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

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Equilibrium

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} (AgCN) = 2.2 \times 1 0^{- 16}

and

K_{a} (HCN) = 6.2 \times 1 0^{- 10}

]

Answer: (D)

At pH 3 nearly every cyanide ion released by the solid is captured as

HCN

, which keeps

[C N^{-}]

low and drives more solid into solution. Writing the solubility as

s

, the silver that dissolves stays as

A g^{+}

while the cyanide is shared between its two forms, so

[A g^{+}] = s

and

[C N^{-}] + [HCN] = s

. The acid dissociation constant fixes the split:

[C N^{-}] = s \frac{K _{a}}{K _{a} + [ H ^{+} ]} \approx s \frac{K _{a}}{[ H ^{+} ]} = s \frac{6.2 \times 1 0 ^{- 10}}{1 0 ^{- 3}} = 6.2 \times 1 0^{- 7} s

Substituting in the solubility product,

K_{s p} = [A g^{+}] [C N^{-}] = 6.2 \times 1 0^{- 7} s^{2} = 2.2 \times 1 0^{- 16}

s^{2} = 3.55 \times 1 0^{- 10}, s = 1.9 \times 1 0^{- 5}

The solubility is

1.9 \times 1 0^{- 5}

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option