For `0.0128 N` solution fo acetic at `25^(@)C` equivalent conductance of the solution is 1.4 mho `cm^(3) eq^(-1)` and `lambda^(oo) = 391` mho `cm^(2)eq^(-1)`. Calculate dissociation constant `(K_(a))` of acetic acid.

Explanation

To calculate the dissociation constant (K_a) of acetic acid from the given data, we can follow these steps:

### Step 1: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:

\[ \alpha = \frac{\Lambda}{\Lambda^\infty} \]

Where:
- \(\Lambda\) is the equivalent conductance of the solution (1.4 mho cm\(^2\) eq\(^{-1}\))
- \(\Lambda^\infty\) is the equivalent conductance at infinite dilution (391 mho cm\(^2\) eq\(^{-1}\))

Substituting the values:

\[ \alpha = \frac{1.4}{391} \]

Calculating this gives:

\[ \alpha \approx 0.00357 \]

### Step 2: Calculate the concentration (C) of acetic acid
Given that the normality (N) of the solution is 0.0128 N, and for acetic acid (which is a weak acid), the concentration (C) in molarity is the same as the normality since it is a monoprotic acid:

\[ C = 0.0128 \, \text{mol/L} \]

### Step 3: Calculate the dissociation constant (K_a)
The dissociation constant \(K_a\) for a weak acid can be calculated using the formula:

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

Substituting the values:

\[ K_a = \frac{0.0128 \cdot (0.00357)^2}{1 - 0.00357} \]

Calculating \(\alpha^2\):

\[ (0.00357)^2 \approx 0.0000127569 \]

Now substituting back into the \(K_a\) formula:

\[ K_a = \frac{0.0128 \cdot 0.0000127569}{0.99643} \]

Calculating the numerator:

\[ 0.0128 \cdot 0.0000127569 \approx 0.000000163 \]

Now calculating \(K_a\):

\[ K_a \approx \frac{0.000000163}{0.99643} \approx 0.000000163 \]

Thus,

\[ K_a \approx 1.63 \times 10^{-7} \, \text{mol/L} \]

### Final Answer:
The dissociation constant \(K_a\) of acetic acid is approximately \(1.63 \times 10^{-7} \, \text{mol/L}\).

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