The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be — Ionic Equilibrium Chemistry Question
Question
The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would be
💡 Solution & Explanation
# Solution: Ionisation Constant of HCN **Given Information:** - Concentration: $C = 0.1$ M - Degree of dissociation: $\alpha = 0.01\% = 0.0001 = 10^{-4}$ **Step 1: Find the concentration of ionized HCN** $$[H^+] = [CN^-] = C \cdot \alpha = 0.1 \times 10^{-4} = 10^{-5} \text{ M}$$ **Step 2: Find the concentration of un-ionized HCN** $$[HCN] = C(1-\alpha) = 0.1(1 - 10^{-4}) \approx 0.1 \text{ M}$$ (Since $\alpha$ is very small, we can approximate $(1-\alpha) \approx 1$) **Step 3: Apply the Ionization Constant Expression** For the equilibrium: $HCN \rightleftharpoons H^+ + CN^-$ $$K_a = \frac{[H^+][CN^-]}{[HCN]} = \frac{(10^{-5})(10^{-5})}{0.1}$$ $$K_a = \frac{10^{-10}}{10^{-1}} = 10^{-9} \text{ M}$$ **Answer:** $K_a = 10^{-9}$ M (or $1 \times 10^{-9}$ M) This matches **option D**.