Question

The molar conductivity of acetic acid at infinite dilution is $387Sc{m}^{2}mo{l}^{-1}$. At the same temperature, $0.001$ M solution of acetic acid, molar conductivity is 55 S $c{m}^{2}mo{l}^{-1}$. $\alpha$ is the degree of dissociation of $0.1$ N acetic acid. What is the value of 600$\alpha$?
Assume $1-\alpha =1$ for 0.1 N acid

Answer 1

For 0.001 M $C{H}_{3}COOH,\alpha =\frac{{\Lambda }_m}{{\Lambda }_{\infty }}$

${\Lambda }_m=55Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{\infty }=387Sc{m}^{2}mo{l}^{-1}$

$\therefore \alpha =\frac{55}{387}=0.142$ or $14.2\%$

Since, $K_a=\left(c{\alpha }^2\right)/(1-\alpha )$

$K_a=\frac{0.001\times (0.142{)}^2}{(1-0.142)}=2.35\times {10}^{-5}$

For 0.1 N acid, $K_a=c{\alpha }^2$

$2.35\times {10}^{-5}=0.1\times {\alpha }^2$

$\therefore \alpha =0.015$ or $1.5\%$

Now,

$600\alpha =600\times 0.015=9.$