The dissociation constant of n-butyric acid is `1.6 xx 10^(-5)` and the molar conductivity at infinite dilution is `380 xx 10^(-4) Sm^(2) mol^(-1)`. The specific conductance of the `0.01M` acid solution is

Options

A. `1.52 xx 10^(-5) Sm^(-1)`
B. `1.52 xx 10^(-2) Sm^(-1)`
C. `1.52 xx 10^(-3) Sm^(-1)`
D. none

Explanation

To find the specific conductance (κ) of a 0.01 M n-butyric acid solution, we can follow these steps:

### Step 1: Understand the given data
- Dissociation constant (K_a) of n-butyric acid = \(1.6 \times 10^{-5}\)
- Molar conductivity at infinite dilution (λ_m) = \(380 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}\)
- Concentration of the acid solution (C) = 0.01 M

### Step 2: Write the expression for the dissociation constant
For a weak acid HA dissociating into H⁺ and A⁻, the dissociation can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

The expression for the dissociation constant (K_a) is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

At equilibrium, if α is the degree of dissociation, we can express the concentrations as:
- \([H^+] = C \cdot \alpha\)
- \([A^-] = C \cdot \alpha\)
- \([HA] = C(1 - \alpha)\)

Substituting these into the K_a expression gives:
\[ K_a = \frac{(C \cdot \alpha)(C \cdot \alpha)}{C(1 - \alpha)} = \frac{C \cdot \alpha^2}{1 - \alpha} \]

### Step 3: Rearrange to find α
Assuming α is small (α < 1), we can approximate \(1 - \alpha \approx 1\):
\[ K_a \approx C \cdot \alpha^2 \]
Thus, we can rearrange to find α:
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

### Step 4: Substitute the values to find α
Substituting the known values:
\[ \alpha = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} = 0.04 \]

### Step 5: Calculate the molar conductivity (λ)
The molar conductivity (λ) of the solution can be calculated using the relation:
\[ \alpha = \frac{\lambda}{\lambda_m} \]
Where λ_m is the molar conductivity at infinite dilution. Rearranging gives:
\[ \lambda = \alpha \cdot \lambda_m \]

### Step 6: Substitute the values to find λ
Substituting the values:
\[ \lambda = 0.04 \cdot (380 \times 10^{-4}) = 15.2 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \]

### Step 7: Convert λ to specific conductance (κ)
The specific conductance (κ) can be calculated using:
\[ \kappa = \frac{\lambda \cdot C}{1000} \]
Where C is in mol/L. Thus:
\[ \kappa = \frac{15.2 \times 10^{-4} \cdot 0.01}{1000} = 1.52 \times 10^{-4} \, \text{S/cm} \]

### Step 8: Convert κ to S/m
To convert S/cm to S/m, we multiply by 100:
\[ \kappa = 1.52 \times 10^{-4} \, \text{S/cm} \times 100 = 1.52 \times 10^{-2} \, \text{S/m} \]

### Final Answer
The specific conductance of the 0.01 M n-butyric acid solution is:
\[ \kappa = 1.52 \times 10^{-2} \, \text{S/m} \]

---