Question

The solubility of AgCN in a buffer solution of pH=3 is x. The value of x is : [Assume : No cyano complex is formed ; $K_{sp}(\text{AgCN}) = 2.2 \times 10^{-16}$ and $K_a(\text{HCN}) = 6.2 \times 10^{-10}$]

• A. $0.625 \times 10^{-6}$
• B. $1.6 \times 10^{-6}$
• C. $2.2 \times 10^{-16}$
• D. $1.9 \times 10^{-5}$

Hint:

Solubility of salts containing anions of weak acids increases as pH decreases because the anion is consumed by $H^+$ ions, shifting the equilibrium forward (Le Chatelier's Principle).

Answer 1

The Correct Option is D

Step 1: In acidic buffer, $CN^-$ reacts with $H^+$ to form $HCN$. Solubility $x = [Ag^+] = [CN^-] + [HCN]$.
Step 2: From $K_a$, $[HCN] = \frac{[H^+][CN^-]}{K_a}$.
Step 3: $x = [CN^-] \left( 1 + \frac{[H^+]}{K_a} \right) \Rightarrow [CN^-] = \frac{x}{1 + \frac{[H^+]}{K_a}}$.
Step 4: $K_{sp} = [Ag^+][CN^-] = x \cdot \left( \frac{x}{1 + \frac{[H^+]}{K_a}} \right)$.
Step 5: $x^2 = K_{sp} \left( 1 + \frac{10^{-3}}{6.2 \times 10^{-10}} \right) \approx K_{sp} \left( \frac{10^{-3}}{6.2 \times 10^{-10}} \right)$.
Step 6: $x^2 = 2.2 \times 10^{-16} \times 1.6 \times 10^6 = 3.52 \times 10^{-10} \Rightarrow x \approx 1.87 \times 10^{-5}$.