Question

At $20^{\circ} C$ the $Ag ^{+}$ion concentration in a saturated solution of $Ag _{2} CrO _{4}$ is $1.5 \times 10^{-4} mol / L$. At $20^{\circ} C$ the solubility product of $Ag _{2} CrO _{4}$ would be

• A. $ 3.3750\times 10^{-12} $
• B. $ 1.6875\times 10^{-10} $
• C. $ 1.6875\times 10^{-12} $
• D. $ 1.6875\times 10^{-11} $

Answer 1

Answer: (C)

Given, $2\left[ Ag ^{+}\right]=1.5 \times 10^{-4} mol / L$
$ \left[ Ag ^{+}\right]=0.75 \times 10^{-4}$
$\left[ Ag ^{+}\right]=\left[ CrO _{4}^{2-}\right]=0.75 \times 10^{-4} mol / L$
$Ag _{2} CrO _{4} \rightleftharpoons 2 Ag ^{+}+ CrO _{4}^{2-}$
$ K _{s p}=\left[ Ag ^{+}\right]^{2} \cdot\left[ CrO _{4}^{2-}\right]$
$K _{s p}=\left(1.5 \times 10^{-4}\right)^{2} \times\left(0.75 \times 10^{-4}\right)$
$K _{s p}=2.25 \times 10^{-8} \times 0.75 \times 10^{-4} $
$K _{s p}=1.6875 \times 10^{-12}$