$\mathrm{HA}(\mathrm{aq})⇌{\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{A}}^{-}(\mathrm{aq})$ The freezing point depression of a $0.1\mathrm{m}$aqueous solution of a monobasic weak acid HA is $0.20^\circ \mathrm{C}$. The dissociation constant for the acid is Given : ${\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_{2}\mathrm{O}\right)=1.8\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1},\mathrm{molality}\equiv \mathrm{molarity}$ [JEE Main 2025, 8 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) $1.38\times {10}^{-3}$. ΔT f = ik f m 0.2 = i × 1.8 × 0.1 $i=\frac{20}{18}=\frac{10}{9}$ $\begin{matrix}For & H{A}_{(aq)} & ⇌ & {H}_{(aq)}^{+} & + & {A}_{(aq)}^{−} \ t=0 & 1 \ t={t}_{eq} & 1−\alpha & & \alpha & & \alpha \end{matrix}$ i = 1 + α $\frac{10}{9}=1+\alpha$ $\alpha =\frac{1}{9}$ ${K}_{eq}=\frac{[{H}^{+}][{A}^{−}]}{[HA]}=\frac{C{\alpha }^{2}}{1−\alpha }$ $=\frac{0.1{(\frac{1}{9})}^{2}}{1−\frac{1}{9}}=\frac{1}{720}$ ${K}_{eq.}=1.38\times {10}^{−3}$

\(\mathrm{HA}(\mathrm{aq})⇌{\mathrm{H}}^{+}(\mathrm{aq})+{\mathrm{A}}^{-}(\mathrm{aq})\) The freezing point depression o…

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