To solve this problem, we need to determine the concentration of hydroxide ions [tex]\(\left[ \text{OH} ^{-}\right]\)[/tex], the pOH, and the pH of the solution given [tex]\(\left[ \text{H}_3\text{O} ^{+}\right]=4.9 \times 10^{-9} M\)[/tex]. Here are the steps to find these values:
1. Calculate the concentration of hydroxide ions [tex]\(\left[ \text{OH} ^{-}\right]\)[/tex]:
We use the ion-product constant for water ([tex]\(K_w\)[/tex]) at 25°C, which is [tex]\(1.0 \times 10^{-14}\)[/tex].
[tex]\[
\left[ \text{OH} ^{-} \right] = \frac{K_w}{\left[ \text{H}_3\text{O} ^{+} \right]}
\][/tex]
Given:
[tex]\[
K_w = 1.0 \times 10^{-14}
\][/tex]
[tex]\[
\left[ \text{H}_3\text{O} ^{+} \right] = 4.9 \times 10^{-9} M
\][/tex]
Substituting the values:
[tex]\[
\left[ \text{OH} ^{-} \right] = \frac{1.0 \times 10^{-14}}{4.9 \times 10^{-9}}
\][/tex]
[tex]\[
\left[ \text{OH} ^{-} \right] \approx 2.04 \times 10^{-6} M
\][/tex]
2. Calculate the pOH of the solution:
The pOH is calculated using the formula:
[tex]\[
\text{pOH} = -\log_{10} \left( \left[ \text{OH} ^{-} \right] \right)
\][/tex]
Substituting the concentration of [tex]\(\left[ \text{OH} ^{-} \right]\)[/tex]:
[tex]\[
\text{pOH} = -\log_{10} \left( 2.04 \times 10^{-6} \right)
\][/tex]
[tex]\[
\text{pOH} \approx 5.69
\][/tex]
3. Calculate the pH of the solution:
The pH is calculated using the formula:
[tex]\[
\text{pH} = -\log_{10} \left( \left[ \text{H}_3\text{O} ^{+} \right] \right)
\][/tex]
Substituting the concentration of [tex]\(\left[ \text{H}_3\text{O} ^{+} \right]\)[/tex]:
[tex]\[
\text{pH} = -\log_{10} \left( 4.9 \times 10^{-9} \right)
\][/tex]
[tex]\[
\text{pH} \approx 8.31
\][/tex]
So, the calculated values are:
[tex]\[
\begin{array}{l}
\left[ \text{OH} ^{-}\right] \approx 2.04 \times 10^{-6} M \\
\text{pOH} \approx 5.69 \\
\text{pH} \approx 8.31
\end{array}
\][/tex]
