Question

The $K_{sp}$ for $Mg(OH{)}_2$ in water is $1.2\times {10}^{-11}$. In an acidic solution, if the $M{g}^{2+}$ concentration is $1.2\times {10}^{-5}mol/L$, what is the pH at which $Mg(OH{)}_2$ just begins to precipitate?

• A. 3
• B. 4
• C. 5
• D. 11
• E. 12

Answer 1

Answer: (D)

11

$K_{sp}=\left[M{g}^{2+}\right]\left[O{H}^{-}\right]$
$K_{sp}=1.2\times {10}^{-11}$
$\left[M{g}^{2+}\right]=1.2\times {10}^{-5}$
$1.2\times {10}^{-11}=1.2\times {10}^{-5}\times [O{H}^{-}{]}^2$
$[O{H}^{-}{]}^2=\frac{1.2\times {10}^{-11}}{1.2\times {10}^{-5}}$

$=1.0\times {10}^{-6}$
$\left[O{H}^{-}\right]=\sqrt{1.0\times {10}^{-6}}$

$=1.0\times {10}^{-3}$
$pOH=-log\left[O{H}^{-}\right]$

$=-log1.0\times {10}^{-3}$

$=3$
$pH=14-POH$

$=14-3$

$=11$