Acid‐Base Calculations – Answers 1. In each of the following solutions, determine: 3
a) 0.030 mol/dm HCl HCl is a strong acid HCl + H2O  H3O+ + Cl‐ i)
ii)
iii)
iv)
iii) pH iv) pOH HBr is a strong acid HCl + H2O  H3O+ + Br‐ HBr completely breaks down, therefore [H3O+] = 0.125 mol/L [OH‐] = 10‐pOH = 8.00 x 10‐14 mol/L pH = ‐log [H3O+] = 0.903 pOH = 14 – pH = 13.097 c) 0.025 mol/dm3 NaOH NaOH is a strong base NaOH  Na+ + OH‐ i) [H3O+] = 10‐pH = 4.00 x 10‐13 mol/L ii) [OH‐] = 0.025 mol/L iii) pH = 14 – pOH = 12.40 iv) pOH = ‐log [OH‐] = 1.60 d) 0.050 mol/dm3 Ca(OH)2 hydronium ion concentration hydroxide ion concentration HCl completely breaks down, therefore [H3O+] = 0.030 mol/L [OH‐] = 10‐pOH = 3.3 x 10‐13 mol/L pH = ‐log [H3O+] = 1.52 pOH = 14 – pH = 12.48 3
b) 0.125 mol/dm HBr i)
ii)
iii)
iv)
i) ii) Ca(OH)2 is a strong base Ca(OH)2  Ca2+ + 2OH‐ i) [H3O+] = 10‐pH = 1.00 x 10‐13 mol/L ii) [OH‐] = 2(0.050 mol/L) = 0.10 mol/L iii) pH = 14 – pOH = 13.00 iv) pOH = ‐log [OH‐] = 1.00 L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 1 of 13 e) 0.100 mol/dm3 H2CO3 H2CO3 is a weak acid: H2CO3 + H2O  H3O+ + HCO3‐ H2CO3 [Initial] 0.100 [Change] ‐x [Equilibrium] 0.100 ‐ x + H2O  H3O+ + HCO3‐
0 0 +x +x x x ‐
Ka =
H3 O+ HCO3
H2 CO3
4.4 x 10‐7 =
x x
0.100‐x
Shortcut check, [HA] ÷ Ka > 500 4.4 x 10‐7 =
x x
0.100
(4.4 x 10‐7)(0.100) = x2 4.4 x 10‐8 = x2 x = 2.1 x 10‐4 i) [H3O+] = 2.1 x 10‐4 mol/L ii) [OH‐] = 10‐pOH = 4.8 x 10‐11 mol/L iii) pH = ‐log [H3O+] = 3.68 iv) pOH = 14 – pH = 10.32 L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 2 of 13 f) 0.250 mol/dm3 HCN HCN is a weak acid: HCN + H2O  H3O+ + CN‐ HCN [Initial] 0.250 [Change] ‐x [Equilibrium] 0.250 ‐ x + H2O  H3O+ + CN‐
0 0 +x +x x x Ka =
H3 O+ CN‐
HCN
6.2 x 10‐10 =
x x
0.250‐x
Shortcut check, [HA] ÷ Ka > 500 6.2 x 10‐10 =
x x
0.250
‐10
2
(6.2 x 10 )(0.250) = x 1.55 x 10‐10 = x2 x = 1.2 x 10‐5 i) [H3O+] = 1.2 x 10‐5 mol/L ii) [OH‐] = 10‐pOH = 8.0 x 10‐10 mol/L iii) pH = ‐log [H3O+] = 4.90 iv) pOH = 14 – pH = 9.10 2. +
–
–
Calculate [H3O ], [ClO4 ] and [OH ] in an aqueous solution that is 0.150 M in HClO4(aq). Is the solution acidic or basic? HClO4 is a strong acid: HClO4 + H2O  H3O+ + ClO4‐ [H3O+] = 0.150 mol/L [ClO4‐] = 0.150 mol/L Kw = [H3O+][OH‐] [OH‐] = Kw ÷ [H3O+] [OH‐] = 1.00 X 10‐14 ÷ (0.150 mol/L) = 6.67 x 10‐14 mol/L The solution is acidic, [H3O+] > [OH‐] L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 3 of 13 3. Calculate , [OH– ], [K+] and [H3O+] in an aqueous solution that is 0.250 M in KOH(aq). Is the solution acidic or basic? KOH is a strong base: 4. KOH  K+ + OH‐ [OH‐] = 0.250 mol/L [K+] = 0.250 mol/L Kw = [H3O+][OH‐] [H3O+] = Kw ÷ [OH‐] [H3O+] = 1.00 X 10‐14 ÷ (0.250 mol/L) = 4.00 x 10‐14 mol/L The solution is basic, [H3O+] < [OH‐] Compute [Ca2+], [OH– ] and [H3O+] for a solution that is prepared by dissolving 0.600 g of Ca(OH)2 in enough water to make 3
0.500 dm of solution. Ca(OH)2 is a strong base: Ca(OH)2  Ca2+ + 2OH‐ [Ca(OH)2] = n ÷ V n = mass ÷ molar mass n = 0.600 g ÷ 74.09268 g/mol = 0.00810 mol [Ca(OH)2] = n ÷ V [Ca(OH)2] = 0.00810 mol ÷ 0.500 L = 0.0162 mol/L [OH‐] = 2(0.0162 mol/L) = 0.0324 mol/L [Ca2+] = 0.0162 mol/L Kw = [H3O+][OH‐] [H3O+] = Kw ÷ [OH‐] [H3O+] = 1.00 X 10‐14 ÷ (0.0324 mol/L) = 3.09 x 10‐13 mol/L The solution is basic, [H3O+] < [OH‐] L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 4 of 13 The value of Ka in water at 25˚C for benzoic acid, HC7H5O2 is 6.46 x 10–5. Calculate the pH of an aqueous solution of 0.0200 M HC7H5O2. 5. HC7H5O2 is a weak acid: HC7H5O2 + H2O  H3O+ + C7H5O2‐ HC7H5O2 [Initial] 0.0200 [Change] ‐x [Equilibrium] 0.0200 ‐ x + H2O  H3O+ + C7H5O2‐
0 0 +x +x x x ‐
Ka =
H3 O+ C7 H5 O2
HC7 H5 O2
6.46 x 10‐5 =
x x
0.0200‐x
Shortcut check, [HA] ÷ Ka < 500, therefore must keep x (6.46 x 10‐5)(0.0200 ‐ x) = x2 1.292 x 10‐6 ‐ 6.46 x 10‐5x = x2 0 = x2 + 6.46 x 10‐5x ‐ 1.292 x 10‐6 6.46 x 10
6.46 x 10
2 1
6.46 x 10
5.172 x 10
2
0.00227
6.46 x 10
2
x = ‐0.001169 or +0.001105 x = [H3O+] = 0.00110 mol/L pH = ‐log(0.00110) = 2.957 4 1
1.292 x 10
L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 5 of 13 6. Sulfamic acid, HO3SNH2, is used as a stabilizer for chlorine in swimming pools. Calculate the pH of a 0.0400 M sulfamic acid solution given that Ka = 0.100. HO3SNH2 is a weak acid: HA + H2O  H3O+ + A‐ HA + H2O  H3O+ + A‐ [Initial] 0.0400 0 0 [Change] ‐x +x +x
[Equilibrium] 0.0400 ‐ x x x Ka =
H3 O+ A‐
HA
0.100 =
x x
0.0400‐x
Shortcut check, [HA] ÷ Ka < 500, therefore must keep x (0.100)(0.0400 ‐ x) = x2 0.00400 – 0.100x = x2 0 = x2 + 0.100x – 0.00400 0.100
0.100
4 1
2 1
0.100
0.0260
2
x = ‐0.131 or +0.0306 x = [H3O+] = 0.00306 mol/L pH = ‐log(0.0306) = 1.514 0.00400
L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 6 of 13 7. Calculate the pH of an aqueous solution of 0.150 M HClO. HClO is a weak acid: HClO + H2O  H3O+ + ClO‐ HClO [Initial] 0.150 [Change] ‐x [Equilibrium] 0.150 ‐ x + H2O  H3O+ + ClO‐
0 0 +x +x x x Ka =
H3 O+ ClO‐
HClO
2.9 x 10‐8 =
x x
0.150‐x
Shortcut check, [HA] ÷ Ka > 500 2.9 x 10‐8 =
x x
0.150
(2.9 x 10‐8)(0.150) = x2 4.35 x 10‐9 = x2 x = 6.6 x 10‐5 [H3O+] = 6.6 x 10‐5 mol/L pH = ‐log [H3O+] = 4.18 L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 7 of 13 8. Suppose two aspirin tablets (1 tablet = 324 mg) are dissolved in enough water to make 500.0 cm3 of solution. Compute the pH of the resulting solution if Ka = 2.75 x 10–5 and the molar mass for aspirin (acetylsalicylic acid) is 180.15 g/mol. 2 tablets = (2)(324 mg) = 648 mg = 0.648 g mol = mass ÷ molar mass mol = (0.648 g) ÷ (180.15 g/mol) = 0.00360 mol aspirin C = n ÷ V C = (0.00360 mol) ÷ (0.5000 L) = 0.00719 mol/L = concentration of aspirin solution aspirin is a weak acid: HA + H2O  H3O+ + A‐ HA + H2O  H3O+ + A‐ [Initial] 0.00719 0 0 [Change] ‐x +x +x
[Equilibrium] 0.00719 ‐ x x x Ka =
H3 O+ A‐
HA
2.75 x 10 =
x x
0.00719‐x
Shortcut check, [HA] ÷ Ka < 500, therefore must keep x (2.75 x 10‐5)(0.00719 ‐ x) = x2 1.98 x 10‐7 – 2.75 x 10‐5x = x2 0 = x2 + 2.75 x 10‐5x – 1.98 x 10‐7 2.75 x 10
2.75 x 10
2 1
2.75 x 10
7.92 x 10
2
2.75 x 10
8.90 x 10
2
x = ‐4.59 x 10‐4 or +4.31 x 10‐4 x = [H3O+] = 4.31 x 10‐4 mol/L pH = ‐log(4.31 x 10‐4) = 3.365 4 1
1.98 x 10
L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 8 of 13 9. The pH of a 0.72 mol/L solution of benzoic acid, HC7H5O2, is 2.68. Calculate the numerical value of the Ka for this acid. HC7H5O2 is a weak acid: HC7H5O2 + H2O  H3O+ + C7H5O2‐ HC7H5O2 [Initial] 0.72 [Change] ‐x [Equilibrium] 0.72 ‐ x + H2O  H3O+ + C7H5O2‐
0 0 +x +x x x H3 O
‐
C7 H5 O2
Ka =
pH = 2.68 Ka =
HC7 H5 O2
K =
x x
0.0200‐x
[H3O+] = 10‐pH = 2.1 x 10‐3 mol/L = x 0.0021 0.0021
0.72‐0.0021
= 6.1 x 10‐6 10. The value of Ka for acetic acid is 1.8 x 10‐5 at 25°C. Calculate the [H3O+] in a 0.10 mol/L solution of acetic acid. What is the percent ionization of the acetic acid? HC2H3O2 is a weak acid: HC2H3O2 + H2O  H3O+ + C2H3O2‐ HC2H3O2 [Initial] 0.10 [Change] ‐x [Equilibrium] 0.10 ‐ x Ka =
+ H2O  H3O+ + C2H3O2‐
0 0 +x +x x x H3 O+ C2 H3 O‐2
HC2 H3 O2
1.8 x 10‐5 =
x x
0.10 ‐ x
Shortcut check, [HA] ÷ Ka > 500 1.8 x 10‐5 =
x x
0.10
(1.8 x 10‐5)(0.10) = x2 1.8 x 10‐6 = x2 x = 1.3 x 10‐3 [H3O+] = 1.3 x 10‐3 mol/L amount ionized
x 100 percent ionization= original amount
percent ionized= H3 O+
HA
x 100 percent ionized = 0.0013
0.10
x 100 = 1.3% ionized L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 9 of 13 11. The ionization constant, Kb, for the weak base ammonia, NH3, is 1.7 x 10‐5. Calculate the [OH‐], pH, and percent ionization of ammonia in a 0.082 mol/L solution of ammonia. NH3 is a weak base: NH3 + H2O  NH4+ + OH‐ NH3 [Initial] 0.082 [Change] ‐x [Equilibrium] 0.082 ‐ x Kb =
+ H2O  NH4+ + OH‐
0 0 +x +x x x NH+4 OH‐
NH3
1.7 x 10‐5 =
x x
0.082 ‐ x
Shortcut check, [B] ÷ Kb > 500 1.7 x 10‐5 =
x x
0.082
(1.7 x 10‐5)(0.082) = x2 1.394 x 10‐6 = x2 x = 1.18 x 10‐3 [OH‐] = 1.2 x 10‐3 mol/L amount ionized
percent ionization= x 100 original amount
percent ionized= OH‐
B
x 100 percent ionized = 0.00118
0.082
x 100 = 14% ionized L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 10 of 13 12. The hydroxide ion concentration in a 0.157 mol/L solution of sodium propanoate, NaC2H5COO(aq), is found to be 1.1 x 10‐5 mol/L. Calculate the base ionization constant for the propanoate ion. The propanoate ion is a weak base: B + H2O  BH+ + OH‐ B [Initial] 0.157 [Change] ‐x [Equilibrium] 0.157 ‐ x + H2O  BH+ + OH‐
0 0 +x +x x x BH+ OH‐
Kb =
B
K =
[OH‐] = 1.1 x 10‐5 mol/L = x Kb =
x x
0.157‐x
1.1 x 1.1 x = 7.7 x 10‐10 0.157 ‐ 1.1 x 13. Aniline, C6H5NH2, is closely related to ammonia and is also a weak base. If the pH of a 0.10 mol/L aniline solution was found to be 8.81, what is its Kb? Alinine is a weak base: C6H5NH2 + H2O  C6H5NH3+ + OH‐ B [Initial] 0.10 [Change] ‐x [Equilibrium] 0.10 ‐ x + H2O  BH+ + OH‐
0 0 +x +x x x BH+ OH‐
Kb =
B
K =
pH = 8.81 Kb =
x x
0.10 ‐ x
6.5 x pOH = 14 – 8.81 = 5.19 6.5 x [OH‐] = 10‐pOH = 10‐5.19 = 6.5 x 10‐6 mol/L = x = 4.2 x 10‐10 0.10 – 6.5 x L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 11 of 13 14. Codeine has a Kb of 1.73 x 10‐6. Calculate the pH of a 0.020 mol/L codeine solution. Codeine is a weak base: B + H2O  BH+ + OH‐ B [Initial] 0.020 [Change] ‐x [Equilibrium] 0.020 ‐ x + H2O  BH+ + OH‐
0 0 +x +x x x BH+ OH‐
B
Kb =
1.73 x 10‐6 =
x x
0.020 ‐ x
Shortcut check, [B] ÷ Kb > 500 1.73 x 10‐6 =
x x
0.020
(1.73 x 10‐6)(0.020) = x2 3.46 x 10‐8 = x2 x = 1.86 x 10‐4 [OH‐] = 1.9 x 10‐4 mol/L pOH = ‐log(1.9 x 10‐4) = 3.73 pH = 14 – pOH = 10.27 15. Codeine (C18H21NO3) is a weak organic base. A 5.0 x 10ˉ3 M solution of codeine has a pH of 9.95. Calculate the value of Kb for this substance under the conditions in this question. Codeine is a weak base: C18H21NO3 + H2O  C18H21NO3H+ + OH‐ B [Initial] 0.0050 [Change] ‐x [Equilibrium] 0.0050 ‐ x + H2O  BH+ + OH‐
0 0 +x +x x x Kb =
BH+ OH‐
B
K =
pH = 9.95 x x
0.0050 ‐ x
Kb = 8.9 x 10‐5
pOH = 14 – 9.95 = 4.05 8.9 x 10‐5
0.0050 – 8.9 x 10‐5
[OH‐] = 10‐pOH = 10‐4.05 = 8.9 x 10‐5 mol/L = x = 1.5 x 10‐6 L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 12 of 13 16. A 0.0135 M solution of a weak base (generic formula = B) has a pH of 8.39. Calculate the Kb for this weak base. Weak Base: B + H2O  BH+ + OH‐ B [Initial] 0.0135 [Change] ‐x [Equilibrium] 0.0135 ‐ x + H2O  BH+ + OH‐
0 0 +x +x x x Kb =
BH+ OH‐
B
K =
pH = 8.39 Kb = x x
0.0135 ‐ x
pOH = 14 – 8.39 = 5.61 2.5 x 10‐6
2.5 x 10‐6
0.0135 – 2.5 x 10
‐6
[OH‐] = 10‐pOH = 10‐5.61 = 2.5 x 10‐6 mol/L = x = 4.5 x 10‐10 17. Calculate the pH in a 3.0 M Pyridine solution with Kb= 1.5 x 10‐9. Pyridine is a weak base: B + H2O  BH+ + OH‐ B [Initial] 3.0 [Change] ‐x [Equilibrium] 3.0 ‐ x Kb =
+ H2O  BH+ + OH‐
0 0 +x +x x x BH+ OH‐
B
1.5 x 10‐9 =
x x
3.0 ‐ x
Shortcut check, [B] ÷ Kb > 500 1.5 x 10‐9 =
x x
3.0
(1.5 x 10‐9)(3.0) = x2 4.5 x 10‐9 = x2 x = 6.7 x 10‐5 [OH‐] = 6.7 x 10‐5mol/L pOH = ‐log(6.7 x 10‐5) = 4.17 pH = 14 – pOH = 9.83 L. Farrell – Chemistry 12 – Acid‐Base Calculations – Answers – Page 13 of 13