Question

The $K_{\text {sp }}$ of $Ag _{2} CrO _{4}, AgCl , AgBr$ 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 $Na _{2} CrO _{4}$ ?

• A. $Ag I$
• B. $AgCl$
• C. $AgBr$
• D. $Ag _{2} CrO _{4}$

Answer 1

Answer: (D)

$\therefore Ag _{2} CrO _{4} \rightleftharpoons \underset{2 S}{2 Ag ^{+}}+ \underset{S}{CrO _{4}^{2-}}$
$\therefore$ Solubility product,
$K_{ sp }=(2 S)^{2} \times S=4 S^{3}, K_{ sp }=\left(1.1 \times 10^{-12}\right)$
$\Rightarrow S=\sqrt[3]{\frac{K_{ sp }}{4}}=0.65 \times 10^{-4}$
Now, $ AgCl \rightleftharpoons \underset{S}{Ag ^{+}}+ \underset{S}{Cl ^{-}}$
$\therefore K_{ sp }=S \times S \left(K_{ sp }=1.8 \times 10^{-10}\right)$
$\Rightarrow S=\sqrt[2]{K_{ sp }}=1.34 \times 10^{-5}$
$AgBr \rightleftharpoons \underset{S}{Ag^{+}}+\underset{S}{ Br ^{-}}$
$\therefore K_{\text {sp }}=S \times S \left(K_{\text {sp }}=5.0 \times 10^{-13}\right)$
$\Rightarrow S=\sqrt{K_{\text {sp }}}=0.71 \times 10^{-6}$
$AgI \rightleftharpoons Ag _{S}^{+}+ I _{S}^{-}$
$\therefore K_{ sp }=S \times S \left(K_{ sp }=8.3 \times 10^{-17}\right)$
$\Rightarrow S=\sqrt{K_{ sp }}=0.9 \times 10^{-8}$
$\because$ Solubility of $Ag _{2} CrO _{4}$ is highest.
Therefore, it is added to the solution.