Part 1: Degree of ionization in 0.02 M solution

The ionization of dimethylamine in water is represented as:

$$(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-$$

Given: $K_b = 5.4 \times 10^{-4}$, $c = 0.02$ M

Using the approximation for weak bases (since $K_b$ is small):

$$\alpha = \sqrt{\frac{K_b}{c}} = \sqrt{\frac{5.4 \times 10^{-4}}{0.02}} = \sqrt{2.7 \times 10^{-2}} = 0.164$$

Part 2: Percentage ionized in 0.1 M NaOH

When the solution is also 0.1 M in NaOH, the concentration of $OH^-$ is approximately $0.1$ M (since NaOH is a strong base and completely ionized, and the contribution from dimethylamine is negligible due to the common ion effect).

Using the equilibrium expression:

$$K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}$$

Let $\alpha$ be the new degree of ionization. Then:

$$[(CH_3)_2NH_2^+] = c\alpha = 0.02\alpha$$ $$[OH^-] = 0.1$$ $$[(CH_3)_2NH] = c(1 - \alpha) \approx 0.02$$

Substituting the values into the $K_b$ expression:

$$5.4 \times 10^{-4} = \frac{(0.02\alpha)(0.1)}{0.02} = 0.1\alpha$$

Solving for $\alpha$:

$$\alpha = \frac{5.4 \times 10^{-4}}{0.1} = 5.4 \times 10^{-3}$$

To find the percentage of dimethylamine ionized: Percentage ionized = $\alpha \times 100 = 5.4 \times 10^{-3} \times 100 = 0.54\%$