Q. The solubility product of $Ag_{2}CrO_{4}$is $32 \times 10^{- 12}$. What is the concentration of $CrO_{4}^{-}$ ions in that solution?
A) $2 \times 10^{- 4}M$
B) $16 \times 10^{- 4}M$
C) $8 \times 10^{- 4}M$
D) $8 \times 10^{- 8}M$
Explanation
(a)
$Ag_{2}CrO_{4} \rightarrow 2Ag^{+} + CrO_{4}^{2 -}$
$s\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2s\ \ \ \ \ \ \ \ \ \ \ \ s$
$K_{sp} = (2s)^{2}s = 4s^{3}$
$s = \left( \frac{K_{sp}}{4} \right)^{1/3} = \left( \frac{32 \times 10^{- 12}}{4} \right)^{1/3} = 2 \times 10^{- 4}M$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.