Answer
$pH = 10.92$
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.
$ 4.4\times 10^{- 4} = \frac{[CH_3N{H_3}^+][H_3O^+]}{[CH_3NH_2]}$
$ 4.4\times 10^{- 4} = \frac{( 0.135 + x )* x}{ 0.255 - x}$
Considering 'x' has a very small value.
$ 4.4\times 10^{- 4} = \frac{ 0.135 * x}{ 0.255}$
$ 4.4\times 10^{- 4} = 0.529x$
$\frac{ 4.4\times 10^{- 4}}{ 0.529} = x$
$x = 8.3\times 10^{- 4}$
Percent dissociation: $\frac{ 8.3\times 10^{- 4}}{ 0.255} \times 100\% = 0.33\%$
Less than $5%$: OK.
x = $[H_3O^+]$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 8.3 \times 10^{- 4})$
$pOH = 3.08$
$pH + pOH = 14$
$pH + 3.08 = 14$
$pH = 10.92$
