The solubility of solid silver chromate $A{g}_{2}Cr{O}_4$ , is determined in three solvents substance $K_{sp}$   of $A{g}_{2}Cr{O}_4=9\times {10}^{-12}$  .

I) Pure water         II) 0.1M $AgN{O}_3$              III) 0.1M  $N{a}_{2}Cr{O}_4$
Predict the relative solubility of  $A{g}_{2}Cr{O}_4$  in I , II and III. 
 

Solubility in water is more than remaining two because, as common ion effect increases solubility decreases
$A{g}_{2}Cr{O}_4\stackrel{ }{\rightleftharpoons }\underset{2s}{2A{g}^{+}}+\underset{s}{Cr{O}_4^{-2}}$ 
(i) 0.1M $A{g}^{+}$  ions is common ion
 $9\times {10}^{-12}={\left(0.1\right)}^2\times s\Rightarrow s=9\times {10}^{-10}M$
(ii) 0.1M  $Cr{O}_4^{-2}$  ions is common ion

$$\begin{gathered} 9\times {10}^{-12}={\left(2s\right)}^2\left(0.1\right) \\ 9\times {10}^{-12}=4s^2\left(0.1\right) \\ s=\sqrt{\frac{9}{4}\times {10}^{-11}}\simeq 4.5\times {10}^{-6}M \end{gathered}$$