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 (NH4CH3COO), we can follow these steps:

### Step 1: Understand the Reaction
Ammonium acetate is formed from the weak base ammonium hydroxide (NH4OH) and the weak acid acetic acid (CH3COOH). The degree of hydrolysis (h) is the fraction of the salt that undergoes hydrolysis in water.

### Step 2: Write the Hydrolysis Equation
The hydrolysis of ammonium acetate can be represented as:
\[ NH_4CH_3COO + H_2O \rightleftharpoons NH_4^+ + CH_3COO^- \]

### Step 3: Use the Formula for Degree of Hydrolysis
The degree of hydrolysis (h) can be calculated using the formula:
\[ h = \sqrt{\frac{K_w}{K_a \cdot K_b}} \]
where:
- \( K_w \) is the ion product of water (\( 1.0 \times 10^{-14} \) at 25°C),
- \( K_a \) is the dissociation constant of acetic acid,
- \( K_b \) is the dissociation constant of ammonium hydroxide.

Given:
- \( K_a = 2.0 \times 10^{-5} \)
- \( K_b = 2.0 \times 10^{-5} \)

### Step 4: Substitute the Values
Substituting the values into the formula:
\[ h = \sqrt{\frac{1.0 \times 10^{-14}}{(2.0 \times 10^{-5}) \cdot (2.0 \times 10^{-5})}} \]

### Step 5: Calculate the Denominator
Calculate the denominator:
\[ (2.0 \times 10^{-5}) \cdot (2.0 \times 10^{-5}) = 4.0 \times 10^{-10} \]

### Step 6: Substitute and Simplify
Now substitute this back into the equation for h:
\[ h = \sqrt{\frac{1.0 \times 10^{-14}}{4.0 \times 10^{-10}}} \]

### Step 7: Simplify the Fraction
\[ h = \sqrt{2.5 \times 10^{-5}} \]

### Step 8: Calculate the Square Root
Now calculate the square root:
\[ h = \sqrt{2.5} \times 10^{-2.5} \]
\[ h \approx 1.58 \times 10^{-3} \]

### Step 9: Express in Scientific Notation
Expressing this in scientific notation gives:
\[ h \approx 5.0 \times 10^{-3} \]

### Final Answer
Thus, the degree of hydrolysis of ammonium acetate is approximately \( 5.0 \times 10^{-3} \).

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