The pH of a `10^(-3)` M HCl solution at `25^(@)C` if it is diluted 1000 times, will be-

Options

A. 3
B. zero
C. 5.98
D. 6.02

Explanation

To find the pH of a `10^(-3)` M HCl solution at `25°C` after it is diluted 1000 times, we can follow these steps:

### Step-by-Step Solution:

1. **Initial Concentration of HCl**:
The initial concentration of HCl is given as `10^(-3)` M. Since HCl is a strong acid, it completely dissociates in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Therefore, the concentration of \(\text{H}^+\) ions will also be `10^(-3)` M.

2. **Dilution Calculation**:
When the solution is diluted 1000 times, the new concentration (\(M_2\)) can be calculated using the dilution formula:
\[ M_1 \times V_1 = M_2 \times V_2 \]
Here, \(M_1 = 10^{-3}\) M and \(V_2 = 1000 \times V_1\). Thus:
\[ M_2 = \frac{M_1 \times V_1}{V_2} = \frac{10^{-3} \times V_1}{1000 \times V_1} = 10^{-3} \div 1000 = 10^{-6} \text{ M} \]

3. **Calculating pH**:
The pH of a solution is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of \(\text{H}^+\):
\[ \text{pH} = -\log(10^{-6}) = 6 \]

4. **Considering Water Contribution**:
However, we must consider that the pH of pure water at `25°C` is around 7, which means that in a very dilute solution, the contribution of \(\text{H}^+\) ions from water (which is \(10^{-7}\) M) will also affect the pH. Therefore, the total concentration of \(\text{H}^+\) ions will be:
\[ [\text{H}^+] = 10^{-6} + 10^{-7} = 1.1 \times 10^{-6} \text{ M} \]

5. **Recalculating pH**:
Now, we recalculate the pH using the new concentration:
\[ \text{pH} = -\log(1.1 \times 10^{-6}) \]
Using logarithmic properties, this can be approximated:
\[ \text{pH} \approx 6 - 0.04 \approx 5.96 \]

### Conclusion:
Thus, the pH of the diluted solution is approximately **5.96**.