Q.The ionization constant of dimethylamine is 5.4 × 10⁻⁴. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Concept: Degree of ionization of a weak base, and its suppression by a common ion.

In pure 0.02 M solution. For $(CH_3)_2NH$ with $K_b = 5.4 \times 10^{-4}$:

Because $\alpha$ is not small compared with 1, the exact quadratic

$c\alpha^2 + K_b\alpha - K_b = 0$ is better; it gives $\alpha = 0.151$ (15.1% ionized).

In 0.1 M NaOH. The NaOH supplies a large common-ion concentration

$[OH^-] \approx 0.1$ M, and the base's own contribution is negligible. Then

$K_b = \frac{(c\alpha')(0.1)}{c} = 0.1\,\alpha' \quad\Rightarrow\quad \alpha' = \frac{K_b}{0.1} = \frac{5.4 \times 10^{-4}}{0.1} = 5.4 \times 10^{-3}$ …

Understanding Weak Base Ionization

Dimethylamine, $(CH_3)_2NH$, is a weak base. When dissolved in water, it accepts a proton from water molecules:

The equilibrium constant for this reaction is the base ionization constant $K_b = 5.4 \times 10^{-4}$. The degree of ionization $\alpha$ tells us what fraction of the original base molecules have accepted a proton at equilibrium. For a weak base, $\alpha$ is small but not negligible, and it depends on both $K_b$ and the initial concentration.

When we add a strong base like NaOH, which completely dissociates to give $OH^-$ ions, we flood the solution with the product of our equilibrium. By Le Chatelier's principle, this shifts the equilibrium backward, suppressing the ionization of dimethylamine—the common-ion effect.

Part 1: Degree of Ionization in Pure 0.02 M Solution

1. Set up the equilibrium expression

Let the initial concentration be $c = 0.02$ M and the degree of ionization be $\alpha$. At equilibrium:

• $(CH_3)_2NH$: concentration = $c(1 - \alpha)$
• $(CH_3)_2NH_2^+$: concentration = $c\alpha$
• $OH^-$: concentration = $c\alpha$

The base ionization constant is:

2. Check if we can use the approximation

For weak electrolytes, if $\alpha \ll 1$, we can approximate $1 - \alpha \approx 1$. Let's check: if $c\alpha^2 \approx K_b$, then $\alpha \approx \sqrt{K_b/c} = \sqrt{5.4 \times 10^{-4}/0.02} = \sqrt{0.027} \approx 0.164$.

Since $\alpha \approx 0.164$ is not negligible compared to 1, we should solve the exact equation.

3. Solve the quadratic equation

Dividing through by 0.02:

Using the quadratic formula:

Taking the positive root:

Percentage ionization = $0.151 \times 100 = 15.1\%$ (the quick approximation $\alpha = \sqrt{K_b/c}$ gives 0.164 — the exact quadratic value is the better one here because $\alpha$ is not small)

Part 2: Degree of Ionization in 0.1 M NaOH

4. Account for the common ion

NaOH is a strong base that dissociates completely:

So the solution already contains $[OH^-] = 0.1$ M from NaOH before any dimethylamine ionizes. …