1. Acid-Base Equilibria The reaction of weak acids with water, OR the reaction of weak bases with water, always results in an equilibrium!! The equilibrium constant for the reaction of a weak acid with water is K a 1 Mar. 25

2. Acid-Base Equilibria eg. HF (aq) + H 2 O (l) º K a = [H 3 O+] [F - ] [HF] H 3 O + (aq) + F - (aq) K eq = ? 2 Mar. 25

3. Acid-Base Equilibria For any weak acid Why is H 2 O (l) omitted from the K a expression? K a = [H 3 O+] [conjugate base] [weak acid] 3 Mar. 25

4. Acid-Base Equilibria the equilibrium constant for the reaction of a weak base with water is K b HS - (aq) + H 2 O (l) º K b = H 2 S (aq) + OH - (aq) 4 Mar. 25

5. Acid-Base Equilibria For any weak base K b = [OH - ] [conjugate acid] [weak base] 5 Mar. 25

6. eg. Write the expression for K b for S 2- (aq) ANSWER: S 2- (aq) + H 2 O (l) º K b = [OH - ] [HS - ] [S 2- ] HS - (aq) + OH - (aq) 6 Mar. 25

7. 5.a) Use K a to find [H 3 O+] for 0.100 mol/L HF (aq) HF (aq) + H 2 O (l) º H 3 O + (aq) + F - (aq) K a = 6.6 x 10 -4 Mar. 25

8. x 2 = (0.100)(6.6 x 10 -4 ) x 2 = 6.6 x 10 -5 x = 8.1 x 10 -3 mol/L 1 st try - Ignore x Mar. 25

9. 2 nd try– Include x x 2 = (0.0919)(6.6 x 10 -4 ) x 2 = 6.0654 x 10 -5 x = 7.8 x 10 -3 mol/L Different than 1 st try: CANNOT IGNORE DISSOCIATION Mar. 28

10. 3 rd try– Include new x x 2 = (0.0922)(6.6 x 10 -4 ) x 2 = 6.0852 x 10 -5 x = 7.8 x 10 -3 mol/L [H 3 O + ] = 7.8 x 10 -3 mol/L Same as 2 nd try: Mar. 28

11. 5.b) find [H 3 O+] for 0.250 mol/L CH 3 COOH (aq CH 3 COOH (aq) + H 2 O (l) º H 3 O + (aq) + CH 3 COO - (aq) K a = 1.8 x 10 -5 Mar. 28

12. x 2 = (0.250)(1.8 x 10 -5 ) x 2 = 4.5 x 10 -6 x = 2.1 x 10 -3 mol/L 1 st try - Ignore x Mar. 28

13. 2 nd try– Include x x 2 = (0.2479)(1.8 x 10 -5 ) x 2 = 4.462 x 10 -6 x = 2.1 x 10 -3 mol/L [H 3 O + ] = 2.1 x 10 -3 mol/L Same as 1 st try: Mar. 28

14. 14 To ignore OR not to ignore: that is the question Mar. 28

15. pH of a weak acid Step #1: Write a balanced equation Step #2: ICE table OR assign variables Step #3: Write the K a expression Step #4: Check (can we ignore dissociation) Step #5: Substitute into Ka expression 15 Mar. 28

16. pH of a weak acid eg. Find pH of 0.100 mol/L HF(aq). Step #1: Write a balanced equation HF (aq) + H 2 O (l) º H 3 O + (aq) + F - (aq) 16 Mar. 30

17. Step #2: Equilibrium Concentrations Let x = [H 3 O + ] at equilibrium [F - ] = x [HF] = 0.100 - x 17 Mar. 30

18. Step #3: Write the K a expression K a = [H 3 O+] [F - ] [HF] 18 Mar. 30

19. Step #4: Check (can we ignore dissociation) dissociation (- x) may be IGNORED = 151 (0.100) 6.6 x 10 -4 Acid dissociation CANNOT be IGNORED in this question. [weak acid] KaKa If > 500 We have to use the – x 19 Mar. 30

20. Step #5: Substitute into Ka expression x 2 = 6.6 x 10 -5 - 6.6 x 10 -4 x x 2 + 6.6 x 10 -4 x - 6.6 x 10 -5 = 0 a = 1b = 6.6 x 10 -4 c = -6.6 x 10 -5 Quadratic Formula!! 20 Mar. 30

21. Ignore negative roots 21 Mar. 30

22. a) Find the [H 3 O + ] in 0.250 mol/L HCN (aq) Check: 4.0 x 10 8 x = 1.24 x 10 -5 [H 3 O + ] = 1.24 x 10 -5 b) Calculate the pH of 0.0300 mol/L HCOOH (aq) Check: 167 x = 2.24 x 10 -3 pH = 2.651 Try these: Mar. 31

23. HCN + H 2 O ⇋ H 3 O + + CN - Let x = [H 3 O + ] x = [CN - ] 0.250 – x = [HCN] Check: 23 K a = [H 3 O + ] [CN - ] [HCN] = 4.0 x 10 8 0.250 6.2 x 10 -10 Quadratic NOT needed Mar. 31

24. 24 x = 1.25 x 10 -5 [H 3 O + ] = 1.25 x 10 -5 mol/L pH = 4.904 x 2 = 1.55 x 10 -10 Mar. 31

25. HCOOH + H 2 O ⇋ H 3 O + + HCOO - 0.0300 0 0 -x +x +x 0.0300 – x x x Check: 25 K a = [H 3 O + ] [HCOO - ] [HCOOH] = 167 0.0300 1.8 x 10 -4 Quadratic needed Mar. 31

26. 26 A = 1 B = 1.8 x 10 -4 C = -5.4 x 10 -6 x = 2.24 x 10 -3 [H 3 O + ] = 2.24 x 10 -3 mol/L pH = 2.651 x 2 = 5.4 x 10 -6 - 1.8 x 10 -4 x x 2 + 1.8 x 10 -4 x - 5.4 x 10 -6 = 0 Mar. 31

27. Practice 1. Formic acid, HCOOH, is present in the sting of certain ants. What is the [H 3 O + ] of a 0.025 mol/L solution of formic acid? (0.00203 mol/L) 2. Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid. ( [H 3 O + ] = 3.87 x 10 -3 pH = 2.413 ) 3. What is the percent dissociation of the vinegar in 2.? % diss = 0.466 % Mar. 31

28. Practice 4. A solution of hydrofluoric acid has a molar concentration of 0.0100 mol/L. What is the pH of this solution? ( [H 3 O + ] = 0.00226 pH = 2.646 ) 5. The word “butter” comes from the Greek butyros. Butanoic acid, C 3 H 7 COOH, gives rancid butter its distinctive odour. Calculate the [H 3 O + ] of a 1.0 × 10 −2 mol/L solution of butanoic acid. (Ka = 1.51 × 10 −5 ) (Ans: 3.89 x 10 -4 mol/L) Mar. 31

29. pH of a weak base  same method as acids  ignore dissociation if  to calculate K b (usually given on the exam) 29 [weak base] KbKb > 500 Apr. 4

30. pH of a weak base Calculate the pH of 0.0100mol/L Na 2 CO 3(aq) 30 Apr. 4

31. CO 3 2- + H 2 O ⇋ HCO 3 - + OH - 0.0100 0 0 -x +x +x 0.0100 – x x x Check: 31 K b = [OH - ] [HCO 3 - ] [CO 3 2- ] = 47 0.0100 2.13 x 10 -4 → Quadratic needed K b = 1.00 x 10 -14 4.7 x 10 -11 = 2.13 x 10 -4 Apr. 4

32. 32 x 2 = 2.13 x 10 -6 - 2.13 x 10 -4 x x 2 + 2.13 x 10 -4 x - 2.13 x 10 -6 = 0 A = 1 B = 2.13 x 10 -4 C = -2.13 x 10 -6 x = 1.36 x 10 -3 [OH - ] = 1.36 x 10 -3 mol/L pOH = ??pH = 11.13 Apr. 4

33. pH of a weak base Calculate the pH of 0.500 mol/L NaNO 2(aq) 33 Apr. 4 Na + NO 2 - H2OH2O

34. NO 2 - + H 2 O ⇋ HNO 2 + OH - 0.500 0 0 -x +x +x 0.500 – x x x Check: 34 K b = [OH - ] [HCO 3 - ] [CO 3 2- ] = 3.6 x 10 10 0.500 1.39 x 10 -11 OK to ignore –x here ie. NO Quadratic K b = 1.00 x 10 -14 7.2 x 10 -4 = 1.39 x 10 -11 Apr. 4

35. 35 x 2 = 6.95 x 10 -12 [OH - ] = 2.6 x 10 -6 mol/L pOH = ?? pH = 8.42 x = 2.6 x 10 -6 Apr. 4

36. Calculating K a from [weak acid] and pH eg.The pH of a 0.072 mol/L solution of benzoic acid, C 6 H 5 COOH, is 2.68. Calculate the numerical value of the Ka for this acid. - Equation - Find [H 3 O + ] from pH - Subtract from [weak acid] - Substitute to find K a See p. 591 #6 & 8 36 Apr. 5

37. C 6 H 5 COOH (aq) + H 2 O (l) º H 3 O + (aq) + C 6 H 5 COO - (aq) [H 3 O + ] = 10 -2.68 = 0.00209 mol/L [C 6 H 5 COOH] = 0.072 – 0.00209 = 0.06991 mol/L Find K a K a = (0.00209)(0.00209) (0.06991) = 6.2 x 10 -5 [C 6 H 5 COO - ] = 0.00209 mol/L 37 Apr. 5

38. Calculating K a from [weak acid] and pH eg.The pH of a 0.072 mol/L solution of benzoic acid, C 6 H 5 COOH, is 2.68. Calculate the % dissociation for this acid. See p. 591 #’s 5 & 6 [H 3 O + ] = 10 -2.68 = 0.00209 mol/L = 2.9 % 38 Apr. 5

39. a)0.250 mol/L chlorous acid, HClO 2(aq) ; pH = 1.31 0.01219.5% b) 0.150 mol/L cyanic acid, HCNO (aq) ; pH = 2.15 0.000354.7% c)0.100 mol/L arsenic acid, H 3 AsO 4(aq) ; pH = 1.70 0.005020% d)0.500 mol/L iodic acid, HIO 3(aq) ; pH = 0.670 0.16042.8% Calculate the acid dissociation constant, K a, and the percent dissociation for each acid: 39 Apr. 5

40. More Practice: Weak Acids: pp. 591, 592 #’s 6 -8 Weak Bases: p. 595 #’s 11 - 16 (K b ’s on p. 592) 40 Apr. 5

41. Acid-Base Stoichiometry Solution Stoichiometry (Review) 1. Write a balanced equation 2. Calculate moles given ( OR n = CV) 3. Mole ratios 4. Calculate required quantity OR OR m = nM 41 Apr. 6

42. 42

43. eg.25.0 mL of 0.100 mol/L H 2 SO 4(aq) was used to neutralize 36.5 mL of NaOH (aq). Calculate the molar concentration of the NaOH solution. H 2 SO 4(aq) + NaOH (aq) → H 2 O (l) + Na 2 SO 4(aq) 22 n H 2 SO 4 = n NaOH = C NaOH = 43 Apr. 6

44. Acid-Base Stoichiometry pp. 600, 601 – Sample Problems p. 602 #’s 17 - 20 44 Apr. 6

45. Dilution Given 3 of the four variables Only one solution C i V i = C f V f Stoichiometry Given 3 of the four variables Two different solutions 4 step method 45

46. Excess Acid or Base To calculate the pH of a solution produced by mixing an acid with a base:  write the B-L equation (NIE)  calculate the moles of H 3 O + and OH -  subtract to determine the moles of excess H 3 O + or OH -  divide by total volume to get concentration  calculate pH 46

47. eg.20.0 mL of 0.0100 M Ca(OH) 2(aq) is mixed with 10.0 mL of 0.00500 M HCl (aq). Determine the pH of the resulting solution. ANSWER: Species present: Ca 2+ OH - H 3 O + Cl - H 2 O SB SA 47

48. C = 0.0200 mol/L V = 0.0200 L C = 0.00500 mol/L V = 0.0100 L NIE:OH - + H 3 O + → 2 H 2 O 4.00 x 10 -4 mol OH - 5.0 x 10 -5 mol H 3 O + 3.5 x 10 -4 mol excess OH - 48 n = CV

49. = 0.01167 mol/L [OH - ] = 0.01167 mol/L pOH = 1.933 pH = 12.067 49

50. Indicators An indicator is a weak acid that changes color with changes in pH HIn is the general formula for an indicator To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color 50 Apr. 13

51. HIn (aq) + H 2 O (l) º H 3 O + (aq) + In - (aq) Colour #1Colour #2 HIn is the acid form of the indicator. Adding H 3 O + Adding OH - 51 causes colour 1 (LCP) removes the H 3 O + & causes colour #2 Apr. 13

52. methyl orange HMo (aq) + H 2 O (l) º H 3 O + (aq) + Mo - (aq) red yellow bromothymol blue HBb (aq) + H 2 O (l) º H 3 O + (aq) + Bb - (aq) yellow blue 52 Apr. 13

53. Indicators: p. 23 1.a) HMv + H 2 O (l) º H 3 O + (aq) + Mv - (aq) b) HBb + H 2 O (l) º H 3 O + (aq) + Bb - (aq) 2. IndicatorpHcolour thymol blue3.0 yellow methyl red7.9 yellow phenolpthalein7.1colourless indigocarmine13.5 yellow 3.a) pH range: 2.8 – 4.5 b) pH range: 8.0 – 8.2 53 Apr. 13

54. Acid-Base Titration (p. 603 → ) A titration is a lab technique used to determine an unknown solution concentration. A standard solution is added to a known volume of solution until the endpoint of the titration is reached. The endpoint occurs when there is a sharp change in colour The equivalence point occurs when the moles of H 3 O + equals the moles of OH - 54 Apr. 13

55. Acid-Base Titration The colour change is caused by the indicator added to the titration flask. An indicator is a chemical that changes color over a given pH range (See indicator table) A buret is used to add the standard solution standard solution - solution of known concentration 55

56. Acid-Base Titration primary standard - a standard solution which can be made by direct weighing of a stable chemical. Data from titrations allows us to calculate an unknown solution concentration. 56

57. Titration Calculation eg. Clem Student performed a titration by adding 0.250 mol/L HCl (aq) to 10.0 mL samples of Ca(OH) 2(aq). Use the data below to determine the molar concentration of Ca(OH) 2(aq). Trial1 2 3 4 Final volume 8.48 mL 15.70 mL 22.91 mL 30.14 mL Initial volume 1.05 mL 8.48 mL 15.70 mL 22.91 mL Volume HCl (aq) used 57 7.43 mL7.23 mL7.21 mL7.22 mL Omit first trial OVERSHOT the endpoint

58. Equation: 2 HCl (aq) + Ca(OH) 2(aq) → CaCl 2(aq) + 2 H 2 O (l) n HCl = n Ca(OH) 2 = C = 58 C = 0.250 mol/L V ave = 0.00722 L C = ? mol/L V ave = 0.0100 L

59. Acid-Base Titration Titration Lab – pp. 606, 607 59

60. Multi-Step Titrations (p. 609 - 611) Polyprotic acids donate their protons one at a time when reacted with a base. eg. Write the equations for the steps that occur when H 3 PO 4(aq) is titrated with NaOH (aq) H 3 PO 4(aq) + OH - (aq) H 2 PO 4 - (aq) + OH - (aq) HPO 4 2- (aq) + OH - (aq) 60

61. Multi-Step Titrations H 3 PO 4(aq) + OH - (aq) → H 2 PO 4 - (aq) + H 2 O (l) H 2 PO 4 - (aq) + OH - (aq) → HPO 4 2- (aq) + H 2 O (l) HPO 4 2- (aq) + OH - (aq) º PO 4 3- (aq) + H 2 O (l) H 3 PO 4(aq) + 3 OH - (aq) º PO 4 3- (aq) + 3 H 2 O (l) 61

62. Multi-Step Titrations Write the balanced net ionic equations, and the overall equation, for the titration of Na 2 S (aq) with HCl (aq). p. 611 #’s 21.b), 22, & 23 LAST TOPIC!! Titration CurvesTitration Curves 62

63. Properties / Operational Definitions Acid-Base Theories and Limitations  Arrhenius – H-X and X-OH  Modified – react with water → hydronium  BLT – proton donor/acceptor (CA and CB) Writing Net Ionic Equations (BLT) Strong vs. Weak pH & pOH calculations Equilibria (K w, K a, K b ) Titrations/Indicators/Titration Curves Dilutions and Excess Reagent questions Acids and Bases 63

64. Step #2: ICE table [ HF] [H 3 O + ] [F - ] I C E 0.100 mol/L0 0 - x + x + x 0.100 - x x x 64

65. CO 3 2- (aq) + H 2 O (l) º HCO 3 - (aq) + OH - (aq) 0.0100mol/L CO 2 3- (aq) [PO 4 3- ] [HPO 4 2- ] [OH - ] I C E 0.0100 mol/L0 0 - x + x + x 0.0100 - x x x 65

66. Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us 66

67. A primary standard is a pure substance that is stable enough to be stored indefinitely without decomposition, can be weighed accurately without special precautions when exposed to air, and will undergo an accurate stoichiometric reaction in a titration. 67

68. 15.pH of 0.297 mol/L HOCl HOCl (aq) + H 2 O (l) º H 3 O + (aq) + OCl - (aq) Let x = [H 3 O + ] at equilibrium [OCl - ] = x [HOCl] = 0.297 - x K a = [H 3 O+] [OCl - ] [HOCl] 68

69. Check: dissociation (- x) may be IGNORED = 1.02 x 10 7 (0.297) 2.9 x 10 -8 X = 9.28 x 10 -5 pH = 4.03 69

70. 0.484 mol/L 0.07000 L 0.125 mol/L 0.02500 L 16. NIE: OH - + H 3 O + → 2 H 2 O 0.03388 mol OH - 0.003125 mol H 3 O + 0.030755 mol excess OH - [OH - ] = 0.3237 mol/LpOH = 0.490 pH = 13.510 70

71. 17. Ignore dissociation [OH - ] = 0.0146 mol/L % diss = 2.92 % 18. V ave = 10.975 mL n NaOH = 0.001262 mol n H 2 SO 4 = 0.000631 mol C = 0.0252 mol/L 19. K b = 3.93 x 10 -4 % diss = 6.27 71

72. c) 2.50 mol/L NaCN (aq) K b = 1.61 x 10 -5 Check: 1.5 x 10 5 x = 6.34 x 10 -3 pOH = 2.20pH = 11.80 d) 0.100 mol/L K 2 S (aq) K b = 0.0769 Check: 1.3 x = 0.0573 pOH = 1.24pH = 12.76 72