Answer
$pH = 10.643$
Work Step by Step
1. Draw the ICE table (It is in the end of this answer)
2. Calculate 'x' using the $K_b$ expression.
$ 4.4\times 10^{- 4} = \frac{[CH_3N{H_3}^+][OH^-]}{[CH_3NH_2]}$
$ 4.4\times 10^{- 4} = \frac{( 0.18 + x )* x}{ 0.18 - x}$
Considering 'x' has a very small value.
$ 4.4\times 10^{- 4} = \frac{ 0.18 * x}{ 0.18}$
$ 4.4\times 10^{- 4} = 1x$
$\frac{ 4.4\times 10^{- 4}}{ 1} = x$
$x = 4.4\times 10^{- 4}$
Percent ionization: $\frac{ 4.4\times 10^{- 4}}{ 0.18} \times 100\% = 0.2444\%$
x = $[OH^-]$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 4.4 \times 10^{- 4})$
$pOH = 3.357$
$pH + pOH = 14$
$pH + 3.357 = 14$
$pH = 10.643$
