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Chemistry · Ch 9 — Electrochemistry

Molar Conductivity

9.1.1

Molar Conductivity

Two solutions at different concentrations contain different numbers of ions per unit volume, and so naturally show different specific conductance — this makes κ a poor basis for comparing how well different electrolytes conduct. Molar conductivity, symbol Λm\Lambda_m, fixes this by standardising the amount of electrolyte considered: it is defined as the conductance of the volume of solution — call it V m³ — that contains exactly one mole of the dissolved electrolyte, measured between electrodes exactly 1 m apart.

Since the specific conductance κ is, by definition, the conductance of 1 m³ of the solution, the conductance of the V m³ containing one mole of electrolyte follows simply as

Λm=κ×V\Lambda_m = \kappa \times V

To turn this into something computable from an ordinary concentration reading, recall that molarity is defined as M=nV(in dm3)M = \dfrac{n}{V(\text{in dm}^3)}, so the volume containing exactly one mole of solute is V=1M dm3 mol−1V = \dfrac{1}{M}\ \text{dm}^3\,\text{mol}^{-1}. Converting this volume-per-mole into SI units (m³ per mol) and substituting into the relation above gives the standard working formula:

Λm (S m2 mol−1)=κ (S m−1)×10−3M (mol L−1)\Lambda_m\ (\text{S m}^2\,\text{mol}^{-1}) = \dfrac{\kappa\ (\text{S m}^{-1}) \times 10^{-3}}{M\ (\text{mol L}^{-1})}

or, in the more commonly used CGS-flavoured form with κ in S cm⁻¹ and M in mol L⁻¹, simply Λm=1000 κM\Lambda_m = \dfrac{1000\,\kappa}{M} (S cm² mol⁻¹). This single equation is the workhorse of every molar-conductivity numerical in the chapter: given any two of κ, M and Λm\Lambda_m, the third follows immediately. …