The $K_{sp}$ of $Ag_2CrO_4$, $AgCl$, $AgBr$ and $AgI$ are respectively, $1.1 \times 10^{-12}$, $1.8 \times…
The $K_{sp}$ of $Ag_2CrO_4$, $AgCl$, $AgBr$ and $AgI$ are respectively, $1.1 \times 10^{-12}$, $1.8 \times 10^{-10}$, $5.0 \times 10^{-13}$, $8.3 \times 10^{-17}$. Which one of the following salts will precipitate last if $AgNO_3$ solution is added to the solution containing equal moles of NaCl, NaI, NaBr and $Na_2CrO_4$?
AgI
AgCl
AgBr
Solution
For, $\mathrm{Ag}_2 \mathrm{CrO}_4 \rightleftharpoons 2 \mathrm{Ag}^{+}+\mathrm{CrO}_4^{2-}$
solubility product
$\begin{aligned}
& \mathrm{K}_{\text {sp }}=(2s)^2 \times s=4 s^3 \\
& \mathrm{K}_{\text {sp }}=1.1 \times 10^{-12} \\
& \mathrm{S}=\sqrt[3]{\frac{K_{\text {sp }}}{4}}=0.65 \times 10^{-4}
\end{aligned}$
For, $\mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
$\begin{aligned}
& \mathrm{K}_{\text {sp }}=S \times S \quad \left(\mathrm{K}_{\mathrm{sp}}=1.8 \times 10^{-10}\right) \\
& \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=1.34 \times 10^{-5}
\end{aligned}$
For, $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=S \times S \left(\mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13}\right) \\
& \mathrm{S}=\sqrt{K_{s p}}=0.71 \times 10^{-6}
\end{aligned}$
$\begin{aligned}
& \text {For, } \mathrm{AgI} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{I}^{-} \\
& \mathrm{K}_{\text {sp }}=S \times S \left(\mathrm{K}_{\text {sp }}=8.3 \times 10^{-17}\right) \\
& \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}=0.9 \times 10^{-8}
\end{aligned}$
therefore, the solubility of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is highest so it is precipitate last.