Degree of dissociation of 0.1 N is (Dissociation constant = 1 × 10 –5) — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution: Degree of Dissociation of Acetic Acid **Given:** - Concentration: $C = 0.1$ N (or 0.1 M) - Dissociation constant: $K_a = 1 \times 10^{-5}$ **Step 1: Set up the dissociation equation** $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$ Let $\alpha$ = degree of dissociation **Step 2: Apply the expression for $K_a$** $$K_a = \frac{C\alpha^2}{1-\alpha}$$ **Step 3: Check if we can use the approximation** Since $K_a = 1 \times 10^{-5}$ is very small, we expect $\alpha \ll 1$, so $(1-\alpha) \approx 1$ $$K_a \approx C\alpha^2$$ **Step 4: Solve for $\alpha$** $$\alpha^2 = \frac{K_a}{C} = \frac{1 \times 10^{-5}}{0.1} = 1 \times 10^{-4}$$ $$\alpha = \sqrt{1 \times 10^{-4}} = 0.01 = 1\%$$ **Verification:** $\alpha = 0.01 \ll 1$ ✓ (approximation valid) **Answer: The degree of dissociation is 0.01 or 1%** (Option D)