If K_{sp} of Ag₂CO₃ is 8 ×10^{-12}, molar solubility of Ag₂CO₃ in 0.1,M,AgNO₃ is

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6-2.Equilibrium-II (Ionic Equilibrium)

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If $K_{sp}$ of $Ag_2CO_3$ is $8 \times10^{-12},$ the molar solubility of $Ag_2CO_3$ in $0.1\,M\,AgNO_3$ is

A

$8\times 10^{-12}\,M$

B

$8\times 10^{-11}\,M$

C

$8\times 10^{-10}\,M$

D

$8\times 10^{-13}\,M$

(JEE MAIN-2019)

Solution

$8 \times {10^{ – 12}} = {(2S' + 0.1)^2}S'$

Or $S' = 8\times 10^{-10}\,M$

Standard 11

Chemistry

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