The solubility product of Ag_{2}CrO_{4} is 32 | vedclass

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(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+} + CrO_{4}^{2-}$.Let the solubility be $S \ M$. Then,$[A

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$2 	imes 10^{-4} \ M$

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(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} ightleftharpoons 2Ag^{+} + CrO_{4}^{2-}$.Let the solubility be $S \ M$. Then,$[Ag^{+}] = 2S$ and $[CrO_{4}^{2-}] = S$.The solubility product expression is $K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}]$.Substituting the values: $K_{sp} = (2S)^{2}(S) = 4S^{3}$.Given $K_{sp} = 32 	imes 10^{-12}$,we have $4S^{3} = 32 	imes 10^{-12}$.$S^{3} = 8 	imes 10^{-12}$.$S = \sqrt[3]{8 	imes 10^{-12}} = 2 	imes 10^{-4} \ M$.Since $[CrO_{4}^{2-}] = S$,the concentration of $CrO_{4}^{2-}$ ions is $2 	imes 10^{-4} \ M$.

The solubility product of $Ag_{2}CrO_{4}$ is $32 \times 10^{-12}$. What is the concentration of $CrO_{4}^{2-}$ ions in that solution?

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