Sure! To calculate the pH of a [tex]$7.8 \times 10^{-8} M$[/tex] aqueous solution of hydrobromic acid [tex]\((HBr)\)[/tex], follow these steps:
1. Understand the relationship:
Hydrobromic acid ([tex]\(HBr\)[/tex]) is a strong acid, which means it dissociates completely in water. Each [tex]\(HBr\)[/tex] molecule dissociates to give one [tex]\(H^+\)[/tex] ion.
Thus, the concentration of hydrogen ions [tex]\([H^+]\)[/tex] in the solution is equal to the concentration of [tex]\(HBr\)[/tex].
Given:
[tex]\[
[H^+] = 7.8 \times 10^{-8} \, M
\][/tex]
2. Use the formula for pH:
The pH of a solution is calculated using the formula:
[tex]\[
\text{pH} = -\log_{10}([H^+])
\][/tex]
3. Substitute the concentration of hydrogen ions into the formula:
[tex]\[
\text{pH} = -\log_{10}(7.8 \times 10^{-8})
\][/tex]
4. Calculate the logarithm:
The logarithm base 10 of [tex]\(7.8 \times 10^{-8}\)[/tex] can be split into:
[tex]\[
\log_{10}(7.8 \times 10^{-8}) = \log_{10}(7.8) + \log_{10}(10^{-8})
\][/tex]
Now break it down:
- [tex]\(\log_{10}(10^{-8}) = -8\)[/tex]
- [tex]\(\log_{10}(7.8)\)[/tex] is approximately [tex]\(0.892\)[/tex]
Therefore:
[tex]\[
\log_{10}(7.8 \times 10^{-8}) = 0.892 - 8 = -7.108
\][/tex]
5. Compute the pH:
[tex]\[
\text{pH} = -(-7.108)
\][/tex]
[tex]\[
\text{pH} = 7.108
\][/tex]
6. Round the answer to two decimal places:
Round 7.108 to two decimal places:
[tex]\[
\text{pH} \approx 7.11
\][/tex]
Hence, the pH of the [tex]\(7.8 \times 10^{-8} \, M\)[/tex] aqueous solution of hydrobromic acid [tex]\((HBr)\)[/tex] is approximately [tex]\(7.11\)[/tex].
