Answer
$pH = 4.55$
Work Step by Step
1. Draw the ICE table (It is in the end of this answer)
2. Calculate 'x' using the $K_a$ expression.
$ 1.8\times 10^{- 5} = \frac{[C_2H_3{O_2}^-][H_3O^+]}{[HC_2H_3O_2]}$
$ 1.8\times 10^{- 5} = \frac{( 0.125 + x )* x}{ 0.195 - x}$
Considering 'x' has a very small value.
$ 1.8\times 10^{- 5} = \frac{ 0.125 * x}{ 0.195}$
$ 1.8\times 10^{- 5} = 0.641x$
$\frac{ 1.8\times 10^{- 5}}{ 0.641} = x$
$x = 2.8\times 10^{- 5}$
Percent dissociation: $\frac{ 2.8\times 10^{- 5}}{ 0.195} \times 100\% = 0.014\%$
- Less than $5\%$, good approximation.
x = $[H_3O^+]$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.8 \times 10^{- 5})$
$pH = 4.55$
