Question

Conductivity of $0.00241M$ acetic acid is $7.896\times {10}^{-5}Sc{m}^{-1}$. Calculate its molar conductivity. If ${\wedge }_m^0$ for acetic acid is $390.5Sc{m}^{2}mo{l}^{-1}$, what is its dissociation constant?

Answer 1

Molar conductivity $\Lambda$ and specific conductivity $\kappa$ are related by the following expression:
$\Lambda =\frac{1000\kappa }{C}$
Here, $C$ is the molar concentration.
$\Lambda =\frac{1000\times 7.896\times {10}^{-5}}{0.00241}=32.76Sc{m}^2/mol$
The relationship between the degree of dissociation $\alpha$, the molar conductivity at concentration $C$, ${\Lambda }^C$ and the molar conductivity at infinite dilution ${\Lambda }^o$ is as follows:
$\alpha =\frac{{\Lambda }^C}{{\Lambda }^0}$

$\alpha =\frac{32.76}{390.5}=8.39\times {10}^{-2}$

The dissociation constant $K=\frac{C{\alpha }^2}{1-\alpha }$

$K=\frac{0.00241\times (8.39\times {10}^{-2}{)}^2}{(1-8.39\times {10}^{-2})}=1.86\times {10}^{-5}$