A saturated solution of ${\text{Ag}}_2{\text{CrO}}_4$ has a solubility product $K_{sp}=4\times 10^{-12}$. Calculate the solubility of ${\text{Ag}}_2{\text{CrO}}_4$ in a $0.1\text{ M}$ solution of ${\text{AgNO}}_3$.

Solution

1. The solubility equilibrium is ${\text{Ag}}_2{\text{CrO}}_4(s)\rightleftharpoons 2{\text{Ag}}^{+}(aq)+{\text{CrO}}_4^{2-}(aq)$.
2. The concentration of ${\text{Ag}}^{+}$ is dominated by ${\text{AgNO}}_3$, so $[{\text{Ag}}^{+}]\approx 0.1\text{ M}$.
3. $$\begin{array}{rl}{K}_{sp}& =[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]\\ & =(0.1{)}^2\times S\\ & =4\times 10^{-12}\end{array}$$.
4. $$\begin{array}{rl}S& =\frac{4\times 10^{-12}}{0.01}\\ & =4\times 10^{-10}\text{ M}\end{array}$$.
5. Hence the answer is (A).