The equivalent conductance of `(M)/(20)`solution of a weak
monobasic acid is 10 mhos `cm^(2)` and at infinite dilution is 200 mhos `cm^(2)`. The dissociation consant of this acid is :-

Options

A. `1.25xx10^(-4)`
B. `1.25xx10^(-5)`
C. `1.25xx10^(-6)`
D. `6.26xx10^(-4)`

Explanation

To find the dissociation constant (Ka) of the weak monobasic acid, we can follow these steps:

### Step 1: Understand the given information
We are given:
- The equivalent conductance (Λ) of a \( \frac{M}{20} \) solution of the acid is 10 mhos \( cm^2 \).
- The equivalent conductance at infinite dilution (Λ∞) is 200 mhos \( cm^2 \).

### Step 2: Convert the concentration
The concentration of the acid solution is:
\[ C = \frac{M}{20} = \frac{1}{20} \text{ M} = 0.05 \text{ M} \]

### Step 3: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_{\infty}} \]
Substituting the values we have:
\[ \alpha = \frac{10}{200} = 0.05 \]

### Step 4: Use the formula for the dissociation constant (Ka)
For a weak monobasic acid, the dissociation constant (Ka) is given by:
\[ K_a = C \cdot \alpha^2 \]
Substituting the values of C and α:
\[ K_a = 0.05 \cdot (0.05)^2 \]
\[ K_a = 0.05 \cdot 0.0025 = 0.000125 \]

### Step 5: Express Ka in scientific notation
\[ K_a = 1.25 \times 10^{-4} \]

### Final Answer
The dissociation constant (Ka) of the weak monobasic acid is:
\[ K_a = 1.25 \times 10^{-4} \]

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