The pH of a solution obtained by mixing 100 mL of a solution pH=3 with 400 mL of a solution of pH=4 is

Options

A. 3- log 2.8
B. 7- log 2.8
C. 4- log 2.8
D. 5- log 2.8

Explanation

To find the pH of a solution obtained by mixing 100 mL of a solution with pH = 3 and 400 mL of a solution with pH = 4, we can follow these steps:

### Step 1: Calculate the concentration of H⁺ ions for each solution.

1. **For the solution with pH = 3:**
\[ \text{pH} = 3 \implies [\text{H}^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} \]

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

### Step 2: Calculate the total moles of H⁺ ions in each solution.

1. **For the 100 mL solution (pH = 3):**
\[ \text{Moles of H}^+ = [\text{H}^+] \times \text{Volume} = 10^{-3} \, \text{M} \times 0.1 \, \text{L} = 10^{-4} \, \text{moles} \]

2. **For the 400 mL solution (pH = 4):**
\[ \text{Moles of H}^+ = [\text{H}^+] \times \text{Volume} = 10^{-4} \, \text{M} \times 0.4 \, \text{L} = 4 \times 10^{-5} \, \text{moles} \]

### Step 3: Calculate the total moles of H⁺ ions in the mixed solution.

\[ \text{Total moles of H}^+ = 10^{-4} + 4 \times 10^{-5} = 10^{-4} + 0.4 \times 10^{-4} = 1.4 \times 10^{-4} \, \text{moles} \]

### Step 4: Calculate the total volume of the mixed solution.

\[ \text{Total Volume} = 100 \, \text{mL} + 400 \, \text{mL} = 500 \, \text{mL} = 0.5 \, \text{L} \]

### Step 5: Calculate the concentration of H⁺ ions in the mixed solution.

\[ [\text{H}^+] = \frac{\text{Total moles of H}^+}{\text{Total Volume}} = \frac{1.4 \times 10^{-4} \, \text{moles}}{0.5 \, \text{L}} = 2.8 \times 10^{-4} \, \text{M} \]

### Step 6: Calculate the pH of the mixed solution.

\[ \text{pH} = -\log([\text{H}^+]) = -\log(2.8 \times 10^{-4}) \]
Using a calculator:
\[ \text{pH} \approx 3.55 \]

### Final Answer:
The pH of the solution obtained by mixing 100 mL of a solution with pH = 3 and 400 mL of a solution with pH = 4 is approximately **3.55**.

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