Question

The $pH$ of $Mg(OH{)}_2$ solution is $10.45$ at ${25}^{\circ }C$ . The solubility product of magnesium hydroxide will be:

• A. $2.24\times {10}^{-11}{M}^3$
• B. $1.12\times {10}^{-11}{M}^3$
• C. $3.36\times {10}^{-11}{M}^3$
• D. $5.60\times {10}^{-12}{M}^3$

Answer 1

The correct option is B $1.12\times {10}^{-11}{M}^3$

The $pH$ of the magnesium hydroxide solution is $10.45$

Hence $pOH=14-10.45$

$=3.55$

$\left[O{H}^{-}\right]=antilog(-3.55)$

$=0.000282$ M.

$\left[M{g}^{2+}\right]=\frac{0.000282}{2}$

$=0.000141$ M.

The expression for the solubility product is $K_{sp}=\left[M{g}^{2+}\right][O{H}^{-}{]}^2$

$=0.000141\times (0.000282{)}^2$

$=1.12\times {10}^{-11}{M}^3$.

Option B is correct.