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 For a weak acid dissociation: $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$ **Step 1: Use the dissociation constant expression** $$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$$ **Step 2: Set up using degree of dissociation ($\alpha$)** For initial concentration $C = 0.1$ N: - $[H^+] = C\alpha = 0.1\alpha$ - $[CH_3COO^-] = C\alpha = 0.1\alpha$ - $[CH_3COOH] = C(1-\alpha) \approx C = 0.1$ (since $\alpha$ is small) **Step 3: Substitute into $K_a$ expression** $$1 \times 10^{-5} = \frac{(0.1\alpha)(0.1\alpha)}{0.1}$$ $$1 \times 10^{-5} = \frac{0.01\alpha^2}{0.1} = 0.1\alpha^2$$ **Step 4: Solve for $\alpha$** $$\alpha^2 = \frac{1 \times 10^{-5}}{0.1} = 1 \times 10^{-4}$$ $$\alpha = \sqrt{1 \times 10^{-4}} = 0.01 = \frac{1}{100}$$ **Answer: The degree of dissociation is 0.01 or 1%** (Option D is likely 0.01 or 1/100)