Answer
$[H_3O^+] = [CH_3COO^-] \approx 1.9 \times 10^{- 3}M $
$[CH_3COOH] \approx 0.2M$
Work Step by Step
$K_a(CH_3COOH) = 1.8 \times 10^{-5}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3COO^-] = x$
-$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.2 - x$
For approximation, we consider: $[CH_3COOH] = 0.2M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.2}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.2}$
$ 3.6 \times 10^{- 6} = x^2$
$x = 1.897 \times 10^{- 3}$
Percent ionization: $\frac{ 1.897 \times 10^{- 3}}{ 0.2} \times 100\% = 0.9487\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3COO^-] = x = 1.897 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.2M$
