Question

The freezing point depression of a $0.1M$ aqueous solution of weak acid $\left(HX\right)$ is $-{0.20}^{\circ }C$. What is the value of equilibrium constant for the reaction:

$HX\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+X^{-}\left(aq\right)$

[Given: $K_f$ for water = $1.8kgmo{l}^{-1}K$. and $\text{Molality = Molarity}$]

• A. $1.46\times {10}^{-4}$
• B. $1.35\times {10}^{-3}$
• C. $1.21\times {10}^{-3}$
• D. $1.35\times {10}^{-4}$

Answer 1

The correct option is B $1.35\times {10}^{-3}$

The expression for the depression in freezing point is $\Delta T_f=k_f\times m\times i$
Substitute values in the above expression
$0.20=1.8\times 0.1\times i$
$i=1.111$

	HX	;$H^{+}$	;$Y^{-}$
Initial moles	1	0	0
Change	-n	n	n
Moles at equilibrium	(1-n)	n	n

Total number of moles at equilibrium $=1+n=0.111$
Hence, $n=0.111$
$K=\frac{0.111\times 0.111}{0.889\times 10}=1.35\ast {10}^{-3}$