Question

Explain the determination of molar conductivity of a weak electrolyte at infinite dilution or zero concentration using Kohlrausch's law.

Answer 1

Consider the molar conductivity (Λ₀) of a weak acid, CH₃COOH at zero concentration.

By Kohlrausch’s law, \[\ce{Λ_{0CH_3COOH} = λ_{CH_3COO^-} + λ^0_H+}\]

where \[\ce{λ^0_{CH_3COO^-}}\] and \[\ce{λ^0_H+}\] are the molar ionic conductivities of CH₃COO⁻ and H⁺ ions respectively.

If \[\ce{Λ_{0CH_3COONa},  Λ_{0HCl} and Λ_{0NaCl}}\] are the molar conductivities of CH₃COONa, HCl and NaCl respectively at zero concentration, then by Kohlrausch’s law,

\[\ce{Λ_{0CH_3COONa} = λ^0_{CH_3COO-} + λ^0_{Na^+}}\]

\[\ce{Λ_{0HCl} = λ^0_{H^+} + λ^0_{Cl^-}}\]

\[\ce{Λ_{0NaCl} = λ^0_{Na^+} + λ^0_{Cl^-}}\]

Now, 

\[\ce{Λ_{0CH_3COONa} + Λ_{0HCl} - Λ_{0NaCl}}\]

= \[\ce{[λ^0_{CH_3COO-} + λ^0_{Na^+} + [λ^0_{H^+} + λ^0_{Cl^-}] - [λ^0_{Na^+} + λ^0_{Cl^-}]}\]

= \[\ce{λ^0_{CH_3COO-} + λ^0_{H^+}}\]

= \[\ce{Λ_{0CH_3COOH}}\]