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Acid-Base Equilibria

Mar. 24. 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. [H 3 O+] [F - ]. K a =. [HF]. Mar. 24.

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Acid-Base Equilibria

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  1. Mar. 24 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 Ka

  2. [H3O+] [F-] Ka = [HF] Mar. 24 Acid-Base Equilibria H3O+(aq) + F-(aq) eg. HF(aq) + H2O(l) º Keq = ?

  3. [H3O+] [conjugate base] Ka = [weak acid] Mar. 24 Acid-Base Equilibria • For any weak acid • Why is H2O(l) omitted from the Ka expression?

  4. Mar. 24 Acid-Base Equilibria • the equilibrium constant for the reaction of a weak base with water is Kb HS-(aq) + H2O(l) º Kb = H2S(aq) + OH-(aq)

  5. [OH-] [conjugate acid] Kb = [weak base] Mar. 24 Acid-Base Equilibria • For any weak base

  6. [OH-] [HS-] Kb = [S2-] Mar. 24 eg. Write the expression for Kb for S2-(aq) ANSWER: S2-(aq) + H2O(l)º Worksheet #5 HS-(aq) + OH-(aq)

  7. Mar. 25 5.a) Use Ka to find [H3O+] for 0.100 mol/L HF(aq) HF(aq) + H2O(l)º H3O+(aq) + F-(aq) Ka = 6.6 x 10-4

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

  9. Mar. 25 2nd 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 1st try: CANNOT IGNORE DISSOCIATION

  10. Mar. 25 3rd 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 [H3O+] = 7.8 x 10-3 mol/L Same as 2nd try:

  11. Mar. 25 5.b) find [H3O+] for 0.250 mol/L CH3COOH(aq CH3COOH(aq) + H2O(l)º H3O+(aq) + CH3COO-(aq) Ka = 1.8 x 10-5

  12. Mar. 25 1st try - Ignore x x 2 = (0.250)(1.8 x 10-5) x 2 = 4.5 x 10-6 x = 2.1 x 10-3 mol/L

  13. Mar. 25 2nd 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 [H3O+] = 2.1 x 10-3 mol/L Same as 1st try:

  14. Mar. 29 To ignore OR not to ignore: that is the question

  15. Mar. 29 pH of a weak acid Step #1: Write a balanced equation Step #2: ICE table OR assign variables Step #3: Write the Ka expression Step #4: Check (can we ignore dissociation) Step #5: Substitute into Ka expression

  16. Mar. 29 pH of a weak acid eg. Find pH of 0.100 mol/L HF(aq). Step #1: Write a balanced equation HF(aq) + H2O(l)º H3O+(aq) + F-(aq)

  17. Mar. 29 Step #2: Equilibrium Concentrations Let x = [H3O+] at equilibrium [F-] = x [HF] = 0.100 - x

  18. [H3O+] [F-] Ka = [HF] Mar. 29 Step #3: Write the Ka expression

  19. If > 500 [weak acid] Ka Mar. 29 Step #4: Check (can we ignore dissociation) dissociation (- x) may be IGNORED =151 (0.100) Acid dissociation CANNOT be IGNORED in this question. 6.6 x 10-4 We have to use the – x

  20. Quadratic Formula!! Mar. 29 Step #5: Substitute into Ka expression x2 = 6.6 x 10-5 - 6.6 x 10-4 x x2 + 6.6 x 10-4 x - 6.6 x 10-5 = 0 a = 1 b = 6.6 x 10-4 c = -6.6 x 10-5

  21. Mar. 29 Ignore negative roots

  22. Mar. 30 Try these: a) Find the [H3O+] in 0.250 mol/L HCN(aq) Check: 4.0 x 108 x = 1.24 x 10-5 [H3O+] = 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

  23. [H3O+] [CN-] Ka = [HCN] 0.250 6.2 x 10-10 =4.0 x 108 Quadratic NOT needed HCN + H2O ⇋ H3O+ + CN- Let x = [H3O+] x = [CN-] 0.250 – x = [HCN] Check:

  24. x2 = 1.55 x 10-10 x = 1.25 x 10-5 pH = 4.904

  25. [H3O+] [HCOO-] Ka = [HCOOH] 0.0300 Quadratic needed 1.8 x 10-4 =167 HCOOH + H2O ⇋ H3O+ + HCOO- 0.0300 0 0 -x +x +x 0.0300 – x x x Check:

  26. x2 = 5.4 x 10-6 - 1.8 x 10-4x x2 + 1.8 x 10-4x - 5.4 x 10-6 = 0 A = 1 B = 1.8 x 10-4 C = -5.4 x 10-6 x = 2.24 x 10-3 pH = 2.651

  27. Mar. 30 Practice Formic acid, HCOOH, is present in the sting of certain ants. What is the [H3O+] of a 0.025 mol/L solution of formic acid? (0.00203 mol/L) Calculate the pH of a sample of vinegar that contains 0.83 mol/L acetic acid. ( [H3O+] = 3.87 x 10-3 pH = 2.413 ) What is the percent dissociation of the vinegar in 2.? % diss = 0.466 %

  28. Mar. 30 Practice Worksheet #6 A solution of hydrofluoric acid has a molar concentration of 0.0100 mol/L. What is the pH of this solution? ( [H3O+] = 0.00226 pH = 2.646 ) The word “butter” comes from the Greek butyros. Butanoic acid, C3H7COOH,gives rancid butter its distinctive odour. Calculate the [H3O+] 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)

  29. > 500 [weak base] Kb Apr. 1 pH of a weak base • same method as acids • ignore dissociation if • to calculate Kb (usually given on the exam)

  30. Apr. 1 pH of a weak base Calculate the pH of 0.0100mol/L Na2CO3(aq)

  31. [OH-] [HCO3-] 1.00 x 10-14 Kb = Kb = [CO32-] 4.7 x 10-11 Apr. 1 0.0100 = 2.13 x 10-4 2.13 x 10-4 = 47 → Quadratic needed CO32- + H2O ⇋ HCO3- + OH- 0.0100 0 0 -x +x +x 0.0100 – x x x Check:

  32. Apr. 1 x2 = 2.13 x 10-6 - 2.13 x 10-4x x2 + 2.13 x 10-4x - 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

  33. Apr. 1 pH of a weak base Calculate the pH of 0.500 mol/L NaNO2(aq) Worksheet #7

  34. [OH-] [HCO3-] 1.00 x 10-14 Kb = Kb = [CO32-] 7.2 x 10-4 Apr. 1 0.500 = 1.39 x 10-11 1.39 x 10-11 = 3.6 x 1010 OK to ignore –x here ie. NO Quadratic NO2- + H2O ⇋ HNO2 + OH- 0.500 0 0 -x +x +x 0.500 – x x x Check:

  35. Apr. 1 x2 = 6.95 x 10-12 x = 2.6 x 10-6 [OH-] = 2.6 x 10-6 mol/L pOH = ?? pH = 8.42

  36. Apr. 4 Calculating Ka from [weak acid] and pH See p. 591 #6 & 8 eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the numerical value of the Ka for this acid. • Equation • Find [H3O+] from pH • Subtract from [weak acid] • Substitute to find Ka

  37. (0.00209)(0.00209) Ka = (0.06991) Apr. 4 C6H5COOH(aq) + H2O(l)º H3O+(aq) + C6H5COO-(aq) [H3O+] = 10-2.68 = 0.00209 mol/L [C6H5COO-] = 0.00209 mol/L [C6H5COOH] = 0.072 – 0.00209 = 0.06991 mol/L Find Ka = 6.2 x 10-5

  38. Apr. 4 Calculating Ka from [weak acid] and pH [H3O+] = 10-2.68 = 0.00209 mol/L See p. 591 #’s 5 & 6 = 2.9 % eg. The pH of a 0.072 mol/L solution of benzoic acid, C6H5COOH, is 2.68. Calculate the % dissociation for this acid.

  39. Apr. 4 Calculate the acid dissociation constant, Ka , and the percent dissociation for each acid: • a) 0.250 mol/L chlorous acid, HClO2(aq); pH = 1.31 • 0.012 19.5% • b) 0.150 mol/L cyanic acid, HCNO(aq); pH = 2.15 • 0.00035 4.7% • c) 0.100 mol/L arsenic acid, H3AsO4(aq); pH = 1.70 • 0.0050 20% • 0.500 mol/L iodic acid, HIO3(aq); pH = 0.670 • 0.160 42.8%

  40. Apr. 4 More Practice: • Weak Acids: pp. 591, 592 #’s 6 -8 • Weak Bases: p. 595 #’s 11 - 16 (Kb’s on p. 592) Worksheet #8

  41. Apr. 5 Acid-Base Stoichiometry Solution Stoichiometry (Review) • Write a balanced equation • Calculate moles given ( OR n = CV) • Mole ratios • Calculate required quantity OR OR m = nM

  42. Apr. 5 eg. 25.0 mL of 0.100 mol/L H2SO4(aq) was used to neutralize 36.5 mL of NaOH(aq). Calculate the molar concentration of the NaOH solution. H2SO4(aq) + NaOH(aq) 2 → H2O(l) + Na2SO4(aq) 2 nH2SO4 = nNaOH = CNaOH =

  43. Apr. 7 Acid-Base Stoichiometry pp. 600, 601 – Sample Problems p. 602 #’s 17 - 20 Worksheet #9

  44. Acids & Bases #9 answers: • 0.210 mol/L • a) 22.5 mL b) 24.7 mL c) 4.8 mL • 31.5 mL • a) 0.0992 mol/L b) 0.269 mol/L c) 0.552 mol/L

  45. Apr. 7 Dilution • Given 3 of the four variables • Only one solution • CiVi = CfVf Stoichiometry • Given 3 of the four variables • Two different solutions • 4 step method

  46. Apr. 8 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 H3O+ and OH- • subtract to determine the moles of excess H3O+ or OH- • divide by total volume to get concentration • calculate pH

  47. Apr. 8 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: Ca2+ OH- H3O+ Cl- H2O SB SA

  48. Apr. 8 NIE: OH- + H3O+ → 2 H2O C = 0.00500 mol/L V = 0.0100 L C = 0.0200 mol/L V = 0.0200 L n = CV n = CV 4.00 x 10-4 mol OH- 5.0 x 10-5 mol H3O+ 3.5 x 10-4 mol excess OH-

  49. Apr. 8 = 0.01167 mol/L [OH-] = 0.01167 mol/L pOH = 1.933 pH = 12.067 Worksheet #10

  50. Apr. 8 Acids % Bases #10 answers: 1. pH = 13.000 2. [H3O+] = 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

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