Question

The freezing point depression of a $0.109Maq.$ solution of formic acid is $-{0.21}^{\circ }C$. The equilibrium constant for the reaction is $1.44\times {10}^{-x}$.
$HCOOH\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+HCO{O}^{\ominus }\left(aq\right)$
$K_f$ for water $=1.86kgmo{l}^{-1}K$
Value of $x$ is:

Answer 1

$\Delta T_f=(1+\alpha )K_f\times m$
$0.21=(1+\alpha )\times 1.86\times 0.109$
$1+\alpha =1.0358$
$\alpha =0.0358$
$K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.109(0.0358{)}^2}{1-0.0358}=1.44\times {10}^{-4}$