Question

The $pH$ of a saturated solution of $Mg{\left(OH\right)}_2$ in water will be :

$(K_{sp}=4\times {10}^{-12})$

• A. $4-\log 2$
• B. $10-\log 2$
• C. $4+\log 2$
• D. $10+\log 2$

Answer 1

The correct option is D $10+\log 2$

$Mg(OH{)}_2\to M{g}^{2+}+2O{H}^{-}$

$1$ $-$ $-$

$1-x$ $x$ $2x$

$K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2=x\times (2x{)}^2=4x^3$

$4x^3=4\times {10}^{-12}$

$x=1\times {10}^{-4}$

$2x=2\times {10}^{-4}$

Also, $pOH=-log\left[O{H}^{-}\right]=4-log2$

$pH=14-pOH=10+log2$