Question

The solubility product of $Mg(OH)_2$ is $1.0\times 10^{-12}$. Concentrated aqueous $NaOH$ solution is added to a $0.01 \,M$ aqueous solution of $MgCl_2$. The $pH$ at which precpitation occurs is

• A. 7.2
• B. 7.8
• C. 8.0
• D. 9.0

Answer 1

Answer: (D)

For the reaction,
$Mg(OH)_2 {\rightleftharpoons} Mg^{2+} + 2OH^{-}$
$K_{sp}=\left[Mg^{2+}\right]\left[OH^{-}\right]^{2}$
$10^{-12}=\left(0.01\right)\left[OH^{-}\right]^{2}$
$\frac{10^{-12}}{10^{-2}}=\left[OH^{-}\right]^{2}$
$\therefore \left[OH^{-}\right]=10^{-5}$
$\Rightarrow pOH=+5$
Also, $pH + pOH = 14$
$pH = 14 - pOH$
$14 - 5 = 9$