Question

Calculate the molar conductivity at infinite dilution for CH₃COOH if the molar conductivity at infinite dilution for NaCl, HCl and CH₃COONa are 126.45, 426.16 and 91.0 ohm⁻¹ cm² mol⁻¹ respectively.

The molar conductivity of NaCl, CH₃COONa and HCl at infinite dilution is 126.45, 91.0 and 426.16 ohm⁻¹ cm² mol⁻¹ respectively. Calculate the molar conductivity `(Lambda_m^infty)` for CH₃COOH at infinite dilution.

Answer 1

Given: \[\ce{\Lambda{^{\circ}_{m}} (HCl)}\] = 426.16 Ω⁻¹ cm² mol⁻¹

\[\ce{\Lambda{^{\circ}_{m}} (CH3COONa)}\] = 91.0 Ω⁻¹ cm² mol⁻¹

\[\ce{\Lambda{^{\circ}_{m}} (NaCl)}\] = 126.45 Ω⁻¹ cm² mol⁻¹

By using Kohlrausch’s Law of Independent Ionic Migration:

\[\ce{\Lambda{^{\circ}_{m}} ​(CH3COOH) = \Lambda{^{\circ}_{m}} ​(HCl) + \Lambda{^{\circ}_{m}} ​(CH3COONa) − \Lambda{^{\circ}_{m}} ​(NaCl)}\]

= 426.16 + 91.0 − 126.45

= 517.16 − 126.45

= 390.71 Ω⁻¹ cm² mol⁻¹