Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)`

Explanation

To calculate the dissociation constant \( K_a \) of acetic acid from the given data, we can follow these steps:

### Step 1: Understand the given data
We are provided with:
- Normality of acetic acid solution, \( N = 0.05 \)
- Equivalent conductance of the solution, \( \Lambda_{eq} = 7.36 \, \text{mho cm}^{-2} \)
- Equivalent conductance at infinite dilution, \( \Lambda_{eq}^{\infty} = 390.70 \, \text{mho cm}^{-2} \)

### Step 2: Calculate the degree of dissociation (\( \alpha \))
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_{eq}}{\Lambda_{eq}^{\infty}} \]
Substituting the values:
\[ \alpha = \frac{7.36}{390.70} \approx 0.0188 \]

### Step 3: Calculate the concentration of acetic acid (\( C \))
Since we are given the normality, we can directly use it as the concentration for weak acids:
\[ C = 0.05 \, \text{N} = 0.05 \, \text{mol/L} \]

### Step 4: Calculate \( K_a \)
For a weak acid, the dissociation constant \( K_a \) can be calculated using the formula:
\[ K_a = C \cdot \alpha^2 \]
Substituting the values:
\[ K_a = 0.05 \cdot (0.0188)^2 \]
Calculating \( (0.0188)^2 \):
\[ (0.0188)^2 \approx 0.00035344 \]
Now substituting this back into the equation for \( K_a \):
\[ K_a = 0.05 \cdot 0.00035344 \approx 1.7672 \times 10^{-5} \, \text{mol/L} \]

### Final Answer:
Thus, the dissociation constant \( K_a \) of acetic acid is approximately:
\[ K_a \approx 1.76 \times 10^{-5} \, \text{mol/L} \]

---