Calculate the pH of a 1.0 × 10^-5M solution of HBr. Be sure your answer has the correct number of significant figures.

1 answer

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:

\[ \boxed{4.99} \]