The molar conductivity of `0.025 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Its degree of dissociation `(alpha)` and dissociation constant. Given `lambda^(@)(H^(+))=349.6 S cm^(-1)` and `lambda^(@)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1)`.

Options

A. 11.4%, `3.67xx10^(-4) "mol L"^(-1)`
B. 22.8%, `1.83xx10^(-4) "mol L"^(-1)`
C. 52.2%, `4.25xx10^(-4) "mol L"^(-1)`
D. 1.14%, `3.67xx10^(-6) "mol L"^(-1)`

Explanation

To solve the problem, we need to find the degree of dissociation (α) and the dissociation constant (Ka) for methanoic acid (HCOOH) given the molar conductivity and other parameters. Here’s a step-by-step solution:

### Step 1: Write down the given data
- Concentration of methanoic acid (C) = 0.025 mol L^(-1)
- Molar conductivity (λm) = 46.1 S cm^2 mol^(-1)
- Limiting molar conductivity of H^+ (λ⁰(H^+)) = 349.6 S cm^2 mol^(-1)
- Limiting molar conductivity of HCOO^− (λ⁰(HCOO^−)) = 54.6 S cm^2 mol^(-1)

### Step 2: Calculate the limiting molar conductivity of methanoic acid (λ⁰(HCOOH))
Using Kohlrausch's Law:
\[ \lambda^0(HCOOH) = \lambda^0(H^+) + \lambda^0(HCOO^−) \]
\[ \lambda^0(HCOOH) = 349.6 + 54.6 = 404.2 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda^0(HCOOH)} \]
Substituting the values:
\[ \alpha = \frac{46.1}{404.2} \]
Calculating α:
\[ \alpha \approx 0.114 \, \text{(or 11.4% when multiplied by 100)} \]

### Step 4: Calculate the dissociation constant (Ka)
The dissociation of methanoic acid can be represented as:
\[ HCOOH \rightleftharpoons H^+ + HCOO^- \]
Using the initial concentration (C) and the degree of dissociation (α):
- Initial concentration of HCOOH = C = 0.025 mol L^(-1)
- Change in concentration = -Cα for HCOOH, +Cα for H^+, and +Cα for HCOO^−

At equilibrium:
- Concentration of HCOOH = C(1 - α)
- Concentration of H^+ = Cα
- Concentration of HCOO^− = Cα

The expression for the dissociation constant (Ka) is:
\[ K_a = \frac{[H^+][HCOO^-]}{[HCOOH]} = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]
Substituting the values:
\[ K_a = \frac{0.025 \cdot (0.114)^2}{1 - 0.114} \]
Calculating:
\[ K_a = \frac{0.025 \cdot 0.012996}{0.886} \approx 3.67 \times 10^{-4} \, \text{mol L}^{-1} \]

### Final Results
- Degree of dissociation (α) = 11.4%
- Dissociation constant (Ka) = \(3.67 \times 10^{-4} \, \text{mol L}^{-1}\)

### Summary
The degree of dissociation of methanoic acid is 11.4%, and the dissociation constant is \(3.67 \times 10^{-4} \, \text{mol L}^{-1}\).

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