# How to separate KCl from NaCl and MgCl₂?

Having mixed powder of $\ce{KCl}$, $\ce{NaCl}$, and $\ce{MgCl2}$ (and, perhaps, some crap), how can the components be separated?

• Do you want to recover all the compounds as separate, pure chlorides, in a quantitative manner? That is probably impossible without specialized equipment/reagents. Why do you need to perform this separation? – Nicolau Saker Neto Dec 26 '13 at 15:28
• Because I need potassium chloride and have the described mix (a dietary salt replacement) is the only available source of it to buy at my location, @NicolauSakerNeto. – Ivan Dec 26 '13 at 15:40
• If separating the three into separate pieces is too complex then I can consider it enough to get rid of the Sodium and leave Magnesium in place. Perhaps there is a way to gather Sodium ions from a water solution? – Ivan Dec 26 '13 at 17:00
• Purity requirements are very liberal, I don't need very pure substances, getting KCl concentration from 30% to, say, 80% or even 70% can be considered better than nothing. – Ivan Dec 26 '13 at 17:11
• By the way, if you're doing this to try and get very low sodium table salt, I should warn you that actual potassium chloride isn't very pleasant. Also, this separation is really difficult because sodium and potassium are chemically extremely similar, so it usually relies on steric effects AFAIK. If you have problems getting hold of pure potassium chloride, you'll probably have problems getting hold of what you need to separate sodium ions and potassium ions. – Aesin Dec 26 '13 at 17:26

As I mentioned in the comments, a quantitative separation of the three compounds will probably not be possible outside of a proper laboratory, however a coarse separation is possible. A quick search on the internet yields this link for the separation of $\ce{NaCl}$ and $\ce{KCl}$. You didn't give us the exact composition of the table salt substitute, so I assume $\ce{MgCl2}$ is present in much smaller amounts than the first two compounds.

The simplest strategy is to use the difference between the solubilities of the salts in water or ethanol. Using water is much more convenient, so we'll stick to that. I'll outline the method to obtain the highest purity $\ce{KCl}$. First, add a very large amount of table salt substitute to a container with boiling water (you'll want to add some 200-250 g per 100 g of water) and let the mixture dissolve as much as possible (though it probably won't even come close to dissolving completely). What you have then is a boiling water solution saturated with both $\ce{NaCl}$ and $\ce{KCl}$. Now, while the water is still very hot, quickly filter off the excess salt (might be doable with a fine sieve, or a funnel with some cotton would work better) and reserve the liquid in another container. Let the liquid cool down (using an ice bath is optional but increases yield and purity), and it will start to precipitate the salts. Now you perform another filtration, but this time you keep the solid retained on the filter, not the liquid. This yields purified $\ce{KCl}$.

This strategy works because it turns out that $\ce{NaCl}$ and $\ce{KCl}$ behave rather differently in how their solubility in water is affected by temperature.

$\ce{NaCl}$ soubility in water:

• 35.6 g/100 g water (0°C)
• 35.9 g/100 g water (20°C)
• 39.0 g/100 g water (100°C)

$\ce{KCl}$ soubility in water:

• 28.0 g/100 g water (0°C)
• 34.2 g/100 g water (20°C)
• 56.3 g/100 g water (100°C)

Evidently, as the filtered liquid cools down, $\ce{NaCl}$ remains almost as soluble in cooler water as in hot water, while $\ce{KCl}$ is only about half as soluble. This means that a large amount of $\ce{KCl}$ will precipitate as the filtered liquid cools, while only a bit of $\ce{NaCl}$ will follow. Hence, ideally, from 100 g of water saturated with both salts at 100°C (and containing nothing else), you will be able to extract 25.0 g of a mixture of salts containing 87.6% $\ce{KCl}$ if you perform the second filtration at 20°C. If you filter at 0°C, then the yield increases to 31.7 g of a mixture containing 89.3% $\ce{KCl}$. If for some reason you don't want to do the second filtration, you can just let the water evaporate completely after the first filtration, producing 95.3 g of a mixture containing 59.1% $\ce{KCl}$ from 100 g of saturated boiling water.

In practice your yield and the purity of the $\ce{KCl}$ will likely be somewhat lower due to losses in manipulation and the presence of other impurities. The procedure may be complicated by the very large initial amount of table salt mixture that you have to dissolve in water. You can reduce the amount of the mixture, but at some point the purification yield will start to go down.

• Thank you very much, @nicolau-saker-neto. For those seeking to use the method described in the answer I must note that although I have approved it (as it looks reasonable) I have not tested it myself yet. – Ivan Dec 27 '13 at 19:03
• @Ivan How would you even test it without a lab? – jeremy Jan 2 '14 at 22:08
• Because the procedure is basically just to boil water... – jheindel Sep 8 '15 at 3:53

Add $\ce{MgSO4}$, until a saturated solution of magnesium salts has been obtained. Upon evaporating down, it will be found that all the potassium salts have been precipitated more or less completely, and with kainite $\ce{KCl. MgSO4.3H2O}$ being the essential potassium-containing salt.

Add the kainite to a boiling solution of $\ce{NaHCO3}$ with moles equal to those of $\ce{Mg^{2+}}$. Then, freeze the mix until $\ce{Na2SO4.10H2O}$ is visible along with the visible $\ce{MgCO3}$ suspension.