An acid solution of `pH=6` is diluted `1000` times, the `pH` of the final solution is

Options

A. `6.99`
B. `6.0`
C. `3.0`
D. `9.0`

Explanation

To solve the problem of finding the pH of an acid solution after it has been diluted 1000 times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial pH**:
The initial pH of the acid solution is given as 6.

2. **Calculate the Initial H⁺ Concentration**:
The concentration of hydrogen ions (H⁺) can be calculated from the pH using the formula:
\[ [H^+] = 10^{-\text{pH}} = 10^{-6} \, \text{M} \]

3. **Account for Water's Contribution**:
Pure water contributes H⁺ ions as well, with a concentration of:
\[ [H^+]_{\text{water}} = 10^{-7} \, \text{M} \]
Since the solution is acidic, we cannot neglect this contribution when calculating the total H⁺ concentration after dilution.

4. **Dilution Factor**:
The solution is diluted 1000 times. Therefore, the new concentration of H⁺ ions from the acid after dilution will be:
\[ [H^+]_{\text{diluted}} = \frac{10^{-6}}{1000} = 10^{-9} \, \text{M} \]

5. **Calculate Total H⁺ Concentration**:
Now, we need to add the contributions from both the diluted acid and water:
\[ [H^+]_{\text{total}} = [H^+]_{\text{diluted}} + [H^+]_{\text{water}} = 10^{-9} + 10^{-7} \]
To add these, we can factor out \(10^{-7}\):
\[ [H^+]_{\text{total}} = 10^{-7} \left(10^{-2} + 1\right) = 10^{-7} \times 1.01 = 1.01 \times 10^{-7} \, \text{M} \]

6. **Calculate the Final pH**:
Finally, we can find the pH of the final solution using the formula:
\[ \text{pH} = -\log[H^+]_{\text{total}} = -\log(1.01 \times 10^{-7}) \]
Using properties of logarithms:
\[ \text{pH} = -\log(1.01) - \log(10^{-7}) = -\log(1.01) + 7 \]
Since \(-\log(1.01)\) is a small positive number (approximately 0.00432), we can estimate:
\[ \text{pH} \approx 7 - 0.00432 \approx 6.99568 \]
Rounding this gives us:
\[ \text{pH} \approx 6.99 \]

### Final Answer:
The pH of the final solution after dilution is approximately **6.99**.