The pH of a buffer solution of 0.1 M `CH_(3)COOH` and 0.1 `MCH_(3)COONa` is
`(pK_(a)CH_(3)COOH =4.745)`

Options

A. 4.745
B. 3.745
C. 5.745
D. 6.745

Explanation

To find the pH of the buffer solution composed of 0.1 M acetic acid (CH₃COOH) and 0.1 M sodium acetate (CH₃COONa), we can use the Henderson-Hasselbalch equation:

### Step-by-Step Solution:

1. **Identify the components of the buffer solution**:
- The acid is acetic acid (CH₃COOH).
- The conjugate base (salt) is sodium acetate (CH₃COONa).

2. **Write the Henderson-Hasselbalch equation**:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

3. **Substitute the known values into the equation**:
- Given:
- \(\text{pK}_a = 4.745\)
- \([\text{Salt}] = 0.1 \, M\) (concentration of CH₃COONa)
- \([\text{Acid}] = 0.1 \, M\) (concentration of CH₃COOH)

\[ \text{pH} = 4.745 + \log \left( \frac{0.1}{0.1} \right) \]

4. **Calculate the logarithmic term**:
- Since \(\frac{0.1}{0.1} = 1\), we have:
\[ \log(1) = 0 \]

5. **Final calculation of pH**:
\[ \text{pH} = 4.745 + 0 = 4.745 \]

### Conclusion:
The pH of the buffer solution is **4.745**.

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