First slide

Kohlraush law

Question

The molar conductivities $λ_{NaOAc}^{0}$ and $λ_{HCl}^{0}$ at infinite dilution in water at 25⁰C and 91.0 and 426.2 S cm² / mol respectively. To caculate $λ_{HOAc}^{0}$
the additional value required is

Easy

A

$\lambda_{NaCl}^0$
B

$\lambda_{{H_2}O}^0$
C

$\lambda_{KCl}^0$
D

$\lambda_{NaOH}^0$

Solution

$\lambda _{C{H_3}COONa}^m + \lambda _{HCl}^m = \lambda _{C{H_3}CO{O^ - }}^m + \lambda _{N{a^ + }}^m + \lambda _{{H^ + }}^m + \lambda _{C{l^ - }}^m = \lambda _{C{H_3}COOH}^m + \lambda _{NaCl}^m$

$λ_{NaCl}^{m}$ is required.

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