Question

The concentration of ${Ag}^{+}$ ion in a saturated solution of ${Ag}_{2}Cr{O}_4$ at ${20}^{o}C$ is $1.5\times {10}^{-4}mol{L}^{-1}$. Determine the solubility product of ${Ag}_{2}Cr{O}_4$ at ${20}^{o}C$.

Answer 1

The equilibrium is:
${Ag}_{2}Cr{O}_4\rightleftharpoons 2{Ag}^{+}+Cr{O}_4^{2-}$
On the basis of this equation, the concentration of $Cr{O}_4^{2-}$ ion will be half of the concentration of ${Ag}^{+}$ ions.
Thus, $\left[{Ag}^{+}\right]=1.5\times {10}^{-4}M$
and $\left[Cr{O}_4^{2-}\right]=0.75\times {10}^{-4}M$
$={\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{2-}\right]=1.6875\times {10}^{-12}$