Answer
$pH = 5.03$
Work Step by Step
1. Drawing the ICE table, we get these concentrations at the equilibrium:
$C_3H_5O_2H(aq) + H_2O(l) \lt -- \gt C_3H_5O_2(aq) + H_3O^+(aq)$
Remember: Reactants at equilibrium = Initial Concentration - x
And Products = Initial Concentration + x
$[C_3H_5O_2H] = 0.2 M - x$
$[C_3H_5O_2] = 0.3M + x$
$[H_3O^+] = 0 + x$
2. Calculate 'x' using the $K_a$ expression.
$ 1.4\times 10^{- 5} = \frac{[C_3H_5O_2][H_3O^+]}{[C_3H_5O_2H]}$
$ 1.4\times 10^{- 5} = \frac{( 0.3 + x )* x}{ 0.2 - x}$
Considering 'x' has a very small value.
$ 1.4\times 10^{- 5} = \frac{ 0.3 * x}{ 0.2}$
$ 1.4\times 10^{- 5} = 1.5x$
$\frac{ 1.4\times 10^{- 5}}{ 1.5} = x$
$x = 9.3 \times 10^{- 6}$
Percent dissociation: $\frac{ 9.3 \times 10^{- 6}}{ 0.2} \times 100\% = 4.7\times 10^{- 3}\%$
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 9.3 \times 10^{- 6})$
$pH = 5.03$
