The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

I need to find the ionization constant ($K_a$) for HCN given the concentration and degree of dissociation. **Given Information:** - Initial concentration: $C = 0.1$ M - Degree of dissociation: $\alpha = 0.01\% = 0.0001$ **Step 1: Find the concentration of $H^+$ ions at equilibrium** $$[H^+] = C \times \alpha = 0.1 \times 0.0001 = 10^{-5} \text{ M}$$ **Step 2: Set up the dissociation equilibrium** $$HCN \rightleftharpoons H^+ + CN^-$$ At equilibrium: - $[H^+] = 10^{-5}$ M - $[CN^-] = 10^{-5}$ M (from stoichiometry) - $[HCN] = 0.1 - 10^{-5} \approx 0.1$ M (since $\alpha$ is very small) **Step 3: Calculate $K_a$** $$K_a = \frac{[H^+][CN^-]}{[HCN]} = \frac{(10^{-5})(10^{-5})}{0.1}$$ $$K_a = \frac{10^{-10}}{10^{-1}} = 10^{-10} \times 10^{1} = 10^{-9} \text{ M}$$ **Answer:** $K_a = 10^{-9}$ M (or $1 \times 10^{-9}$ M) This matches option **D** and is consistent with the known $K_a$ value for HCN, a weak acid.