At 20 , ^circ C,the Ag^+ ion concentration in | vedclass

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(C) The dissociation of silver chromate is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$The solubility product expression is:

Vedclass · Accepted Answer

$1.6875 	imes 10^{-12}$

Vedclass · Answer

(C) The dissociation of silver chromate is given by: $Ag_2CrO_4(s) ightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$The solubility product expression is: $K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$Given that $[Ag^+] = 1.5 	imes 10^{-4} \, M$,the concentration of chromate ions is half the concentration of silver ions due to stoichiometry:$[CrO_4^{2-}] = \frac{[Ag^+]}{2} = \frac{1.5 	imes 10^{-4}}{2} = 0.75 	imes 10^{-4} \, M$Substituting these values into the $K_{sp}$ expression:$K_{sp} = (1.5 	imes 10^{-4})^2 	imes (0.75 	imes 10^{-4})$$K_{sp} = (2.25 	imes 10^{-8}) 	imes (0.75 	imes 10^{-4})$$K_{sp} = 1.6875 	imes 10^{-12}$

At $20 \, ^\circ C$,the $Ag^+$ ion concentration in a saturated solution of $Ag_2CrO_4$ is $1.5 \times 10^{-4} \, mol/L$. At $20 \, ^\circ C$,the solubility product of $Ag_2CrO_4$ would be:

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