Answer
(a) 1.095%
(b) 3.464%
Work Step by Step
(a)
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3COO^-] = x$
-$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.15 - x$
For approximation, we consider: $[CH_3COOH] = 0.15M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.15}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.15}$
$ 2.7 \times 10^{- 6} = x^2$
$x = 1.643 \times 10^{- 3}$
Percent ionization: $\frac{ 1.643 \times 10^{- 3}}{ 0.15} \times 100\% = 1.095\%$
%ionization < 5% : Right approximation.
(b)
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3COO^-] = x$
-$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.015 - x$
For approximation, we consider: $[CH_3COOH] = 0.015M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.015}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.015}$
$ 2.7 \times 10^{- 7} = x^2$
$x = 5.196 \times 10^{- 4}$
Percent ionization: $\frac{ 5.196 \times 10^{- 4}}{ 0.015} \times 100\% = 3.464\%$
%ionization < 5% : Right approximation.
