An analyst wants to convert $1 \mathrm{L} \mathrm{HCl}$ of $\mathrm{pH}=1$ to a solution of $\mathrm{HCl}$ of $\mathrm{pH}=2$. The volume of water needed to do this dilution is _____ $\mathrm{mL}$. (Nearest integer)

The concentration of ${\mathrm{H}}^{+}$ or say $\mathrm{HCl}$ in both the given solution is ${10}^{–1}$ and ${10}^{–2}$ mole ${\mathrm{L}}^{–1}$ respectively.

For the given amount of $\mathrm{HCl}$ the concentration depends over dilution hence

$$\begin{gathered} {\mathrm{M}}_1{\mathrm{V}}_1={\mathrm{M}}_2{\mathrm{V}}_2 \\ \frac{{\mathrm{M}}_1}{{\mathrm{M}}_2}=\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1} \end{gathered}$$

$\frac{{10}^{-1}}{{10}^{-2}}=\frac{{\mathrm{V}}_2}{1\mathrm{L}}$

${\mathrm{V}}_2=10 \mathrm{L}=10000 \mathrm{mL}$

$∆\mathrm{V}=10000-1000 \mathrm{mL}=9000 \mathrm{mL}$