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 sodium acetate (CH3COONa) with a concentration of C M, we can follow these steps:

### Step 1: Understand the nature of sodium acetate
Sodium acetate is a salt formed from a weak acid (acetic acid, CH3COOH) and a strong base (sodium hydroxide, NaOH). In an aqueous solution, it undergoes hydrolysis.

### Step 2: Write the hydrolysis reaction
When sodium acetate is dissolved in water, it dissociates into sodium ions (Na⁺) and acetate ions (CH3COO⁻). The acetate ions can react with water as follows:
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 3: Determine the pH expression
Since acetate is the conjugate base of acetic acid, we can use the following relationship to find the pH:
\[ \text{pH} = \frac{1}{2} \text{pK}_w + \frac{1}{2} \text{pK}_a + \frac{1}{2} \log C \]
Where:
- \( \text{pK}_w \) is the ion product of water (approximately 14 at 25°C).
- \( \text{pK}_a \) is the dissociation constant of acetic acid.
- \( C \) is the concentration of the sodium acetate solution.

### Step 4: Substitute known values
Assuming we know the values of \( \text{pK}_w \) and \( \text{pK}_a \) for acetic acid, we can plug these into the equation. For acetic acid, \( \text{pK}_a \) is approximately 4.76.

### Step 5: Final expression for pH
Thus, the final expression for the pH of the sodium acetate solution can be summarized as:
\[ \text{pH} = \frac{1}{2} \times 14 + \frac{1}{2} \times 4.76 + \frac{1}{2} \log C \]

### Step 6: Conclusion
The pH of an aqueous solution of CH3COONa of concentration C(M) is given by:
\[ \text{pH} = \frac{1}{2} \text{pK}_w + \frac{1}{2} \text{pK}_a + \frac{1}{2} \log C \]