Answer
(a) $pH = 2.57$
(b) $pH = 4.44$
Work Step by Step
(a)
1. We have those concentrations at equilibrium:
-$[H_3O^+] = [CH_3COO^-] = x$
-$[CH_3COOH] = [CH_3COOH]_{initial} - x = 0.40 - x$
For approximation, we consider: $[CH_3COOH] = 0.40M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3COO^-]}{ [CH_3COOH]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.4}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.4}$
$ 7.2 \times 10^{- 6} = x^2$
$x = 2.7 \times 10^{- 3}$
Percent dissociation: $\frac{ 2.7 \times 10^{- 3}}{ 0.4} \times 100\% = 0.67\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3COO^-] = x = 2.7 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3COOH] \approx 0.4M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.7 \times 10^{- 3})$
$pH = 2.57$
(b)
1. Drawing the ICE table we get these concentrations at the equilibrium:
$CH_3COOH(aq) + H_2O(l) \lt -- \gt CH_3COO^-(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[CH_3COOH] = 0.40 M - x$
$[CH_3COO^-] = 0.20M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 1.8\times 10^{- 5} = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$
$ 1.8\times 10^{- 5} = \frac{( 0.2 + x )* x}{ 0.4 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.2 * x}{ 0.4}$
$ 1.8\times 10^{- 5} = 0.5x$
$\frac{ 1.8\times 10^{- 5}}{ 0.5} = x$
$x = 3.6\times 10^{- 5}$
Percent dissociation: $\frac{ 3.6\times 10^{- 5}}{ 0.4} \times 100\% = 9\times 10^{- 3}\%$
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 3.6 \times 10^{- 5})$
$pH = 4.44$
