The pH of 10^(-7) M NaOH is: A) 7.01 B) Between 7 and 8 C) Between 8 and 9 D) Greater than 10

To find the pH of a \({10^{-7}}\, \text{M NaOH}\) solution, we need to consider the following points:

1. **Understanding the concentration**: A \({10^{-7}}\, \text{M}\) NaOH solution is a very dilute base. NaOH is a strong base, so it dissociates completely in water:

\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Therefore, the concentration of \(\text{OH}^-\) ions will be \(10^{-7}\, \text{M}\) in this solution.

2. **Water's contribution**: In pure water, the concentration of \(\text{OH}^-\) ions is \(10^{-7}\, \text{M}\) at 25°C. So, in the presence of \({10^{-7}}\, \text{M}\) NaOH, the total \(\text{OH}^-\) concentration is:

\[ [\text{OH}^-]_{\text{total}} = 10^{-7}\, \text{M} + 10^{-7}\, \text{M} = 2 \times 10^{-7}\, \text{M} \]

3. **Calculating pOH**: The pOH of the solution is given by:

\[ \text{pOH} = -\log [\text{OH}^-] \]

Plugging in the total \(\text{OH}^-\) concentration:

\[ \text{pOH} = -\log (2 \times 10^{-7}) \]

To simplify this, we use the logarithm properties:

\[ \text{pOH} = -\log (2) - \log (10^{-7}) \]
\[ \text{pOH} = -\log (2) + 7 \]

The value of \(-\log (2)\) is approximately \(-0.3\):

\[ \text{pOH} \approx 7 - 0.3 = 6.7 \]

4. **Finding pH**: The relationship between pH and pOH is:

\[ \text{pH} + \text{pOH} = 14 \]

Therefore:

\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 6.7 = 7.3 \]

So, the pH of \({10^{-7}}\, \text{M NaOH}\) is approximately \(7.3\), which falls between 7 and 8.

**Answer: B) Between 7 and 8**