Sure, let's solve this problem step by step:
1. Determine the concentration of hydroxide ions ([tex]\(OH^-\)[/tex]):
Since NaOH is a strong base, it dissociates completely in water. Therefore, the concentration of hydroxide ions ([tex]\(OH^-\)[/tex]) will be equal to the concentration of the NaOH solution.
Given:
[tex]\[
\text{{Concentration of NaOH}} = 9.67 \times 10^{-3} \text{{ M}}
\][/tex]
Therefore,
[tex]\[
\text{{Concentration of }} OH^- = 9.67 \times 10^{-3} \text{{ M}}
\][/tex]
2. Calculate the pOH of the solution:
The pOH is calculated using the formula:
[tex]\[
\text{{pOH}} = -\log[\text{{OH}}^-]
\][/tex]
Substituting the concentration of [tex]\(OH^-\)[/tex]:
[tex]\[
\text{{pOH}} = -\log(9.67 \times 10^{-3}) \approx 2.0146
\][/tex]
3. Calculate the pH of the solution:
The relationship between pH and pOH is given by:
[tex]\[
\text{{pH}} + \text{{pOH}} = 14
\][/tex]
Solving for pH:
[tex]\[
\text{{pH}} = 14 - \text{{pOH}} = 14 - 2.0146 \approx 11.9854
\][/tex]
Therefore, the pH of the NaOH solution with a concentration of [tex]\(9.67 \times 10^{-3}\)[/tex] M is approximately:
[tex]\[
\boxed{11.99}
\][/tex]
