The solubility product of Ag_2CO_3 and FeCO_3 | vedclass

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(B) For $Ag_2CO_3$: The dissociation is $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.Let solubility be $S_1$. Then $K_{sp1} = (2S_1)^2(S

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$20 : 1$

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(B) For $Ag_2CO_3$: The dissociation is $Ag_2CO_3(s) ightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.Let solubility be $S_1$. Then $K_{sp1} = (2S_1)^2(S_1) = 4S_1^3$.Given $K_{sp1} = 4 	imes 10^{-12}$,so $4S_1^3 = 4 	imes 10^{-12} \implies S_1^3 = 10^{-12} \implies S_1 = 10^{-4} \, M$.For $FeCO_3$: The dissociation is $FeCO_3(s) ightleftharpoons Fe^{2+}(aq) + CO_3^{2-}(aq)$.Let solubility be $S_2$. Then $K_{sp2} = (S_2)(S_2) = S_2^2$.Given $K_{sp2} = 2.5 	imes 10^{-11} = 25 	imes 10^{-12}$,so $S_2^2 = 25 	imes 10^{-12} \implies S_2 = 5 	imes 10^{-6} \, M$.Now,$S_1/S_2 = (10^{-4}) / (5 	imes 10^{-6}) = 100 / 5 = 20$.Thus,$S_1/S_2 = 20 : 1$.

The solubility product of $Ag_2CO_3$ and $FeCO_3$ in water at $25 \, ^\circ C$ are $4 \times 10^{-12}$ and $2.5 \times 10^{-11}$ respectively. If $S_1$ and $S_2$ are their solubilities in water respectively when dissolved separately,then $S_1/S_2$ is

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