How many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2?

First, recall the definition of pH: $$\text{pH}= -\log_{10}[H^+]$$, where $$[H^+]$$ is the molar concentration of hydrogen ions in $$\text{mol L}^{-1}$$.

We have an initial aqueous solution of HCl whose pH is $$1$$. Using the definition, the hydrogen-ion concentration is obtained by inverting the logarithm:

$$[H^+]_{\text{initial}} = 10^{-\,\text{pH}} = 10^{-1} = 0.1\ \text{mol L}^{-1}.$$

The volume of this solution is given as $$1\ \text{L}$$. Concentration multiplied by volume gives the actual amount (moles) of hydrogen ions present. Hence,

$$n_{H^+}= [H^+]_{\text{initial}}\times V_{\text{initial}} = 0.1\ \text{mol L}^{-1} \times 1\ \text{L} = 0.1\ \text{mol}.$$

Because only pure water will be added, the number of moles of $$H^+$$ will not change; only the volume will increase. The target pH of the new solution is $$2$$, so its desired hydrogen-ion concentration is

$$[H^+]_{\text{final}} = 10^{-\,2} = 0.01\ \text{mol L}^{-1}.$$

If the final volume after dilution is $$V_{\text{final}}$$ (in litres), the unchanged amount of $$H^+$$ must equal the new concentration multiplied by this new volume. Therefore,

$$n_{H^+}=0.1\ \text{mol}= [H^+]_{\text{final}}\times V_{\text{final}} = 0.01\ \text{mol L}^{-1} \times V_{\text{final}}.$$

Solving for $$V_{\text{final}}$$ gives

$$V_{\text{final}} = \frac{0.1\ \text{mol}}{0.01\ \text{mol L}^{-1}} = 10\ \text{L}.$$

Originally we had $$1\ \text{L}$$. The volume of water that must be added is the difference between the required total volume and the initial volume:

$$V_{\text{water added}} = V_{\text{final}} - V_{\text{initial}} = 10\ \text{L} - 1\ \text{L} = 9\ \text{L}.$$

Hence, the correct answer is Option B.