$\mathrm{H} A_{(a q)} ightleftharpoons \mathrm{H}_{(a q)}^{+}+A_{(a q)}^{-}$\ The freezing point

Question

$\mathrm{H} A_{(a q)} ightleftharpoons \mathrm{H}_{(a q)}^{+}+A_{(a q)}^{-}$\ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid $\mathrm{H} A$ is $0.20^{\circ} \mathrm{C}$. The dissociation constant for the acid is\ Given : $K_{f}\left(\mathrm{H}_{2} \mathrm{O}ight)=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molality $\equiv$ molarity\

Professor PC Thomas Classes · Answer

(c) : Given : $m=0.1 \mathrm{~m}$,\ $K_{f}=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$\ Molality $\equiv$ Molarity, $\Delta T_{f}=0.20^{\circ} \mathrm{C}$\ We know, $\Delta T_{f}=i K_{f} \mathrm{~m}$\ $0.20=i 	imes 1.8 	imes 0.1$\ $i=\frac{0.20}{1.8 	imes 0.1}=\frac{20}{18}=\frac{10}{9}$\ Given reaction,       $\begin{array}{lccc} & \mathrm{H} A_{(a q)} ightleftharpoons \mathrm{H}_{(a q)}^{+}+A_{(a q)}^{-} \ t=0 ; & 1 & 0 & 0 \ t=t_{	ext {eq }} ; & 1-\alpha & \alpha & \alpha\end{array}$\ Total moles after dissociation $=1-\alpha+ \alpha+\alpha=1+\alpha$\ $i=1+\alpha ; \frac{10}{9}=1+\alpha ; \alpha=\frac{10}{9}-1=\frac{1}{9}$\ $K_{a}=\frac{\left[\mathrm{H}^{+}ight]\left[A^{-}ight]}{[\mathrm{H} A]}=\frac{C \alpha^{2}}{1-\alpha}=\frac{0.1(1 / 9)^{2}}{1-1 / 9}=\frac{1}{720}$\ $K_{a}=1.38 	imes 10^{-3}$\