Question

The $K_{\text{sp }}$ of $A{g}_{2}Cr{O}_4,AgCl,AgBr$ and $AgI$ 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 $N{a}_{2}Cr{O}_4$ ?

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

Answer 1

Answer: (D)

$\therefore A{g}_{2}Cr{O}_4\rightleftharpoons \underset{2S}{2A{g}^{+}}+\underset{S}{Cr{O}_4^{2-}}$
$\therefore$ Solubility product,
$K_{sp}=(2S{)}^2\times S=4S^3,K_{sp}=\left(1.1\times 10^{-12}\right)$
$\Rightarrow S=\sqrt[3]{\frac{K_{sp}}{4}}=0.65\times 10^{-4}$
Now, $AgCl\rightleftharpoons \underset{S}{A{g}^{+}}+\underset{S}{C{l}^{-}}$
$\therefore K_{sp}=S\times S\left(K_{sp}=1.8\times 10^{-10}\right)$
$\Rightarrow S=\sqrt[2]{K_{sp}}=1.34\times 10^{-5}$
$AgBr\rightleftharpoons \underset{S}{A{g}^{+}}+\underset{S}{B{r}^{-}}$
$\therefore K_{\text{sp }}=S\times S\left(K_{\text{sp }}=5.0\times 10^{-13}\right)$
$\Rightarrow S=\sqrt{K_{\text{sp }}}=0.71\times 10^{-6}$
$AgI\rightleftharpoons A{g}_S^{+}+I_S^{-}$
$\therefore K_{sp}=S\times S\left(K_{sp}=8.3\times 10^{-17}\right)$
$\Rightarrow S=\sqrt{K_{sp}}=0.9\times 10^{-8}$
$\because$ Solubility of $A{g}_{2}Cr{O}_4$ is highest.
Therefore, it is added to the solution.