The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 " mho cm"^(2)` and at infinite dilution is `400 mho cm^(2)`. The dissociation constant of this acid is :

Options

A. `6.25 xx 10^(-4)`
B. `1.25 xx 10^(-6)`
C. `6.25 xx 10^(-5)`
D. `1.25 xx 10^(-5)`

Explanation

To find the dissociation constant (K) of a weak monobasic acid given its equivalent conductance at a certain concentration and at infinite dilution, we can follow these steps:

### Step 1: Understand the given data
- Equivalent conductance (Λ) of M/32 solution = 8.0 mho cm²
- Equivalent conductance at infinite dilution (Λ0) = 400 mho cm²

### Step 2: Calculate the degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]
Substituting the values:
\[ \alpha = \frac{8.0}{400} = 0.02 \]

### Step 3: Determine the concentration (C)
The concentration of the solution is given as M/32. Therefore, we can express the concentration in terms of molarity:
\[ C = \frac{1}{32} \text{ M} \]

### Step 4: Calculate the dissociation constant (K)
The dissociation constant (K) for a weak monobasic acid can be calculated using the formula:
\[ K = C \cdot \alpha^2 \]
Substituting the values we have:
\[ K = \left(\frac{1}{32}\right) \cdot (0.02)^2 \]
Calculating \( (0.02)^2 \):
\[ (0.02)^2 = 0.0004 \]
Now substituting back into the equation for K:
\[ K = \left(\frac{1}{32}\right) \cdot 0.0004 = \frac{0.0004}{32} = 0.0000125 \]
Converting this to scientific notation:
\[ K = 1.25 \times 10^{-5} \]

### Conclusion
The dissociation constant (K) of the weak monobasic acid is:
\[ K = 1.25 \times 10^{-5} \]