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

To find the dissociation constant \(K_a\) for the weak acid HA, we need to use the principle of freezing point depression. Here are the steps for the calculation:

1. The freezing point depression (\(\Delta T_f\)) is given as \(0.20 \, ^\circ \text{C}\).
2. We can use the formula for freezing point depression: \(\Delta T_f = i \cdot K_f \cdot m\), where
   • \(i\) is the van 't Hoff factor.
   • \(K_f = 1.8 \, \text{K kg mol}^{-1}\) (cryoscopic constant of water).
   • \(m = 0.1 \, \text{mol kg}^{-1}\) (molality, given as 0.1 m).
3. For a monobasic weak acid dissociating as \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\), the van 't Hoff factor \(i\) is given by: \(i = 1 + \alpha\), where \(\alpha\) is the degree of dissociation.
4. Substitute the known values into the depression formula: \(0.20 = (1 + \alpha) \cdot 1.8 \cdot 0.1\)
5. Solving the equation, we get: \(0.20 = 0.18 + 0.18\alpha\)
6. This simplifies to: \(0.18\alpha = 0.02\), hence \(\alpha = \frac{0.02}{0.18} = \frac{1}{9}\).
7. The dissociation constant \(K_a\) is related to \(\alpha\) by: \(K_a = \frac{c \alpha^2}{1 - \alpha}\), where \(c = 0.1 \, \text{mol L}^{-1}\).
8. Substituting the values, we get: \(K_a = \frac{0.1 \cdot \left(\frac{1}{9}\right)^2}{1 - \frac{1}{9}}\)
9. This yields: \(K_a = \frac{0.1 \cdot \frac{1}{81}}{\frac{8}{9}}\)
10. Simplifying: \(K_a = \frac{0.1 \cdot 1}{81} \times \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 \times 10^{-3}\)
11. Thus, the dissociation constant \(K_a\) for the acid is \(1.38 \times 10^{-3}\).

Therefore, the correct answer is \( 1.38 \times 10^{-3} \).

Answer 2

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 \((K_a)\) \[ K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.1 \times (0.11)^2}{1 - 0.11} = \frac{0.1 \times 0.0121}{0.89} = 1.36 \times 10^{-3} \approx 1.38 \times 10^{-3} \]