Calculate the degree of hydrolysis of a decimolar solution of ammonium acetate at `25^@C`. `K_a(CH_3COOH)=1.75xx10^(-5), K_b(NH_4OH)=1.81xx10^(-5)`

Explanation

To calculate the degree of hydrolysis (h) of a decimolar (0.1 M) solution of ammonium acetate at 25°C, we can use the formula for the degree of hydrolysis for a salt formed from a weak acid and a weak base:

\[ h = \frac{K_w}{K_a \cdot K_b} \]

### Step-by-Step Solution:

1. **Identify the Constants**:
- Given:
- \( K_a \) (for acetic acid, \( CH_3COOH \)) = \( 1.75 \times 10^{-5} \)
- \( K_b \) (for ammonium hydroxide, \( NH_4OH \)) = \( 1.81 \times 10^{-5} \)
- \( K_w \) (ion product of water at 25°C) = \( 1.0 \times 10^{-14} \)

2. **Substitute the Values into the Formula**:
- Using the formula for degree of hydrolysis:
\[ h = \frac{K_w}{K_a \cdot K_b} \]
- Substitute the values:
\[ h = \frac{1.0 \times 10^{-14}}{(1.75 \times 10^{-5}) \cdot (1.81 \times 10^{-5})} \]

3. **Calculate \( K_a \cdot K_b \)**:
- First, calculate the product of \( K_a \) and \( K_b \):
\[ K_a \cdot K_b = (1.75 \times 10^{-5}) \cdot (1.81 \times 10^{-5}) = 3.1775 \times 10^{-10} \]

4. **Calculate Degree of Hydrolysis (h)**:
- Now substitute back into the equation for \( h \):
\[ h = \frac{1.0 \times 10^{-14}}{3.1775 \times 10^{-10}} \approx 3.15 \times 10^{-5} \]

5. **Final Result**:
- The degree of hydrolysis of the decimolar solution of ammonium acetate at 25°C is approximately \( 3.15 \times 10^{-5} \).