Explanation

(a)
$Mg(OH)_{2} \rightleftharpoons {Mg}^{2 +} + 2{OH}^{-}$
$s$ $2s$
$K_{sp}Mg(OH)_{2} = \left\lbrack {Mg}^{2 +} \right\rbrack\left\lbrack {OH}^{-} \right\rbrack^{2}$
$\Rightarrow$ $K_{sp}\ Mg(OH)_{2} = 4S^{3}$
$1.96 \times 10^{- 11} = 4S^{3}$
or $S = \left\lbrack \frac{1.96 \times 10^{- 11}}{4} \right\rbrack^{\frac{1}{3}}$
or $S = \left( 4.9 \times 10^{- 12} \right)^{\frac{1}{3}}$
$\therefore$ $S = 1.96 \times 10^{- 4}$
So, concentration of $\left\lbrack OH^{-} \right\rbrack = 2S$
$\therefore$ $\left\lbrack OH^{-} \right\rbrack = 3.38 \times 10^{- 4}$
$\Rightarrow$ $pOH = - \log\left\lbrack {OH}^{-} \right\rbrack$
$= - \log\left\lbrack 3.38 \times 10^{- 4} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ pOH = 3.471$
$pH = 14 - pOH$
$= 14 - 3.471$
$\therefore$ pH=10.529