To determine the pH of a \(1 \times 10^{-4}\) mol HCl solution, we first note that HCl is a strong acid and will dissociate completely in solution. Therefore, the concentration of hydrogen ions \([H^+]\) will also be \(1 \times 10^{-4} , \text{M}\).
The formula for calculating pH is:
\[ \text{pH} = -\log[H^+] \]
Substituting the value we have:
\[ \text{pH} = -\log(1 \times 10^{-4}) = 4 \]
However, we need to consider that the solution also contains a small amount of \(H^+\) ions from the water itself (from the autoionization of water). Pure water has a \([H^+]\) concentration of \(1 \times 10^{-7} , \text{M}\). When calculating the pH of very dilute acid solutions, we must account for this contribution.
This results in the total \([H^+]\):
\[ [H^+]_{\text{total}} = 1 \times 10^{-4} + 1 \times 10^{-7} \approx 1 \times 10^{-4} , (\text{since } 1 \times 10^{-4} \text{ is much larger than } 1 \times 10^{-7}) \]
Thus, the estimated pH remains close to 4.
Therefore, the correct answer to the pH of a \(1 \times 10^{-4}\) mol HCl solution is:
4