The pH of a solution obtained by mixing equal volume of solution having pH = 3 and pH = 4.
`[log5.5=0.7404]`

Options

A. 3.26
B. 3.5
C. `4.0`
D. 3.42

Answer

To find the pH of a solution obtained by mixing equal volumes of solutions with pH 3 and pH 4, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Concentration of H⁺ Ions:**
- For a solution with pH = 3:
\[ [H^+]_1 = 10^{-3} \, \text{M} \]
- For a solution with pH = 4:
\[ [H^+]_2 = 10^{-4} \, \text{M} \]

2. **Calculate Moles of H⁺ Ions:**
- Assume we mix equal volumes (V) of both solutions.
- Moles of H⁺ from the first solution:
\[ \text{Moles}_1 = [H^+]_1 \times V = 10^{-3} \times V \]
- Moles of H⁺ from the second solution:
\[ \text{Moles}_2 = [H^+]_2 \times V = 10^{-4} \times V \]

3. **Total Moles of H⁺ Ions:**
- Total moles of H⁺ after mixing:
\[ \text{Total Moles} = \text{Moles}_1 + \text{Moles}_2 = 10^{-3}V + 10^{-4}V = (10^{-3} + 10^{-4})V \]
- Simplifying:
\[ \text{Total Moles} = (0.001 + 0.0001)V = 0.0011V \]

4. **Calculate the Total Volume:**
- Since we mixed equal volumes of both solutions:
\[ \text{Total Volume} = V + V = 2V \]

5. **Calculate the Concentration of H⁺ Ions in the Mixture:**
- The concentration of H⁺ ions in the mixed solution:
\[ [H^+]_{\text{final}} = \frac{\text{Total Moles}}{\text{Total Volume}} = \frac{0.0011V}{2V} = \frac{0.0011}{2} = 0.00055 \, \text{M} \]

6. **Calculate the pH of the Final Solution:**
- Using the formula for pH:
\[ \text{pH} = -\log[H^+]_{\text{final}} = -\log(0.00055) \]
- We can express this as:
\[ \text{pH} = -\log(5.5 \times 10^{-4}) = -(\log 5.5 + \log 10^{-4}) = -\log 5.5 + 4 \]
- Given that \(\log 5.5 = 0.7404\):
\[ \text{pH} = -0.7404 + 4 = 3.2596 \]

7. **Final Result:**
- Rounding to two decimal places, the final pH is:
\[ \text{pH} \approx 3.26 \]