Dissociation constants of `CH_(3)COOH and NH_(4)OH` are `1.8xx10^(5)` each at `25^(@)C`. The equilibrium constant for the reaction of `CH_(3)COOH and NH_(4)OH` will be -

Options

A. `(1.8xx1.8)/(10^(4))`
B. `(1.8)/(10^(-9))`
C. `1.8xx1.8xx10^(4)`
D. `3.24xx10^(-10)`

Answer

To find the equilibrium constant for the reaction between acetic acid (CH₃COOH) and ammonium hydroxide (NH₄OH), we can follow these steps:

### Step 1: Write the dissociation reactions
1. **Dissociation of Acetic Acid (CH₃COOH)**:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
The dissociation constant \( K_a \) for this reaction is given as:
\[ K_a = 1.8 \times 10^{-5} \]

2. **Dissociation of Ammonium Hydroxide (NH₄OH)**:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
The dissociation constant \( K_b \) for this reaction is also given as:
\[ K_b = 1.8 \times 10^{-5} \]

### Step 2: Write the overall reaction
The overall reaction between acetic acid and ammonium hydroxide can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NH}_4\text{OH} \rightleftharpoons \text{CH}_3\text{COONH}_4 + \text{H}_2\text{O} \]

### Step 3: Relate the equilibrium constant to \( K_a \) and \( K_b \)
The equilibrium constant \( K_{eq} \) for the overall reaction can be expressed in terms of the dissociation constants:
\[ K_{eq} = \frac{K_a \cdot K_b}{K_w} \]
Where \( K_w \) is the ion product of water, which at 25°C is:
\[ K_w = 1.0 \times 10^{-14} \]

### Step 4: Substitute the values
Now substituting the values of \( K_a \), \( K_b \), and \( K_w \):
\[ K_{eq} = \frac{(1.8 \times 10^{-5}) \cdot (1.8 \times 10^{-5})}{1.0 \times 10^{-14}} \]

### Step 5: Calculate \( K_{eq} \)
Calculating the numerator:
\[ (1.8 \times 10^{-5}) \cdot (1.8 \times 10^{-5}) = 3.24 \times 10^{-10} \]
Now substituting into the equation for \( K_{eq} \):
\[ K_{eq} = \frac{3.24 \times 10^{-10}}{1.0 \times 10^{-14}} = 3.24 \times 10^{4} \]

### Final Answer
Thus, the equilibrium constant for the reaction is:
\[ K_{eq} = 3.24 \times 10^{4} \]