mL L 4.25x10 -3 g 100mL soln 303.3g PbSO 4 mol PbSO 4 1.96x10 -8

74. Equilibria of Slightly Soluble Ionic Compounds Determining K sp from Solubility continued (b) PbF 2 (s) Pb 2+ (aq) + 2F - (aq) K sp = [Pb 2+ ][F - ] 2 = 2.6x10 -3 M K sp = (2.6x10 -3 )(5.2x10 -3 ) 2 = 0.64g L soln245.2g PbF 2 mol PbF 2 7.0x10 -8

75. Equilibria of Slightly Soluble Ionic Compounds Solubility-Product Constants (K sp ) of Selected Ionic Compounds at 25 0 C Name, FormulaK sp Aluminum hydroxide, Al(OH) 3 Cobalt ( II ) carbonate, CoCO 3 Iron ( II ) hydroxide, Fe(OH) 2 Lead ( II ) fluoride, PbF 2 Lead ( II ) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn( I O 3 ) 2 3 x 10 -34 1.0 x 10 -10 4.1 x 10 -15 3.6 x 10 -8 1.6 x 10 -8 4.7 x 10 -29 8 x 10 -48 Mercury ( I ) iodide, Hg 2 I 2 3.9 x 10 -6

76. Equilibria of Slightly Soluble Ionic Compounds Determining Solubility from K sp PROBLEM:Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH) 2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH) 2 in water if the K sp is 6.5x10 -6. PLAN:Write out a dissociation equation and K sp expression; Find the molar solubility (S) using a table. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq ) K sp = [Ca 2+ ][OH - ] 2 -Initial Change Equilibrium - - 00 +S+ 2S S2S K sp = (S)(2S) 2 S == 1.2x10x -2 M Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration (M)

77. Equilibria of Slightly Soluble Ionic Compounds Relationship Between K sp and Solubility at 25 0 C No. of IonsFormulaCation:AnionK sp Solubility (M) 2MgCO 3 1:13.5 x 10 -8 1.9 x 10 -4 2PbSO 4 1:11.6 x 10 -8 1.3 x 10 -4 2BaCrO 4 1:12.1 x 10 -10 1.4 x 10 -5 3Ca(OH) 2 1:25.5 x 10 -6 1.2 x 10 -2 3BaF 2 1:21.5 x 10 -6 7.2 x 10 -3 3CaF 2 1:23.2 x 10 -11 2.0 x 10 -4 3Ag 2 CrO 4 2:12.6 x 10 -12 8.7 x 10 -5

78. Equilibria of Slightly Soluble Ionic Compounds The effect of a common ion on solubility PbCrO 4 ( s ) Pb 2+ ( aq ) + CrO 4 2- ( aq ) CrO 4 2- added

79. Equilibria of Slightly Soluble Ionic Compounds Calculating the Effect of a Common Ion on Solubility PROBLEM:In Sample Problem 19.6, we calculated the solubility of Ca(OH) 2 in water. What is its solubility in 0.10M Ca(NO 3 ) 2 ? K sp of Ca(OH) 2 is 6.5x10 -6. PLAN:Set up a reaction equation and table for the dissolution of Ca(OH) 2. The Ca(NO 3 ) 2 will supply extra [Ca 2+ ] and will relate to the molar solubility of the ions involved. SOLUTION:Ca(OH) 2 ( s ) Ca 2+ ( aq ) + 2OH - ( aq )Concentration(M) Initial Change Equilibrium - - - 0.100 +S+2S 0.10 + S2S K sp = 6.5x10 -6 = (0.10 + S)(2S) 2 = (0.10)(2S) 2 S << 0.10 S = = 4.0x10 -3 Check the assumption: 4.0% 0.10M 4.0x10 -3 x 100 =

80. Equilibria of Slightly Soluble Ionic Compounds Predicting the Effect on Solubility of Adding Strong Acid PROBLEM:Write balanced equations to explain whether addition of H 3 O + from a strong acid affects the solubility of these ionic compounds: (a) Lead ( II ) bromide(b) Copper ( II ) hydroxide(c) Iron ( II ) sulfide PLAN:Write dissolution equations and consider how strong acid would affect the anion component. No effect. SOLUTION:(a) PbBr 2 ( s ) Pb 2+ ( aq ) + 2Br - ( aq ) (b) Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) OH - is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility. (c) FeS( s ) Fe 2+ ( aq ) + S 2- ( aq )S 2- is the anion of a weak acid and will react with water to produce OH -. Both weak acids serve to increase the solubility of FeS. FeS( s ) + H 2 O( l ) Fe 2+ ( aq ) + HS - ( aq ) + OH - ( aq )

81. Equilibria of Slightly Soluble Ionic Compounds Predicting Whether a Precipitate Will Form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO 3 ) 2 is mixed with 0.200L of 0.060M NaF? PLAN:Write out a reaction equation to see which salt would be formed. Look up the K sp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION:CaF 2 (s) Ca 2+ (aq) + 2F - (aq) K sp = 3.2x10 -11 mol Ca 2+ = 0.100L(0.30mol/L) = 0.030mol[Ca 2+ ] = 0.030mol/0.300L = 0.10M mol F - = 0.200L(0.060mol/L) = 0.012mol[F - ] = 0.012mol/0.300L = 0.040M Q = [Ca 2+ ][F - ] 2 =(0.10)(0.040) 2 = 1.6x10 -4 Q is >> K sp and the CaF 2 WILL precipitate.

82. Ionic Equilibria in Chemical Analysis Separating Ions by Selective Precipitation SOLUTION: PROBLEM:A solution consists of 0.20M MgCl 2 and 0.10M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x10 -10 ; K sp of Cu(OH) 2 is 2.2x10 -20. PLAN:Both precipitates are of the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. It is obvious that Cu(OH) 2 will precipitate first so we calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This should ensure that we do not precipitate Mg(OH) 2. Then we can check how much Cu 2+ remains in solution. Mg(OH) 2 ( s ) Mg 2+ ( aq ) + 2OH - ( aq ) K sp = 6.3x10 -10 Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) K sp = 2.2x10 -20 [OH - ] needed for a saturated Mg(OH) 2 solution = = 5.6x10 -5 M

83. Ionic Equilibria in Chemical Analysis Separating Ions by Selective Precipitation continued Use the K sp for Cu(OH) 2 to find the amount of Cu remaining. [Cu 2+ ] = K sp /[OH - ] 2 = 2.2x10 -20 /(5.6x10 -5 ) 2 =7.0x10 -12 M Since the solution was 0.10M CuCl 2, virtually none of the Cu 2+ remains in solution.

84. Ionic Equilibria in Chemical Analysis The general procedure for separating ions in qualitative analysis. Add precipitating ion Centrifuge Add precipitating ion Centrifuge

85. Ionic Equilibria in Chemical Analysis A qualitative analysis scheme for separating cations into five ion groups. Add 6M HCl Centrifuge Acidify to pH 0.5; add H 2 S Centrifuge Add NH 3 /NH 4 + buffer(pH 8) Centrifuge Add (NH 4 ) 2 HPO 4 Centrifuge

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