The pH of an aqueous solution of CH3COONa of concentration C(M) is given by :

Options

A. `7-1/2 pK_(a) -1/2 logC`
B. `1/2 pK_(w) + 1/2 pK_(b) + 1/2 logC`
C. `1/2 pK_(w) - 1/2 pK_(b) + 1/2 logC`
D. `1/2 pK_(w) + 1/2 pK_(a) + 1/2 logC`

Explanation

To determine the pH of an aqueous solution of CH3COONa (sodium acetate) with a concentration of C M, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of the Salt**:
- Sodium acetate (CH3COONa) is composed of the acetate ion (CH3COO⁻) and sodium ion (Na⁺).
- The acetate ion is the conjugate base of acetic acid (CH3COOH), which is a weak acid.

2. **Understand the Hydrolysis of the Salt**:
- When CH3COONa is dissolved in water, it dissociates into Na⁺ and CH3COO⁻ ions.
- The acetate ion (CH3COO⁻) can hydrolyze in water to produce hydroxide ions (OH⁻):
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

3. **Determine the Nature of the Solution**:
- Since CH3COO⁻ is a weak base (conjugate base of a weak acid), the solution will be basic due to the production of OH⁻ ions.

4. **Calculate the pOH**:
- The equilibrium constant for the hydrolysis of the acetate ion can be expressed as:
\[ K_b = \frac{K_w}{K_a} \]
- Where \( K_w \) is the ion product of water (1.0 x 10⁻¹⁴ at 25°C) and \( K_a \) is the acid dissociation constant for acetic acid.

5. **Use the Henderson-Hasselbalch Equation**:
- For a weak base and its conjugate acid, the pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
- In this case, the base is CH3COO⁻ and the acid is CH3COOH. The concentration of the base is C M.

6. **Substituting Values**:
- Since we have a weak base (CH3COO⁻) and its conjugate acid (CH3COOH), we can substitute into the equation:
\[ \text{pH} = 14 - \frac{1}{2} \text{pK}_a - \frac{1}{2} \log C \]
- Here, \( \text{pK}_a \) is the negative logarithm of \( K_a \) for acetic acid.

7. **Final Expression**:
- The final expression for the pH of the solution is:
\[ \text{pH} = 7 + \frac{1}{2} \text{pK}_a + \frac{1}{2} \log C \]