Q. At $25℃$, the solubility product of $Mg(OH)_{2}$ is $1.0\ \ 10^{\ 11}$. At which pH, will $Mg^{2}$ ions start precipitating in the form of $Mg(OH)_{2}$ from a solution of $0.001\ M\ Mg^{2}$ ions?
A) 9
B) 10
C) 11
D) 8
Explanation
(b)
$Mg(OH)_{2} \rightleftharpoons {Mg}^{2 +} + 2{OH}^{-}$
$K_{sp} = \left\lbrack {Mg}^{2 +} \right\rbrack\left\lbrack {OH}^{-} \right\rbrack^{2}$
${\lbrack OH}^{-}\rbrack = \sqrt{\frac{K_{sp}}{\left\lbrack {Mg}^{2 +} \right\rbrack}}$
$\ {\lbrack OH}^{-}\rbrack = \sqrt{\frac{1.0 \times 10^{- 11}}{0.001}} = 10^{- 4}$
$pOH$=4 and pH=10
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.