An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would
(Given log 2 = 0.30)

For an aqueous solution, $$\text{pH} = -\log_{10}[H^{+}]$$.

Initial pH of the HCl solution is given as $$1.0$$.
Therefore the initial hydrogen-ion concentration is $$[H^{+}]_1 = 10^{-1}\,\text{M} = 0.1\,\text{M}$$.

The solution is now diluted by adding an equal volume of water.
For any solute, doubling the total volume (while keeping the amount of solute the same) halves its concentration:

$$[H^{+}]_2 = \frac{[H^{+}]_1}{2} = \frac{0.1}{2} = 0.05\,\text{M}$$.

Calculate the new pH:

$$\text{pH}_2 = -\log_{10}(0.05)$$.

Write $$0.05$$ as $$5 \times 10^{-2}$$, then use the logarithm law $$\log(ab) = \log a + \log b$$:

$$\text{pH}_2 = -\left[\log_{10}(5) + \log_{10}(10^{-2})\right]$$.

Since $$\log_{10}(10^{-2}) = -2$$, we get

$$\text{pH}_2 = -\left[\log_{10}(5) - 2\right] = 2 - \log_{10}(5).$$

Using $$\log_{10}(5) \approx 0.699$$,

$$\text{pH}_2 \approx 2 - 0.699 = 1.301 \approx 1.3.$$

Thus the pH increases from $$1.0$$ to approximately $$1.3$$.

Hence, the correct option is Option B: increase to 1.3.