Question

The $K_{sp}$ of $A{g}_{2}Cr{O}_4,AgCl,AgBr$ and $Agl$ 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 $AgN{O}_3$ solution is added to the solution containing equal moles of $NaCl,NaBr,NaI$ and $N{a}_{2}Cr{O}_4$?

• A. $AgBr$
• B. $A{g}_{2}Cr{O}_4$
• C. $AgI$
• D. $AgCl$

Answer 1

Answer: (B)

$A{g}_{2}Cr{O}_4$

The $K_{sp}$ of $A{g}_{2}Cr{O}_4,AgCl,AgBr$ and $Agl$ are respectively, $1.1\times {10}^{-12},1.8\times {10}^{-10},5.0\times {10}^{-13},8.3\times {10}^{-17}$.
Assume that $1\times {10}^{-5}$ M solutions of $NaCl,NaBr,NaI$ and $N{a}_{2}Cr{O}_4$ are added.
Calculate the molarity of $AgN{O}_3$ solution (that should be reached in the solution) to affect the precipitation.

For $A{g}_{2}Cr{O}_4$, $\left[A{g}^{+}\right]=\sqrt{\frac{K_{sp}}{\left[Cr{O}_4^{2-}\right]}}=\sqrt{\frac{1.1\times {10}^{-12}}{{10}^{-5}}}=3.3\times {10}^{-4}$ M.

For $AgCl$, $\left[A{g}^{+}\right]=\frac{K_{sp}}{\left[C{l}^{-}\right]}=\frac{1.8\times {10}^{-10}}{{10}^{-5}}=1.8\times {10}^{-5}$ M.

For $AgBr$, $\left[A{g}^{+}\right]=\frac{K_{sp}}{\left[B{r}^{-}\right]}=\frac{5.0\times {10}^{-13}}{{10}^{-5}}=5.0\times {10}^{-8}$ M.

For $AgI$, $\left[A{g}^{+}\right]=\frac{K_{sp}}{\left[I^{-}\right]}=\frac{8.3\times {10}^{-17}}{{10}^{-5}}=8.3\times {10}^{-12}$ M.

Here concentration of $A{g}^{+}$ in $A{g}_{2}Cr{O}_4$ is highest.
Hence, $A{g}_{2}Cr{O}_4$ will precipitate last.