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$0.023 \mathrm{~g}$ of sodium metal is reacted with $100 \mathrm{~cm}^{3}$ of water. The $\mathrm{pH}$ of the resulting solution is
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Verified Answer
The correct answer is:
12
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
$$
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } &=\frac{1 \times 10^{-3} \times 1000}{100}=1 \times 10^{-2} \\
\mathrm{pOH} &=-\log \left[\mathrm{OH}^{-}\right] \\
&=-\log \left(1 \times 10^{-2}\right) \\
&=2 \\
\mathrm{pH} &=14-2=12
\end{aligned}
$$
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