At `25^(@)C` the equivalent conductance of butanoic acid at infinite dilution is `386.6 ohm^(-1)cm^(2)eq^(-1)`. If the ionization consatant is `1.4xx10^(-5)`, calculate equivalent conductance of `0.05` N butanoic acid solution at `25^(@)C(ohm^(-1)cm^(2)eq^(-1))`?

Options

A. `3.87`
B. `6.46`
C. `6.94`
D. `4.38`

Explanation

To solve the problem, we need to calculate the equivalent conductance of a 0.05 N butanoic acid solution at 25°C using the given information about its equivalent conductance at infinite dilution and its ionization constant.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Equivalent conductance at infinite dilution (Λ°) = 386.6 ohm⁻¹ cm² eq⁻¹
- Ionization constant (Ka) = 1.4 × 10⁻⁵
- Normality of the solution (N) = 0.05 N

2. **Calculate the Degree of Ionization (α):**
The degree of ionization (α) can be calculated using the formula:
\[ K_a = \frac{C \cdot \alpha^2}{1 - \alpha} \]
where C is the concentration of the acid. Since we are given the normality (0.05 N), we can assume C = 0.05 N for a monoprotic acid.

Rearranging the formula for small α (assuming α is much less than 1):
\[ K_a \approx C \cdot \alpha^2 \]
Thus,
\[ \alpha^2 = \frac{K_a}{C} = \frac{1.4 \times 10^{-5}}{0.05} \]
\[ \alpha^2 = 2.8 \times 10^{-4} \]
\[ \alpha = \sqrt{2.8 \times 10^{-4}} \approx 0.0167 \]

3. **Calculate the Equivalent Conductance (Λ) of the Solution:**
The equivalent conductance of the solution (Λ) can be calculated using the formula:
\[ \Lambda = \Lambda^\circ \cdot \alpha \]
Substituting the values:
\[ \Lambda = 386.6 \cdot 0.0167 \approx 6.44 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

4. **Final Result:**
The equivalent conductance of the 0.05 N butanoic acid solution at 25°C is approximately **6.44 ohm⁻¹ cm² eq⁻¹**.