Question

An aqueous solution of 0.1 M HA shows depression in freezing point of 0.2°C. If $ K_f $ (H$_2$O) = 1.86 K kg mol$^{-1}$ and assuming molarity = molality, find the dissociation constant of HA.

• A. \( 4.50 \times 10^{-5} \)
• B. \( 6.25 \times 10^{-3} \)
• C. \( 5.625 \times 10^{-4} \)
• D. \( 2.65 \times 10^{-4} \)

Hint:

In the calculation of depression in freezing point, always remember to account for the dissociation factor \( i \), which helps in determining the degree of dissociation.

Answer 1

The Correct Option is C

We can use the formula for depression in freezing point: \[ \Delta T_f = K_f \cdot m \cdot i \] where:
\( \Delta T_f = 0.2 \, \text{°C} \),
\( K_f = 1.86 \, \text{K kg mol}^{-1} \),
\( m = 0.1 \, \text{mol/kg} \) (molarity = molality),
\( i \) is the van't Hoff factor (dissociation constant).
Step 1: Apply the formula for depression in freezing point:
\[ 0.2 = 1.86 \times 0.1 \times i \] Step 2: Solve for \( i \): \[ i = \frac{0.2}{1.86 \times 0.1} = \frac{0.2}{0.186} \approx 1.075 \] Step 3: Calculate the dissociation constant (K):
For the dissociation constant, \( i = 1 + \alpha \) (where \( \alpha \) is the degree of dissociation), so: \[ i = 1 + \alpha = 1.075 \Rightarrow \alpha = 0.075 \] Thus, the dissociation constant is approximately \( 5.625 \times 10^{-4} \).