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

The dissociation constant for the weak acid HA, denoted as \(K_a\), is determined using the principle of freezing point depression. The calculation proceeds as follows:

1. The observed freezing point depression is \(\Delta T_f = 0.20 \, ^\circ \text{C}\).
2. The freezing point depression is calculated using the formula: \(\Delta T_f = i \cdot K_f \cdot m\), where:
   • \(i\) represents the van 't Hoff factor.
   • \(K_f = 1.8 \, \text{K kg mol}^{-1}\) is the cryoscopic constant of water.
   • \(m = 0.1 \, \text{mol kg}^{-1}\) is the molality, given as 0.1 m.
3. For a monobasic weak acid with the dissociation equilibrium \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), the van 't Hoff factor is \(i = 1 + \alpha\), where \(\alpha\) is the degree of dissociation.
4. Substituting the known values into the depression formula: \(0.20 = (1 + \alpha) \cdot 1.8 \cdot 0.1\).
5. Solving this equation yields: \(0.20 = 0.18 + 0.18\alpha\).
6. This simplifies to: \(0.18\alpha = 0.02\), which gives \(\alpha = \frac{0.02}{0.18} = \frac{1}{9}\).
7. The dissociation constant \(K_a\) is related to \(\alpha\) by the expression: \(K_a = \frac{c \alpha^2}{1 - \alpha}\), where \(c = 0.1 \, \text{mol L}^{-1}\) is the concentration.
8. Substituting the calculated values: \(K_a = \frac{0.1 \cdot \left(\frac{1}{9}\right)^2}{1 - \frac{1}{9}}\).
9. This results in: \(K_a = \frac{0.1 \cdot \frac{1}{81}}{\frac{8}{9}}\).
10. Further simplification: \(K_a = \frac{0.1 \cdot 1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 \times 10^{-3}\).
11. Consequently, the dissociation constant \(K_a\) for the acid is \(1.38 \times 10^{-3}\).

The dissociation constant \(K_a\) for the acid is \(1.38 \times 10^{-3}\).