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

Options

A. 3.29
B. 3.5
C. `4.0`
D. 3.42

Answer

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

### Step 1: Determine the concentration of H⁺ ions in each solution.
- For the solution with pH = 3:
\[ [\text{H}^+]_1 = 10^{-3} \, \text{M} \]

- For the solution with pH = 4:
\[ [\text{H}^+]_2 = 10^{-4} \, \text{M} \]

### Step 2: Calculate the number of moles of H⁺ ions in each solution.
Assuming we mix equal volumes \( V \) of each solution:
- Number of moles of H⁺ in solution 1:
\[ \text{Moles}_1 = [\text{H}^+]_1 \times V = 10^{-3} \times V \]

- Number of moles of H⁺ in solution 2:
\[ \text{Moles}_2 = [\text{H}^+]_2 \times V = 10^{-4} \times V \]

### Step 3: Calculate the total number of moles of H⁺ ions in the mixed solution.
\[ \text{Total moles of H}^+ = \text{Moles}_1 + \text{Moles}_2 = (10^{-3} \times V) + (10^{-4} \times V) = (10^{-3} + 10^{-4}) \times V \]
\[ = (10^{-3} + 0.1 \times 10^{-3}) \times V = (1.1 \times 10^{-3}) \times V \]

### Step 4: Calculate the total volume of the mixed solution.
Since we mixed equal volumes of both solutions:
\[ \text{Total volume} = V + V = 2V \]

### Step 5: Calculate the concentration of H⁺ ions in the mixed solution.
\[ [\text{H}^+]_{\text{mix}} = \frac{\text{Total moles of H}^+}{\text{Total volume}} = \frac{(1.1 \times 10^{-3}) \times V}{2V} = \frac{1.1 \times 10^{-3}}{2} = 0.55 \times 10^{-3} \, \text{M} \]

### Step 6: Calculate the pH of the mixed solution.
\[ \text{pH}_{\text{mix}} = -\log([\text{H}^+]_{\text{mix}}) = -\log(0.55 \times 10^{-3}) \]
Using the logarithmic property:
\[ \text{pH}_{\text{mix}} = -\log(0.55) - \log(10^{-3}) = -\log(0.55) + 3 \]
Using the given value \( \log(5.5) = 0.7404 \):
\[ \log(0.55) = \log\left(\frac{5.5}{10}\right) = \log(5.5) - 1 = 0.7404 - 1 = -0.2596 \]
Thus,
\[ \text{pH}_{\text{mix}} = -(-0.2596) + 3 = 0.2596 + 3 = 3.2596 \approx 3.29 \]

### Final Answer:
The pH of the mixed solution is approximately **3.29**.

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