Calculate the extent of hydrolysis & the pH of 0.02 M `CH_3COONH_4`.
`[K_(b) (NH_(3))=1.8 xx 10^(-5), K_(a) (CH_(3)COOH)=1.8 xx 10^(-5)]`

Answer

To solve the problem of calculating the extent of hydrolysis and the pH of a 0.02 M solution of ammonium acetate (CH₃COONH₄), we will follow these steps:

### Step 1: Understand the Hydrolysis Reaction
Ammonium acetate dissociates in water to form ammonium ions (NH₄⁺) and acetate ions (CH₃COO⁻). The hydrolysis of these ions can be represented as:
\[ \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+ \]
\[ \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \]

### Step 2: Determine the Equilibrium Constants
We know:
- \( K_a \) for acetic acid (CH₃COOH) = \( 1.8 \times 10^{-5} \)
- \( K_b \) for ammonia (NH₃) = \( 1.8 \times 10^{-5} \)

### Step 3: Calculate the Hydrolysis Constant (K_h)
Using the relationship:
\[ K_h = \frac{K_w}{K_a \cdot K_b} \]
Where \( K_w = 1.0 \times 10^{-14} \) at 25°C.

Substituting the values:
\[ K_h = \frac{1.0 \times 10^{-14}}{(1.8 \times 10^{-5}) \cdot (1.8 \times 10^{-5})} \]
\[ K_h = \frac{1.0 \times 10^{-14}}{3.24 \times 10^{-10}} \]
\[ K_h \approx 3.09 \times 10^{-5} \]

### Step 4: Set Up the Hydrolysis Equation
Let \( x \) be the extent of hydrolysis. The initial concentration of ammonium acetate is 0.02 M. At equilibrium:
- Concentration of NH₄⁺ = \( 0.02 - x \)
- Concentration of CH₃COO⁻ = \( 0.02 - x \)
- Concentration of NH₃ = \( x \)
- Concentration of H₃O⁺ = \( x \)

Using the hydrolysis constant:
\[ K_h = \frac{[NH_3][H_3O^+]}{[NH_4^+]} \]
\[ K_h = \frac{x^2}{0.02 - x} \]

### Step 5: Solve for x
Assuming \( x \) is small compared to 0.02, we can approximate:
\[ K_h \approx \frac{x^2}{0.02} \]
Thus,
\[ x^2 = K_h \cdot 0.02 \]
\[ x^2 = 3.09 \times 10^{-5} \cdot 0.02 \]
\[ x^2 = 6.18 \times 10^{-7} \]
\[ x = \sqrt{6.18 \times 10^{-7}} \]
\[ x \approx 0.000785 \, \text{M} \]

### Step 6: Calculate the Percentage Hydrolysis
Percentage hydrolysis = \( \frac{x}{\text{initial concentration}} \times 100 \)
\[ \text{Percentage hydrolysis} = \frac{0.000785}{0.02} \times 100 \approx 3.93\% \]

### Step 7: Calculate the pH
To find the pH, we can use the concentration of H₃O⁺:
\[ [H_3O^+] = x \approx 0.000785 \]
\[ pH = -\log[H_3O^+] \]
\[ pH = -\log(0.000785) \approx 3.104 \]

### Final Answer
- Extent of hydrolysis: \( 0.000785 \, \text{M} \)
- Percentage hydrolysis: \( 3.93\% \)
- pH: \( 3.10 \)