The ionization constant of dimethylamine is $5.4 \times 10^{-4}$. Calculate its degree of ionization in its $0.02 \ M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1 \ M$ in $NaOH$?

(N/A) $K_{b} = 5.4 \times 10^{-4}$
$c = 0.02 \ M$
For a weak base,the degree of ionization $\alpha$ is given by $\alpha = \sqrt{\frac{K_{b}}{c}}$.
$\alpha = \sqrt{\frac{5.4 \times 10^{-4}}{0.02}} = \sqrt{270 \times 10^{-4}} = 0.1643$.
When $0.1 \ M \ NaOH$ is added,it acts as a strong base and provides $0.1 \ M \ OH^{-}$ ions.
$(CH_{3})_{2}NH + H_{2}O \longleftrightarrow (CH_{3})_{2}NH_{2}^{+} + OH^{-}$
Initial: $0.02 \ M \quad 0 \quad 0.1 \ M$
Equilibrium: $(0.02 - x) \ M \quad x \ M \quad (0.1 + x) \ M$
Since $K_{b}$ is small,$x$ is negligible compared to $0.02$ and $0.1$.
$K_{b} = \frac{x(0.1)}{0.02} = 5.4 \times 10^{-4}$
$x = \frac{5.4 \times 10^{-4} \times 0.02}{0.1} = 1.08 \times 10^{-4} \ M$.
Percentage ionization $= \frac{x}{c} \times 100 = \frac{1.08 \times 10^{-4}}{0.02} \times 100 = 0.54 \%$.