HBr is a strong acid, which means it completely dissociates in solution. Therefore, the concentration of hydrogen ions \([H^+]\) in a 1.0 × 10^-5 M solution of HBr is also 1.0 × 10^-5 M.
To calculate the pH, we can use the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting in the concentration of hydrogen ions:
\[ \text{pH} = -\log(1.0 \times 10^{-5}) \]
Calculating this:
\[ \text{pH} = -(-5) = 5.0 \]
However, we need to consider that in very dilute solutions of strong acids like HBr, the concentration of \([H^+]\) from the water itself (which contributes \(1.0 \times 10^{-7} M\)) can also be significant. In such cases, we should include this contribution:
The total \([H^+]\) would then be:
\[ \text{Total } [H^+] = 1.0 \times 10^{-5} + 1.0 \times 10^{-7} \approx 1.0 \times 10^{-5} \]
(because \(1.0 \times 10^{-5}\) is much larger than \(1.0 \times 10^{-7}\)). Thus, we can assume:
\[ [H^+] \approx 1.0 \times 10^{-5} M \]
Now we recalculate the pH:
\[ \text{pH} \approx -\log(1.0 \times 10^{-5}) = 5.0 \]
But the total concentration including the water's contribution would need to be calculated more precisely:
The total concentration before taking log is considered as:
\[ [H^+] \approx 1.0 \times 10^{-5} + 1.0 \times 10^{-7} = 1.01 \times 10^{-5} \]
Calculating the pH with this value gives:
\[ \text{pH} = -\log(1.01 \times 10^{-5}) \approx 4.99 \]
Finally, rounding to two significant figures for the \(1.0 \times 10^{-5}\) concentration:
\[ \text{pH} = 4.99 \]
Thus, the final answer for the pH of a 1.0 × 10^-5 M solution of HBr is:
\[ \boxed{4.99} \]