Equal volumes of two solutions of a strong acid having `pH 3` and `pH 4` are mixed together. The `pH` of the resulting solution will then be equal to

Options

A. `3.5`
B. `3.26`
C. `7`
D. `1.0`

Explanation

To determine the pH of the resulting solution when equal volumes of two strong acid solutions with pH 3 and pH 4 are mixed, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions in each solution.
- For the solution with pH 3:
\[ [H^+] = 10^{-pH} = 10^{-3} \, \text{M} \]
- For the solution with pH 4:
\[ [H^+] = 10^{-pH} = 10^{-4} \, \text{M} \]

### Step 2: Mix the two solutions.
Since we are mixing equal volumes of both solutions, we can denote the volume of each solution as V. Therefore, the total volume after mixing will be:
\[ V_{\text{total}} = V + V = 2V \]

### Step 3: Calculate the total moles of H⁺ ions in the mixed solution.
- Moles of H⁺ from the first solution:
\[ \text{Moles from pH 3} = [H^+] \times V = 10^{-3} \times V \]
- Moles of H⁺ from the second solution:
\[ \text{Moles from pH 4} = [H^+] \times V = 10^{-4} \times V \]
- Total moles of H⁺ in the mixed solution:
\[ \text{Total moles} = 10^{-3}V + 10^{-4}V = (10^{-3} + 10^{-4})V = (0.001 + 0.0001)V = 0.0011V \]

### Step 4: Calculate the concentration of H⁺ ions in the mixed solution.
The concentration of H⁺ ions in the resulting solution is given by:
\[ [H^+]_{\text{mixed}} = \frac{\text{Total moles}}{V_{\text{total}}} = \frac{0.0011V}{2V} = \frac{0.0011}{2} = 0.00055 \, \text{M} \]

### Step 5: Calculate the pH of the resulting solution.
To find the pH:
\[ pH = -\log[H^+]_{\text{mixed}} = -\log(0.00055) \]
Calculating this gives:
\[ pH \approx 3.26 \]

### Final Answer:
The pH of the resulting solution when equal volumes of the two strong acid solutions are mixed is approximately **3.26**.

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