1. Initial pH and hydrogen ion concentration

Given: $\text{pH} = 1.0$

Using $\text{pH} = -\log[H^+]$ we get $[H^+] = 10^{-1} = 0.1\,\text{M}$

Since HCl is a strong acid, it dissociates completely, so this is the acid concentration as well.

2. Effect of adding equal volume of water

If equal volume of water is added, the total volume becomes double.

Hence concentration becomes half: $[H^+]_{\text{new}} = \frac{0.1}{2} = 0.05\,\text{M}$

3. Calculate new pH

$\text{pH}_{\text{new}} = -\log(0.05)$

Now, $0.05 = 5 \times 10^{-2}$ so $\log(0.05) = \log 5 + \log 10^{-2}$

Using $\log 5 = \log\left(\frac{10}{2}\right) = 1 - \log 2 = 1 - 0.30 = 0.70$

Therefore, $\log(0.05) = 0.70 - 2 = -1.30$

Hence, $\text{pH}_{\text{new}} = -(-1.30) = 1.30$

4. Conclusion

The pH increases from $1.0$ to $1.3$.

So the correct option is: $\boxed{\text{A: increase to }1.3}$

5. Comparison with stored correct answer

Stored correct answer: A

This matches the derived answer.