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 weak acid approximation** Since $K_a$ is very small, we check if $C >> K_a$: $$\frac{K_a}{C} = \frac{1 \times 10^{-5}}{0.1} = 1 \times 10^{-4} << 1$$ The weak acid approximation is valid. **Step 3: Use the dissociation constant formula** $$K_a = \frac{[\text{products}]}{[\text{reactants}]} = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$$ where $\alpha$ is the degree of dissociation. **Step 4: Solve for α** $$\alpha^2 = \frac{K_a}{C} = \frac{1 \times 10^{-5}}{0.1} = 1 \times 10^{-4}$$ $$\alpha = \sqrt{1 \times 10^{-4}} = 0.01 = 1\%$$ **Answer: The degree of dissociation is 0.01 or 1%** (Option D)