An excess of Ag_{2}CrO_{4(s)} is added to a 5 | vedclass

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(B) The dissolution of $Ag_{2}CrO_{4}$ is given by the equilibrium: $Ag_{2}CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + Cr{O_{4}}^{2-}_{(aq)}$.The sol

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$1.5 	imes 10^{-5} \ M$

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(B) The dissolution of $Ag_{2}CrO_{4}$ is given by the equilibrium: $Ag_{2}CrO_{4(s)} ightleftharpoons 2Ag^+_{(aq)} + Cr{O_{4}}^{2-}_{(aq)}$.The solubility product expression is $K_{sp} = [Ag^+]^2 [CrO_{4}^{2-}]$.Given $K_{sp} = 1.1 	imes 10^{-12}$ and the common ion concentration $[CrO_{4}^{2-}] = 5 	imes 10^{-3} \ M$.Substituting the values: $1.1 	imes 10^{-12} = [Ag^+]^2 (5 	imes 10^{-3})$.$[Ag^+]^2 = \frac{1.1 	imes 10^{-12}}{5 	imes 10^{-3}} = 0.22 	imes 10^{-9} = 2.2 	imes 10^{-10}$.$[Ag^+] = \sqrt{2.2 	imes 10^{-10}} \approx 1.48 	imes 10^{-5} \ M$.This is closest to $1.5 	imes 10^{-5} \ M$.

An excess of $Ag_{2}CrO_{4(s)}$ is added to a $5 \times 10^{-3} \ M$ $K_{2}CrO_{4}$ solution. The concentration of $Ag^{+}$ in the solution is closest to
[ Solubility product for $Ag_{2}CrO_{4} = 1.1 \times 10^{-12}$ ]

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An excess of $Ag_{2}CrO_{4(s)}$ is added to a $5 \times 10^{-3} \ M$ $K_{2}CrO_{4}$ solution. The concentration of $Ag^{+}$ in the solution is closest to[ Solubility product for $Ag_{2}CrO_{4} = 1.1 \times 10^{-12}$ ]