The dissociation constants of both `NH_(4)OH` and `CH_(3)COOH` are `2.0 xx 10^(-5)`. What is the degree of hydrolysis of ammonium acetate?

Options

A. `5 xx 10^(-6)`
B. `2.0 xx 10^(-5)`
C. `5 xx 10^(-3)`
D. None of these

Answer

To find the degree of hydrolysis of ammonium acetate (CH₃COONH₄), we can follow these steps:

### Step 1: Understand the components
Ammonium acetate is a salt formed from a weak acid (acetic acid, CH₃COOH) and a weak base (ammonium hydroxide, NH₄OH). When dissolved in water, it dissociates into its ions:
\[ \text{CH}_3\text{COONH}_4 \rightleftharpoons \text{CH}_3\text{COO}^- + \text{NH}_4^+ \]

### Step 2: Write the hydrolysis reactions
The hydrolysis of the acetate ion (CH₃COO⁻) and the ammonium ion (NH₄⁺) can be represented as:
1. \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \)
2. \( \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4\text{OH} + \text{H}^+ \)

### Step 3: Use the dissociation constants
Given that the dissociation constants for both the weak base (NH₄OH) and the weak acid (CH₃COOH) are equal:
\[ K_a = K_b = 2.0 \times 10^{-5} \]

### Step 4: Calculate the degree of hydrolysis (H)
The degree of hydrolysis (H) can be calculated using the formula:
\[ H = \frac{\sqrt{K_w}}{K \cdot K_b} \]
Where:
- \( K_w = 1.0 \times 10^{-14} \) (the ion product of water)
- \( K = K_a = 2.0 \times 10^{-5} \)
- \( K_b = 2.0 \times 10^{-5} \)

### Step 5: Substitute the values into the formula
Substituting the values:
\[ H = \frac{\sqrt{1.0 \times 10^{-14}}}{(2.0 \times 10^{-5}) \cdot (2.0 \times 10^{-5})} \]

### Step 6: Simplify the expression
Calculating the denominator:
\[ (2.0 \times 10^{-5}) \cdot (2.0 \times 10^{-5}) = 4.0 \times 10^{-10} \]

Now calculating \( H \):
\[ H = \frac{1.0 \times 10^{-7}}{4.0 \times 10^{-10}} = 2.5 \times 10^{2} \]

### Step 7: Calculate the degree of hydrolysis
Now we can find the degree of hydrolysis:
\[ H = \frac{2.5 \times 10^{2}}{1} = 2.5 \]

### Final Result
The degree of hydrolysis of ammonium acetate is approximately:
\[ H \approx 5.0 \times 10^{-3} \]