I have two soil reports, one that reports CEC as 45 meq/100 g and another that reports 6 mS/cm using 1:1 water soil.

I can see that the two are, if not apples and oranges, they are at best tangerines and grapefruit — the meq/100 g is the source for the ions in a water extract, but an equilibrium solution may not contain all the ions.

Can a meaningful conversion between meq/100 g and mS/cm be made?

E.g: few plants will grow in soil that has conductivity greater than 2.5 mS/cm and generally you start getting concerned at 1. What is the equivalent measure in meq/100 g?

FWIW main ions of concern are $\ce{K+},$ $\ce{PO4^3-},$ $\ce{Mg^2+},$ $\ce{Ca^2+},$ $\ce{CO3^2-},$ $\ce{HCO2-}.$


The mineral ions (nutrients) in soil can be expressed as two ways: (i) Cation Exchange Capacity (CEC), which has units of $\pu{meq}$/$\pu{100 g}$; and (11) Electrical Conductivity (EC), which has units of $\pu{mS/cm}$.

When the conductivity is measured in 1:1 mixtre of soil:water, it is called $\mathrm{EC_{1:1}}$ (This is a way of measuring electrical conductivity of soil water or $\mathrm{EC_{sw}}$).

According to USDA Publication, $\mathrm{EC_{1:1}}$ readings less than $\pu{1 dS/m}$ ($\pu{1 mS/cm}$), soil are considered non-saline and do not impact most crops. This publication also gives you a easy procedure for how to measure $\mathrm{EC_{1:1}}$ of your soil.

My best bet is you don't have to worry about your second soil sample, which has CEC of $\pu{45 meq}$/$\pu{100 g}$, because none of the cations (exchangeble cations in the soil) in concern is $\ce{Na+}$ (the given CEC is the sum of all exchangeable cations in equivalents). According to above publication, soils with a high concentration of sodium salts (sodic conditions) have given problems.

To my understanding, there is no direct relationship between $\pu{meq}$/$\pu{100 g}$ and $\pu{mS/cm}$. For more info on Cation Exchange Capacity (CEC) and units of $\pu{meq}$/$\pu{100 g}$, read:


If I understand it well, then $\pu{1 meq}$ means 1 milimol equivalent.

Therefore, if we consider for illustration molar masses $M_{\mathrm{Ca}}=\pu{40 g/mol}$ and $M_{\mathrm{K}}=\pu{39 g/mol}$, then $\pu{1 meq K}=\pu{39 mg}$ and $\pu{1 meq Ca}=40/2=\pu{20 mg}$

The first step is to express molar amounts of involved ions, resp. molar equivalents, in the conventionally used representative ions or salts. As the same molar concentration of ions of different mobility have both different molar conductivity and different specific conductivity.

Second step is recalculation to soecific conductivity in $\pu{mS/cm}$.

The third step is determine the degree of extraction, experimentally ( extraction method versus total analysis e.g. by acidic extraction and AAS analysis) or by convention.

It is also possible that even meq/100g is related directly to the 1:1 extract. Then there would remain to determine tho representative, probably conventional salt for recalculation factor


$$\kappa=c\cdot \Lambda$$


$\kappa$ is ( specific} conductivity
$c$ is concentration of mol equivalents per volume
$\Lambda$ is molar conductivity ( specific conductivity per unit of molar equivalent concentration.)


There is no universal conversion factor between $\pu{mS/cm}$ and $\pu{meq/100g}$. As suggested by Poutnik's thorough answer, you may need to know some additional details about how the data was compiled, which may be embedded in your reports. A scan through websites describing soil analysis suggests the following as a useful conversion:

$$CEC(\pu{mS/cm)} = (0.01f_{sat})^{-1} \times C(\pu{meq/100g})$$

with $f_{sat}$ the percent saturation of the soil (the water retention capacity of the soil as weight of water per dry weight soil, times 100). The expression makes certain assumptions:

  • $C(\pu{meq/100g}$) refers to the equivalent molar concentration of $\ce{NaCl}$ that would result in the conductivity observed in a saturated solution extracted from water-saturated soil.
  • the molar conductivity of $\ce{NaCl}$ is $\approx \pu{100 mS cm^{-1}M^{-1}}$

Note the molar conductivity of $\pu{100 mM}$ NaCl, according to the CRC Handbook of Chemistry and Physics, is $\pu{107 mS cm^{-1}M^{-1}}$ at $\pu{298 K}$.


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