What is the solubility of Ag_2CO_3 in 0.1 , M | vedclass

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(A) The solubility equilibrium for $Ag_2CO_3$ is: $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.Given $K_{sp} = [Ag^+]^2 [CO_3^{2-}] = 4

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$10^{-6}$

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(A) The solubility equilibrium for $Ag_2CO_3$ is: $Ag_2CO_3(s) ightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.Given $K_{sp} = [Ag^+]^2 [CO_3^{2-}] = 4 	imes 10^{-13}$.In the presence of $0.1 \, M \, Na_2CO_3$,the concentration of $CO_3^{2-}$ is approximately $0.1 \, M$ due to the common ion effect.Let $s$ be the solubility of $Ag_2CO_3$. Then $[Ag^+] = 2s$ and $[CO_3^{2-}] \approx 0.1$.Substituting these into the $K_{sp}$ expression: $4 	imes 10^{-13} = (2s)^2 (0.1)$.$4 	imes 10^{-13} = 4s^2 	imes 0.1$.$s^2 = \frac{4 	imes 10^{-13}}{4 	imes 0.1} = 10^{-12}$.$s = \sqrt{10^{-12}} = 10^{-6} \, M$.

What is the solubility of $Ag_2CO_3$ in $0.1 \, M \, Na_2CO_3$? The solubility product $(K_{sp})$ of $Ag_2CO_3$ at the same temperature is $4 \times 10^{-13}$.

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