HA_{(aq)} rightleftharpoons H^{+}_{(aq)} + A^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The formula for freezing point depression is $\Delta T_{f} = i K_{f} m$.Given $\Delta T_{f} = 0.20^{\circ} C$,$K_{f} = 1.8 \ K \ kg \ mol^{-1}$,an

Vedclass · Accepted Answer

$1.38 	imes 10^{-3}$

Vedclass · Answer

(A) The formula for freezing point depression is $\Delta T_{f} = i K_{f} m$.Given $\Delta T_{f} = 0.20^{\circ} C$,$K_{f} = 1.8 \ K \ kg \ mol^{-1}$,and $m = 0.1 \ m$.$0.20 = i 	imes 1.8 	imes 0.1$$i = \frac{0.20}{0.18} = \frac{10}{9}$.For the dissociation $HA ightleftharpoons H^{+} + A^{-}$,the van't Hoff factor $i = 1 + \alpha$,where $\alpha$ is the degree of dissociation.$1 + \alpha = \frac{10}{9} \implies \alpha = \frac{1}{9}$.The dissociation constant $K_{a} = \frac{C \alpha^2}{1 - \alpha}$.Substituting $C = 0.1$ and $\alpha = \frac{1}{9}$:$K_{a} = \frac{0.1 	imes (1/9)^2}{1 - 1/9} = \frac{0.1 	imes (1/81)}{8/9} = \frac{0.1}{81} 	imes \frac{9}{8} = \frac{0.1}{72} = \frac{1}{720} \approx 1.38 	imes 10^{-3}$.

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.

Similar Questions

The experimental depression in freezing point of a dilute solution is $0.025 \ K$. If the van't Hoff factor $(i)$ is $2.0$,the calculated depression in freezing point (in $K$) is

If a $0.02 \, M \, Pb(NO_3)_2$ solution is isotonic with a $0.05 \, M$ aqueous glucose solution,the degree of dissociation of $Pb(NO_3)_2$ will be ..........

When acetic acid is dissolved in benzene,its molecular mass:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.