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 of the weak monobasic acid, we will follow these steps:

### Step 1: Calculate the Degree of Dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]
where:
- \(\Lambda\) = equivalent conductance of the solution = 6.0 mho cm² eq⁻¹
- \(\Lambda_0\) = equivalent conductance at infinite dilution = 200 mho cm² eq⁻¹

Substituting the values:
\[ \alpha = \frac{6.0}{200} = 0.03 \]

### Step 2: Use Ostwald's Dilution Law
According to Ostwald's dilution law, the dissociation constant (K) for a weak acid can be expressed as:
\[ K = \frac{C \alpha^2}{1 - \alpha} \]
where:
- \(C\) = concentration of the acid solution
- \(\alpha\) = degree of dissociation

### Step 3: Determine the Concentration (C)
The concentration of the solution is given as \(M/32\). Since we are dealing with a weak monobasic acid, we can express the concentration in molarity:
\[ C = \frac{1}{32} \text{ M} = 0.03125 \text{ M} \]

### Step 4: Substitute Values into the Dissociation Constant Formula
Now we can substitute the values into the dissociation constant formula. Since \(\alpha\) is much less than 1, we can approximate \(1 - \alpha \approx 1\):
\[ K = C \alpha^2 \]
Substituting \(C\) and \(\alpha\):
\[ K = 0.03125 \times (0.03)^2 \]
Calculating \((0.03)^2\):
\[ (0.03)^2 = 0.0009 \]
Now substituting this back:
\[ K = 0.03125 \times 0.0009 = 0.000028125 \]
Converting to scientific notation:
\[ K = 2.8125 \times 10^{-5} \]

### Step 5: Final Result
Thus, the dissociation constant of the weak monobasic acid is:
\[ K \approx 2.81 \times 10^{-5} \]