The `pH` of a solution at `25^(@)C` containing `0.10 M` sodium acetate and `0.03 M` acetic acid is (`pK_(a)` for `CH_(3)COOH=4.57`)

Options

A. 4.09
B. 6.09
C. 5.09
D. 7.09

Explanation

To find the pH of a solution containing 0.10 M sodium acetate and 0.03 M acetic acid at 25°C, we can use the Henderson-Hasselbalch equation, which is given by:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Where:
- \([\text{A}^-]\) is the concentration of the base (sodium acetate in this case).
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid in this case).
- \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant.

### Step-by-step Solution:

1. **Identify the components**:
- Sodium acetate (CH₃COONa) acts as the base, so \([\text{A}^-] = 0.10 \, \text{M}\).
- Acetic acid (CH₃COOH) is the weak acid, so \([\text{HA}] = 0.03 \, \text{M}\).
- Given \(\text{pK}_a\) for acetic acid is 4.57.

2. **Substitute the values into the Henderson-Hasselbalch equation**:
\[ \text{pH} = 4.57 + \log \left( \frac{0.10}{0.03} \right) \]

3. **Calculate the ratio**:
\[ \frac{0.10}{0.03} = \frac{10}{3} \approx 3.33 \]

4. **Calculate the logarithm**:
\[ \log(3.33) \approx 0.522 \]

5. **Add the values**:
\[ \text{pH} = 4.57 + 0.522 = 5.092 \]

6. **Final result**:
The pH of the solution is approximately 5.09.