An acid solution of `pH =6` is diluted `100` times. The `pH` of solution becomes

Options

A. `6.95`
B. `6`
C. `4`
D. `8`

Explanation

To find the pH of a diluted acid solution, we can follow these steps:

### Step 1: Determine the initial concentration of hydrogen ions
Given that the pH of the initial acid solution is 6, we can find the concentration of hydrogen ions \([H^+]\) using the formula:
\[ \text{pH} = -\log[H^+] \]
Rearranging this gives:
\[ [H^+] = 10^{-\text{pH}} = 10^{-6} \text{ M} \]

### Step 2: Calculate the concentration after dilution
The solution is diluted 100 times. When a solution is diluted, the concentration of the solute decreases. Thus, the new concentration of hydrogen ions after dilution is:
\[ [H^+]_{\text{diluted}} = \frac{[H^+]_{\text{initial}}}{100} = \frac{10^{-6}}{100} = 10^{-8} \text{ M} \]

### Step 3: Consider the contribution of water
In pure water at 25°C, the concentration of hydrogen ions is \(10^{-7} \text{ M}\). When we dilute the solution, we need to consider the contribution of hydrogen ions from water as well. Therefore, the total concentration of hydrogen ions in the diluted solution is:
\[ [H^+]_{\text{total}} = [H^+]_{\text{diluted}} + [H^+]_{\text{water}} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7} \text{ M} \]

### Step 4: Calculate the new pH
Now, we can calculate the new pH using the total concentration of hydrogen ions:
\[ \text{pH} = -\log[H^+]_{\text{total}} = -\log(1.1 \times 10^{-7}) \]
Using logarithmic properties, this can be approximated as:
\[ \text{pH} \approx 7 - \log(1.1) \]
Since \(\log(1.1) \approx 0.041\), we find:
\[ \text{pH} \approx 7 - 0.041 \approx 6.96 \]

### Final Answer
Thus, the pH of the diluted solution is approximately **6.95**.

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