Question

$0.01(M)$ solution an acid HA freezes at-$0.0205^{\circ}C$ If $K_{f}$ for water is $1.86\,K\,kg\,mol^{-1}$, the ionization constant of the conjugate base of the acid will be (consider molarity $\simeq$ molality)

• A. $1.1 \times 10^{-4}$
• B. $1.1 \times 10^{-3}$
• C. $9\times 10^{-11}$
• D. $9\times 10^{-12}$

Answer 1

Answer: (C)

$\Delta T_{f} =k_{f} \times m =1.86\times0.01=0.0186$
$i=\frac{0.0205}{0.0186}=1.10+1+\alpha or \alpha=0.1$
$K_{a}=\frac{C\alpha^{2}}{1-\alpha}$
$=\frac{0.01\times\left(0.1\right)^{2}}{1-0.1}=\frac{1}{9}\times10^{-3}$
or $K_{b}=9\times10^{-11}$