Question

The freezing point of a $0.08$ molal aqueous solution of $NaHSO _{4}$ is $-0.372^{\circ} C$. The dissociation constant for the following reaction is: $\left(K_{f}\right.$ for $\left.H _{2} O =1.86\, K . kg / mol \right)$
$HSO _{4}^{-} \longrightarrow H ^{+}+ SO _{4}^{2-}$

• 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,
$K=\frac{0.08 \alpha \times 0.08 \alpha}{0.08(1-\alpha)}$
$=\frac{0.08 \times 0.5 \times 0.08 \times 0.5}{0.08 \times 0.5}=0.04$