Question

The specific conductance of $0.01M$ solution of acetic was found to be $0.0163S{m}^{-1}$ at ${25}^{0}C$. Calculate the degree of dissociation of the acid.
Molar conductance of acetic acid at infinite dilution is $390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}$ at ${25}^{o}C$.

Answer 1

Given:
$\text{Conductivity,}\kappa =0.0163S{m}^{-1}$

Also, we know
$1L={10}^{-3}{m}^3$
$1L^{-1}={10}^3{m}^{-3}$
$\text{Molar concentration,}\left(C\right)=0.01mol{L}^{-1}=10mol{m}^{-3}$

$\text{Molar conductance,}{\Lambda }_m=\frac{\kappa }{C}$
${\Lambda }_m=\frac{0.0163S{m}^{-1}}{10mol{m}^{-3}}=16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Given,
${\Lambda }_m^0=390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$\text{Degree of dissociation,}\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}=\frac{16.3\times {10}^{-4}S{m}^{2}mo{l}^{-1}}{390.7\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=0.0417$