0.01 mole of $AgN{O}_3$ is added to 1 litre of a solution which is 0.1 M in $N{a}_{2}Cr{O}_4$ and 0.005 M in 0.005 M in $NaI{O}_3.$ Calculate the millimoles of precipitate formed at equilibrium. ($K_{sp}$

 values of $A{g}_{2}Cr{O}_4$ and $AgI{O}_3$ are ${10}^{-8}$ and ${10}^{-13}$ respectively).

The $K_{sp}$ values of $A{g}_{2}Cr{O}_4$ and $AgI{O}_3$ reveals that $Cr{O}_4^{2-}$  and $I{O}_3^{-}$ will be precipitated on addition of $AgN{O}_3$ as : 

$\left[A{g}^{+}\right]\left[I{O}_3^{-}\right]={10}^{-13}$

${\left[A{g}^{+}\right]}_{needed}=\frac{{10}^{-13}}{\left[0.005\right]}=2\times {10}^{-11}$

${\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{2-}\right]={10}^{-8}$

${\left[A{g}^{+}\right]}_{needed}=\sqrt{\frac{{10}^{-8}}{0.1}}=3.16\times {10}^{-4}$

Thus,$AgI{O}_3$ will be precipitate first.

         $\underset{\begin{array}{l}0.01\\ \\ 0.005\end{array}}{AgN{O}_3}+\underset{\begin{array}{l}0.005\\ \\ 0\end{array}}{NaI{O}_3}\stackrel{ }{\to }\underset{\begin{array}{l}0\\ \\ 0.005\end{array}}{AgI{O}_3}+\underset{\begin{array}{l}0\\ \\ 0.005\end{array}}{NaN{O}_3}$

The left mole of $AgN{O}_3$ are now used to precipitate $A{g}_{2}Cr{O}_4$

$\underset{\begin{array}{l}0.005\\ \\ 0\end{array}}{2AgN{O}_3}+\underset{\begin{array}{l}0.1\\ \\ 0.0975\end{array}}{N{a}_{2}Cr{O}_4}\stackrel{ }{\to }\underset{\begin{array}{l}0\\ \\ 0.0025\end{array}}{A{g}_{2}Cr{O}_4}+\underset{\begin{array}{l}0\\ \\ 0.005\end{array}}{2NaN{O}_3}$

a total of moles formed is 0.0075 moles