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

💡 Solution & Explanation

# Solution: Ionization Constant of HCN **Given Data:** - Concentration: $C = 0.1$ M - Degree of dissociation: $\alpha = 0.01\% = 0.0001$ **Step 1: Find the ionized concentration** $$[\text{H}^+] = [\text{CN}^-] = C \cdot \alpha = 0.1 \times 0.0001 = 10^{-5} \text{ M}$$ **Step 2: Find the equilibrium concentration of undissociated HCN** $$[\text{HCN}] = C(1-\alpha) = 0.1(1-0.0001) \approx 0.1 \text{ M}$$ Since $\alpha$ is very small, $(1-\alpha) \approx 1$ **Step 3: Apply the ionization constant expression** For the equilibrium: $\text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^-$ $$K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} = \frac{(10^{-5})(10^{-5})}{0.1}$$ $$K_a = \frac{10^{-10}}{10^{-1}} = 10^{-9}$$ **Answer: $K_a = 10^{-9}$ or $1 \times 10^{-9}$ M** (Option D)