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. 0.86
B. 1.26
C. 1.76
D. 2.26

Answer

To solve the problem of finding the pH of a mixture of two solutions with different pH values, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the pH Concept**:
The pH of a solution is defined as:
\[ \text{pH} = -\log_{10}[\text{H}^+] \]
where \([\text{H}^+]\) is the concentration of hydrogen ions in moles per liter.

2. **Calculate \([\text{H}^+]\) for Each Solution**:
- For the first solution with pH = 1:
\[ \text{pH} = 1 \implies [\text{H}^+] = 10^{-1} = 0.1 \, \text{M} \]
- For the second solution with pH = 2:
\[ \text{pH} = 2 \implies [\text{H}^+] = 10^{-2} = 0.01 \, \text{M} \]

3. **Determine the Volume of Each Solution**:
Both solutions have a volume of 50 mL, which is equivalent to 0.050 L.

4. **Calculate the Total Moles of \([\text{H}^+]\)**:
- Moles of \([\text{H}^+]\) from the first solution:
\[ \text{Moles} = [\text{H}^+] \times \text{Volume} = 0.1 \, \text{M} \times 0.050 \, \text{L} = 0.005 \, \text{moles} \]
- Moles of \([\text{H}^+]\) from the second solution:
\[ \text{Moles} = [\text{H}^+] \times \text{Volume} = 0.01 \, \text{M} \times 0.050 \, \text{L} = 0.0005 \, \text{moles} \]

5. **Calculate the Total Moles of \([\text{H}^+]\)**:
\[ \text{Total moles of } [\text{H}^+] = 0.005 + 0.0005 = 0.0055 \, \text{moles} \]

6. **Calculate the Total Volume of the Mixture**:
The total volume of the mixture is:
\[ \text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L} \]

7. **Calculate the Concentration of \([\text{H}^+]\) in the Mixture**:
\[ [\text{H}^+]_{\text{mixture}} = \frac{\text{Total moles of } [\text{H}^+]}{\text{Total Volume}} = \frac{0.0055 \, \text{moles}}{0.1 \, \text{L}} = 0.055 \, \text{M} \]

8. **Calculate the pH of the Mixture**:
\[ \text{pH}_{\text{mixture}} = -\log_{10}(0.055) \approx 1.26 \]

Thus, the pH of the mixture will be nearly **1.26**.