At T (K), the molar conductivity of 0.04 M acetic acid is 7.8 S `cm^(2) mol^(-1)`. If the limiting molar conductivities of `H^(+)` and `CH_(3)COO^(-)` at T (K) are 349 and 41 S `cm^(2) mol^(-1)` repsectively, the dissociation constant of acetic acid is :

Explanation

To find the dissociation constant of acetic acid (CH₃COOH), we can follow these steps:

### Step 1: Calculate the limiting molar conductivity (Λₘ⁰) of acetic acid
The limiting molar conductivity of acetic acid can be calculated using the limiting molar conductivities of its ions:
\[ \Lambda_m^0 = \Lambda_{H^+} + \Lambda_{CH_3COO^-} \]
Given:
- \(\Lambda_{H^+} = 349 \, \text{S cm}^{-2} \text{mol}^{-1}\)
- \(\Lambda_{CH_3COO^-} = 41 \, \text{S cm}^{-2} \text{mol}^{-1}\)

Calculating:
\[ \Lambda_m^0 = 349 + 41 = 390 \, \text{S cm}^{-2} \text{mol}^{-1} \]

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \]
Where:
- \(\Lambda_m = 7.8 \, \text{S cm}^{-2} \text{mol}^{-1}\)
- \(\Lambda_m^0 = 390 \, \text{S cm}^{-2} \text{mol}^{-1}\)

Calculating:
\[ \alpha = \frac{7.8}{390} \approx 0.02 \]

### Step 3: Calculate the dissociation constant (K_d)
The dissociation constant (K_d) for acetic acid can be calculated using the formula:
\[ K_d = \frac{C \alpha^2}{1 - \alpha} \]
Where:
- \(C = 0.04 \, \text{mol/L}\)
- \(\alpha \approx 0.02\)

Calculating:
\[ K_d = \frac{0.04 \times (0.02)^2}{1 - 0.02} \]
\[ = \frac{0.04 \times 0.0004}{0.98} \]
\[ = \frac{0.000016}{0.98} \approx 1.63 \times 10^{-5} \]

### Final Answer
The dissociation constant of acetic acid (K_d) is approximately:
\[ K_d \approx 1.63 \times 10^{-5} \]

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