The conductivity of 0.001 M acetic acid is `5xx10^(-5)S cm^(-1)` and `^^^(@)` is 390.5 `S cm^(2) "mol"^(-1)` then the calculated value of dissociation constant of acetic acid would be

Options

A. `81.78xx10^(-4)`
B. `81.78xx10^(-5)`
C. `18.78xx10^(-6)`
D. `18.78xx10^(-5)`

Answer

To calculate the dissociation constant of acetic acid from the given data, we will follow these steps:

### Step 1: Calculate Molar Conductivity (λM)
The molar conductivity (λM) can be calculated using the formula:

\[ \lambda_M = \frac{\text{Conductivity} \times 1000}{\text{Concentration}} \]

Given:
- Conductivity = \(5 \times 10^{-5} \, S \, cm^{-1}\)
- Concentration = \(0.001 \, M\)

Substituting the values:

\[ \lambda_M = \frac{5 \times 10^{-5} \times 1000}{0.001} = 50 \, S \, cm^{2} \, mol^{-1} \]

### Step 2: Calculate Degree of Dissociation (α)
The degree of dissociation (α) can be calculated using the formula:

\[ \alpha = \frac{\lambda_M}{\lambda_{M}^{\circ}} \]

Where:
- \(\lambda_{M}^{\circ}\) (molar conductivity at infinite dilution) = \(390.5 \, S \, cm^{2} \, mol^{-1}\)

Substituting the values:

\[ \alpha = \frac{50}{390.5} \approx 0.128 \]

### Step 3: Write the Expression for Dissociation Constant (K_a)
The dissociation of acetic acid can be represented as:

\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]

Let the initial concentration of acetic acid be \(C = 0.001 \, M\). The concentration of dissociated ions at equilibrium will be:

- Concentration of \(CH_3COOH\) at equilibrium = \(C(1 - \alpha)\)
- Concentration of \(CH_3COO^-\) and \(H^+\) at equilibrium = \(C\alpha\)

The expression for the dissociation constant \(K_a\) is given by:

\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 4: Simplify the Expression
Since \(\alpha\) is small, we can approximate \(1 - \alpha \approx 1\):

\[ K_a \approx \frac{C\alpha^2}{1} = C\alpha^2 \]

### Step 5: Substitute Values to Calculate \(K_a\)
Substituting \(C = 0.001 \, M\) and \(\alpha = 0.128\):

\[ K_a \approx 0.001 \times (0.128)^2 \]

Calculating:

\[ K_a \approx 0.001 \times 0.016384 = 1.6384 \times 10^{-5} \, M \]

### Final Result
Thus, the calculated value of the dissociation constant \(K_a\) of acetic acid is approximately:

\[ K_a \approx 16.38 \times 10^{-6} \, M \]