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:** - Concentration: $C = 0.1$ M - Degree of dissociation: $\alpha = 0.01\% = 0.0001$ **Step 1: Find the concentration of ionized HCN** $$[\text{H}^+] = [\text{CN}^-] = C \cdot \alpha = 0.1 \times 0.0001 = 10^{-5} \text{ M}$$ **Step 2: Find the concentration of unionized 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 formula** $$K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} = \frac{(10^{-5})(10^{-5})}{0.1}$$ $$K_a = \frac{10^{-10}}{10^{-1}} = 10^{-9} \text{ M}$$ **Answer:** The ionization constant is **$K_a = 10^{-9}$ M** (or **$10^{-9}$ mol/L**) This matches the weak acid nature of HCN, consistent with its very low degree of dissociation.