Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L...

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Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-2 Chapter 8 Review, pages 576–581 Knowledge 1. (b) 2. (b) 3. (a) 4. (b) 5. (d) 6. (c) 7. (c) 8. (d) 9. (c) 10. (b) 11. (d) 12. (b) 13. (b) 14. False. A strong acid completely ionizes in solution. 15. True 16. False. The hydrogen ion concentration of a solution with a pH of 3.0 is 1 × 10 3 mol/L. 17. False. A weak acid exists primarily in the form of molecules in solution. 18. True 19. False. The reaction of a strong acid with a strong base produces a neutral salt. 20. True 21. False. The endpoint of a strong acid–strong base titration occurs at a pH near 7. 22. True 23. False. A buffer solution can absorb a small amount of acid or base without a significant change in pH. 24. True 25. (a) (vi) (b) (vii) (c) (iii) (d) (xi) (e) (iv) (f) (ix) (g) (x) (h) (viii) (i) (i) (j) (ii) (k) (v) 26. Arrhenius acids form hydrogen ions in aqueous solutions and bases form hydroxide ions in aqueous solution. 27. (a) Water behaves as a base. (b) Water behaves as an acid. (c) Water behaves as a base. (d) Water behaves as an acid.

Transcript of Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L...

Page 1: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-2

Chapter 8 Review, pages 576–581 Knowledge 1. (b) 2. (b) 3. (a) 4. (b) 5. (d) 6. (c) 7. (c) 8. (d) 9. (c) 10. (b) 11. (d) 12. (b) 13. (b) 14. False. A strong acid completely ionizes in solution. 15. True 16. False. The hydrogen ion concentration of a solution with a pH of 3.0 is 1 × 10−3 mol/L. 17. False. A weak acid exists primarily in the form of molecules in solution. 18. True 19. False. The reaction of a strong acid with a strong base produces a neutral salt. 20. True 21. False. The endpoint of a strong acid–strong base titration occurs at a pH near 7. 22. True 23. False. A buffer solution can absorb a small amount of acid or base without a significant change in pH. 24. True 25. (a) (vi) (b) (vii) (c) (iii) (d) (xi) (e) (iv) (f) (ix) (g) (x) (h) (viii) (i) (i) (j) (ii) (k) (v) 26. Arrhenius acids form hydrogen ions in aqueous solutions and bases form hydroxide ions in aqueous solution. 27. (a) Water behaves as a base. (b) Water behaves as an acid. (c) Water behaves as a base. (d) Water behaves as an acid.

Page 2: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-3

28. (a) acid: H3PO4(aq); base: H2O(l); conjugate acid: H3O+(aq); conjugate base: H2PO4

−(aq) (b) acid: H2O(l); base: NH3(aq); conjugate acid: NH4

+(aq); conjugate base: OH−(aq) (c) acid: HBr(aq); base: H2O(l); conjugate acid: H3O+(aq); conjugate base: Br−(aq) (d) acid: H2O(l); base: CN−(aq); conjugate acid: HCN(aq); conjugate base: OH−(aq) 29. (a) The conjugate base of HF is fluoride ion; F–. (b) The conjugate acid of H2O is hydronium ion; H3O+. (c) The conjugate base of H2O is hydroxide ion; OH–. (d) The conjugate acid of NH3 is ammonium ion; NH4

+. 30. HSO4

–(aq) + H2O(l) ⇌ H3O+(aq) + SO42–(aq)

HSO4–(aq) +HNO3(aq) ⇌ H2SO4(aq) + NO3

–(aq) 31. pH = −log[H+] pOH= −log[OH–] pH + pOH = 14 (a) [H+(aq)] = 0.004 mol/L pH = 2.4 (b) The added hydroxide ions are significantly less than the [OH] due to autoionization of water; therefore, pH = 7 (c) [H+(aq)] = 7.3 × 10−6 mol/L pH = 5.14 (d) [H+(aq)] = 0.25 mol/L pH = 0.60 (e) [OH–(aq)] = 0.45 mol/L pOH = 0.35 pH = 13.65

(f)

[OH! (aq)]= 2" 6.1 "10!5 molL

[OH! (aq)]= 1.22"10!4 molL

pOH= ! log [OH! (aq)]= !log [1.22"10!4]

pOH = 3.91pH = 14.00-3.91

=10.09

pH = 10.09 (g) [OH–(aq)] = 0.006 mol/L pOH = 2.2 pH = 11.8 32. pOH = 14.00 – pH 33. H2O(l)

! "!!# !!! H+(aq) + OH–(aq)

Page 3: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-4

34. Given: pH = 8.0 Required: [OH–(aq)] Solution: [H+(aq)] = 10−pH [H+(aq)] = 10−8.0 = 1.0 × 10−8 mol/L

[OH! (aq)]= 1.0"10!14

1.0"10!8

[OH–(aq)] = 1.0 × 10−6 mol/L 35. A strong acid dissociates completely in water, but a weak acid exists with molecular and ionic species in equilibrium. 36. Hydrochloric acid is a strong acid; it will have a large Ka since the ionic products are present in much higher concentration than the molecular reactant.

37. a

3 3

a 3

5a

[H (aq)][A (aq)][HA(aq)]

(1.34 10 )(1.34 10 )0.1000 1.34 10

1.80 10

+ −

− −

=

× ×=− ×

= ×

K

K

K

38. Solution pH pOH [H+(aq)] [OH–] Acidic,

basic, or neutral?

A 6.88 7.12 1.3 × 10−7 mol/L 7.7 × 10−8 mol/L acidic

B 0.92 13.08 0.12 mol/L 8.4 × 10−14 mol/L acidic

C 10.89 3.11 1.3 × 10−11 mol/L 7.8 × 10−4 mol/L basic

D 7.00 7.00 1.0 × 10−7 mol/L 1.0 × 10−7 mol/L neutral

39. pH = 14.0 − pOH pH = 6.4

40. [OH! (aq)]= 1.0"10!14

[H+ (aq)]

(a) [OH–(aq)] = 1.0 × 10–7 mol/L; neutral (b) [OH–(aq)] = 1.2 × 10–9 mol/L; acidic (c) [OH–(aq)] = 1.0 × 10–2 mol/L; basic (d) [OH–(aq)] = 1.9 × 10–10 mol/L; acidic

Page 4: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-5

41. Given: pH = 4.0 Required: [OH–(aq)] Solution: [H+(aq)] = 10−pH [H+(aq)] = 10−4.0 = 1.0 × 10−4 mol/L

14

4

1.0 10[OH (aq)]1.0 10

−−

×=×

[OH–(aq)] = 1 × 10−10 mol/L 42. (a) The ionization of H2SO4 favours the products. (b) The ionization of HCO3

– favours the reactants. (c) The ionization of HF favours the reactants. 43. (a) The autoionization of water is endothermic because the equilibrium shifts toward ionization as the temperature is increased. (b) Kw = [H+(aq)] [OH–(aq)] [H+(aq)] = [OH–(aq]

5.47 !10"14 = [H+ (aq)]2

[H+ (aq)]= 5.47 !10"14

[H+(aq)] = 2.34 × 10–7 mol/L [OH–(aq)] = 2.34 × 10–7 mol/L 44. (a) The solution would be a moderately weak base because one mole of hydroxide ions are produced per mole of conjugate base; HCO3

–(aq) (b) The solution would be a strong acid that produces two moles of hydrogen ions per mole of acid; H2SO4 (aq) 45. Given: [HF(aq)] = 0.050 mol/L; Ka = 6.6 × 10−4

Required: percent ionization Analysis:

HF(aq) ⇌ H+(aq) + F−(aq) I 0.050 0 0 C − x + x + x E 0.050 – x x x

( )

+

a[H (aq)][F (aq)][HF aq ]

K−

=

( )percentage ionization [H (aq)]

100 [HF aq ]

+

=

Page 5: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-6

Solution: 4 ( )( )6.6 10(0.050 – )x x

x−× =

4 ( )( )6.6 100.050x x−× ≈

x2 ≈ 3.3 × 10−5 x = [H+(aq)] ≈ 5.7 × 10−3 mol/L

percentage ionization 0.0057100 0.050

=

percentage ionization = 11 % 46. Given: [H2CO3(aq)] = 0.050 mol/L; Ka1 = 4.4 × 10−7; Ka2 = 4.7 × 10−11

Required: [HCO3–(aq)], [CO3

2–(aq)], [H+ (aq)] Analysis:

H2CO3(aq) ⇌ H+(aq) + HCO3−(aq)

I 0.050 0 0 C − x + x + x E 0.050 – x x x

+

3a

2 3

[H (aq)][HCO (aq)][H CO (aq)]

K−

=

Solution: 7 ( )( )4.4 100.050 –x x

x−× =

7 ( )( )4.4 100.050x x−× ≈

x2 ≈ 2.2 × 10−8 x = [H+(aq)] = [HCO3

−(aq)] ≈ 1.5 × 10−4 mol/L

HCO3

− (aq) ⇌ H+(aq) + CO32−(aq)

I 0.00015 0.00015 0 C − x + x + x E 0.00015 – x 0.00015 + x x + 2

3a

3

[H (aq)][CO (aq)][HCO (aq)]

−=K

11 (0.00015 )( )4.7 100.00015 –

x xx

− +× =

11 (0.00015 )( )4.7 100.00015

x−× ≈ x = [HCO3

−(aq)] ≈ 4.7 × 10−11 mol/L

Page 6: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-7

Statement: [HCO3–(aq)] = 1.5 × 10–4 mol/L; [CO3

2–(aq)] = 4.7 × 10–11 mol/L; [H+(aq)] = 1.5 × 10–4 mol/L

47. Given: [HCO2H(aq)] = 0.100 mol/L; [H+(aq)] = 1.34 × 10–3 mol/L Required: Ka Solution: HCO2H(aq) ⇌ H+(aq) + HCO2

−(aq)

( )+

2a

2

[H (aq)][HCO (aq)][HCO H aq (aq)]

K−

=

–3 –3

a(1.34 10 )(1.34 10 )

(0.100)K × ×=

Ka = 1.80 × 10−5 48. Given: [HC2H3O2 (aq)] = 0.10 mol/L; Ka = 1.8 × 10−5

Required: pH Analysis:

HC2H3O2 (aq) ⇌ H+(aq) + C2H3O2−(aq)

I 0.10 0 0 C − x + x + x E 0.10 – x x x

( )

( )+

2 3 2a

2 3 2

[H (aq)][C H O aq ][HC H O aq ]

=K

Solution: 5 ( )( )1.8 10 0.10 –x x

x−× =

5 ( )( )1.8 10 0.10x x−× ≈

x2 ≈ 1.8 × 10−6 x = [H+(aq)] ≈ 1.34 × 10−3 mol/L pH = −log(1.34 × 10−3) pH = 2.87 49. (a) An aqueous solution of sodium nitrate will be neutral. (b) An aqueous solution of ammonium bromide will be acidic. (c) An aqueous solution of sodium fluoride will be basic. 50. The sample is a weak acid and the titrant is a strong base

Page 7: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-8

51.

52. (a) In a solution with a pH of 4.8, phenolphthalein will be colourless. (b) In a solution with a pH of 4.8, bromocresol green will be green. (c) In a solution with a pH of 4.8, phenol red will be yellow. 53. Given: [HC3H5O3(aq)] = 0.75 mol/L; [NaC3H5O3 (aq)] = 0.25 mol/L; Ka = 1.4 × 10−4

Required: pH Analysis:

HC3H5O3(aq) ⇌ C3H5O3−(aq) + H+(aq)

0.75 0.25 0 − x + x + x 0.75 – x 0.25 + x x

( ) ( )( )

3 5 3a

3 5 3

[C H O aq ][H aq ][HC H O aq ]

K− +

=

Solution: 4 ( )(0.25 )1.4 10(0.75 – )x x

x− +× =

4 ( )(0.25)1.4 100.75x−× ≈

x ≈ 4.2 × 10−4 x = [H+(aq)] ≈ 4.2 × 10−4 mol/L pH = −log(4.2 × 10−4) pH = 3.38

Page 8: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-9

54. Given: [H2CO2(aq)] = 0.10 mol/L; 2 2H CO 1.00 LV = ; [NaHCO2(aq)] = 0.10 mol/L

pH = 3.5; Ka = 3.75 Required:

2NaHCO (aq)V Solution: pH =3.5 = –log[H+(aq)] [H+(aq)] = 10−3.5 [H+(aq)] = 3.16 × 10−4 mol/L

( ) ( )( )

2a

2 2

[HCO aq ][H aq ][H CO aq ]

K− +

=

( ) 42[HCO aq ](3.16 10 )

3.750.10

− −×=

( )

( )

4

2

42

(3.16 10 )[HCO aq ](3.75)(0.10)

[HCO aq ] 8.4 10 mol/L

−−

− −

×=

= ×

For 1.00 L, 8.4 × 10−4 mol must be added.

nNaHCO2 (aq) =VNaHCO2 (aq) ! cNaHCO2 (aq)

VNaHCO2 (aq) =nNaHCO2 (aq)

cNaHCO2 (aq)

= 8.4 ! 10"4 mol0.10 mol / L

VNaHCO2 (aq) = 8.4 ! 10"3 L

VNaHCO2 (aq) = 8.4 mL

55. (a) NH3(aq) + H2O(l) ⇌ NH4+ (aq) + OH−(aq)

(b) The equilibrium shifts toward the right because the acid reduces [OH−(aq)]. 56. The conjugate base of citric acid is citrate ion. 57. Answers may vary. Sample answer: Citric acid is used for flavouring foods, stabilizing pH, and deterring the growth of bacteria. Understanding 58. Nitric acid produces hydrogen ions in aqueous solution. 59. Ammonia is a Brønsted–Lowry base because it accepts protons, but it is not an Arrhenius base because it does not ionize to form hydroxide ions. 60. The concentration of liquid water does not change; so it is part of the constant, K. 61. The acid with the smallest Ka ionizes the least and therefore adds the least amount of hydrogen ion to the solution. Therefore, the acid with a Ka of 1 × 10−6 has the higher pH. 62. When the concentration of a hydrochloric acid solution is doubled, the pH of the solution does not decrease by a factor of two because pH is a logarithmic scale, not a linear scale.

Page 9: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-10

63. pH = −log[H+(aq)] log(1.0 × 10−7) = −7 −log(1.0 × 10−7) = 7 64. A weak acid is an acid that has a low dissociation. Concentration refers to the amount of material in solution, not dissociation; therefore, an acid can be both weak and concentrated. 65. According to Le Châtelier’s principle, increased pressure shifts the equilibrium toward fewer molecules of gas. In this synthesis, there is a shift toward the product. 66. Hydrofluoric acid has the higher percentage ionization because its Ka value is larger. 67. The order is Ka1 > Ka2 > Ka3 because it is harder to remove additional protons as the negative charge on the ion increases. 68. Water does not act as a diprotic acid because the hydroxide ion is such a weak acid that it does not donate a proton in aqueous solutions. 69. (a) The conjugate base is a stronger base than water. (b) [H+(aq)] > [HA(aq)] 70. The pH of the aqueous salt depends on the ability of the ions to hydrolyze and change the pH of the solution. 71. Answers may vary. Flow charts or concept maps should include the following information:

Type of salt Examples Comment pH of solution Cation of a Group 1 or Group 2 element, other than Be; anion is from strong acid

KCl(aq), NaCl(aq), NaNO3(aq)

Neither of the ions acts as an acid or a base

neutral

Cation of a Group 1 or Group 2 element, other than Be; anion is from weak acid

NaC2H3O2(aq), KCN(aq), NaF(aq)

Anion acts as a base; cation has no effect on pH

basic

Cation is conjugate acid of weak base; anion is from strong acid

NH4Cl(aq), NH4NO3(aq)

Cation acts as an acid; anion has no effect on pH

acidic

Cation is conjugate acid of weak base; anion is conjugate base of weak acid

NH4C2H3O2(aq), NH4CN(aq)

Cation acts as an acid; anion acts as a base

acidic if Ka > Kb basic if Kb > Ka neutral if Ka = Kb

Cation is highly charged metal ion; anion is from strong acid

Al(NO3)3(aq), FeCl3(aq)

Hydrated cation acts as an acid; anion has no effect on pH

acidic

Page 10: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-11

72. Given: VHC2H3O2(aq) = 100.00 mL; [HC2H3O2(aq)] = 0.200 mol/L (a) Required: pH before titration begins Solution:

HC2H3O2 (aq) ⇌ H+(aq) + C2H3O2–(aq)

I 0.200 0 0 C – x + x + x E 0.200 – x + x + x

( )

( )

–2 3 2

a2 3 2

–2 3 25

2 3 2

25

25

2 5

3

[H (aq)][C H O aq ][HC H O (aq)]

[H (aq)][C H O aq ]1.8 10

[HC H O (aq)]

1.8 100.200

1.8 100.2003.6 106.0 10

+

+−

=

× =

× =−

× ≈

≈ ×≈ ×

K

xx

x

xx

pH = –log[H+(aq)] = –log [6.0 × 10–3] pH = 2.22 Statement: The pH before titration begins is 2.22.

(b) Required: pH after 50.0 mL of 0.100 mol/L KOH is added Solution: Before titration begins: n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq) = (0.200 mmol/mL)(100.00 mL) n HC2H3O2(aq) = 20.00 mmol Amount of KOH(aq) added: nKOH(aq) =[KOH(aq)] × VKOH(aq) = (0.100 mmol/mL)(50.00 mL) nKOH(aq) = 5.00 mmol Since KOH(aq) reacts completely with ethanoic acid: Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq) = 20.00 mmol – 5.00 mmol Unreacted ethanoic acid = 15.0 mmol Since 50.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the total volume is now 150.00 mL.

Page 11: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-12

The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined: ( )

( )

2 3 2

2 3 2

HC H O aq2 3 2

HC H O aq

2 3 2

[HC H O (aq)]

(15.00 mmol)=(150.00 mL)

[HC H O (aq)] 0.100 mol/L

=

=

nV

Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2

–(aq)], is determined:

( ) ( )

( )

( )

–2 3 2

–2 3 2

C H O aq–2 3 2

C H O aq

–2 3 2

[C H O aq ]

(5.00 mmol)=(150.00 mL)

[C H O aq ] 0.033 mol/L

n

V=

=

For the ethanoic acid/ethanoate solution:

HC2H3O2 (aq)  ⇌ H+(aq) + C2H3O2–(aq)

I 0.100 0 0.033 C – x + x + x E 0.100 – x + x 0.033 + x

The values can now be substituted into the equilibrium equation for the ionization of a weak acid, where Ka = 1.8 × 10–5.

( )+ –2 3 2

2 3 2

5

[H (aq)][C H O aq ][HC H O (aq)]

( )(0.033 )1.8 10(0.200 )

=

+× =−

aK

x xx

Using simplifying assumptions, 0.033 + x ≈ 0.033 and 0.100 – x ≈ 0.100 (valid by the hundred rule),

5

5

5

(0.033)1.8 10 ( )(0.100)

(1.8 10 )(0.100)(0.033)

5.45 10

× ≈

×≈

≈ ×

x

x

x

Therefore [H+(aq)] = x ≈ 5.45 × 10–5 and the pH can be determined: pH = –log[H+(aq)] ≈ –log(5.45 × 10–5) pH ≈ 4.26 Statement: The pH of the solution is 4.26.

Page 12: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-13

(c) Required: pH after 100.0 mL of 0.100 mol/L KOH is added Solution: Before titration begins: n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq) = (0.200 mmol/mL)(100.00 mL) n HC2H3O2(aq) = 20.00 mmol Amount of KOH(aq) added: nKOH(aq) =[KOH(aq)] × VKOH(aq) = (0.100 mmol/mL)(100.00 mL) nKOH(aq) = 10.00 mmol Since KOH(aq) reacts completely with ethanoic acid: Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq) = 20.00 mmol – 10.00 mmol Unreacted ethanoic acid = 10.0 mmol Since 100.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the total volume is now 200.00 mL. The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined:

( )

( )

2 3 2

2 3 2

HC H O aq2 3 2

HC H O aq

2 3 2

[HC H O (aq)]

(10.00 mmol)=(200.00 mL)

[HC H O (aq)] 0.050 mol/L

=

=

nV

Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2

–(aq)], is determined:

( ) ( )

( )

( )

–2 3 2

–2 3 2

C H O aq–2 3 2

C H O aq

–2 3 2

[C H O aq ]

(10.00 mmol)=(200.00 mL)

[C H O aq ] 0.050 mol/L

n

V=

=

For the ethanoic acid/ethanoate solution:

HC2H3O2 (aq) ⇌ H+(aq) + C2H3O2–(aq)

I 0.050 0 0.050 C – x + x + x E 0.050 – x + x 0.050 + x

Page 13: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-14

The values can now be substituted into the equilibrium equation for the ionization of a weak acid, where Ka = 1.8 × 10–5.

( )+ –2 3 2

2 3 2

5

[H (aq)][C H O aq ][HC H O (aq)]

( )(0.050 )1.8 10(0.050 )

=

+× =−

aK

x xx

Using simplifying assumptions, 0.050 + x ≈ 0.050 and 0.050 – x ≈ 0.050 (valid by the hundred rule),

5

5

(0.050)1.8 10 ( )(0.050)

1.8 10

x

x

× ≈

≈ ×

Therefore [H+(aq)] = x ≈ 1.8 × 10–5 and the pH can be determined: pH = –log[H+(aq)] ≈ –log(1.8 × 10–5) pH ≈ 4.74 Statement: The pH of the solution is 4.74

(d) Required: pH after 150.0 mL of 0.100 mol/L KOH(aq) is added Solution: Before titration begins: n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq) = (0.200 mmol/mL)(100.00 mL) n HC2H3O2(aq) = 20.00 mmol Amount of KOH(aq) added: nKOH(aq) =[KOH(aq)] × VKOH(aq) = (0.100 mmol/mL)(150.00 mL) nKOH(aq) = 15.00 mmol Since KOH(aq) reacts completely with ethanoic acid: Unreacted ethanoic acid = n HC2H3O2(aq) – nKOH(aq) = 20.00 mmol – 15.00 mmol Unreacted ethanoic acid = 5.0 mmol Since 150.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the total volume is now 250.00 mL. The concentration of unreacted ethanoic acid, [HC2H3O2(aq)], can be determined:

2 3 2

2 3 2

HC H O (aq)2 3 2

HC H O (aq)

2 3 2

[HC H O (aq)]

(5.00 mmol)=(250.00 mL)

[HC H O (aq)] 0.020 mol/L

=

=

nV

Page 14: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-15

Similarly, the concentration of the conjugate base of ethanoic acid, [C2H3O2–(aq)], is

determined:

( ) ( )

( )

( )

–2 3 2

–2 3 2

C H O aq–2 3 2

C H O aq

–2 3 2

[C H O aq ]

(15.00 mmol)=(250.00 mL)

[C H O aq ] 0.060 mol/L

n

V=

=

For the ethanoic acid/ethanoate solution:

HC2H3O2 (aq) ⇌ H+(aq) + C2H3O2–(aq)

I 0.020 0 0.060 C – x + x + x E 0.020 – x + x 0.060 + x

The values can now be substituted into the equilibrium equation for the ionization of a weak acid, where Ka = 1.8 × 10–5.

( )+ –2 3 2

2 3 2

5

[H (aq)][C H O aq ][HC H O (aq)]

( )(0.060 )1.8 10(0.020 )

=

+× =−

aK

x xx

Using simplifying assumptions, 0.020 + x ≈ 0.020 and 0.060 – x ≈ 0.060 (valid by the hundred rule),

5

5

6

( )(0.060)1.8 10(0.020)(1.8 10 )(0.020)

(0.060)6.000 10

x

x

x

× ≈

×≈

≈ ×

Therefore [H+(aq)] = x ≈ 6.000 × 10–6 and the pH can be determined: pH = –log[H+(aq)] ≈ –log(6.000 x 10–6) pH ≈ 5.22 Statement: The pH of the solution is 5.22

(e) Required: pH after 200.0 mL of 0.100 mol/L KOH is added Solution: Before titration begins: n HC2H3O2(aq) = [HC2H3O2(aq)] × VHC2H3O2(aq) = (0.200 mmol/mL)(100.00 mL) n HC2H3O2(aq) = 20.00 mmol Amount of KOH(aq) added: nKOH =[KOH(aq)] × VKOH(aq) = (0.100 mmol/mL)(200.00 mL) nKOH = 20.00 mmol

Page 15: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-16

Since KOH(aq) reacts completely with ethanoic acid: Unreacted ethanoic acid = n HC2H3O2 – nKOH = 20.00 mmol – 20.00 mmol Unreacted ethanoic acid = 0.00 mmol Since 200.00 mL of KOH(aq) was added to 100.00 mL of ethanoic acid solution, the total volume is now 300.00 mL. The pH of the resulting solution is determined by the hydrolysis of the ethanoate ion.

( ) ( )

( )

2 3 2 initial2 3 2

2 3 2

HC H O aqC H O aq

C H O aq

20.00 mmol

=

=

n n

n

( )( )

( )

( )

2 3 2

2 3 2

2 3 2

2 3 2

C H O aq

C H O aqC H O aq

2C H O aq

20.00 mmol300.0 mL

6.667 10 mol/L

−−

=

=

= ×

nc

V

c

C2H3O2

– (aq) + H2O(l) ⇌ HC2H3O2 (aq) + OH−(aq)

I 6.667 × 10–2 − 0 0 C − x − + x + x E 6.667 × 10–2– x − x x

( )

( )2 3 2

b2 3 2

14

5

10

[HC H O aq ][OH (aq)][C H O aq ]

1.0 101.8 105.556 10 (2 extra digits carried)

=

×=×

= ×

K

5.556!10"10 = (x)(x)(6.667 !10"2 – x)

5.556!10"10 # (x)(x)6.667 !10"2

x2 ≈ 3.706 × 10−11 x = [OH−(aq)] ≈ 6.087 × 10−6 mol/L pOH = −log(6.087 × 10−6)

pOH = 5.22 pH = 14.00 – 5.22 pH = 8.78 The pH of the resulting solution is 8.78.

Page 16: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-17

(f) Required: pH after 250.0 mL of 0.100 mol/L KOH is added. Solution:

Since 200.0 mL of 0.100 mol/L KOH were required for neutralization, 50.0 mL remain unreacted. nKOH =[KOH(aq)] × VKOH(aq) = (0.100 mmol/mL)(50.00 mL) nKOH = 5.00 mmol Total volume of solution = 350.0 mL

( )( )

( )

( )

KOH aqKOH aq

KOH aq

2KOH aq

5.00 mmol350.0 mL

1.4286 10 mol/L (2 extra digits carried)−

=

=

= ×

nc

V

c

pOH = –log[OH–(aq)] = –log( 21.4286 10−× ) = 1.85 pH = 14 – pOH = 14 – 1.85 pH = 12.15 Statement: The pH after 250 mL of 0.100 mol/L KOH(aq) is added is 12.15.

73. Most indicators can be used for this type of titration because the pH changes from very low to very high close to the equivalence point. 74. Methyl yellow is a better indicator because the equivalence point has a pH greater than 7 and thymol blue changes colour at a very low pH. 75. Buffer systems can be either acidic, basic or neutral. An acidic buffer consists of a weak acid in equilibrium with its conjugate base: HA(aq) ⇌ H+(aq) + A−(aq) When an acid is added to this system, the additional hydrogen ions react with the conjugate base, A–1(aq) to form HA(aq). As a result, the system shifts to the left. Conversely, when a base is added to this system, the hydroxide ions from the base react with the hydrogen ions from the equilibrium, causing the system to shift to the right. A basic buffer consists of a weak base, B(aq), in equilibrium with its conjugate acid, BH+(aq) : B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq) When an acid is added to this system, the additional hydrogen ions react with the hydroxide ions from the equilibrium. As a result, the system shifts to the left. Conversely, when a base is added to this system, the hydroxide ions from the base react with the conjugate acid, causing the system to shift to the right.

Page 17: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-18

76. This solution would not make a good buffer because hydrochloric acid and sodium chloride are essentially completely ionized; no equilibrium is established to form a buffer. 77. (a) Given: [H2NNH2(aq)] = 0.40 mol/L; [H2NNH3NO3(aq)] = 0.80 mol/L; Kb = 1.7 × 10−6

Required: pH Analysis:

H2NNH2(aq) + H2O(l) ⇌ H2NNH3

+(aq) + OH−(aq) I 0.40 − 0.80 0 C − x − + x + x E 0.40 – x − x + 0.80 x

( ) ( )( )

2 3b

2 2

[H NNH aq ][OH aq ][H NNH aq ]

K+ −

=

Solution: 6 ( 0.80)( )1.7 10(0.40 – )x x

x− +× =

6 0.801.7 100.40x−× ≈

x = [OH− (aq)] ≈ 8.5 × 10−7 mol/L pOH = −log(8.5 × 10−7) pOH = 6.07 pH = 14.00 – 6.07 pH = 7.93

Analysis and Application 78. (a) In the lungs, Hb(O2)4 is favoured; in cells, HbH4

4+(aq) is favoured. (b) As the carbon dioxide level falls, the equilibrium of the carbonic acid-hydrogen carbonate buffer shifts toward carbonic acid, which decreases pH. Breathing into a bag increases carbon dioxide, which raises the pH and releases oxygen. (c) Sodium hydrogen carbonate increases pH and releases oxygen in the blood. 79. (a) Given: [C6H8O6(aq)] = 2.00 g/L; Ka1 = 7.9 × 10−5; Ka2 = 1.6 × 10−12

Required: [H+], pH Analysis:

6 8 62.00 g C H O6 8 6

6 8 6

1.00 mol C H OL 176 g C H O

× 0.0114 mol/L =

C6H8O6(aq) ⇌ H+(aq) + C6H7O6

−(aq) I 0.0114 0 0 C − x + x + x E 0.0114 – x x x

Page 18: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-19

( )( )

+6 7 6

a16 8 6

[H (aq)][C H O aq ][C H O aq ]

K−

=

Ka1 is much greater than Ka2 ; so Ka2 has no significant contribution to equilibrium.

Solution: 5 ( )( )7.9 10(0.0114 – )

x xx

−× =

5 ( )( )7.9 100.0114x x−× ≈

x2 ≈ 9.006 × 10−7 (2 extra digits carried) x = [H+(aq)] ≈ 9.4899 × 10−4 mol/L pH = −log (9.4899 × 10–4) pH = 3.02 Statement: [H+(aq)] ≈ 9.5 × 10−4 mol/L

(b) Given: [C6H8O6(aq)] = 0.0114 mol/L; Ka1 = 7.9 × 10−5; pH = 1.0 Required: percentage ionization Analysis: pH=1.0 so [H+] = 1.00 × 10−1 mol/L

( )( )

+6 7 6

a16 8 6

[H (aq)][C H O aq ][C H O aq ]

K−

=

Solution:

( )( )

+6 7 6

a16 8 6

[H (aq)][C H O aq ][C H O aq ]

K−

=

( )( )

5 6 7 6

6 8 6

(0.1)[C H O aq ]7.9 10

[C H O aq ]

−−× =

percentage ionization ( )( )

6 7 6

6 8 6

[C H O aq ]100 %

[C H O aq ]

= ×

percentage ionization = 7.9 × 10−4 × 100 % percentage ionization = 0.079 %

80. (a) Both solutions are acidic: 3 NO2(g) + H2O(l) ⇌ NO(g) + 2 HNO3(aq) SO2(g) + H2O(l) ⇌ H2SO3(aq) (b) Lakes on limestone bedrock are less affected by these acidic solutions because calcium carbonate in the limestone reacts with H2SO3 to neutralize the acid. 81. (a) Approximately 200 mL of sodium hydroxide has been added at the second equivalence point. (b)(i) When 0 mL NaOH(aq) is added, the major entities present are H2A(aq) and H2O(l). (ii) When between 0 and100 mL NaOH(aq) is added, the major entities present are H2A(aq), HA–(aq), H2O(l), and Na+(aq). (iii) When 100 mL NaOH(aq) is added, the major entities present are HA–(aq), H2O(l), and Na+(aq).

Page 19: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-20

(iv) When between 100 and 200 mL NaOH(aq) is added, the major entities present are HA–(aq), A2–(aq), H2O(l), and Na+(aq). (v) When 200 mL NaOH(aq) is added, the major entities present are A2–(aq), H2O(l), and Na+(aq). (vi) When more than 200 mL NaOH is added, the major entities present are A2–(aq), H2O(l), Na+(aq), and OH–(l). (c) Given: pH = 4; titration is half completed

Required: Ka1 Solution: At 50 mL, 50 % of H2A(aq) has reacted so [H2A(aq)] = [HA−(aq)] [H+(aq)] = 10–pH [H+(aq)] = 1.0 × 10−4

a12

4

a1

[H (aq)][HA (aq)][H A(aq)]

[1.0 10 ][HA (aq)]

+ −

− −

=

×=

K

K[HA (aq)]−

4a1 1.0 10−= ×K

Given: pH = 8; second titration is half completed Required: Ka2 Solution: At 50 mL, 50 % of HA− has reacted so [HA−] = [A2−] [H+] = 10–pH [H+] = 1.0 × 10−8

2

a2

8 2

a2

[H ][A ][HA ]

[1.0 10 ][A ]

+ −

− −

=

×=

K

K2[A ]−

8a2 1.0 10−= ×K

82. Answers may vary. Sample answer: Prepare buffer solutions with known pH across the entire pH range. Add the indicator to each solution and record the colour. 83. Answers may vary. Sample answer: Prepare buffer solutions with known pH across the entire pH range. Prepare a solution of a weak acid with pH = 4. Add methyl orange indicator to make an orange-coloured solution. Prepare a solution of a weak base at the same concentration with pH = 10. Add thymolphthalein to make a blue solution. Combine equal amounts of the two coloured solutions to make a colourless solution. 84. Answers may vary. Sample answer: Choose an acid or base whose Ka is close to the desired [H+]. Calculate the concentrations of the acid or base and its salt needed to produce a buffer at the pH. Prepare the solution. Determine the pH and adjust by adding the acid or base or its salt to get to the exact pH required. 85. Answers may vary. Sample answer: Titrate samples of the buffer with a strong acid and with a strong base to determine the amount of hydrogen ions and the amount of hydroxide ions that can be added without substantial change to the pH. 86. (a) To prepare a buffer with pH 3.0, I would use phosphoric acid and potassium dihydrogen phosphate.

Page 20: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-21

(b) To prepare a buffer with pH 4.0, I would use hydrofluoric acid and potassium fluoride. (c) To prepare a buffer with pH 5.0, I would use ethanoic acid and sodium ethanoate. (d) To prepare a buffer with pH 7.0, I would use benzoic acid and sodium benzoate. 87. Answers will vary. Concept maps should include Arrhenius acids and bases, Brønsted–Lowry acids and bases, proton donors, proton acceptors, the formation of salts and H2O, and conjugate acids and bases. Evaluation 88. Answers may vary. Spreadsheets should include a table with the following comparative information:

Antacid brand Active ingredient mol of active ingredient per package

Cost per package Cost per mole of antacid

89. Phenolphthalein is not an appropriate indicator for this titration because it does not change colour at a pH near the equivalence point. Methyl red is a better indicator for this titration. 90. A titration of a weak acid with a weak base is not feasible because there would be a buffering effect around the endpoint, and so the curve would not have a sharp break. Reflect on Your Learning 91. Answers will vary. Sample answer: There may be many examples of acids and bases in foods, cleaning solutions, medicines, and other everyday substances. A pH meter or pH indicator strips could be used to search for examples. 92. Answers will vary. Sample answer: Many materials may act as indicators if they are acidic or basic and have a different colour than their salt. A stain remover could work by changing the pH of the coloured substance to a form that does not have colour. 93. Answers will vary based on student experience. Research 94. Answers will vary. Sample answer: Because acid forms as substances in the paper break down, books that are printed on acidic paper are stored at cool temperatures and low humidity to slow the reactions. Valuable books can often be treated with a slightly basic buffer solution to neutralize the acid and react with the additional acid that forms. 95. Answers will vary. Sample answer: Arab scientists learned to make alkali hydroxides at least 1200 years ago. The first alkali hydroxides to be isolated, sodium hydroxides and potassium hydroxide, were produced from the ashes of burnt wood. The word alkali comes from the Arab word for ashes.

Page 21: Chapter 8 Review, pages 576–581 Knowledge · 2018. 8. 30. · A 6.88 7.12 −1.3 × 10−7 mol/L 7.7 × 108 mol/L acidic B 0.92 13.08 −0.12 mol/L 8.4 × 10 14 mol/L acidic C −10.89

Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-22

96.

97. Answers will vary. Students should trace acid rain through the pH change in lakes and streams, and connect this information to the presence of nitrogen and sulfur compounds in the atmosphere. By relating higher concentrations of these compounds to power plant locations and population centres, students can point to the combustion of fuels as the original source. 98. Answers will vary. Sample answer: The main goal of food preservation is to kill and prevent growth of microorganisms such as bacteria, yeasts, and fungi. Most organisms exist in a particular pH range. Acidic foods, such as tomatoes, naturally resist spoiling when they are canned. Other foods can be preserved by pickling in vinegar, which decreases its pH, or sometimes by treating with lye to increase the pH. 99. Answers may vary. Sample answer: Limestone is made of calcium carbonate, a weak base, which can neutralize the acid in acid rain and act as a buffer to maintain lake water at an optimum pH. Limestone is used for treating lakes and streams because it is a natural material, is mined in many locations around the world, and dissolves slowly, producing a long-lasting treatment. 100. Answers may vary. Sample answer: Svante Arrhenius was a research chemist and professor in Sweden during the late 1800s and early 1900s. His contributions to chemistry include the observation that pure water does not conduct electric current but a salt solution does conduct because salts form ions in solution. He expanded on the theory of ions to include a definition of acids and bases as substances that form hydrogen ions and hydroxide ions in solution. In addition to his studies of ions, he developed the concept of activation energy in chemical reactions, one of the basic concepts of collision theory. Arrhenius was the first scientist to relate temperature on Earth to the concentration of carbon dioxide in the atmosphere and also to predict global warming as a result of burning fossil fuels. 101. Answers may vary. Sample answer: Acid indigestion and acid reflux occur when stomach acids leak into the esophagus and damage tissues. The simplest medical treatment is an antacid, which is a basic compound that reduces the acidity of the stomach contents. Other treatments reduce the amount of acid by causing cells in the digestive system to reduce the production of acidic compounds.

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Copyright © 2012 Nelson Education Ltd. Chapter 8: Acid-Base Equilibrium 8-23

102. Answers may vary. Sample answer: Buffers are used in medicines to control the effect of pH changes on the pharmaceutical compound and to match the pH of the medicine to that of the body system where it is used. Most body fluids contain buffering compounds that control pH. Matching the medicine to the pH of the body increases the effectiveness of delivering the medicine and protects tissues from possibly harmful pH changes. 103. Answers may vary. Sample answer: Most toothpastes for controlling tooth decay use fluoride salts such as stannous fluoride. The fluoride ion replaces the hydroxide ion in hydroxyapatite, forming a surface coating that has a lower reactivity with acids. 104. Answers may vary. Sample answer: Occupations in which people would monitor pH regularly include manufacturing chemists and technicians who use pH probes and titrations to measure acids and bases during chemical reactions, environmental technicians who use probes to monitor and control acidity in streams and lakes, medical lab technicians who measure the pH of body fluids, and food chemists who use information about pH to process foods safely. 105. Answers may vary. Sample answer: Household products use a wide variety of acids and bases. Acids used at home include the acetic acid in vinegar, which is a component of many cleaning compounds, and citric acid, used to preserve foods. Household bases include ammonia (base), used in cleaning and disinfecting solutions and in treatments for insect bites and stings, and sodium carbonate, a component of laundry detergents. 106. Answers may vary. Sample answer: Most plants grow best within a particular pH range. Depending on the type of plant and its specific needs for elements and compounds in soil, the optimum pH range can vary from about 6 to 7.5. Farmers adjust soil pH by adding mineral elements and compounds to the soil. The most common mineral soil conditioners are lime, composed of calcium carbonate and/or magnesium carbonate, which is used to increase soil pH, and sulfur, used to decrease soil pH.