A buffer solution is prepared by mixing `10ml` of `1.0 M` acetic acid & `20 ml` of `0.5 M` sodium acetate and then diluted to `100ml` with distilled water. If the `pK_(a)` of `CH_(3)COOH` is `4.76`. What is the pH of the buffer solution prepared?

Options

A. `3.84`
B. `4.76`
C. `4.34`
D. `5.21`

Explanation

To find the pH of the buffer solution prepared by mixing acetic acid and sodium acetate, we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step 1: Calculate the moles of acetic acid and sodium acetate

1. **Moles of acetic acid (CH₃COOH)**:
- Volume = 10 mL = 0.010 L
- Concentration = 1.0 M
- Moles = Volume × Concentration = \(0.010 \, \text{L} \times 1.0 \, \text{mol/L} = 0.010 \, \text{mol}\)

2. **Moles of sodium acetate (CH₃COONa)**:
- Volume = 20 mL = 0.020 L
- Concentration = 0.5 M
- Moles = Volume × Concentration = \(0.020 \, \text{L} \times 0.5 \, \text{mol/L} = 0.010 \, \text{mol}\)

### Step 2: Calculate the total volume of the buffer solution

- Total volume after dilution = 100 mL = 0.100 L

### Step 3: Calculate the concentrations of acetic acid and sodium acetate in the final solution

1. **Concentration of acetic acid**:
\[ [\text{Acid}] = \frac{\text{Moles of Acid}}{\text{Total Volume}} = \frac{0.010 \, \text{mol}}{0.100 \, \text{L}} = 0.1 \, \text{M} \]

2. **Concentration of sodium acetate**:
\[ [\text{Salt}] = \frac{\text{Moles of Salt}}{\text{Total Volume}} = \frac{0.010 \, \text{mol}}{0.100 \, \text{L}} = 0.1 \, \text{M} \]

### Step 4: Substitute the values into the Henderson-Hasselbalch equation

- Given \( \text{pK}_a = 4.76 \)

\[ \text{pH} = 4.76 + \log \left( \frac{0.1}{0.1} \right) \]

Since \( \log(1) = 0 \):

\[ \text{pH} = 4.76 + 0 = 4.76 \]

### Final Answer

The pH of the buffer solution is **4.76**.

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