Explanation:
Given that,
$\begin{aligned}
& \mathrm{pH}=2,\left[\mathrm{H}^{+}\right]=10^{-2} \mathrm{M}, \mathrm{V}_1=1 \text { litre } \\
& \mathrm{pH}=4,\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{M}, \mathrm{V}_2=\text { ? } \\
& \mathrm{M}_1 \mathrm{~V}_1=\mathrm{M}_2 \mathrm{~V}_2 \\
& 10^{-2} \times 1=10^{-4} \times \mathrm{V}_2 \\
& \\
& \mathrm{~V}_2=\frac{10^{-2}}{10^{-4}}=100 \text { litre }
\end{aligned}$
Therefore, 99 litre of water is added.