Q. The solubility in water of a sparingly soluble salt $AB_{2}$ is $1 \times 10^{- 5}\ mol\ L^{- 1}$. Its solubility product number will be
A) $4 \times 10^{- 15}$
B) $4 \times 10^{- 10}$
C) $1 \times 10^{- 15}$
D) $1 \times 10^{- 10}$
Explanation
(a)
$AB_{2} \rightleftharpoons A^{2 +} + 2B^{-}$
$s$ $2s$
$K_{sp} = \left\lbrack A^{2 +} \right\rbrack\left\lbrack B^{-} \right\rbrack^{2}$
$= (S)(2S)^{2} = 4S^{3}$
$= 4\left( 1 \times 10^{- 5} \right)^{3}$
$= 4 \times 10^{- 15}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.