# 14.3 Relative strengths of acids and bases  (Page 13/18)

 Page 13 / 18

From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) CH 3 CO 2 H: $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 1.34 $×$ 10 −3 M ;
$\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]$ = 1.34 $×$ 10 −3 M ;

[CH 3 CO 2 H] = 9.866 $×$ 10 −2 M ;

(b) ClO : [OH ] = 4.0 $×$ 10 −4 M ;

[HClO] = 2.38 $×$ 10 −5 M ;

[ClO ] = 0.273 M ;

(c) HCO 2 H: [HCO 2 H] = 0.524 M ;
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 9.8 $×$ 10 −3 M ;
$\left[{\text{HCO}}_{2}{}^{\text{−}}\right]$ = 9.8 $×$ 10 −3 M ;

(d) ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}:$ $\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$ = 0.233 M ;

[C 6 H 5 NH 2 ] = 2.3 $×$ 10 −3 M ;
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 2.3 $×$ 10 −3 M

From the equilibrium concentrations given, calculate K a for each of the weak acids and K b for each of the weak bases.

(a) NH 3 : [OH ] = 3.1 $×$ 10 −3 M ;
$\left[{\text{NH}}_{4}{}^{\text{+}}\right]$ = 3.1 $×$ 10 −3 M ;

[NH 3 ] = 0.533 M ;

(b) HNO 2 : $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 0.011 M ;
$\left[{\text{NO}}_{2}{}^{\text{−}}\right]$ = 0.0438 M ;

[HNO 2 ] = 1.07 M ;

(c) (CH 3 ) 3 N: [(CH 3 ) 3 N] = 0.25 M ;
[(CH 3 ) 3 NH + ] = 4.3 $×$ 10 −3 M ;

[OH ] = 4.3 $×$ 10 −3 M ;

(d) ${\text{NH}}_{4}{}^{\text{+}}:$ $\left[{\text{NH}}_{4}{}^{\text{+}}\right]$ = 0.100 M ;

[NH 3 ] = 7.5 $×$ 10 −6 M ;
[H 3 O + ] = 7.5 $×$ 10 −6 M

(a) ${K}_{\text{b}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5};$
(b) ${K}_{\text{a}}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4};$
(c) ${K}_{\text{b}}=7.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5};$
(d) ${K}_{\text{a}}=5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Determine K b for the nitrite ion, ${\text{NO}}_{2}{}^{\text{−}}.$ In a 0.10- M solution this base is 0.0015% ionized.

Determine K a for hydrogen sulfate ion, ${\text{HSO}}_{4}{}^{\text{−}}.$ In a 0.10- M solution the acid is 29% ionized.

${K}_{\text{a}}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) F

(b) ${\text{NH}}_{4}{}^{\text{+}}$

(c) ${\text{AsO}}_{4}{}^{3-}$

(d) ${\left({\text{CH}}_{3}\right)}_{2}{\text{NH}}_{2}{}^{\text{+}}$

(e) ${\text{NO}}_{2}{}^{\text{−}}$

(f) ${\text{HC}}_{2}{\text{O}}_{4}{}^{\text{−}}$ (as a base)

Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) HTe (as a base)

(b) ${\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{\text{+}}$

(c) ${\text{HAsO}}_{4}{}^{3-}$ (as a base)

(d) ${\text{HO}}_{2}{}^{\text{−}}$ (as a base)

(e) ${\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}$

(f) ${\text{HSO}}_{3}{}^{\text{−}}$ (as a base)

(a) ${K}_{\text{b}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12};$
(b) ${K}_{\text{a}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8};$
(c) ${K}_{\text{b}}=5.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7};$
(d) ${K}_{\text{b}}=4.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3};$
(e) ${K}_{\text{b}}=2.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3};$
(f) ${K}_{\text{b}}=6.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

For which of the following solutions must we consider the ionization of water when calculating the pH or pOH?

(a) 3 $×$ 10 −8 M HNO 3

(b) 0.10 g HCl in 1.0 L of solution

(c) 0.00080 g NaOH in 0.50 L of solution

(d) 1 $×$ 10 −7 M Ca(OH) 2

(e) 0.0245 M KNO 3

Even though both NH 3 and C 6 H 5 NH 2 are weak bases, NH 3 is a much stronger acid than C 6 H 5 NH 2 . Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH 3 and 0.10 M in C 6 H 5 NH 2 ?

(a) $\left[{\text{OH}}^{\text{−}}\right]=\left[{\text{NH}}_{4}{}^{\text{+}}\right]$

(b) $\left[{\text{NH}}_{4}{}^{\text{+}}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$

(c) $\left[{\text{OH}}^{\text{−}}\right]=\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$

(d) [NH 3 ] = [C 6 H 5 NH 2 ]

(e) both a and b are correct

(a) is the correct statement.

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.25 M in HCO 2 H and 0.10 M in HClO.

Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is 0.134 M in HNO 2 and 0.120 M in HBrO.

[H 3 O + ] = 7.5 $×$ 10 −3 M
[HNO 2 ] = 0.127
[OH ] = 1.3 $×$ 10 −12 M
[BrO ] = 4.5 $×$ 10 −8 M
[HBrO] = 0.120 M

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.25 M in CH 3 NH 2 and 0.10 M in C 5 H 5 N ( K b = 1.7 $×$ 10 −9 ).

Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH 3 and 0.100 M in C 6 H 5 NH 2 .

[OH ] = $\left[{\text{NO}}_{4}{}^{\text{+}}\right]$ = 0.0014 M
[NH 3 ] = 0.144 M
[H 3 O + ] = 6.9 $×$ 10 −12 M
$\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]$ = 3.9 $×$ 10 −8 M
[C 6 H 5 NH 2 ] = 0.100 M

Using the K a value of 1.4 $×$ 10 −5 , place $\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{3+}$ in the correct location in [link] .

Calculate the concentration of all solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I .

(a) 0.0092 M HClO, a weak acid

(b) 0.0784 M C 6 H 5 NH 2 , a weak base

(c) 0.0810 M HCN, a weak acid

(d) 0.11 M (CH 3 ) 3 N, a weak base

(e) 0.120 M $\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}$ a weak acid, K a = 1.6 $×$ 10 −7

(a) $\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{ClO}}^{\text{−}}\right]}{\left[\text{HClO}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0092-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0092}\phantom{\rule{0.2em}{0ex}}=2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$
Solving for x gives 1.63 $×$ 10 −5 M . This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [ClO]= 5.8 $×$ 10 −5 M
[HClO] = 0.00092 M
[OH ] = 6.1 $×$ 10 −10 M ;
(b) $\frac{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{3}{}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{C}}_{6}{\text{H}}_{5}{\text{NH}}_{2}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0784-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0784}\phantom{\rule{0.2em}{0ex}}=4.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
Solving for x gives 5.81 $×$ 10 −6 M . This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
$\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]$ = [OH ] = 5.8 $×$ 10 −6 M
[C 6 H 5 NH 2 ] = 0.00784
[H 3 O + ] = 1.7 $×$ 10 −9 M ;
(c) $\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CN}}^{\text{−}}\right]}{\left[\text{HCN}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.0810-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.0810}\phantom{\rule{0.2em}{0ex}}=4.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$
Solving for x gives 6.30 $×$ 10 −6 M . This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H 3 O + ] = [CN ] = 6.3 $×$ 10 −6 M
[HCN] = 0.0810 M
[OH ] = 1.6 $×$ 10 −9 M ;
(d) $\frac{\left[{\left({\text{CH}}_{3}\right)}_{3}{\text{NH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\left({\text{CH}}_{3}\right)}_{3}\text{N}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.11-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.11}\phantom{\rule{0.2em}{0ex}}=6.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$
Solving for x gives 2.63 $×$ 10 −3 M . This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH 3 ) 3 NH + ] = [OH ] = 2.6 $×$ 10 −3 M
[(CH 3 ) 3 N] = 0.11 M
[H 3 O + ] = 3.8 $×$ 10 −12 M ;
(e) $\frac{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left({\text{OH}\right)}^{\text{+}}\right]\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{\left(0.120-x\right)}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{\left(x\right)\left(x\right)}{0.120}\phantom{\rule{0.2em}{0ex}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}$
Solving for x gives 1.39 $×$ 10 −4 M . This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H 2 O) 5 (OH) + ] = [H 3 O + ] = 1.4 $×$ 10 −4 M
$\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}\right]$ = 0.120 M
[OH ] = 7.2 $×$ 10 −11 M

Propionic acid, C 2 H 5 CO 2 H ( K a = 1.34 $×$ 10 −5 ), is used in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698- M solution of C 2 H 5 CO 2 H?

White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegar is 1.007 g/cm 3 , what is the pH?

pH = 2.41

The ionization constant of lactic acid, CH 3 CH(OH)CO 2 H, an acid found in the blood after strenuous exercise, is 1.36 $×$ 10 −4 . If 20.0 g of lactic acid is used to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?

Nicotine, C 10 H 14 N 2 , is a base that will accept two protons ( K 1 = 7 $×$ 10 −7 , K 2 = 1.4 $×$ 10 −11 ). What is the concentration of each species present in a 0.050- M solution of nicotine?

[C 10 H 14 N 2 ] = 0.049 M
[C 10 H 14 N 2 H + ] = 1.9 $×$ 10 −4 M
$\left[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_{2}{\text{H}}_{2}{}^{2+}\right]$ = 1.4 $×$ 10 −11 M
[OH ] = 1.9 $×$ 10 −4 M
[H 3 O + ] = 5.3 $×$ 10 −11 M

The pH of a 0.20- M solution of HF is 1.92. Determine K a for HF from these data.

The pH of a 0.15- M solution of ${\text{HSO}}_{4}{}^{\text{−}}$ is 1.43. Determine K a for ${\text{HSO}}_{4}{}^{\text{−}}$ from these data.

${K}_{\text{a}}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}$

The pH of a 0.10- M solution of caffeine is 11.16. Determine K b for caffeine from these data:
${\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{10}{\text{N}}_{4}{\text{O}}_{2}{\text{H}}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)$

The pH of a solution of household ammonia, a 0.950 M solution of NH 3, is 11.612. Determine K b for NH 3 from these data.

${K}_{\text{b}}=1.77\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

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