Question

HA $ (aq) \rightleftharpoons H^+ (aq) + A^- (aq) $ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: $ K_f(H_2O) = 1.8 \, \text{K kg mol}^{-1} $, molality ≡ molarity

• A. \( 1.1 \times 10^{-2} \)
• B. \( 1.38 \times 10^{-3} \)
• C. \( 1.90 \times 10^{-3} \)
• D. \( 1.89 \times 10^{-1} \)

Hint:

- For weak electrolytes: \( i = 1 + (n-1)\alpha \) (n = ions produced) - Freezing point depression: \( \Delta T_f = iK_fm \) - \( K_a \) calculation for weak acid: \( K_a = \frac{C\alpha^2}{1-\alpha} \) - Approximation valid when \( \alpha < 0.1 \), otherwise use exact formula

Answer 1

The Correct Option is B

Step 1: Calculate van't Hoff factor (i) \[ \Delta T_f = i \cdot K_f \cdot m 0.20 = i \times 1.8 \times 0.1 i = \frac{0.20}{0.18} = 1.11 \] 
Step 2: Relate i to degree of dissociation \((\alpha)\) For weak acid dissociation: \[ i = 1 + \alpha 1.11 = 1 + \alpha \alpha = 0.11 \] 
Step 3: Calculate dissociation constant \((K_a)\) \[ K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.1 \times (0.11)^2}{1 - 0.11} = \frac{0.1 \times 0.0121}{0.89} = 1.36 \times 10^{-3} \approx 1.38 \times 10^{-3} \]