Question

The solubility of $A{g}_{2}Cr{O}_4$ at ${25}^{\circ }C$ is $8.0\times {10}^{-5}$ moles/litre. Calculate its solubility product

Answer 1

$A{g}_{2}Cr{O}_4\rightleftharpoons \underset{2S}{2A{g}^{+}}+\underset{S}{Cr{O}_4^{2-}}$
$K_{sp\left(A{g}_{2}Cr{O}_4\right)}=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{2-}]$
$=\left[2S{]}^2\right[S]$
$=4S^3$
$=4\times (8.0\times {10}^{-5}{)}^3$
$K_{sp}=2.048\times {10}^{-12}$
$K_{sp}$ for $K_{2}Cr{O}_4$ at ${25}^{\circ }C$ will be $2.048\times {10}^{-12}$.