Answer
$[HC_3H_5O_2] \approx 0.1M$
$[H_3O^+] = [C_3H_5{O_2}^-] = x = 1.14 \times 10^{- 3}M$
$pH = 2.943$
Percent dissociation: $ = 1.14\%$
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.
-$[H_3O^+] = [C_3H_5{O_2}^-] = x$
-$[HC_3H_5O_2] = [HC_3H_5O_2]_{initial} - x = 0.1 - x$
For approximation, we consider: $[HC_3H_5O_2] = 0.1M$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_3H_5{O_2}^-]}{ [HC_3H_5O_2]}$
$Ka = 1.3 \times 10^{- 5}= \frac{x * x}{ 0.1}$
$Ka = 1.3 \times 10^{- 5}= \frac{x^2}{ 0.1}$
$ 1.3 \times 10^{- 6} = x^2$
$x = 1.14 \times 10^{- 3}$
Percent dissociation: $\frac{ 1.14 \times 10^{- 3}}{ 0.1} \times 100\% = 1.14\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [C_3H_5{O_2}^-] = x = 1.14 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HC_3H_5O_2] \approx 0.1M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.14 \times 10^{- 3})$
$pH = 2.943$
