The ${\mathrm{K}}_{\mathrm{sp}}\text{ of }{\mathrm{Ag}}_2{\mathrm{CrO}}_4,\mathrm{AgCl},\mathrm{AgBr}\text{ and }\mathrm{AgI}$ are respectively, $1.1\times {10}^{-12},1.8\times {10}^{-10},5.0\times {10}^{-23} 8.3\times {10}^{-17}$. Which one solution is added to the solution containing equal moles of NaCl, NaI and ${\mathrm{Na}}_2{\mathrm{CrO}}_4$?

${\mathrm{Ag}}_2{\mathrm{CrO}}_4\rightleftharpoons 2{\mathrm{Ag}}^{+}+{\mathrm{CrO}}_4^{2-}$

Solubility product,

${\mathrm{K}}_{\mathrm{sp}}=(2\mathrm{S}{)}^2\times \mathrm{S}=4{\mathrm{S}}^3,{\mathrm{K}}_{\mathrm{sp}}=\left(1.1\times {10}^{-12}\right)$

$\mathrm{S}=\sqrt[3]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{4}}=0.65\times {10}^{-4}$

$\mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}$

${\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}\times \mathrm{S} \left({\mathrm{K}}_{\mathrm{sp}}=1.8\times {10}^{-10}\right)$

$\mathrm{S}=\sqrt[3]{{\mathrm{K}}_{\mathrm{sp}}}=1.34\times {10}^{-5}$

$\mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Br}}^{-}$

${\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}\times \mathrm{S}$

$\mathrm{S}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=0.71\times {10}^{-6}$

$\mathrm{AgI}\rightleftharpoons {\mathrm{Ag}}^{+}+\mathrm{I}$

${\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}\times \mathrm{S} \left({\mathrm{K}}_{\mathrm{sp}}=83\times {10}^{-17}\right)$

$\text{S}\sqrt{{\mathrm{K}}_{\mathrm{sp}}}=0.9\times {10}^{-8}$

$\because \text{ Solubility of }{\mathrm{Ag}}_2{\mathrm{CrU}}_4\text{ is highest. }$

So, it will precipitate last.