Temperature Compensation for Conductivity
The electrical conductivity (EC) of an aqueous solution increases with temperature significantly: about 2 per degree Celsius. In practice, there are several empirical formulas in use to convert measured EC values to the reference temperature at 25. But what is the physical idea behind all these equations?
The answer to this question lies in the fundamental relationship between electrical conductivity, diffusion coefficients, and the viscosity of water.
EC, Diffusion Coefficients, and Viscosity of Water
The two equations that interrelate the three physical quantities (EC, diffusion coefficient D, and viscosity η of water) are:
The Nernst-Einstein equation states the proportionality between the EC and the diffusion coefficients Di of dissolved ions:
where F denotes Faraday’s constant, R is the gas constant, T the temperature in Kelvin, z the electrical charge, and c the molar concentration of the dissolved ion(s).
The Stokes-Einstein equation describes the relation between the diffusion coefficient D and the viscosity η:
with kB as the Boltzmann constant and r as the ‘hydraulic radius’ of the diffusing ion. The parameter r itself is irrelevant to our considerations, since it cancels out if we consider D and η at two temperatures T1 and T2:
On the other hand, according to 1, EC is directly proportional to D/T which yields:
If T1 refers to the water temperature T, and T2 to the reference temperature 25, then – after rearranging of 4 – the temperature compensation formula is expressed as a ratio of the viscosity of water at T and at 25:
where EC and η (without the subscript) refer to the water temperature T.
Viscosity of Water as a Function of Temperature
The viscosity of pure water decreases with temperature. Typical values are:
||η20 = 1.003·10-3 kg m-1 s-1
||at 20 °C
||η25 = 0.891·10-3 kg m-1 s-1
||at 25 °C
A nonlinear parameterization of the dynamical viscosity is presented in the classical textbook of Atkins’ Physical Chemistry:
with the two parameters:
||A = 1.37023 (t – 20) + 8.36·10-4 (t – 20)2
||B = 109 + t and t in °C
It describes the viscosity of water over its entire liquid range (0 to 100) with less than 1 error. [Please note that this equation refers to 20 and not to 25.]
Nonlinear Temperature Compensation of EC
Inserting 6 into 5 yields:
The numerical value of the conversion from 20 to 25 can be easily calculated: (η25/η20) = 0.889. Thus we obtain the final formula:
This nonlinear equation is used by the program aqion to convert EC25 to the EC at the given water temperature. The EC25 itself is calculated by the general approach based on diffusion coefficients and molar concentrations of the dissolved ions.
[Note: Although it is not obvious at first sight, 8 provides the right normalization, i.e., the ratio EC/EC25 becomes exactly 1 for 25.]
Instead of the general approach in 8, linear formulas are in widespread use. The most common type of a linear expression is obtained from 8 by a Taylor series expansion (in which higher-order terms in (t – 25) are neglected):
||EC = [ 1 + a (t – 25) ] EC25
||with a = 0.020
The mathematical derivation is presented in the appendix.
Empirical values of the parameter range between a = 0.01 and 0.03. For example, Hayashi deduced a compensation factor of 0.019 from the examination of natural waters. This is in close agreement with the theoretical value:
||a = 0.020 °C-1
||a = 0.019 °C-1
Comparison of the Nonlinear and Linear Model
It is quite instructive to compare the general and linear compensation formulas. The following diagram presents EC/EC25 in the temperature range between 0 and 100:
||– general approach in 8
|linear approx. (a=0.020)
||– 9 with a = 0.020
|linear approx. (a=0.019)
||– 9 with a = 0.019 [Hayashi]
Obviously, at t = 25 the correction is exactly 1. This is valid for all methods.
Appendix: Linearization of the General Equation
We start with 8 and use the Taylor series expansion, ex = 1 + x + … . It yields:
Now we simplify the ratio A/B, that is based on 6a and (6b). Ignoring all quadratic and higher terms in θ = t – 25 leads to:
That is, we get A/B · ln 10 = 2.302 · A/B ≈ 0.023 (θ + 5). Inserting this into A1 yields
Returning from θ = t – 25 to t and ignoring the small temperature offset of 0.5, gives the final expression:
This is the linear approximation used in 9.
Remarks and References
[last modified: 2018-04-25]