Q. The solubility product of a salt having general formula $MX_{2}$, in water is $4 \times 10^{- 12}$. The concentration of $M^{2 +}$ ions in the aqueous solution of the salt is
A) $4.0 \times 10^{- 10}$ M
B) $1.6 \times 10^{- 4}$ M
C) $1.0 \times 10^{- 4}$ M
D) $2.0 \times 10^{- 6}$ M
Explanation
(c)
$MX_{2} \rightleftharpoons M^{2 +} + 2X^{-}$
$s$ $2s$
$K_{sp} = \left\lbrack M^{2 +} \right\rbrack\left\lbrack X^{-} \right\rbrack^{2}$
If solubility be $s$ then
$K_{sp} = (s)(2s)^{2} = 4s^{3}$
$4s^{3} = 4 \times 10^{- 12}$
$\therefore$ $s = 1 \times 10^{- 4}$ M
$\therefore$ $M^{2 +} = s = 1 \times 10^{- 4}$M
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.