`10^(-6) M NaOH` is diluted by `100` times. The `pH` of diluted base is

Options

A. between 7 and 8
B. between 5 and 6
C. between 6 and 7
D. between 10 and 11

Explanation

To solve the problem of finding the pH of a diluted solution of NaOH, we will follow these steps:

### Step-by-Step Solution:

1. **Determine the initial concentration of NaOH**:
The initial concentration of NaOH is given as \(10^{-6} \, M\).

2. **Calculate the concentration after dilution**:
Since the solution is diluted by 100 times, we calculate the new concentration of NaOH:
\[ \text{Concentration of NaOH after dilution} = \frac{10^{-6}}{100} = 10^{-8} \, M \]

3. **Determine the concentration of hydroxide ions (\(OH^-\))**:
In a dilute solution, the concentration of hydroxide ions from NaOH is \(10^{-8} \, M\). However, we must also consider the contribution of \(OH^-\) ions from water:
- The concentration of \(OH^-\) ions from water at 25°C is \(10^{-7} \, M\).

4. **Calculate the total concentration of \(OH^-\) ions**:
The total concentration of \(OH^-\) ions in the solution is the sum of the \(OH^-\) from NaOH and from water:
\[ \text{Total } [OH^-] = 10^{-8} + 10^{-7} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \, M \]

5. **Calculate the pOH**:
To find the pOH, we use the formula:
\[ \text{pOH} = -\log[OH^-] \]
Substituting the total concentration:
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \]

6. **Calculate the pH**:
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
First, we calculate the pOH:
\[ \text{pOH} \approx 7 - \log(1.1) \quad (\text{since } -\log(10^{-7}) = 7) \]
The value of \(\log(1.1)\) is approximately \(0.0414\):
\[ \text{pOH} \approx 7 - 0.0414 = 6.9586 \]
Now, using the relation to find pH:
\[ \text{pH} = 14 - \text{pOH} \approx 14 - 6.9586 = 7.0414 \]

### Final Answer:
The pH of the diluted NaOH solution is approximately **7.0414**.