Question

Conductivity of 0.00241 M acetic acid s 7.896 x 10 ${}^{-5}$ S cm, calculate its molar conductivity. If $\lambda$^{0}${}_m$ for acetic acid is 390.5 S cm${}^2$ mol ${}^{-1}$, what is its dissociation constant?

Answer 1

Given: $K=7.896\times {10}^{-5}Sc{m}^{-1}$ $C=M=0.00241mol{L}^{-1}$

$\therefore$ Molar conductivity

$\wedge m=(K\times 1000)/M=(7.896\times {10}^{-5}\times 1000)/0.00241$

$=32.76Sc{m}^{2}mo{l}^{-1}$

Degree of dissociation

$\alpha =\frac{\lambda m}{{\lambda }^{o}m}=\frac{32.76}{390.5}=0.084$

$\therefore$ Dissociation constant

$\Rightarrow K=\frac{C\alpha }{1-\alpha }$

$=\frac{0.00241\times 0.084}{(1-0.084)}$

$=1.86\times {10}^{-5}mol{L}^{-1}$