Question

The freezing point of a $0.08$ molal aqueous solution of $NaHS{O}_4$ is $-0.372^{\circ }C$. The dissociation constant for the following reaction is: $(K_f$ for $H_{2}O=1.86K.kg/mol)$
$HS{O}_4^{-}⟶H^{+}+S{O}_4^{2-}$

• A. $0.2$
• B. $0.02$
• C. $0.01$
• D. $0.04$

Answer 1

Answer: (D)

$i=\frac{0.08+0.08(1-\alpha )+0.08\alpha +0.08\alpha }{0.08}=2+\alpha$
$\Delta T=i\times K_f\times m$
$0.372=i\times 1.86\times 0.08$
$i=2.5$
$\therefore 2+\alpha =2.5$
or $\alpha =2.5$
Dissociation constant,
$K=\frac{0.08\alpha \times 0.08\alpha }{0.08(1-\alpha )}$
$=\frac{0.08\times 0.5\times 0.08\times 0.5}{0.08\times 0.5}=0.04$