pH of a solution of salt of weak acid and weak base is :
`pH=1/2pK_w+1/2pK_a-1/2pK_b`
and that of weak acid and strong base is
`pH=1/2pK_w+1/2pK_a+1/2logc`
pH of 0.1 M solution of ammonium cyanide (`pK_a`=9.02 and `pK_b` =4.76 ) is

Options

A. 13.78
B. 2.26
C. 9.13
D. 8.13

Explanation

To find the pH of a 0.1 M solution of ammonium cyanide (NH4CN), we will use the formula for the pH of a salt formed from a weak acid and a weak base. The formula is:

\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b \]

### Step-by-Step Solution:

1. **Identify the Constants**:
- Given:
- \( pK_a = 9.02 \) (for HCN)
- \( pK_b = 4.76 \) (for NH4OH)
- \( pK_w = 14 \) (at 25°C)

2. **Substitute the Values into the Formula**:
- Plugging in the values into the formula:
\[ \text{pH} = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b \]
\[ = \frac{1}{2} (14) + \frac{1}{2} (9.02) - \frac{1}{2} (4.76) \]

3. **Calculate Each Term**:
- Calculate \( \frac{1}{2} pK_w \):
\[ \frac{1}{2} (14) = 7 \]
- Calculate \( \frac{1}{2} pK_a \):
\[ \frac{1}{2} (9.02) = 4.51 \]
- Calculate \( \frac{1}{2} pK_b \):
\[ \frac{1}{2} (4.76) = 2.38 \]

4. **Combine the Results**:
- Now combine these results into the pH formula:
\[ \text{pH} = 7 + 4.51 - 2.38 \]
\[ = 7 + 4.51 - 2.38 = 9.13 \]

5. **Final Result**:
- Thus, the pH of the 0.1 M solution of ammonium cyanide is:
\[ \text{pH} = 9.13 \]