Question

Conductivity of $2.5\times {10}^{-4}$ M methanoic acid is $5.25\times {10}^{-5}Sc{m}^{-1}$. Calculate it molar conductivity and degree of dissociation.
(Given: ${\lambda }^0\left(H^{+}\right)=349.5sc{m}^{2}mo{l}^{-1}and{\lambda }^0\left(HCO{O}^{.}\right)=50.5Sc{m}^{2}mo{l}^{-1}$)

Answer 1

We know molar conductivity, $\left({\lambda }_m\right)=\frac{1000\times conductivity\left(k\right)}{concentration\left(c\right)}$
$\therefore {\lambda }_m=\frac{1000\times 5.25\times {10}^{-5}}{2.5\times {10}^{-4}}=210Sc{m}^{2}mo{l}^{-1}$
${\lambda }_{CHCOOH}^0={\lambda }_{H^{+}}^0+{\lambda }_{\left(HCO{O}^{-}\right)}^0=(349.5+50.5)=400Sc{m}^{2}mo{l}^{-1}$
$\therefore \alpha =\frac{{\lambda }_m}{{\lambda }_0}=\frac{210}{400}=0.52$
or, $\alpha =52.5$ %