Explanation

(d)
$2Na\ \ \ + 2H_{2}O\ \rightarrow 2NaOH + H_{2}$
2 mol 2 mol 2 mol
Given, $\frac{0.023}{23}\ mol\ \frac{100}{22400}\ mol$
$= 1 \times 10^{- 3}mol = 4.46 \times 10^{- 3}\ mol$
Thus, $Na$ is the limiting reagent and decide the amount of $NaOH$ formed.
$\because 1\ mole\ Na\ give\ NaOH = 1\ mol$
$\therefore 1 \times 10^{- 3}mole\ Na\ will\ give\ NaOH$
$\ \ \ \ \ \ = 1 \times 10^{- 3}\ mol$
Concentration of
$\lbrack OH^{-}\rbrack = \frac{1 \times 10^{- 3} \times 1000}{100} = 1 \times 10^{- 2}$
$pOH = - \log{\lbrack OH^{-}\rbrack}$
$= - \log{(1 \times 10^{- 2})}$
$= 2$
$pH = 14 - 2 = 12$