Question

The freezing point of an aqueous solution of $NaHS{O}_4$ is $-0.372{}^{o}C$. The dissociation constant for the reaction
$HS{O}_4^{-}\to H^{+}+S{O}_4^{-2}$ is 0.4. Find the concentration of $NaHS{O}_4$ in mol/kg.

[$K_{f}for{H}_{2}O=1.86Kkgmo{l}^{–1}$]
(Assume dilute solution and take molality = molarity)

Answer 1

Let the molality of $NaHS{O}_4$ solution be m, then assuming molarity = molality.
$HS{O}_4^{-}⟶H^{+}+S{O}_4^{-2}$
$m(1-\alpha )m\alpha m\alpha$

$i=1+1+\alpha =2+\alpha$

Effective molality $=\left[N{a}^{+}\right]+i\left[HS{O}_4^{-}\right]$
$=m+m(1+\alpha )=m(2+\alpha )$
$△T_f=K_{f}im$
$\Rightarrow 0.372=1.86\times m(2+\alpha )$
$\Rightarrow 0.2=m(2+\alpha )$

Also, $K_a=\frac{m{\alpha }^2}{1-\alpha }=0.4$
$\Rightarrow \frac{0.2}{2+\alpha }\times \frac{{\alpha }^2}{1-\alpha }=0.4$
$\Rightarrow {\alpha }^2=0.2(2+\alpha )(1-\alpha )$
So,
$1.2{\alpha }^2+0.2\alpha -0.4=0$
$\Rightarrow 6{\alpha }^2+\alpha -2=0$
$\Rightarrow \alpha =0.5$
So, $m=\frac{0.2}{2.5}=0.08mol/kg$