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

Given:

K_sp (AgCN) = 2.2 × 10⁻¹⁶
K_a (HCN) = 6.2 × 10⁻¹⁰
pH = 3
[H⁺] = 10⁻³ M

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Step 1: Write equilibrium reactions

AgCN(s) ⇌ Ag⁺ + CN⁻

K_sp = [Ag⁺][CN⁻]

CN⁻ + H⁺ ⇌ HCN

K_a = [H⁺][CN⁻] / [HCN]

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Step 2: Relation between total solubility and CN⁻

Let solubility = s

Total cyanide produced = s

In acidic medium, most CN⁻ converts to HCN.

From K_a expression:

[CN⁻] = K_a[HCN] / [H⁺]

Since almost all dissolved salt gives HCN,

[HCN] ≈ s

Therefore:

[CN⁻] = (K_a s) / [H⁺]

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Step 3: Substitute in K_sp

K_sp = [Ag⁺][CN⁻]

= s × (K_a s / [H⁺])

K_sp = (K_a s²) / [H⁺]

Rearranging:

s² = (K_sp [H⁺]) / K_a

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Step 4: Substitute numerical values

s² = (2.2 × 10⁻¹⁶ × 10⁻³) / (6.2 × 10⁻¹⁰)

s² = 2.2 × 10⁻¹⁹ / 6.2 × 10⁻¹⁰

s² = 3.55 × 10⁻¹⁰

s = √(3.55 × 10⁻¹⁰)

s = 1.9 × 10⁻⁵ M

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Final Answer:

Solubility of AgCN at pH 3
= 1.9 × 10⁻⁵ M