Answer
(a) 0.00152%
(b) 0.00228%
(c) 0.85%
Work Step by Step
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_6H_5O^-] = x$
-$[C_6H_5OH] = [C_6H_5OH]_{initial} - x$
For approximation, we consider: $[C_6H_5OH] = [C_6H_5OH]_{initial}$
(a)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$
$Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 5.6 \times 10^{- 1}}$
$Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 5.6 \times 10^{- 1}}$
$ 7.27 \times 10^{- 11} = x^2$
$x = 8.53 \times 10^{- 6}$
5% test: $\frac{ 8.53 \times 10^{- 6}}{ 5.6 \times 10^{- 1}} \times 100\% = 0.00152\%$
%ionization < 5% : Right approximation.
(b)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$
$Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 2.5 \times 10^{- 1}}$
$Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 2.5 \times 10^{- 1}}$
$ 3.25 \times 10^{- 11} = x^2$
$x = 5.7 \times 10^{- 6}$
5% test: $\frac{ 5.7 \times 10^{- 6}}{ 2.5 \times 10^{- 1}} \times 100\% = 0.00228\%$
%ionization < 5% : Right approximation.
(c)
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$
$Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 1.8 \times 10^{- 6}}$
$Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 1.8 \times 10^{- 6}}$
$ 2.33 \times 10^{- 16} = x^2$
$x = 1.52 \times 10^{- 8}$
5% test: $\frac{ 1.52 \times 10^{- 8}}{ 1.8 \times 10^{- 6}} \times 100\% = 0.85\%$
%ionization < 5% : Right approximation.
