At 20^{circ}C,the concentration of Ag^{+} ion | 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 be $S \ mol/L$. Then

Vedclass · Accepted Answer

$1.6875 	imes 10^{-12}$

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 be $S \ mol/L$. Then $[Ag^+] = 2S$ and $[CrO_4^{2-}] = S$.Given $[Ag^+] = 2S = 1.5 	imes 10^{-4} \ mol/L$.Therefore,$S = \frac{1.5 	imes 10^{-4}}{2} = 0.75 	imes 10^{-4} \ mol/L$.The solubility product expression is $K_{sp} = [Ag^+]^2 [CrO_4^{2-}] = (2S)^2 (S) = 4S^3$.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 $20^{\circ}C$,the concentration of $Ag^{+}$ ions in a saturated solution of $Ag_2CrO_4$ is $1.5 \times 10^{-4} \ mol/L$. The solubility product $(K_{sp})$ of $Ag_2CrO_4$ at $20^{\circ}C$ is:

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