When `1 ml` of `0.1 N HCI` is added to `1` litre of a solution of `pH` value `4`, the pH will be nearly

Options

A. 5
B. 4.477
C. 3
D. 3.699

Explanation

To determine the pH of the solution after adding 1 ml of 0.1 N HCl to 1 liter of a solution with a pH of 4, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the initial concentration of H⁺ ions in the pH 4 solution:**
- The pH is given as 4.
- The concentration of H⁺ ions can be calculated using the formula:
\[ [H^+] = 10^{-pH} = 10^{-4} \, \text{M} \]

2. **Calculate the number of moles of H⁺ ions in the 1 liter solution:**
- Since the concentration is \(10^{-4} \, \text{M}\) and the volume is 1 liter:
\[ \text{Moles of H}^+ = [H^+] \times \text{Volume} = 10^{-4} \times 1 = 10^{-4} \, \text{moles} \]

3. **Calculate the number of moles of H⁺ ions from the added HCl:**
- The concentration of HCl is 0.1 N, and we are adding 1 ml (which is 0.001 L):
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \times 0.001 = 0.0001 \, \text{moles} = 0.1 \, \text{mmoles} \]

4. **Convert the moles of H⁺ from the added HCl to millimoles:**
- Since 1 mole = 1000 millimoles:
\[ \text{HCl in millimoles} = 0.1 \, \text{mmoles} \]

5. **Calculate the total moles of H⁺ ions after mixing:**
- Total moles of H⁺ ions = moles from the original solution + moles from HCl:
\[ \text{Total moles of H}^+ = 10^{-4} + 0.1 \, \text{mmoles} = 0.1 + 0.0001 = 0.1001 \, \text{mmoles} \]

6. **Calculate the total volume of the solution after adding HCl:**
- Initial volume = 1000 ml, added volume = 1 ml:
\[ \text{Total volume} = 1000 + 1 = 1001 \, \text{ml} \]

7. **Calculate the new concentration of H⁺ ions in the resultant solution:**
- Convert total moles to moles per liter:
\[ [H^+] = \frac{0.1001 \, \text{mmoles}}{1001 \, \text{ml}} = \frac{0.1001}{1001} \, \text{moles/L} \approx 2 \times 10^{-4} \, \text{M} \]

8. **Calculate the new pH of the solution:**
- Use the formula for pH:
\[ \text{pH} = -\log([H^+]) = -\log(2 \times 10^{-4}) \]
- Solving this gives:
\[ \text{pH} \approx 3.699 \]

### Final Answer:
The pH of the resultant solution will be approximately **3.699**.