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** $$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$ **Step 2: Apply the dissociation constant formula** For weak acids, when degree of dissociation is $\alpha$: $$K_a = \frac{C\alpha^2}{1-\alpha}$$ Since $K_a$ is small, assume $\alpha \ll 1$, so $1-\alpha \approx 1$: $$K_a \approx C\alpha^2$$ **Step 3: Solve for $\alpha$** $$\alpha^2 = \frac{K_a}{C} = \frac{1 \times 10^{-5}}{0.1} = \frac{10^{-5}}{10^{-1}} = 10^{-4}$$ $$\alpha = \sqrt{10^{-4}} = 10^{-2} = 0.01$$ **Step 4: Verify the approximation** Since $\alpha = 0.01 \ll 1$, our assumption is valid. **Answer:** The degree of dissociation is **0.01 or 1%** (or **$10^{-2}$** depending on how option D is presented).