Hydrofluoric acid is a weak acid. At `25^(@)C`, the molar conductivity of 0.002 M HF is `176.2 Omega^(-1)cm^(2) mol^(-1)`. If its `Lambda_(m)^(@)=405.2Omega^(-1)cm^(2)mol^(-1)`. Equilibrium constant at the given concentration is

Options

A. `6.7xx10^(-4)M`
B. `3.2xx10^(-4)M`
C. `6.7xx10^(-5)M`
D. `3.2xx10^(-5)M`

Explanation

To solve the problem, we need to find the equilibrium constant (K) for the dissociation of hydrofluoric acid (HF) at a concentration of 0.002 M. Here are the steps to arrive at the solution:

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

\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \]

Where:
- \(\Lambda_m\) = Molar conductivity of the solution = 176.2 Ω⁻¹ cm² mol⁻¹
- \(\Lambda_m^0\) = Limiting molar conductivity = 405.2 Ω⁻¹ cm² mol⁻¹

Substituting the values:

\[ \alpha = \frac{176.2}{405.2} \approx 0.435 \]

### Step 2: Set up the dissociation equation
The dissociation of hydrofluoric acid can be represented as:

\[ HF \rightleftharpoons H^+ + F^- \]

Let the initial concentration of HF be \(C = 0.002 \, \text{M}\). At equilibrium:
- Concentration of \(HF\) = \(C(1 - \alpha)\)
- Concentration of \(H^+\) = \(C\alpha\)
- Concentration of \(F^-\) = \(C\alpha\)

### Step 3: Write the expression for the equilibrium constant (K)
The equilibrium constant (K) for the dissociation of HF can be expressed as:

\[ K = \frac{[H^+][F^-]}{[HF]} \]

Substituting the equilibrium concentrations:

\[ K = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 4: Substitute the values into the equilibrium constant expression
Now substituting \(C = 0.002 \, \text{M}\) and \(\alpha = 0.435\):

\[ K = \frac{0.002 \times (0.435)^2}{1 - 0.435} \]

Calculating \((0.435)^2\):

\[ (0.435)^2 \approx 0.189225 \]

Now substituting this back into the equation for K:

\[ K = \frac{0.002 \times 0.189225}{1 - 0.435} = \frac{0.00037845}{0.565} \]

Calculating the final value:

\[ K \approx 0.000669 \, \text{M} \approx 6.69 \times 10^{-4} \, \text{M} \]

### Final Answer
Thus, the equilibrium constant \(K\) is approximately:

\[ K \approx 6.7 \times 10^{-4} \, \text{M} \]

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