Q. Solubility of $Ca(OH)_{2}$ is a $s\ mol\ L^{- 1}$. The solubility product $(K_{sp})$ under the same condition is
A) $4s^{3}$
B) $3s^{4}$
C) $4s^{2}$
D) $s^{3}$
Explanation
(a)
$Ca(OH)_{2} \rightleftharpoons {Ca}^{2 +} + 2{OH}^{-}$
$K_{sp} = \left\lbrack {Ca}^{2 +} \right\rbrack\left\lbrack {OH}^{-} \right\rbrack^{2}$
$= (s)(2s)^{2} = 4s^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.