The values of limiting ionic conductance of `H^(+)` and `HCOO^(-)` ions are respectively 347 and 53 S `cm^(2) mol^(-1)` at 298 K, If the molar conductance of 0.025 M methanoic acid at 298 K is 40 `S cm^(2) mol^(-1)` , the dissociation constant of methanoic acid at 298 K is

Options

A. `1 xx 10^(-5)`
B. `2 xx 10^(-5)`
C. `1.5 xx 10^(-4)`
D. `2.5 xx 10^(-4)`

Explanation

To find the dissociation constant \( K_a \) of methanoic acid, we can follow these steps:

### Step 1: Determine the limiting molar conductance (\( \lambda_{\infty} \))
The limiting molar conductance for methanoic acid can be calculated using the limiting conductance values of its ions:
\[ \lambda_{\infty} = \lambda_{H^+} + \lambda_{HCOO^-} \]
Given:
- \( \lambda_{H^+} = 347 \, S \, cm^2 \, mol^{-1} \)
- \( \lambda_{HCOO^-} = 53 \, S \, cm^2 \, mol^{-1} \)

Calculating:
\[ \lambda_{\infty} = 347 + 53 = 400 \, S \, cm^2 \, mol^{-1} \]

### Step 2: Calculate the degree of dissociation (\( \alpha \))
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda_{\infty}} \]
Where:
- \( \lambda_m = 40 \, S \, cm^2 \, mol^{-1} \) (given)

Calculating:
\[ \alpha = \frac{40}{400} = 0.1 \]

### Step 3: Use the degree of dissociation to find \( K_a \)
The dissociation constant \( K_a \) can be calculated using the formula:
\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]
Where:
- \( C = 0.025 \, mol \, L^{-1} \) (concentration of methanoic acid)
- \( \alpha = 0.1 \)

Substituting the values:
\[ K_a = \frac{0.025 \times (0.1)^2}{1 - 0.1} \]
Calculating:
\[ K_a = \frac{0.025 \times 0.01}{0.9} = \frac{0.00025}{0.9} \approx 2.78 \times 10^{-4} \, mol \, L^{-1} \]

### Final Result
Thus, the dissociation constant \( K_a \) of methanoic acid at 298 K is approximately:
\[ K_a \approx 2.78 \times 10^{-4} \, mol \, L^{-1} \]