\(0.023 \mathrm{~g}\) of sodium metal is reacted with \(100 \mathrm{~cm}^{3}\) of water. The \(\mathrm{pH}\) of the resulting solution is

Given, \(\frac{0.023}{23}\) mol \(\frac{100}{22400} \mathrm{~mol}\) \(=1 \times 10^{-3} \mathrm{~mol}=4.46 \times 10^{-3} \mathrm{~mol}\)
Thus, Na is the limiting reagent and decide the amount of \(\mathrm{NaOH}\) formed.
\(\because 1\) mole Na give \(\mathrm{NaOH}=1 \mathrm{~mol}\)
\(\therefore 1 \times 10^{-3}\) mole Na will give \(\mathrm{NaOH}\) \(=1 \times 10^{-3} \mathrm{~mol}\)
Concentration of
\(
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } &=\frac{1 \times 10^{-3} \times 1000}{100}=1 \times 10^{-2} \\
\mathrm{pOH} &=-\log \left[\mathrm{OH}^{-}\right] \\
&=-\log \left(1 \times 10^{-2}\right) \\
&=2 \\
\mathrm{pH} &=14-2=12
\end{aligned}
\)