A 0.1 m aqueous solution of a weak acid HX i | vedclass

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Question

(D) For the dissociation of weak acid $HX$: $HX \rightleftharpoons H^+ + X^-$Initial moles: $1 \quad 0 \quad 0$At equilibrium: $(1-\alpha) \quad \alph

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$- 0.24$

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(D) For the dissociation of weak acid $HX$: $HX ightleftharpoons H^+ + X^-$Initial moles: $1 \quad 0 \quad 0$At equilibrium: $(1-\alpha) \quad \alpha \quad \alpha$Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$Van't Hoff factor $i = 1 + \alpha = 1 + 0.3 = 1.3$Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m$$\Delta T_f = 1.3 	imes 1.86 	imes 0.1 = 0.2418 \ ^{\circ}C$Freezing point of solution $T_f = T_f^{\circ} - \Delta T_f = 0 - 0.2418 \ ^{\circ}C = - 0.24 \ ^{\circ}C$

$A$ $0.1 \ m$ aqueous solution of a weak acid $HX$ is $30\%$ ionized. If $K_f$ for water is $1.86 \ ^{\circ}C/m,$ the freezing point of the solution will be ......... $^{\circ}C$.

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