At `25^(@)C`, the ionic mobility of `CH_(3)COO^(-)`, `H^(+)` are respectively `4.1xx10^(-4),3.63xx10^(-3)`cm/sec. The conductivity of `0.01"M CH"_(3)COOH` is `5xx10^(-5)" S.cm"^(-1)`. Dissociation constant of `CH_(3)COOH` is

Options

A. `1.34xx10^(-4)`
B. `3xx10^(-4)`
C. `3xx10^(-5)`
D. `3xx10^(-6)`

Explanation

To find the dissociation constant \( K_a \) of acetic acid (\( CH_3COOH \)), we can follow these steps:

### Step 1: Calculate the total ionic mobility
The total ionic mobility \( \mu \) of the ions produced when acetic acid dissociates can be calculated by summing the ionic mobilities of the ions involved:

\[ \mu = \mu_{CH_3COO^-} + \mu_{H^+} \]

Given:
- \( \mu_{CH_3COO^-} = 4.1 \times 10^{-4} \, \text{cm}^2/\text{sec} \)
- \( \mu_{H^+} = 3.63 \times 10^{-3} \, \text{cm}^2/\text{sec} \)

Calculating \( \mu \):

\[ \mu = 4.1 \times 10^{-4} + 3.63 \times 10^{-3} = 4.04 \times 10^{-3} \, \text{cm}^2/\text{sec} \]

### Step 2: Calculate the degree of dissociation \( \alpha \)
The degree of dissociation \( \alpha \) can be calculated using the formula for conductivity \( \kappa \):

\[ \kappa = C \cdot F \cdot \mu \]

Where:
- \( \kappa = 5 \times 10^{-5} \, \text{S/cm} \) (conductivity)
- \( C = 0.01 \, \text{M} \) (concentration)
- \( F \) is the Faraday constant (approximately \( 96500 \, \text{C/mol} \))

Rearranging the formula to solve for \( \alpha \):

\[ \alpha = \frac{\kappa}{C \cdot \mu} \]

Substituting the values:

\[ \alpha = \frac{5 \times 10^{-5}}{0.01 \cdot 4.04 \times 10^{-3}} = 0.012 \]

### Step 3: Calculate the dissociation constant \( K_a \)
The dissociation constant \( K_a \) for the reaction:

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

is given by:

\[ K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \]

At equilibrium:
- \( [CH_3COO^-] = \alpha C \)
- \( [H^+] = \alpha C \)
- \( [CH_3COOH] = C(1 - \alpha) \)

Substituting these into the \( K_a \) expression:

\[ K_a = \frac{(\alpha C)(\alpha C)}{C(1 - \alpha)} = \frac{\alpha^2 C}{1 - \alpha} \]

Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):

\[ K_a \approx \alpha^2 C \]

Substituting the values:

\[ K_a \approx (0.012)^2 \cdot 0.01 = 1.44 \times 10^{-6} \]

### Final Answer
The dissociation constant \( K_a \) of acetic acid is approximately \( 1.44 \times 10^{-6} \).

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