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 solve the problem, we need to find the equilibrium constant for the reaction between acetic acid (CH₃COOH) and ammonium hydroxide (NH₄OH). Here are the steps to arrive at the solution:

### 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 \( K_1 \) for this reaction is given as \( 1.8 \times 10^{-5} \).

2. **Dissociation of Ammonium Hydroxide (NH₄OH)**:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]
The dissociation constant \( K_2 \) for this reaction is also given as \( 1.8 \times 10^{-5} \).

### Step 2: Write the neutralization reaction

When acetic acid reacts with ammonium hydroxide, the reaction can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NH}_4\text{OH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{NH}_4^+ + \text{H}_2\text{O} \]
This is a neutralization reaction that forms a salt (CH₃COONH₄) and water.

### Step 3: Relate the equilibrium constants

According to the principle of equilibrium constants, when you add two reactions, their equilibrium constants multiply. Therefore, the equilibrium constant \( K_3 \) for the neutralization reaction can be expressed as:
\[ K_3 = K_1 \times K_2 \]

### Step 4: Substitute the values of \( K_1 \) and \( K_2 \)

Since \( K_1 = K_2 = 1.8 \times 10^{-5} \), we can substitute:
\[ K_3 = (1.8 \times 10^{-5}) \times (1.8 \times 10^{-5}) = (1.8 \times 10^{-5})^2 \]

### Step 5: Calculate \( K_3 \)

Now we calculate \( K_3 \):
\[ K_3 = 3.24 \times 10^{-10} \]

### Conclusion

Thus, the equilibrium constant for the reaction of acetic acid and ammonium hydroxide is:
\[ K_3 = 3.24 \times 10^{-10} \]