The pH of a buffer solution prepared by adding 10 mL of `0.1 ` M `CH_(3) COOH` and 20 mL `0.1` M sodium acetate will be ( given : `pK_(a) ` of `CH_(3)COOH = 4.74` )

Options

A. `4.05`
B. `3.04`
C. `5.04`
D. `3.05`

Explanation

To find the pH of the buffer solution prepared by mixing 10 mL of 0.1 M acetic acid (CH₃COOH) and 20 mL of 0.1 M sodium acetate (CH₃COONa), we can use the Henderson-Hasselbalch equation:

### Step 1: Write the Henderson-Hasselbalch equation
The equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 2: Identify the components
- **Acid**: Acetic acid (CH₃COOH)
- **Salt**: Sodium acetate (CH₃COONa)
- Given: \( \text{pK}_a \) of acetic acid = 4.74

### Step 3: Calculate moles of acetic acid and sodium acetate
1. **Moles of Acetic Acid**:
\[ \text{Moles of CH}_3\text{COOH} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.010 \, \text{L} = 0.001 \, \text{mol} \]

2. **Moles of Sodium Acetate**:
\[ \text{Moles of CH}_3\text{COONa} = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times 0.020 \, \text{L} = 0.002 \, \text{mol} \]

### Step 4: Calculate concentrations in the final solution
The total volume of the buffer solution after mixing is:
\[ \text{Total Volume} = 10 \, \text{mL} + 20 \, \text{mL} = 30 \, \text{mL} = 0.030 \, \text{L} \]

1. **Concentration of Acetic Acid**:
\[ [\text{CH}_3\text{COOH}] = \frac{0.001 \, \text{mol}}{0.030 \, \text{L}} = 0.0333 \, \text{M} \]

2. **Concentration of Sodium Acetate**:
\[ [\text{CH}_3\text{COONa}] = \frac{0.002 \, \text{mol}}{0.030 \, \text{L}} = 0.0667 \, \text{M} \]

### Step 5: Substitute values into the Henderson-Hasselbalch equation
Now we can substitute the values into the equation:
\[ \text{pH} = 4.74 + \log\left(\frac{0.0667}{0.0333}\right) \]

### Step 6: Calculate the logarithm
\[ \frac{0.0667}{0.0333} = 2 \]
Thus,
\[ \log(2) \approx 0.301 \]

### Step 7: Final calculation of pH
\[ \text{pH} = 4.74 + 0.301 = 5.041 \]

### Conclusion
The pH of the buffer solution is approximately **5.04**.

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