Equal volumes of two `HCl` solutions of `pH=3` and `pH=5` were mixed. What is the `Ph` of the resulting solution ?

Options

A. `3.5`
B. `4.0`
C. `4.5`
D. `3.3`

Explanation

To solve the problem of finding the pH of the resulting solution when equal volumes of two HCl solutions with pH 3 and pH 5 are mixed, we can follow these steps:

### Step-by-Step Solution:

1. **Convert pH to H⁺ Concentration:**
- For the first solution (pH = 3):
\[ [H^+]_1 = 10^{-3} \, \text{M} \]
- For the second solution (pH = 5):
\[ [H^+]_2 = 10^{-5} \, \text{M} \]

2. **Assume Equal Volumes:**
- Let the volume of each solution be \( V \). Therefore, when mixed, the total volume becomes \( 2V \).

3. **Calculate Total Moles of H⁺:**
- 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:
\[ \text{Moles}_2 = [H^+]_2 \times V = 10^{-5} \times V \]
- Total moles of H⁺ in the mixed solution:
\[ \text{Total Moles} = \text{Moles}_1 + \text{Moles}_2 = 10^{-3}V + 10^{-5}V \]

4. **Calculate the Concentration of H⁺ in the Mixed Solution:**
- Total moles of H⁺:
\[ \text{Total Moles} = (10^{-3} + 10^{-5})V = (10^{-3} + 0.01 \times 10^{-3})V = (1 + 0.01) \times 10^{-3}V = 1.01 \times 10^{-3}V \]
- Concentration of H⁺ in the total volume \( 2V \):
\[ [H^+]_{\text{final}} = \frac{1.01 \times 10^{-3}V}{2V} = \frac{1.01 \times 10^{-3}}{2} = 0.505 \times 10^{-3} \, \text{M} \]

5. **Calculate the pH of the Resulting Solution:**
- Using the concentration to find pH:
\[ \text{pH} = -\log([H^+]_{\text{final}}) = -\log(0.505 \times 10^{-3}) = -\log(0.505) - \log(10^{-3}) \]
- Since \( -\log(10^{-3}) = 3 \), we need to calculate \( -\log(0.505) \):
- \( \log(0.505) \approx -0.30 \) (using logarithmic properties)
- Therefore:
\[ \text{pH} \approx 3 - 0.30 = 3.30 \]

### Final Answer:
The pH of the resulting solution is approximately **3.30**.

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