Q. The solubility in water of a sparingly soluble salt $A_{2}B$is $1.0 \times 10^{- 3}mol\ L^{- 1}$. Its solubility product will be
A) $4 \times 10^{- 9}$
B) $4 \times 10^{9}$
C) $1 \times 10^{9}$
D) $1 \times 10^{- 9}$
Explanation
(a)
$AB_{2} \rightleftharpoons \ A^{+} + \ \ \ \ \ 2B^{-}$
$1 \times 10^{- 3}\ \ 2 \times 10^{- 3}$
$K_{sp} = \left\lbrack 1 \times 10^{- 3} \right\rbrack\left\lbrack 2 \times 10^{- 3} \right\rbrack^{2} = {4 \times 10}^{- 9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.