Answer
$[OH^-]$ = 0.01025M
$pOH = 1.989$
$pH = 12.011$
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.
-$[OH^-] = [CH_3N{H_3}^+] = x$
-$[CH_3NH_2] = [CH_3NH_2]_{initial} - x = 0.25 - x$
For approximation, we consider: $[CH_3NH_2] = 0.25M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][CH_3N{H_3}^+]}{ [CH_3NH_2]}$
$Ka = 4.2 \times 10^{- 4}= \frac{x * x}{ 0.25}$
$Ka = 4.2 \times 10^{- 4}= \frac{x^2}{ 0.25}$
$ 1.05 \times 10^{- 4} = x^2$
$x = 1.025 \times 10^{- 2}$
Percent ionization: $\frac{ 1.025 \times 10^{- 2}}{ 0.25} \times 100\% = 4.099\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [CH_3N{H_3}^+] = x = 1.025 \times 10^{- 2}M $
3. Calculate the pOH and the pH.
$pOH = -log[OH^-]$
$pOH = -log( 0.01025)$
$pOH = 1.989$
$pH + pOH = 14$
$pH + 1.989 = 14$
$pH = 12.011$
