I'm looking for a simple explanation how the Karl Fischer method to determine the content of water works. I'm not a chemist but an engineer from an electronic-physical field and my chemistry courses are a while ago, meanwhile. That's why I can't understand what's going on but I don't really need a detail explanation, just low-level enough that I can say "Yep, makes sense" and high-level enough, that I can understand it.

Hence, could someone please describe roughly what's going on?


1 Answer 1


Attempting an almost non-chemical description, you need to know that the concentration of a sample, $c$, equates the number of moles $n$ per unit of volume $V$; expressed as $c = n/V$.

For once, a known volume $V$ of your liquid to be analyzed for its concentration $c$ of water into is injected into a chamber equipped with a stirrer and electrodes. There is a stock of reagent solution for which you know ahead of the analysis its concentration. You equally know ahead how many molecules of the reagent will react with how many water molecules to yield a balanced reaction equation (stoichiometry). The reagent solution shall react with water only.

The addition of the reagent solution to the analyte is monitored; for one, you know the volume of the reagent solution added. Because of this known volume added, and the known concentration of the stock of reagent solution, you know how many molecules of the reagent were added. The mixing and stirring ensures that each molecule of the reagent quickly encounters a molecule of water to react with.

For two, during the addition of the reagent solution to the analyte, the electric potential between the electrodes dipped into the analyte is monitored. Initially, the addition of a little volume of reagent solution changes this potential only a little. However, continuing the addition of small volumes of reagent solution, there is one point where this change of potential suddenly is large. This is the point when 1) basically all the water molecules reacted with the reagent, 2) the electrodes now monitor an excess of the reagent in respect of the water molecules, and 3) the titration may be stopped.

Because you know the volume of the reagent (of known concentration) added and the balanced reaction equations used in this analysis, you may calculate the number $n$ of water molecules consumed to reach this point of sudden change of the potential. However, because you know both the number of water molecules consumed as well as the volume $V$ of analyte initially injected, you may compute the concentration of water molecules in the analyte, too ($c = n/V$).

The Karl-Fischer analysis offers a rapid characterization of organic liquids (minute scale), provided they do not react differently with the reagent. Samples containing alcohol, for example, may differ in the underlying reaction equations and stoichiometries from samples containing water. Generally speaking, to work well, typical samples submitted do not contain more than 1 mol/L of water (about 18 g/L).

  • $\begingroup$ Awesome, that was the type of explanation I was looking for! I think I got the idea, I just wonder how it can be that precise as I have to add liquid/material continously. Furthermore, I wonder how it can be that fast as these added ingredients have to react (resp. reach an equilibrium?!)? $\endgroup$
    – Ben
    Aug 13, 2020 at 13:50
  • 1
    $\begingroup$ #1 Karl-Fischer titration is a batch-wise analysis (e.g. publicity video by Metrohm, youtube.com/watch?v=W7wxPpADsRw); one-time addition of the analyte, step-wise incremental addition of the reagent, e.g. by a piston driven by a stepper motor such the reactions still advance much faster than the addition. #2 Energetically speaking, the underlying reactions run this much downhill (like an apple falling from a table) that even if the redox reactions could be reversed, the typical reaction conditions render the backward reactions (reformation of water and reagent) insignificant. $\endgroup$
    – Buttonwood
    Aug 13, 2020 at 14:50
  • $\begingroup$ very good, thanks a lot! $\endgroup$
    – Ben
    Aug 14, 2020 at 6:04

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