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

Options

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

Explanation

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

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

### Step 2: Calculate the concentration after dilution
The solution is diluted by 100 times. Therefore, the concentration after dilution can be calculated as follows:
\[ \text{Concentration after dilution} = \frac{10^{-6} \, M}{100} = 10^{-8} \, M \]

### Step 3: Consider the contribution of water
In pure water, the concentration of hydroxide ions \([OH^-]\) is \(10^{-7} \, M\) at 25°C. Since we are diluting NaOH, we need to consider the contribution of hydroxide ions from both the NaOH and the water:
\[ \text{Total } [OH^-] = [OH^-]_{\text{NaOH}} + [OH^-]_{\text{water}} = 10^{-8} \, M + 10^{-7} \, M = 1.1 \times 10^{-7} \, M \]

### Step 4: Calculate the pOH
To find the pOH, we use the formula:
\[ \text{pOH} = -\log[OH^-] \]
Substituting the total hydroxide ion concentration:
\[ \text{pOH} = -\log(1.1 \times 10^{-7}) \]

### Step 5: Calculate the pH
Since \(pH + pOH = 14\), we can find the pH:
\[ pH = 14 - pOH \]

### Step 6: Calculate the logarithm
Using the logarithm properties, we can calculate:
\[ \log(1.1 \times 10^{-7}) = \log(1.1) + \log(10^{-7}) = \log(1.1) - 7 \]
Using \(\log(1.1) \approx 0.0414\):
\[ \text{pOH} \approx - (0.0414 - 7) = 7 - 0.0414 = 6.9586 \]

### Step 7: Final pH calculation
Thus, we can find the pH:
\[ pH = 14 - 6.9586 \approx 7.0414 \]

### Conclusion
The pH of the diluted NaOH solution is approximately **7.04**.

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