Answer
Hydroxide ion concentration $[OH^-] = 1.2 \times 10^{-2}$;
pH = 12.08.
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CH_3NH_2 ]& [ CH_3N{H_3}^{+} ]& [ OH^- ]\\
Initial& 0.33 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.33 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ CH_3N{H_3}^{+} ][ H^+ ]}{[ CH_3NH_2 ]}$$
$$K_b = \frac{(x)(x)}{[ CH_3NH_2 ]_{initial} - x}$$
3. Assuming $ 0.33 \gt\gt x$:
$$K_b = \frac{x^2}{[ CH_3NH_2 ]_{initial}}$$
$$x = \sqrt{K_b \times [ CH_3NH_2 ]_{initial}} = \sqrt{ 4.4 \times 10^{-4} \times 0.33 }$$
$x = 0.012 $
4. Test if the assumption was correct:
$$\frac{ 0.012 }{ 0.33 } \times 100\% = 3.6 \%$$
5. Thus, it is correct to say that $x = 0.012 $
6. $[OH^-] = x = 0.012 = 1.2 \times 10^{-2} \space M$
7. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.012 } = 8.3 \times 10^{-13} \space M$$
$$pH = -log[H_3O^+] = -log( 8.3 \times 10^{-13} ) = 12.08 $$
