$\mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q)$

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: $K_f$(H₂O) = 1.8 K kg mol⁻¹, molality ≡ molarity

<p><strong>Freezing Point Depression:</strong></p> <p>$ \Delta T_f = i \times K_f \times m $</p> <p>Given:</p> <p>$ \Delta T_f = 0.2 \, \text{°C}, \quad K_f = 1.8 \, \text{K kg mol}^{-1}, \quad m = 0.1 \, \text{m} $</p> <p>Substituting the given values:</p> <p>$ 0.2 = i \times 1.8 \times 0.1 $</p> <p>Solving for $i$:</p> <p>$ i = \frac{0.2}{1.8 \times 0.1} = \frac{20}{18} = \frac{10}{9} $</p></p> <p><p><strong>Degree of Dissociation ($\alpha$):</strong></p> <p>For the reaction $\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$:</p> <p>$ i = 1 + \alpha $</p> <p>Given $i = \frac{10}{9}$:</p> <p>$ \frac{10}{9} = 1 + \alpha $</p> <p>$ \alpha = \frac{1}{9} $</p></p> <p><p><strong>Dissociation Constant ($K_{eq}$):</strong></p> <p>$ \mathrm{K}_{eq} = \frac{[H^+][A^-]}{[HA]} $</p> <p>At equilibrium:</p> <p>$ [H^+] = [A^-] = \alpha \times C = \frac{1}{9} \times 0.1 $</p> <p>$ [HA] = 0.1 \times (1 - \alpha) = 0.1 \times \left(1 - \frac{1}{9}\right) $</p> <p>Substituting these into $K_{eq}$:</p> <p>$ \mathrm{K}_{eq} = \frac{(0.1 \times \frac{1}{9})^2}{0.1 \times \left(1 - \frac{1}{9}\right)} $</p> <p>Simplifying:</p> <p>$ \mathrm{K}_{eq} = \frac{0.1 \times \left(\frac{1}{81}\right)}{0.1 \times \frac{8}{9}} = \frac{1}{720} $</p> <p>Therefore:</p> <p>$ \mathrm{K}_{eq} = 1.38 \times 10^{-3} $</p></p>