At 20,^oC,the concentration of Ag^{+} in a sa | vedclass

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(C) The dissociation of $Ag_2CrO_4$ is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$.Let the solubility of $Ag_2CrO_4$ be $

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$1.6875 	imes 10^{-12}$

Vedclass · Answer

(C) The dissociation of $Ag_2CrO_4$ is given by: $Ag_2CrO_4(s) ightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$.Let the solubility of $Ag_2CrO_4$ be $s$. Then $[Ag^{+}] = 2s$ and $[CrO_4^{2-}] = s$.Given $[Ag^{+}] = 1.5 	imes 10^{-4}\,mol\,L^{-1}$.Since $[Ag^{+}] = 2s$,we have $s = \frac{1.5 	imes 10^{-4}}{2} = 0.75 	imes 10^{-4}\,mol\,L^{-1}$.The solubility product $K_{sp}$ is defined as: $K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}] = (2s)^2 (s) = 4s^3$.Substituting the values: $K_{sp} = (1.5 	imes 10^{-4})^2 	imes (0.75 	imes 10^{-4})$.$K_{sp} = (2.25 	imes 10^{-8}) 	imes (0.75 	imes 10^{-4}) = 1.6875 	imes 10^{-12}$.

At $20\,^oC$,the concentration of $Ag^{+}$ in a saturated solution of $Ag_2CrO_4$ is $1.5 \times 10^{-4}\,mol\,L^{-1}$. The solubility product of $Ag_2CrO_4$ at $20\,^oC$ is .......

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