Question

Solubility product of ${Mg\left(OH\right)}_2$ at ordinary temperature is $1.96\times {10}^{-11}$. $pH$ of a saturated solution of ${Mg\left(OH\right)}_2$ will be

• A. 10.53
• B. 8.47
• C. 6.94
• D. 3.47

Answer 1

The correct option is A 10.53

${Mg\left(OH\right)}_2\rightleftharpoons M{\underset{S}{g}}^{2+}+2{\underset{2S}{OH}}^{-}$

$K_{sp}{Mg\left(OH\right)}_2=\left[{Mg}^{2+}\right]{\left[{OH}^{-}\right]}^2$

$\Rightarrow K_{sp}{Mg\left(OH\right)}_2=S\times {2S}^2=4S^3$

$1.96\times {10}^{-11}=4S^3$

or $S={\left[\frac{1.96\times {10}^{-11}}{4}\right]}^{1/3}$

or $S={(4.9\times {10}^{-12})}^{1/3}$

$\therefore S=1.69\times {10}^{-4}$

So, concentration of $\left[{OH}^{-}\right]=2S$

$\therefore \left[{OH}^{-}\right]=3.38\times {10}^{-4}$

$\Rightarrow pOH=-\log \left[{OH}^{-}\right]$

$=-\log [3.38\times {10}^{-4}]$

$\therefore pOH=3.471$

$pH=14-pOH$

$pH=14-3.471$

$\therefore pH=10.529$