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### Chapter16

2. 2. Contents and Concepts Solutions of a Weak Acid or Base 1. Acid−Ionization Equilibria 2. Polyprotic Acids 3. Base−Ionization Equilibria 4. Acid–Base Properties of Salt Solutions Solutions of a Weak Acid or Base with Another Solute 5. Common−Ion Effect 6. Buffers 7. Acid–Base Titration Curves Copyright © Cengage Learning. All rights reserved. 16 | 2
3. 3. Learning Objectives Solutions of a Weak Acid or Base Acid−Ionization Equilibria a. Write the chemical equation for a weak acid undergoing acid ionization in aqueous solution. b. Define acid−ionization constant and degree of ionization. c. Determine Ka from the solution pH. d. Calculate concentrations of species in a weak acid solution using Ka (approximation method). Copyright © Cengage Learning. All rights reserved. 16 | 3
4. 4. 1. Acid−Ionization Equilibria (cont.) a. State the assumption that allows for using approximations when solving problems. b. Calculate concentrations of species in a weak acid solution using Ka (quadratic formula). 2. Polyprotic Acids a. State the general trend in the ionization constants of a polyprotic acid. b. Calculate concentrations of species in a solution of a diprotic acid. Copyright © Cengage Learning. All rights reserved. 16 | 4
5. 5. 3. Base−Ionization Equilibria a. Write the chemical equation for a weak base undergoing ionization in aqueous solution. b. Define base−ionization constant. c. Calculate concentrations of species in a weak base solution using Kb. Copyright © Cengage Learning. All rights reserved. 16 | 5
6. 6. 4. Acid–Base Properties of Salt Solutions a. Write the hydrolysis reaction of an ion to form an acidic solution. b. Write the hydrolysis reaction of an ion to form a basic solution. c. Predict whether a salt solution is acidic, basic, or neutral. d. Obtain Ka from Kb or Kb from Ka. e. Calculating concentrations of species in a salt solution. Copyright © Cengage Learning. All rights reserved. 16 | 6
7. 7. Solutions of a Weak Acid or Base with Another Solute 5. Common−Ion Effect a. Explain the common−ion effect. b. Calculate the common−ion effect on acid ionization (effect of a strong acid). c. Calculate the common−ion effect on acid ionization (effect of a conjugate base). Copyright © Cengage Learning. All rights reserved. 16 | 7
8. 8. 6. Buffers a. Define buffer and buffer capacity. b. Describe the pH change of a buffer solution with the addition of acid or base. c. Calculate the pH of a buffer from given volumes of solution. d. Calculate the pH of a buffer when a strong acid or a strong base is added. e. Learn the Henderson–Hasselbalch equation. f. State when the Henderson–Hasselbalch equation can be applied. Copyright © Cengage Learning. All rights reserved. 16 | 8
9. 9. 7. Acid–Base Titration Curves a. Define equivalence point. b. Describe the curve for the titration of a strong acid by a strong base. c. Calculate the pH of a solution of a strong acid and a strong base. d. Describe the curve for the titration of a weak acid by a strong base. e. Calculate the pH at the equivalence point in the titration of a weak acid by a strong base. Copyright © Cengage Learning. All rights reserved. 16 | 9
10. 10. 7. Acid–Base Titration Curves (cont) f. Describe the curve for the titration of a weak base by a strong acid. g. Calculate the pH of a solution at several points of a titration of a weak base by a strong acid. Copyright © Cengage Learning. All rights reserved. 16 | 10
11. 11. The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water. We can write an acid equilibrium reaction for the generic acid, HA. HA(aq) + H2O(l)  H3O+(aq) + A−(aq) Copyright © Cengage Learning. All rights reserved. 16 | 11
12. 12. Acetic acid is a weak acid. It reacts with water as follows: HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2−(aq) Here A from the previous slide = C2H3O2−(aq) Copyright © Cengage Learning. All rights reserved. 16 | 12
13. 13. The equilibrium constant for the reaction of a weak acid with water is called the acid−ionization constant (or acid−dissociation constant), Ka. Ka [H O ] [ A ] = 3 + − [HA ] Liquid water is not included in the equilibrium constant expression. Copyright © Cengage Learning. All rights reserved. 16 | 13
16. 16. Concept Check 16.1 You have prepared dilute solutions of equal molar concentrations of HC2H3O2 (acetic acid), HNO2, HF, and HCN. Rank the solutions from the highest pH to the lowest pH. (Refer to Table 16.1.) In order of decreasing Ka: HF HNO2 HC2H3O2 HCN Ka 6.8 × 10−4 4.5 × 10−4 1.7 × 10−5 4.9 × 10−10 Copyright © Cengage Learning. All rights reserved. Highest pH HCN HC2H3O2 HNO2 HF Lowest pH 16 | 16
17. 17. Calculations with Ka Given the value of Ka and the initial concentration of HA, you can calculate the equilibrium concentration of all species. Given the value of Ka and the initial concentration of HA, you can calculate the degree of ionization and the percent ionization. Given the pH of the final solution and the initial concentration of HA, you can find the value of Ka and the percent ionization. Copyright © Cengage Learning. All rights reserved. 16 | 17
18. 18. We can be given pH, percent or degree ionization, initial concentration, and Ka. From pH, we can find [H3O+]. From percent or degree ionization, we can find Ka. Using what is given, we can find the other quantities. Copyright © Cengage Learning. All rights reserved. 16 | 18
19. 19. ? Sore−throat medications sometimes contain the weak acid phenol, HC6H5O. A 0.10 M solution of phenol has a pH of 5.43 at 25°C. a. What is the acid−ionization constant, Ka, for phenol at 25°C? b. What is the degree of ionization? Copyright © Cengage Learning. All rights reserved. 16 | 19
20. 20. HC6H5O(aq) + H2O(l)  Initial Change Equilibrium H3O+(aq) + C6H5O−(aq) 0.10 0 0 –x +x +x 0.10 – x x x We were told that pH = 5.43. That allows us to find [H3O+] = 3.7 × 10−6 M = x = [C6H5O−]. Now we find [HC6H5O] = 0.10 – x = 0.10 M. Copyright © Cengage Learning. All rights reserved. 16 | 20
21. 21. Finally, we write the expression for Ka and substitute the concentrations we now know. HC6H5O + H2O  H3O+ + C6H5O− [H3O+] = [C6H5O−] = 3.7 × 10−6 M [HC6H5O] = 0.10 M H3 O + C 6 H5 O − Ka = [HC 6H5 O] [ ][ ] (3.7 × 10 −6 )2 Ka = 0.10 Ka = 1.4 × 10−10 Copyright © Cengage Learning. All rights reserved. 16 | 21
22. 22. The degree of ionization is the ratio of ionized concentration to original concentration: x 3.7 × 10−6 Degree of ionization = = 0.10 0.10 Degree of ionization = 3.7 × 10−5 Percent ionization is the degree of ionization × 100%: Percent ionization = 3.7 × 10−3% or 0.0037% Copyright © Cengage Learning. All rights reserved. 16 | 22
23. 23. Simplifying Assumption for Acid and Base Ionizations The equilibrium concentration of the acid is most often ([HA]0 – x). If x is much, much less than [HA]0, we can assume that subtracting x makes no difference to [HA]: ([HA]0 – x) = [HA]0 This is a valid assumption when the ratio of [HA] 0 to Ka is > 103. If it is not valid, you must use the quadratic equation to solve the problem. Copyright © Cengage Learning. All rights reserved. 16 | 23
24. 24. ? Para−hydroxybenzoic acid is used to make certain dyes. What are the concentrations of this acid, of hydronium ion, and of para−hydroxybenzoate ion in a 0.200 M aqueous solution at 25°C? What is the pH of the solution and the degree of ionization of the acid? The Ka of this acid is 2.6 × 10−5. We will use the generic formula HA for para−hydroxybenzoic acid and the following equilibrium: HA + H2O  H3O+ + A− Copyright © Cengage Learning. All rights reserved. 16 | 24
25. 25. HA(aq) + Initial Change Equilibrium H2O(l)  H3O+(aq) + A−(aq) 0.200 0 0 –x +x +x 0.200 – x x x [H O ] [ A ] = [HA ] + Ka 2.6 × 10 − 3 −5 x2 = 0.200 − x 0.200 > 1000, so 0.200 − x ≈ 0.200 −5 2.6 × 10 Copyright © Cengage Learning. All rights reserved. 16 | 25
26. 26. x2 Ka = 0.200 5.2 × 10 −6 = x 2 x = 2.3 × 10 −3 = [H3 O + ] pH = − log [H3O+ ] = − log (2.3 × 10−3 ) pH = 2.64 Copyright © Cengage Learning. All rights reserved. 16 | 26
27. 27. Polyprotic Acids A polyprotic acid has more than one acidic proton —for example, H2SO4, H2SO3, H2CO3, H3PO4. These acids have successive ionization reactions with Ka1, Ka2, . . . The next example illustrates how to do calculations for a polyprotic acid. Copyright © Cengage Learning. All rights reserved. 16 | 27
28. 28. ? Tartaric acid, H2C4H4O6, is a diprotic acid used in food products. What is the pH of a 0.10 M solution? What is the concentration of the C 4H4O62− ion in the same solution? Ka1 = 9.2 × 10−4; Ka2 = 4.3 × 10−5. First, we will use the first acid−ionization equilibrium to find [H+] and [HC4H4O6−]. In these calculations, we will use the generic formula H2A for the acid. Next, we will use the second acid−ionization equilibrium to find [C4H4O62−]. Copyright © Cengage Learning. All rights reserved. 16 | 28
29. 29. H2A(aq) + Initial H3O+(aq) + HA−(aq) 0.10 0 0 –x +x +x 0.10 – x x x Change Equilibrium K a1 H2O(l)  [H O ] [HA ] = 3 + − [H2 A ] 9.2 × 10 −4 2 x = 0.10 - x 0.10 < 1000 −4 9.2 × 10 We cannot use the simplifying assumption . Copyright © Cengage Learning. All rights reserved. 16 | 29
30. 30. 9.2 × 10 −4 (0.10 − x ) = x 2 9.2 × 10 −5 − 9.2 × 10 −4 x = x 2 x 2 + 9.2 × 10−4 x − 9.2 × 10−5 = 0 a =1 b = 9.2 × 10 −4 c = − 9.2 × 10 −5 − b ± b 2 − 4ac x= 2a −(9.2 × 10−4 ) ± (9.2 × 10−4 )2 − 4(1)( −9.2 × 10−5 ) x= 2(1) − (9.2 × 10 −4 ) ± (1.92 × 10 −2 ) −4 −3 x= = −4.6 × 10 ± 9.6 × 10 2 Copyright © Cengage Learning. All rights reserved. 16 | 30
31. 31. x = 9.1× 10 −3 and x = −1.0 × 10 −2 We eliminate the negative value because it is physically impossible to have a negative concentration. At the end of the first acid ionization equilibrium, the concentrations are [H3O+] = 9.1 × 10−3 M [HA−] = 9.1 × 10−3 M [H2A] = 9.0 × 10−4 M Now we use these for the second acid−ionization equilibrium. Copyright © Cengage Learning. All rights reserved. 16 | 31
32. 32. HA−(aq) + Initial Change Equilibrium Ka2 H2O(l)  H3O+(aq) + A2−(aq) 0.0091 0.0091 0 –x +x +x 0.0091 – x 0.0091 + x x H3O+   A 2−    = HA −    4.3 × 10 −5 (0.0091 + x ) x = (0.0091 − x ) We can assume that 0.0091 + x ≈ 0.0091 and 0.0091 − x ≈ 0.0091 Copyright © Cengage Learning. All rights reserved. 16 | 32
33. 33. 4.3 × 10 −5 0.0091 x = 0.0091 x = 4.3 × 10 −5 M Our assumption was valid. [C4H4O62− ] = 4.3 × 10−5 M [HC4H4O6− ] = 9.1× 10−3 M [H3O + ] = 9.1 × 10 −3 M [H2C 4H4O6 ] = 9.0 × 10 −4 M pH = −log (9.1× 10 −3 ) Copyright © Cengage Learning. All rights reserved. pH = 2.04 16 | 33
34. 34. Base−Ionization Equilibrium The simplest acid–base equilibria are those in which a weak acid or a weak base reacts with water. We can write a base equilibrium reaction for the generic base, B. B(aq) + H2O(l)  HB+(aq) + OH−(aq) Copyright © Cengage Learning. All rights reserved. 16 | 34
35. 35. Ammonia is a weak base. It reacts with water as follows: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) Here the generic B from the previous slide is NH3(aq). Copyright © Cengage Learning. All rights reserved. 16 | 35
36. 36. The equilibrium constant for the reaction of a weak base with water is called the base−ionization constant, Kb. Kb [HB ][OH ] = + − [ B] Liquid water is not included in the equilibrium constant expression. Copyright © Cengage Learning. All rights reserved. 16 | 36
38. 38. Writing Kb Reactions The bases in Table 16.2 are nitrogen bases; that is, the proton they accept adds to a nitrogen atom. Next we’ll practice writing the Kb reactions. Ammonium becomes ammonium ion: NH3+ H2O → NH4+ + OH− Ethylamine becomes ethyl ammonium ion: C2H5NH2 + H2O → C2H5NH3+ + OH− Copyright © Cengage Learning. All rights reserved. 16 | 38
39. 39. Dimethylamine becomes dimethylammonium ion: (CH3)2NH2 + H2O → (CH3)2NH3+ + OH− Pyridine becomes pyridinium ion: C5H5N+ H2O → C5H5NH+ + OH− Hydrazine becomes hydrazinium ion: N2H4+ H2O → N2H5+ + OH− Copyright © Cengage Learning. All rights reserved. 16 | 39
40. 40. We can be given pH, initial concentration, and Kb. From the pH, we can find first [H3O+] and then [OH−]. Using what is given, we can find the other quantities. We can also use a simplifying assumption: When [B]0 / Kb > 103, the expression ([B]0 – x) = [B]0. Copyright © Cengage Learning. All rights reserved. 16 | 40
41. 41. ? Aniline, C6H5NH, is used in the manufacture of some perfumes. What is the pH of a 0.035 M solution of aniline at 25°C? Kb= 4.2 × 10−10 at 25°C. We will construct an ICE chart and solve for x. Copyright © Cengage Learning. All rights reserved. 16 | 41
42. 42. C6H5H(aq) + Initial Change Equilibrium H2O(l)  C6H5NH+(aq) + OH−(aq) 0.035 0 0 –x +x +x 0.035 – x x x We are told that Kb = 4.2 × 10−10. That allows us to substitute into the Kb expression to solve for x. Copyright © Cengage Learning. All rights reserved. 16 | 42
43. 43. C6H5N + H2O C6H5NH+ + OH− [B] = 0.200 – x and [B+] = [OH−] = x [C H NH ] [OH ] = + Kb 0.200 4.2 × 10−10 6 − 5 [ C6H5N] x2 −10 4.2 × 10 = (0.200 − x ) > 1000, so we can assume that 0.200 − x = x x2 4.2 × 10−10 = 0.200 x 2 = 8.4 × 10−11 x = 9.2 × 10−6 M Copyright © Cengage Learning. All rights reserved. 16 | 43
44. 44. The question asks for the pH: pH = 14.00 = pOH = 14.00 − log [OH− ] pH = 14.00 − log (9.2 × 10−6 ) pH = 14.00 − 5.04 pH = 8.96 Copyright © Cengage Learning. All rights reserved. 16 | 44
45. 45. Salt Solutions We will look at the cation and the anion separately, and then combine the result to determine whether the solution is acidic, basic, or neutral. Copyright © Cengage Learning. All rights reserved. 16 | 45
46. 46. The conjugate acid of a strong base is very weak and does not react with water. It is therefore considered to be neutral. Na+ + H2O → No Reaction (NR) The conjugate base of a strong acid is very weak and does not react with water. It is therefore considered to be neutral. Cl− + H2O → NR Copyright © Cengage Learning. All rights reserved. 16 | 46
47. 47. The conjugate acid of a weak base reacts with water to form an acidic solution: NH4+ + H2O  NH3 + H3O+ The conjugate base of a weak acid reacts with water to form a basic solution: F− + H2O  HF + OH− Copyright © Cengage Learning. All rights reserved. 16 | 47
48. 48. We classify each salt by examining its cation and its anion, and then combining the result. NaBr Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is neutral. Br− is the conjugate base of HBr, a strong acid. It does not react with water, so is neutral. The cation is neutral; the anion is neutral. NaBr is neutral. Copyright © Cengage Learning. All rights reserved. 16 | 48
49. 49. NaC2H3O2 Na+ is the conjugate acid of NaOH, a strong base. It does not react with water, so it is neutral. C2H3O2− is the conjugate base of HC2H3O2, a weak acid. It reacts with water to give a basic solution. The cation is neutral; the anion is basic. NaC2H3O2 is basic. Copyright © Cengage Learning. All rights reserved. 16 | 49
50. 50. NH4Cl NH4+ is the conjugate acid of NH3, a weak base. It reacts with water to give an acidic solution. Cl− is the conjugate base of HCl, a strong acid. It does not react with water, so it is neutral. The cation is acidic; the anion is neutral. NH4Cl is acidic. Copyright © Cengage Learning. All rights reserved. 16 | 50
51. 51. NH4F NH4+ is the conjugate acid of NH3, a weak base. It reacts with water to give an acidic solution. F− is the conjugate base of HF, a weak acid. It does not react with water, so it is basic. The cation is acidic; the anion is basic. We need more information in this case. We compare Ka for HF to Kb for NH3. Copyright © Cengage Learning. All rights reserved. 16 | 51
52. 52. If Ka > Kb, the solution is acidic. If Ka < Kb, the solution is basic. If Ka = Kb, the solution is neutral. K w 1.0 × 10 −14 + For NH4 : K a = = = 5.6 × 10 −10 K b 1.8 × 10 −5 K w 1.0 × 10 −14 For F − : K b = = = 1.5 × 10 −11 K a 6.8 × 10 − 4 + K a for NH 4 > K b for F − NH4F is acidic. Copyright © Cengage Learning. All rights reserved. 16 | 52
53. 53. ? Ammonium nitrate, NH4NO3, is administered as an intravenous solution to patients whose blood pH has deviated from the normal value of 7.40. Would this substance be used for acidosis (blood pH < 7.40) or alkalosis (blood pH > 7.40)? Copyright © Cengage Learning. All rights reserved. 16 | 53
54. 54. NH4+ is the conjugate acid of a NH3, a weak base. NH4+ is acidic. NO3− is the conjugate base of HNO3, a strong acid. NO3− is neutral. NH4NO3 is acidic, so it could be used for alkalosis. Copyright © Cengage Learning. All rights reserved. 16 | 54
55. 55. The hydrolysis equilibrium constant can be used in problems to determine the pH of a salt solution. To use the hydrolysis equilibrium, we need to compute the K value for it. Copyright © Cengage Learning. All rights reserved. 16 | 55
56. 56. We can determine K for the reaction of the ion with water (hydrolysis reaction). NH4+ + H2O <==> NH3 + H3O+ is the sum of NH4+ + OH− <==> NH3 + H2O K1 = 1/Kb 2H2O <==> H3O+ + OH− K2 = Kw Overall reaction NH4+ + H2O <==> NH3 + H3O+ Kw K a = K1 K 2 = Kb Copyright © Cengage Learning. All rights reserved. 16 | 56
57. 57. ? What is Kb for the F− ion, the ion added to the public water supply to protect teeth? For HF, Ka = 6.8 × 10−4. Copyright © Cengage Learning. All rights reserved. 16 | 57
58. 58. F− + H2O <==> HF + OH− Is the sum of H3O+ + F− <==> HF + H2O; 2H2O <==> H3O+ + OH−; K1 = 1/Ka K2 = Kw Overall reaction: F− + H2O <==> HF + OH− K w 1.0 × 10 −14 K b = K1 K 2 = = -10 K a 6.8 × 10 For F- : Kb = 1.5 × 10−5 Copyright © Cengage Learning. All rights reserved. 16 | 58
59. 59. ? Household bleach is a 5% solution of sodium hypochlorite, NaClO. This corresponds to a molar concentration of about 0.70 M NaClO. What is the [OH−] and the pH of the solution? For HClO, Ka = 3.5 × 10−8. Copyright © Cengage Learning. All rights reserved. 16 | 59
60. 60. ClO−(aq) + Initial Change Equilibrium H2O(l)  HClO(aq) + OH−(aq) 0.70 0 0 –x +x +x 0.70 – x x x We are told that Ka = 3.5 × 10−8. That means that Kb= 2.9 × 10−7. This allows us to substitute into the Kb expression to solve for x. Copyright © Cengage Learning. All rights reserved. 16 | 60
61. 61. ClO− + H2O  HClO + OH− [H3O+] = [A−] = x and [HA] = 0.200 – x 0.70 2.9 × 10−7 HClO] OH−  [   Kb = ClO−    2 x −7 2.9 × 10 = (0.70 − x ) > 1000, so we can assume that 070 − x = 0.70. x2 2.9 × 10−7 = 0.70 x 2 = 2.0 × 10 -7 x = 4.5 × 10 -4 M Copyright © Cengage Learning. All rights reserved. 16 | 61
62. 62. The question asks for [OH−] and the pH: [OH− ] = x = 4.5 × 10−4 M pH = 14.00 = pOH = 14.00 − log [OH− ] −4 pH = 14.00 − log (4.5 × 10 ) pH = 14.00 − 3.35 pH = 10.65 Copyright © Cengage Learning. All rights reserved. 16 | 62
63. 63. Common−Ion Effect The common−ion effect is the shift in an ionic equilibrium caused by the addition of a solute that takes part in the equilibrium. Copyright © Cengage Learning. All rights reserved. 16 | 63
64. 64. Buffers A buffer solution is characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. A buffer is made by combining a weak acid with its conjugate base or a weak base with its conjugate acid. Copyright © Cengage Learning. All rights reserved. 16 | 64
65. 65. ? What is the pH of a buffer made by mixing 1.00 L of 0.020 M benzoic acid, HC7H5O2, with 3.00 L of 0.060 M sodium benzoate, NaC7H5O2? Ka for benzoic acid is 6.3 × 10−5. This problem involves dilution first. Once we know the concentrations of benzoic acid and benzoate ion, we can use the acid equilibrium to solve for x. We will use HBz to represent benzoic acid and Bz − to represent benzoate ion. Copyright © Cengage Learning. All rights reserved. 16 | 65
66. 66. (0.020 M)(1.0 L) [HBz] = = 0.0050 M (1.0 + 3.0) L (0.060 M)(3.0 L) [Bz ] = = 0.045 M (1.0 + 3.0) L − HBz(aq) + H2O(l)  Initial Change Equilibrium H3O+(aq) + Bz−(aq) 0.0050 0 0.045 –x +x +x 0.0050 – x x 0.045 + x Copyright © Cengage Learning. All rights reserved. 16 | 66
67. 67. H3O+  Bz−    Ka =  [ HBz] 6.3 × 10 −5 x (0.045 + x ) = (0.0050 − x ) We can assume that 0.045 + x ≈ 0.045 and 6.3 × 10 −5 0.0050 − x ≈ 0.0050 0.045 x = 0.0050 x = 7.0 × 10 −6 M Our assumption was valid. Copyright © Cengage Learning. All rights reserved. 16 | 67
68. 68. [H3O + ] = x = 7.0 × 10 −6 M + −6 pH = − log [H3O ] = − log (7.0 × 10 ) pH = 5.15 Copyright © Cengage Learning. All rights reserved. 16 | 68
69. 69. ? Calculate the pH of a 0.10 M HF solution to which sufficient sodium fluoride has been added to make the concentration 0.20 M NaF. Ka for HF is 6.8 × 10−4. We will use the acid equilibrium for HF. NaF provides the conjugate base, F−, so [F−] = 0.20 M. Copyright © Cengage Learning. All rights reserved. 16 | 69
70. 70. [HF] = 0.10 M HF(aq) + H2O(l)  Initial H3O+(aq) + F−(aq) 0.10 0 0.20 –x +x +x 0.10 – x x 0.20 + x Change Equilibrium [F − ] = 0.20 M x (0.20 + x ) [H3O ] [F ] −4 6.8 × 10 = Ka = (0.10 - x ) [HF] We can assume that 0.20 + x ≈ 0.20 and 0.10 − x ≈ 0.10 + − Copyright © Cengage Learning. All rights reserved. 16 | 70
71. 71. 6.8 × 10 −4 0.20x = 0.10 x = 3.4 × 10 −4 M Our assumption was valid. [H3O + ] = x = 3.4 × 10 −4 M pH = − log [H3O + ] = − log (3.4 × 10 −4 ) pH = 3.47 Copyright © Cengage Learning. All rights reserved. 16 | 71
72. 72. Henderson—Hasselbalch Equation Buffers at a specific pH can be prepared using the Henderson−Hasselbalch Equation. Buffers are prepared from a conjugate acid−base pair in which the ionization is approximately equal to the desired H3O+ concentration. Copyright © Cengage Learning. All rights reserved. 16 | 72
73. 73. Consider a buffer made up of a weak acid HA and its conjugate base A−: HA(aq) + H2O(l)  H3O+(aq) + A−(aq) The acid−ionization constant is [H3O + ][A − ] Ka = [HA] Rearranging to solve for [H3O+]  [HA]  [H3O ] = Ka  − ÷  [A ]  + Copyright © Cengage Learning. All rights reserved. 16 | 73
74. 74. The preceding equation can be used to derive an equation for the pH of the buffer. First, we take the negative logarithm of both sides of the equation.  [HA]  − log[H3O ] = −log K a x −   [A ]    +  [HA]  −log[H3O ] = −log Ka − log  − ÷  [A ]  + Copyright © Cengage Learning. All rights reserved. 16 | 74
75. 75. The pKa of a weak acid is defined in a manner similar to pH and pOH pKa = − log Ka We can now express the equation as  [HA]  pH = pKa − log  − ÷  [A ]  This is equivalent to  [A − ]  pH = pK a + log   [HA]     Copyright © Cengage Learning. All rights reserved. 16 | 75
76. 76. This equation is generally shown as: [base] pH = pK a + log [acid] This equation relates buffer pH for different concentrations of conjugate acid and base and is known as the Henderson−Hasselbalch equation. Copyright © Cengage Learning. All rights reserved. 16 | 76
77. 77. ? What is the [H3O+] for a buffer solution that is 0.250 M in acid and 0.600 M in the corresponding salt if the weak acid Ka = 5.80 x 10−7? Use as example the following equation of a conjugate acid−base pair. HA(aq) + H2O(l)  H3O+(aq) + A−(aq) First, we use the value of Ka to find pKa: pKa = −log(5.80 x 10−7) pKa = −(−6.237) pKa = 6.237 Copyright © Cengage Learning. All rights reserved. 16 | 77
78. 78. [base] pH = pK a + log [acid] pKa = 6.237 [base] = 0.600 M [acid] = 0.250 M Substituting: [0.600] pH = 6.237 + log [0.250] pH = 6.237 + log 2.40 pH = 6.237 + 0.380 pH = 6.617 Copyright © Cengage Learning. All rights reserved. 16 | 78
79. 79. Using the pH, we can now find [H3O+]: [H3O+] = 10−pH [H3O+] = 10−6.617 [H3O+] = 2.42 × 10−7 M Copyright © Cengage Learning. All rights reserved. 16 | 79
80. 80. Acid–Base Titration An acid–base titration is a procedure for determining the amount of acid (or base) in a solution by determining the volume of base (or acid) of known concentration that will completely react with it. Copyright © Cengage Learning. All rights reserved. 16 | 80
81. 81. An acid–base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use the titration curve to choose an indicator that will show when the titration is complete. Copyright © Cengage Learning. All rights reserved. 16 | 81
82. 82. This titration plot shows the titration of a strong acid with a strong base. Note that the pH at the equivalence point is 7.0. Copyright © Cengage Learning. All rights reserved. 16 | 82
83. 83. The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added. The indicator must change color near the pH at the equivalence point. Copyright © Cengage Learning. All rights reserved. 16 | 83
84. 84. When solving problems in which an acid reacts with a base, we first need to determine the limiting reactant. This requires that we use moles in the calculation. The next step is to review what remains at the end of the reaction. In the case of a strong acid and a strong base, the solution is neutral. When a weak acid or weak base is involved, the product is a salt. After finding the salt concentration, we use a hydrolysis equilibrium to find the pH. Copyright © Cengage Learning. All rights reserved. 16 | 84
85. 85. ? Calculate the pOH and the pH of a solution in which 10.0 mL of 0.100 M HCl is added to 25.0 mL of 0.100 M NaOH. First we calculate the moles of HCl and NaOH. Then we determine what remains after the reaction. Copyright © Cengage Learning. All rights reserved. 16 | 85
86. 86. The overall reaction is HCl + NaOH → H2O + NaCl Moles HCl = (0.100 M)(10.0 × 10−3 L) = 1.00 × 10−3 mol Moles NaOH = (0.100 M)(25.0 × 10−3 L) = 2.50 × 10−3 mol All of the HCl reacts, leaving 1.50 × 10−3 mol NaOH. The new volume is 10.0 mL + 25.0 mL = 35.0 mL. Copyright © Cengage Learning. All rights reserved. 16 | 86
87. 87. 1.50 × 10−3 mol [OH− ] = = 4.29 × 10−2 M 35.0 × 10−3 L − pH = 14.00 − pOH = 14.00 − ( −log[OH ]) pH = 14.00 − 1.368 pH = 12.632 Copyright © Cengage Learning. All rights reserved. 16 | 87
88. 88. ? Calculate the [OH−] and the pH at the equivalence point for the titration of 500. mL of 0.10 M propionic acid with 0.050 M calcium hydroxide. (This can be used to prepare a preservative for bread.) Ka = 1.3 × 10−5 First we need to find the moles of propionic acid to find the volume of calcium hydroxide. Then we will examine the salt that is produced and use the hydrolysis equilibrium to find the pH. We will use HPr to represent propionic acid and Pr − to represent propionate ion. Copyright © Cengage Learning. All rights reserved. 16 | 88
89. 89. Moles propionic acid = (0.10 M)(500. × 10−3 L) = 0.050 mol HPr To find the moles of calcium hydroxide, we need the balanced chemical equation: 2HPr + Ca(OH)2 → Ca(Pr)2 + 2H2O Moles calcium hydroxide = 1 mol Ca(OH)2 0.050 mol HPr = 0.025 mol Ca(OH)2 2 mol HPr Copyright © Cengage Learning. All rights reserved. 16 | 89
90. 90. Now we find the volume of Ca(OH)2 required to give 0.025 mol: 0.025 mol Volume Ca(OH)2 = = 0.50 L mol 0.050 L The new volume is 500. × 10−3 L + 0.50 L = 1.00 L The concentration of Pr− is 2(0.025 M) = 0.050 M 0.050 mol [Pr ] = = 5.0 × 10−2 M = 0.050 M 1.0 L − Copyright © Cengage Learning. All rights reserved. 16 | 90
91. 91. The hydrolysis reaction is Pr− + H2O  HPr + OH− K w 1.00 × 10−14 −10 Kb = = = 7.7 × 10 −5 Ka 1.3 × 10 Pr−(aq) + Initial Change Equilibrium H2O(l)  HPr(aq) + OH−(aq) 0.050 0 0 –x +x +x 0.050 – x x x Copyright © Cengage Learning. All rights reserved. 16 | 91
92. 92. − [HPr] [OH ] Ka = [Pr − ] 7.7 × 10−10 x2 = (0.050 − x ) 0.050 > 1000 −10 7.7 × 10 We can assume that 0.050 − x ≈ 0.050. x2 −10 7.7 × 10 = 0.050 x 2 = 3.85 × 10 −11 x = 6.20 × 10 -6 M [OH- ] = x = 6.20 × 10−6 M pOH = 5.21 Copyright © Cengage Learning. All rights reserved. pH = 8.79 16 | 92