Question

A $0.0128N$ solution of acetic acid has $\Lambda =14mho{equiv}^{-1}$ and ${\Lambda }_{\infty }=391mho{eq}^{-1}$ at ${25}^{o}C$. Calculate the dissociation constant of the acid.

Answer 1

Degree of dissocation, $\alpha =\frac{{\wedge }_{\infty }}{{\wedge }_o}$

$\alpha =\frac{14}{391}=3.58{\times 10}^{-2}$

Applying Ostwald's dilution law

$K_a=\frac{{C\alpha }^2}{1-\alpha }$

It is given that $C=0.0128N$ and $\alpha =3.58{\times 10}^{-2}$

So, $K_a=1.64\times {10}^{-5}$