Answer
$pH = 11.949$
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.
-$[OH^-] = [CH_3N{H_3}^+] = x$
-$[CH_3NH_2] = [CH_3NH_2]_{initial} - x = 0.18 - x$
For approximation, we consider: $[CH_3NH_2] = 0.18M$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][CH_3N{H_3}^+]}{ [CH_3NH_2]}$
$Kb = 4.4 \times 10^{- 4}= \frac{x * x}{ 0.18}$
$Kb = 4.4 \times 10^{- 4}= \frac{x^2}{ 0.18}$
$ 7.92 \times 10^{- 5} = x^2$
$x = 8.899 \times 10^{- 3}$
Percent ionization: $\frac{ 8.899 \times 10^{- 3}}{ 0.18} \times 100\% = 4.944\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3N{H_3}^+] = x = 8.899 \times 10^{- 3}M $
$[CH_3NH_2] \approx 0.18M$
3. Calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 8.899 \times 10^{- 3})$
$pOH = 2.051$
$pH + pOH = 14$
$pH + 2.051 = 14$
$pH = 11.949$
