A `50 ml` solution of `pH=1` is mixed with a `50 ml` solution of `pH=2`. The `pH` of the mixture will be nearly

Options

A. `0.76`
B. `1.26`
C. `1.76`
D. `2.26`

Explanation

To find the pH of the mixture when a 50 ml solution with pH 1 is mixed with a 50 ml solution with pH 2, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Concentration of H⁺ Ions**:
- For the solution with pH 1:
\[ [H^+]_1 = 10^{-1} \, \text{M} = 0.1 \, \text{M} \]
- For the solution with pH 2:
\[ [H^+]_2 = 10^{-2} \, \text{M} = 0.01 \, \text{M} \]

2. **Calculate the Total Moles of H⁺ Ions**:
- Volume of each solution = 50 ml = 0.050 L
- Moles of H⁺ from solution A (pH 1):
\[ \text{Moles}_1 = [H^+]_1 \times \text{Volume}_1 = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]
- Moles of H⁺ from solution B (pH 2):
\[ \text{Moles}_2 = [H^+]_2 \times \text{Volume}_2 = 0.01 \, \text{M} \times 0.050 \, \text{L} = 0.0005 \, \text{moles} \]

3. **Calculate the Total Moles of H⁺ Ions in the Mixture**:
\[ \text{Total Moles} = \text{Moles}_1 + \text{Moles}_2 = 0.005 + 0.0005 = 0.0055 \, \text{moles} \]

4. **Calculate the Total Volume of the Mixture**:
\[ \text{Total Volume} = 50 \, \text{ml} + 50 \, \text{ml} = 100 \, \text{ml} = 0.1 \, \text{L} \]

5. **Calculate the Final Concentration of H⁺ Ions**:
\[ [H^+]_{\text{final}} = \frac{\text{Total Moles}}{\text{Total Volume}} = \frac{0.0055 \, \text{moles}}{0.1 \, \text{L}} = 0.055 \, \text{M} \]

6. **Calculate the pH of the Mixture**:
\[ \text{pH} = -\log([H^+]_{\text{final}}) = -\log(0.055) \approx 1.26 \]

### Final Answer:
The pH of the mixture will be approximately **1.26**.