Question

What is the molar conductivity of acetic acid?

Answer 1

1. Molar conductivity is defined as the conducting power produced by all the ions when one mole of electrolyte is dissolved in a solution. It can also be defined as the ionic strength of a solution.
2. we can find the molar conductivity of any solution by applying Kohlrausch's law which states that the sum of ionic conductivities of cations and anions is the molar conductivity of any electrolyte at infinite dilution. Mathematically, $$\begin{gathered} {\mathrm{\lambda m}}^{\infty }={\mathrm{x\lambda a}}^{\infty }+{\mathrm{y\lambda c}}^{\infty } \\ \mathrm{where}{\mathrm{\lambda m}}^{\infty }=\mathrm{molar}\mathrm{conductivity}\mathrm{of}\mathrm{an}\mathrm{electrolyte} \\ {\mathrm{\lambda a}}^{\infty }=\mathrm{ionic}\mathrm{conductivity}\mathrm{of}\mathrm{anion} \\ {\mathrm{\lambda c}}^{\infty }=\mathrm{ionic}\mathrm{conductivity}\mathrm{of}\mathrm{cation} \\ \mathrm{x}=\mathrm{number}\mathrm{of}\mathrm{anions} \\ \mathrm{y}=\mathrm{number}\mathrm{of}\mathrm{cations} \end{gathered}$$
3. So, by applying the formula we can find the molar conductivity of acetic acid$\left({\mathrm{CH}}_3\mathrm{COOH}\right)$.
4. The molar conductivity of acetic acid $\left({\mathrm{CH}}_3\mathrm{COOH}\right)$can be calculated by the molar conductivity of acetate ions ${\left({\mathrm{CH}}_3\mathrm{COO}\right)}^{-}$and the molar conductivity of Hydrogen ion${\left(\mathrm{H}\right)}^{+}$.
5. $$\begin{gathered} 426.1={\mathrm{\lambda }}_{\left({\mathrm{H}}^{+}\right)}^{°}+{\mathrm{\lambda }}_{\left({\mathrm{Cl}}^{-}\right)}^{°}...\left(\mathrm{i}\right) \\ 91.0={\mathrm{\lambda }}_{\left({\mathrm{Na}}^{+}\right)}^{°}+{\mathrm{\lambda }}_{\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)}^{°}...\left(\mathrm{ii}\right) \\ 126.5={\mathrm{\lambda }}_{\left({\mathrm{Na}}^{+}\right)}^{°}+{\mathrm{\lambda }}_{\left({\mathrm{Cl}}^{-}\right)}^{°}....\left(\mathrm{iii}\right) \\ \mathrm{Adding}\mathrm{eqn}.\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{and}\mathrm{subtracting}\mathrm{eqn}.\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get} \\ {\mathrm{\lambda }}_{\left({\mathrm{H}}^{+}\right)}^{°}+{\mathrm{\lambda }}_{\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)}^{°}=426.1+91.0-126.5=390.6{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1} \end{gathered}$$
6. By adding the molar conductivity of both the ion we get $390.7{\mathrm{ohm}}^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$ an infinite dilution.
7. Infinite dilution is a state where we can add as much as a solvent but it has zero effect on concentration.