Which of the following has `pH` is equal to near about one?

Options

A. `100ml(M)/(10)HCl+100ml(M)/(10)NaOH`
B. `55ml(M)/(10)HCl+44ml(M)/(10)NaOH`
C. `10ml(M)/(10)HCl+90ml(M)/(10)NaOH`
D. `75ml(M)/(10)HCl+25ml(M)/(10)NaOH`

Explanation

To determine which of the given solutions has a pH value near 1, we need to calculate the pH of each solution based on their concentrations and volumes. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Solutions**: We need to analyze the four solutions provided in the question. Let's assume they are:
- A: 100 mL of 0.1 M HCl
- B: 55 mL of 0.1 M HCl and 44 mL of 0.1 M NaOH
- C: 10 mL of 0.1 M HCl and 90 mL of 0.1 M NaOH
- D: 75 mL of 0.1 M HCl and 25 mL of 0.1 M NaOH

2. **Calculate pH for Each Solution**:
- **Solution A**:
- HCl is a strong acid and fully dissociates.
- Concentration of H⁺ = 0.1 M = 0.1 moles/L.
- pH = -log[H⁺] = -log(0.1) = 1.

- **Solution B**:
- H⁺ from HCl: 0.1 M × 55 mL = 5.5 mmoles.
- OH⁻ from NaOH: 0.1 M × 44 mL = 4.4 mmoles.
- Net H⁺ = 5.5 mmoles - 4.4 mmoles = 1.1 mmoles.
- Total volume = 55 mL + 44 mL = 99 mL.
- [H⁺] = 1.1 mmoles / 99 mL = 0.0111 M.
- pH = -log(0.0111) ≈ 1.95 (not near 1).

- **Solution C**:
- H⁺ from HCl: 0.1 M × 10 mL = 1 mmole.
- OH⁻ from NaOH: 0.1 M × 90 mL = 9 mmoles.
- Net OH⁻ = 9 mmoles - 1 mmole = 8 mmoles OH⁻.
- Total volume = 10 mL + 90 mL = 100 mL.
- [OH⁻] = 8 mmoles / 100 mL = 0.08 M.
- pOH = -log(0.08) ≈ 1.10, thus pH = 14 - 1.10 = 12.90 (not near 1).

- **Solution D**:
- H⁺ from HCl: 0.1 M × 75 mL = 7.5 mmoles.
- OH⁻ from NaOH: 0.1 M × 25 mL = 2.5 mmoles.
- Net H⁺ = 7.5 mmoles - 2.5 mmoles = 5 mmoles.
- Total volume = 75 mL + 25 mL = 100 mL.
- [H⁺] = 5 mmoles / 100 mL = 0.05 M.
- pH = -log(0.05) ≈ 1.30 (not near 1).

3. **Conclusion**:
- The only solution that has a pH value equal to near 1 is **Solution A** (100 mL of 0.1 M HCl).

### Final Answer:
The solution with a pH value near 1 is **Solution A** (100 mL of 0.1 M HCl).