pH of a solution of salt of weak acid and weak base is :
`pH=1/2pK_w+1/2pK_a-pK_b`
and that of weak acid and strong base is
`pH=1/2pK_w+1/2pK_a+1/2logc`
The pH of 0.1 M sodium acetate (`K_a` for `CH_3COOH=1.8xx10^(-5)` ) is

Options

A. 8.87
B. 9.36
C. 4.86
D. 7.2

Explanation

To find the pH of a 0.1 M sodium acetate solution, we will use the formula for the pH of a salt derived from a weak acid and a strong base. The formula is:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} pK_a + \frac{1}{2} \log C \]

Where:
- \( pK_w \) is the ion product of water, which is 14 at 25°C.
- \( pK_a \) is the dissociation constant of the weak acid (acetic acid in this case).
- \( C \) is the concentration of the salt (sodium acetate).

### Step 1: Calculate \( pK_a \)

Given:
- \( K_a \) for acetic acid (\( CH_3COOH \)) = \( 1.8 \times 10^{-5} \)

We can calculate \( pK_a \) using the formula:

\[ pK_a = -\log(K_a) \]

Calculating \( pK_a \):

\[ pK_a = -\log(1.8 \times 10^{-5}) \]

Using a calculator:

\[ pK_a \approx 4.745 \]

### Step 2: Calculate \( pK_w \)

Since \( pK_w \) is a constant at 25°C:

\[ pK_w = 14 \]

### Step 3: Calculate \( \log C \)

Given:
- Concentration \( C = 0.1 \, M \)

Calculating \( \log C \):

\[ \log C = \log(0.1) = -1 \]

### Step 4: Substitute values into the pH formula

Now we can substitute the values into the pH formula:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} pK_a + \frac{1}{2} \log C \]

Substituting the values we found:

\[ \text{pH} = \frac{1}{2} (14) + \frac{1}{2} (4.745) + \frac{1}{2} (-1) \]

Calculating each term:

1. \( \frac{1}{2} (14) = 7 \)
2. \( \frac{1}{2} (4.745) \approx 2.3725 \)
3. \( \frac{1}{2} (-1) = -0.5 \)

Now summing these values:

\[ \text{pH} = 7 + 2.3725 - 0.5 \]

\[ \text{pH} = 9.8725 \]

### Final Result

The pH of the 0.1 M sodium acetate solution is approximately:

\[ \text{pH} \approx 9.87 \]