The pH of a solution at `25°C` containing 0.20 M sodium acetate and 0.06 M acetic acid is `(pK_a for CH_3COOH = 4.74 and log 3 = 0.477)`

Options

A. 4.36
B. 5.26
C. 5.84
D. 6.32

Answer

To find the pH of a solution containing sodium acetate and acetic acid, we can use the Henderson-Hasselbalch equation, which is given by:

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

### Step-by-step solution:

1. **Identify the components:**
- We have sodium acetate (the salt) and acetic acid (the weak acid).
- The concentrations are:
- \([\text{Sodium Acetate}] = 0.20 \, M\)
- \([\text{Acetic Acid}] = 0.06 \, M\)

2. **Use the given pK_a value:**
- We are given that \(\text{pK}_a\) for acetic acid (CH₃COOH) is \(4.74\).

3. **Substitute the values into the Henderson-Hasselbalch equation:**
\[ \text{pH} = 4.74 + \log\left(\frac{0.20}{0.06}\right) \]

4. **Calculate the ratio:**
\[ \frac{0.20}{0.06} = 3.33 \]

5. **Calculate the logarithm:**
- Using the given value \(\log 3 \approx 0.477\), we can calculate:
\[ \log(3.33) \approx \log(3) + \log(1.11) \approx 0.477 + 0.045 = 0.522 \quad (\text{since } \log(1.11) \text{ is small}) \]

6. **Substitute back into the pH equation:**
\[ \text{pH} = 4.74 + 0.522 = 5.262 \]

7. **Final rounding:**
- Rounding to two decimal places, we get:
\[ \text{pH} \approx 5.26 \]

### Conclusion:
The pH of the solution is approximately **5.26**.