An aqueous solution of HCl initially has a pH of 1.0, meaning the hydrogen ion concentration is $[\text{H}^+]=10^{-1} \, \text{M}$.

When we dilute the solution by adding an equal volume of water, the concentration of hydrogen ions will be halved:

$ [\text{H}^+]_{\text{new}} = \frac{10^{-1}}{2} = 5 \times 10^{-2} \, \text{M} $

To find the new pH, we use the pH formula:

$ \text{pH} = -\log[\text{H}^+] $

Substituting the new hydrogen ion concentration gives:

$ \text{pH} = -\log(5 \times 10^{-2}) $

Using the logarithm property $\log(a \times b) = \log a + \log b$:

$ \text{pH} = -(\log 5 + \log 10^{-2}) = -\log 5 + 2 $

Given $\log 2 = 0.30$, we need $\log 5$, which can be calculated as:

$ \log 10 = \log(2 \times 5) = \log 2 + \log 5 \Rightarrow \log 5 = \log 10 - \log 2 = 1 - 0.30 = 0.70 $

Thus:

$ \text{pH} = -(0.70) + 2 = 1.3 $

Therefore, the pH of the solution increases to 1.3.