The equivalent conductance of `M//32` solution of a weak monobasic acid is `6.0` mho `cm^(2) eq^(–1)` and at infinite dilution is 200 mho `cm^(2) eq^(–1)`. The dissociation constant of this acid is:

Options

A. `1.25xx10^(-5)`
B. `1.25xx10^(-6)`
C. `2.81xx10^(-5)`
D. `1.25xx10^(-4)`

Answer

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

### Step 1: Understand the given data
- Equivalent conductance (Λ) of the M/32 solution = 6.0 mho cm² eq⁻¹
- Equivalent conductance at infinite dilution (Λ₀) = 200 mho cm² eq⁻¹
- Molarity of the solution (C) = 1/32 M

### 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 given values:

\[ \alpha = \frac{6.0}{200} \]

Calculating this:

\[ \alpha = 0.03 \]

### Step 3: Calculate the dissociation constant (Ka)
The dissociation constant (Ka) for a weak monobasic acid can be expressed as:

\[ K_a = C \cdot \alpha^2 \]

Where:
- C = concentration of the acid = 1/32 M = 0.03125 M
- α = degree of dissociation = 0.03

Now substituting these values into the equation:

\[ K_a = 0.03125 \cdot (0.03)^2 \]

Calculating (0.03)²:

\[ (0.03)^2 = 0.0009 \]

Now substituting this back into the equation for Ka:

\[ K_a = 0.03125 \cdot 0.0009 \]

Calculating this:

\[ K_a = 2.8125 \times 10^{-5} \]

### Final Answer:
The dissociation constant (Ka) of the weak monobasic acid is approximately:

\[ K_a \approx 2.81 \times 10^{-5} \]

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