Given pH of a solution A is 3 and it is mixed with another solution B having pH 2. If both mixed, then resultant pH of the solution will be

Options

A. 3.2
B. 1.9
C. 3.4
D. 3.5

Answer

To find the resultant pH when mixing two solutions A and B with given pH values, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the pH values**:
- Solution A has a pH of 3.
- Solution B has a pH of 2.

2. **Calculate the concentration of H⁺ ions**:
- The concentration of H⁺ ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} \]
- For solution A:
\[ [H^+]_A = 10^{-3} \, \text{M} \]
- For solution B:
\[ [H^+]_B = 10^{-2} \, \text{M} \]

3. **Assume equal volumes for both solutions**:
- Let’s assume we mix V liters of solution A with V liters of solution B.
- Therefore, the total volume after mixing will be:
\[ V_{total} = V + V = 2V \]

4. **Calculate total moles of H⁺ ions**:
- Moles of H⁺ from solution A:
\[ \text{Moles of } H^+_A = [H^+]_A \times V = (10^{-3}) \times V \]
- Moles of H⁺ from solution B:
\[ \text{Moles of } H^+_B = [H^+]_B \times V = (10^{-2}) \times V \]
- Total moles of H⁺ ions:
\[ \text{Total moles of } H^+ = (10^{-3}V) + (10^{-2}V) = (10^{-3} + 10^{-2})V = (0.001 + 0.01)V = 0.011V \]

5. **Calculate the concentration of H⁺ ions in the mixed solution**:
- The concentration of H⁺ ions in the total volume (2V):
\[ [H^+]_{total} = \frac{\text{Total moles of } H^+}{V_{total}} = \frac{0.011V}{2V} = \frac{0.011}{2} = 0.0055 \, \text{M} \]

6. **Calculate the resultant pH**:
- Use the formula for pH:
\[ \text{pH} = -\log[H^+] \]
- Substitute the concentration of H⁺:
\[ \text{pH} = -\log(0.0055) \]
- Calculate the pH:
\[ \text{pH} \approx 2.26 \]

### Final Result:
The resultant pH of the mixed solution is approximately **2.26**.