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

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

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

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

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

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

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 7

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 8

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 9

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 10

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 11

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 12

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 13

14. 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 14

15. 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) 15

16. Step #2: Equilibrium Concentrations Let x = [H 3 O + ] at equilibrium [F - ] = x [HF] = 0.100 - x 16

17. Step #3: K a expression K a = [H 3 O+] [F - ] [HF] 17

18. 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 18

19. 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!! 19

20. Ignore negative roots 20

21. 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: 21

22. 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 % 22

23. 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) 23

24. pH of a weak base  same method as acids  calculate K b  ignore dissociation if 24

25. pH of a weak base Calculate the pH of 0.0100mol/L Na 2 CO 3(aq) 25

26. pH of a weak base Calculate the pH of 0.500 mol/L NaNO 2(aq) 26

27. 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 27

28. 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 28

29. 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 % 29

30. 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: 30

31. More Practice: Weak Acids: pp. 591, 592 #’s 6 -8 Weak Bases: p. 595 #’s 11 - 16 (K b ’s on p. 595) 31

32. 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 32

33. 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 = 33

34. Acid-Base Stoichiometry pp. 600, 601 – Sample Problems p. 602 #’s 17 - 20 34

35. 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 35

36. 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 36

37. 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 37

38. 0.0200 mol/L 0.0200 L 0.00500 mol/L 0.0100 L NIE:OH - + H 3 O + → 2 H 2 O n = CV 4.00 x 10 -4 mol OH - 5.0 x 10 -5 mol H 3 O + 3.5 x 10 -4 mol excess OH - 38

39. = 0.01167 mol/L [OH - ] = 0.01167 mol/L pOH = 1.933 pH = 12.067 39

40. Indicators An indicator is a weak acid that changes color with changes in pH To choose an indicator for a titration, the pH of the endpoint must be within the pH range over which the indicator changes color 40

41. 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 + causes colour 1 (LCP) Adding OH - removes the H 3 O + & causes colour #2 41

42. 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 42

43. 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. 43

44. Acid-Base Titration The endpoint occurs when there is a sharp change in colour The equivalence point occurs when the moles of hydronium equals the moles of hydroxide The colour change is caused by the indicator added to the titration flask. 44

45. Acid-Base Titration An indicator is a chemical that changes colour over a given pH range (See indicator table) A buret is used to deliver the standard solution 45

46. Acid-Base Titration standard solution - solution of known concentration primary standard - a standard solution which can be made by direct weighing of a stable chemical. Titration Lab 46

47. 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) 47

48. 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) 48

49. 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 49

50. 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 50

51. 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 51

52. 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 52

53. 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: Ca(OH) 2(aq) + 2 HCl (aq) → 2 H 2 O (l) + CaCl 2(aq) n base = 0.0002 mol → needs 0.004 mol HCl n acid = 0.00005 mol → needs 0.000025 mol Ca(OH)2 Limiting reactant Excess reactant 53

54. Acid-Base Stoichiometry Solution stoichiometry (4 question sheet) Excess reagent problems (use NIE) Titrations Titration curves Indicators STSE: Acids Around Us 54

55. WorkSheet #9 answers: 1. 0.210 mol/L 2. a)22.5 mL b) 24.7 mL c) 4.8 mL 3. 31.5 mL 4. a)0.0992 mol/L b)0.269 mol/L c)0.552 mol/L 55

56. WorkSheet #10 answers: 1.pH = 13.000 2.[H 3 O + ] = 4.12 x 10 -2 mol/L [OH - ] = 2.43 x 10 -13 mol/L 3.pH = 13.125 4. pH = 7 p. 586 #’s 1 – 4 56

57. 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. 57

58. 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] 58

59. 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 59

60. 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 60

61. 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 61