The ${\mathrm{K}}_{\mathrm{sp}} \mathrm{of} {\mathrm{Ag}}_2{\mathrm{CrO}}_4, \mathrm{AgCl}, \mathrm{AgBr}$ 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 of the following salts will precipitate last if ${\mathrm{AgNO}}_3$ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and ${\mathrm{Na}}_2{\mathrm{CrO}}_4$

${\mathrm{Ag}}_2{\mathrm{CrO}}_4\rightleftharpoons 2{\mathrm{Ag}}^{+}+{\mathrm{CrO}}_4^{2-}$
t = 0 s 0 0
At equilibrium – 2s s
$\begin{array}{l}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{C}}_2{\mathrm{O}}_4-2\right]\\ {\mathrm{K}}_{\mathrm{sp}}=\left(2\mathrm{s}{)}^2\right(\mathrm{s})\\ {\mathrm{K}}_{\mathrm{sp}}=4{\mathrm{s}}^3\\ \mathrm{l}.1\times {10}^{-12}=4{\mathrm{s}}^3\\ \mathrm{s}=6.5\times {10}^{-5}\\ \mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}\\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{s}}^2\\ \mathrm{s}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}}\\ \mathrm{s}=\sqrt{1.8\times {10}^{-10}}\\ \mathrm{s}=1.34\times {10}^{-5}\end{array}$
Similarly, for AgBr and AgI, solubility is 7.1 x 10^–7 and 9 x 10^–9, respectively. Since solubility of Ag₂CrO₄ is highest, it will precipitate last.