Q. The solubility product of $Ag_{2}CrO_{4}$ in water at 298 K is $3.2 \times 10^{- 11}$. What will be the concentration of $CrO_{4}^{2 -}$ ions in the saturated solution of $Ag_{2}CrO_{4}$?
A) $2 \times 10^{- 4}$ M
B) $5.7 \times 10^{- 5}$ M
C) $5.7 \times 10^{- 6}$ M
D) $3.2 \times 10^{- 11}\$ M
Explanation
(a)
For saturated solution of $Ag_{2}CrO_{4}$, if solubility is ‘s'$\ mol\ L^{- 1}$. Then
$Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+}(aq) + CrO_{4}^{2 -}(aq)$
$s$ $2s$ $s$
$K_{sp} = (2s)^{2}(s) = 4s^{3}$
$K_{sp} = 3.2 \times 10^{- 11}$ (given)
$\therefore$ $3.2 \times 10^{- 11} = 4s^{3}$
$s^{3} = \frac{3.2 \times 10^{- 11}}{4} = 8 \times 10^{- 12}$
$\therefore$ $s = \sqrt{8 \times 10^{- 12}} = 2 \times 10^{- 4}$ M
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.