If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_{2} \mathrm{CO}_{3}$ in $0.1 \mathrm{M} \mathrm{AgNO}_{3}$ is :

(C) $8 \times 10^{-10} \mathrm{M}$

(A) $8 \times 10^{-12} \mathrm{M}$

(B) $8 \times 10^{-11} \mathrm{M}$

(D) $8 \times 10^{-13} \mathrm{M}$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Equilibrium

K_{sp}

Ag_{2} CO_{3}

8 \times 1 0^{- 12}

, the molar solubility of

Ag_{2} CO_{3}

0.1 M AgNO_{3}

is :

Answer: (C)

For

Ag X_{2} CO X_{3} 2 Ag X^{+} + CO X_{3} X^{2 -}

, dissolving in

0.1

AgNO X_{3}

fixes

[Ag X^{+}] \approx 0.1

M by the common-ion effect. Then

K_{s p} = [Ag X^{+}]^{2} [CO X_{3} X^{2 -}] = (0.1)^{2} S^{'}

, so

8 \times 1 0^{- 12} = 0.01 S^{'}

, giving a molar solubility

S^{'} = 8 \times 1 0^{- 10}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option