What is the molar solubility of Ag_{2}CO_{3} | vedclass

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(A) Let the solubility of $Ag_{2}CO_{3}$ in the presence of $0.1 \ M \ Na_{2}CO_{3}$ be $x \ M$.The dissociation equilibrium is: $Ag_{2}CO_{3(s)} \rig

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$10^{-6}$

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(A) Let the solubility of $Ag_{2}CO_{3}$ in the presence of $0.1 \ M \ Na_{2}CO_{3}$ be $x \ M$.The dissociation equilibrium is: $Ag_{2}CO_{3(s)} ightleftharpoons 2 Ag^{+}_{(aq)} + C{O_{3}}^{2-}_{(aq)}$.At equilibrium,the concentration of $Ag^{+}$ is $2x \ M$ and the concentration of $CO_{3}^{2-}$ is $(x + 0.1) \ M$.Since $x$ is very small compared to $0.1$,we can approximate $[CO_{3}^{2-}] \approx 0.1 \ M$.The solubility product expression is: $K_{sp} = [Ag^{+}]^{2} [CO_{3}^{2-}]$.Substituting the values: $4 	imes 10^{-13} = (2x)^{2} 	imes 0.1$.$4 	imes 10^{-13} = 4x^{2} 	imes 10^{-1}$.$x^{2} = 10^{-12}$.$x = 10^{-6} \ M$.

What is the molar solubility of $Ag_{2}CO_{3}$ $(K_{sp} = 4 \times 10^{-13})$ in $0.1 \ M \ Na_{2}CO_{3}$ solution?

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