Assume that you have an aqueous solution of sodium chloride and potassium hydroxide. Is it possible to isolate potassium chloride from this?

$$\ce{Na+(aq) + Cl-(aq) + K+(aq) + OH-(aq) -> Na+(aq) + OH-(aq) + KCl(s)}$$

Since all involved salts/salt combinations are highly water-soluble, I'm not sure how one could isolate the potassium chloride. Is there maybe a way to selectively precipitate the sodium via a non-water soluble salt?

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Apr 10, 2022 at 2:54

2 Answers 2


NaCl is about three times as soluble as KCl in methanol at room temp. That said, while it might barely be feasible to separate their salt that way, assuming nearly 100% solvent recovery, and near-zero energy cost (perhaps waste geothermal?), the most efficient commercial method to separate various salts is with evaporation ponds.

That said, this is not an answer for your question on separating a few liters of solution, but I've written it as one to add the links. If others think this should be placed as a comment, I'll try to do so.

  • $\begingroup$ I was thinking about methanol too, in the deleted comment, but decided against it. $\endgroup$
    – Poutnik
    Apr 10, 2022 at 3:19

I suggest to buy $\ce{KCl}$, or to prepare it by neutralization of $\ce{KOH}$ and $\ce{HCl}$.

If it happens you have mixed solution of $\ce{NaCl}$ and $\ce{KOH}$, neutralize it first by $\ce{HCl}$ (pH meter or pH indicator paper strips).

From Wikipedia solubility table one can conclude that while $\ce{NaCl}$ solubility almost does not depend on temperature, solubility of $\ce{KCl}$ is almost twice at 100 °C, compared 0 °C.

My suggestion is fractional crystallization:

  1. Partially evaporate the solution until significant mass of salts crystallizes.

  2. Heat the mixture to near boiling and mix well - part of $\ce{KCl}$ would dissolve.

  3. Separate the precipitate enriched by $\ce{NaCl}$.

  4. Cool down solution and let crystalyze salt enriched by $\ce{KCl}$.

  5. With both precipitates and residual solution, you can recursively repeat the procedure.

  6. You can set a system of N batches, where $\ce{NaCl}$ enriched precipitate would go to direction of the batch more enriched by $\ce{NaCl}$ and vice versa.

  7. If it looks complicated to you, buy $\ce{KCl}$ or remember the Pierre and Marie Curie struggles with the isolation of traces of $\ce{RaCl2}$ from $\ce{BaCl2}$, where their solubilities are much closer.

Another option is using about 3 times better solubility of $\ce{NaCl}$ in methanol, compared to $\ce{KCl}$. But as it is like $\pu{5 g/L}\ \ce{KCl}$ and $\pu{15 g/L}\ \ce{NaCl}$, there would be extensive solvent manipulation with price and safety issues.

As useful info, here is the question/answer about the solubility diagram of mixtures of both salts

  • $\begingroup$ The solubility of a salt like $\ce{NaCl}$ or $\ce{KCl}$ is not the same in pure water and in a solution of another salt. Please look at solubility diagrams $\ce{NaCl - KCl}$ ! $\endgroup$
    – Maurice
    Apr 10, 2022 at 8:18
  • $\begingroup$ @Maurice I am well aware of it. :-) In fact, I have answered some question with such a diagram last year. The solubility in pure water is just a guidance, there is not totally messed up relation of their temperature dependencies.(seen at the diagram as well). $\endgroup$
    – Poutnik
    Apr 10, 2022 at 9:17
  • $\begingroup$ Ion exchange chromatography could be an option, too. A couple of (late) transition metals are separated from each other this way; the value attributed to them justifies the investment. It very well may be a different story for NaCl, in economic perspective. $\endgroup$
    – Buttonwood
    Mar 16 at 9:27