0.023 g of sodium metal is reacted with 100 cm³ of water. The pH of the resulting solution is

$$\begin{gathered} 2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}⟶2\mathrm{NaOH}+{\mathrm{H}}_2 \\ 2 \mathrm{mol} 2 \mathrm{mol} 2 \mathrm{mol} \end{gathered}$$
$$\begin{gathered} \text{ For }\mathrm{Na},\mathrm{n}=\frac{0.023}{23}={10}^{-3}\mathrm{mol} \\ \because therefore the no.of moles of NaOH is also {10}^{-3} \\ molarity = \frac{no.of moles of solute}{volume of solution in litres} \\ {P}^{OH}=2, hence P^H= 12 \end{gathered}$$