Equal volumes of two solutions of `Hc1` are mixed. One solution has a `pH = 1`, while the other has a `pH = 5`. The `pH` of the resulting solution is

Options

A. `lt 1`
B. Between `1` and `2`
C. `3`
D. Between `4` and `5`

Explanation

To find the pH of the resulting solution when equal volumes of two hydrochloric acid (HCl) solutions with different pH values are mixed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pH Values**:
- The first solution has a pH of 1.
- The second solution has a pH of 5.

2. **Calculate the Concentration of H⁺ Ions**:
- For the first solution (pH = 1):
\[ [H^+]_1 = 10^{-1} \text{ M} = 0.1 \text{ M} \]
- For the second solution (pH = 5):
\[ [H^+]_2 = 10^{-5} \text{ M} = 0.00001 \text{ M} \]

3. **Mix the Solutions**:
- Since equal volumes of both solutions are mixed, the total volume doubles. Therefore, the concentrations of H⁺ ions will be halved in the resulting solution.

4. **Calculate the Total H⁺ Concentration**:
- The total concentration of H⁺ ions after mixing can be calculated as:
\[ [H^+]_{\text{final}} = \frac{[H^+]_1 + [H^+]_2}{2} \]
- Substituting the values:
\[ [H^+]_{\text{final}} = \frac{0.1 + 0.00001}{2} = \frac{0.10001}{2} = 0.050005 \text{ M} \]

5. **Calculate the pH of the Resulting Solution**:
- The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
- Substituting the final concentration:
\[ \text{pH} = -\log(0.050005) \]
- This can be approximated as:
\[ \text{pH} \approx -\log(5 \times 10^{-2}) = -\log(5) - \log(10^{-2}) = -\log(5) + 2 \]
- The value of \(-\log(5)\) is approximately 0.7, so:
\[ \text{pH} \approx 2 - 0.7 = 1.3 \]

6. **Conclusion**:
- The pH of the resulting solution after mixing the two solutions is approximately **1.3**.