To determine the pH of a 1 x 10^-7 M solution of NaOH, we first need to consider that NaOH is a strong base that completely dissociates in solution:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Thus, a 1 x 10^-7 M solution of NaOH will produce an equal concentration of hydroxide ions (OH⁻), which is also 1 x 10^-7 M.
However, in aqueous solutions, pure water has an autoionization equilibrium where:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
At room temperature (25°C), the concentration of hydrogen ions ([H⁺]) in pure water is also 1 x 10^-7 M, leading to a neutral pH of 7.
When you add NaOH to the solution, the total [OH⁻] becomes the sum of the OH⁻ from NaOH and the OH⁻ from water:
Total [OH⁻] = [OH⁻ from NaOH] + [OH⁻ from water] = 1 x 10^-7 + 1 x 10^-7 = 2 x 10^-7 M.
Now, we can calculate the pOH of the solution using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the total hydroxide ion concentration:
\[ \text{pOH} = -\log(2 \times 10^{-7}) \approx 6.70 \]
To find the pH, we can use the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} \approx 14 - 6.70 = 7.30 \]
Therefore, the pH of a 1 x 10^-7 M solution of NaOH is approximately 7.30.