0.01 mole of ${\mathrm{AgNO}}_3$ is added to 1 litre of a solution which is 0.1 M in ${\mathrm{Na}}_2{\mathrm{CrO}}_4$ and 0.005 M in${\mathrm{NaIO}}_3$. Calculate the millimol of precipitate formed at equilibrium.

$\left(K_{sp}\text{ values of }{\mathrm{Ag}}_2{\mathrm{CrO}}_4\text{ and }{\mathrm{AgIO}}_3\text{ are }{10}^{-8}\text{ and }{10}^{-13}\right.\text{ respectively). }$

$\text{ The }{K}_{sp}\text{ values of }{\mathrm{Ag}}_2{\mathrm{CrO}}_4\text{ and }{\mathrm{AgIO}}_3\text{ reveals that }{\mathrm{CrO}}_4^{2-}\text{ and }I{O}_3^{-}\text{will be }$$\text{ precipitated on addition of }{\mathrm{AgNO}}_3\text{ 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, ${\mathrm{AgIO}}_3$ will be precipitate first.
Now, in order to precipitate ${\mathrm{AgNO}}_3$ one can show:

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

The left mole of ${\mathrm{AgNO}}_3$ are now used to precipitate ${\mathrm{Ag}}_2{\mathrm{CrO}}_4$

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