boonga
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Extracting Potassium Carbonate
Hey guys,
after the following reaction has occurred:
Na2CO3(aq) + K2SO4(aq) ---> Na2S04(aq) + K2C03(aq)
what is the easiest method to extract the potassium carbonate (K2C03) as a powder form? Using simple lab equipment and basic chemicals, seeing as I'm
a very amateur chemist.
Any help would be much appreciated.
Boonga 
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Retard-3000
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Fractional Crystallization will work, the solubility of sodium sulphate is 4.7g/100ml at 0C whereas potassium carbonates solubility is 112g/100ml at
20C.
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boonga
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"If a mixture of two or more substances in solution is allowed to crystallize, for example by allowing the temperature of the solution to decrease,
the precipitate will contain more of the least soluble substance." The sodium sulfate will crystallize out of solution but some will remain in
solution as impurity? is it possible to crystallize it all out without loosing too much potassium carbonate at the same time(as it will crystallize as
temperature decreases as well?
Correct me if i'm wrong, still learning
Cheers,
boonga
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Retard-3000
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You're correct, some will still remain in solution but will be possible to remove by doing another crystallization. You will end up losing a small
amount of potassium carbonate if you want it to be quite pure.
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bbartlog
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Potassium and sodium sulfate form double salts, e.g. K<sub>3</sub>Na(SO4)<sub>2</sub>. Depends on concentrations of the ions
and such. But I am skeptical that fractional crystallization will work here. Also, the reaction that is initially posited does not really occur; what
you have is a bunch of anions and cations in solution. The solid compounds do not exist until something drives them out of solution. And as far as I
know the first thing that will drop out at low temperatures is the aforementioned double salt.
[Edited on 12-8-2011 by bbartlog]
[Edited on 12-8-2011 by bbartlog]
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Otter
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Are there ANY "easy" ways to separate the sodium and potassium salts of one given acid? I.E. separate sodium carbonate from potassium carbonate?
*Otter sniffs the puddle.*
*Licks a tiny bit of it up.*
*Pukes out dinner.*
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not_important
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All depends on the anion. In some cases the Na and K salts have significantly differing solubilities, and don't readily form mixed salts. In a few
cases the solubilities in some non-aqueous solvent are significantly different.
But Na and K aren't that different, so in many cases it takes multiple recrystallisations to do the job. Using ion exchange resins might be easier -
saturate the resin using KCl, then run Na2CO3 solution through it with there being a large excess of the potassium charged resin' finish off with
recrystallisation.
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Aqua_Fortis_100%
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Just adding a little bit of sand in n_i's help:
Quote: Originally posted by not_important  | | All depends on the anion. In some cases the Na and K salts have significantly differing solubilities, and don't readily form mixed salts.
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i.e.
NaNO3 + KCl ----> NaCl + KNO3
Many amateur pyrotechnicians use this reaction to produce KNO3, and it works because KNO3 has extremely differing solubilities in hot and cold water
and NaCl solubility is almost constant in hot and cold water.
ps: Depending on how much you have, maybe will be easier to you react your K2SO4 with Ca(OH)2, get KOH and react with CO2 to get your K2CO3..
K2SO4 + Ca(OH)2 ---> 2KOH + CaSO4(s)
2 KOH + CO2 --> K2CO3 + H2O
excess CO2 will make KHCO3, but I dont know if this is stable like NaHCO3 (which isnt much anyway).
[Edited on 13-8-2011 by Aqua_Fortis_100%]
"The secret of freedom lies in educating people, whereas the secret of tyranny is in keeping them ignorant."
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not_important
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The question was " separate the sodium and potassium salts of one]/b] given acid?" so the KCl+NaNO3 example is not a proper fit.
CaSO4 and its hydrates have solubilities of a couple of grams per liter of water, so a bit of them will end up in the KOH. Upon adding CO2 you'll get
CaCO3 with a much lower solubility and precipitate out, if you don't overdo the CO2 and boil the resulting solution to decompose any calcium
bicarbonate.
KHCO3 is reasonably stable, it is the basis of "purple K" used in dry powder fire extinguishers, breaking down at temperatures above 100 C into K2CO3,
CO2, and H2O.
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Aqua_Fortis_100%
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Thanks for correcting me.. Not payed too much attention to this little detail. It just instantly appeared first on my mind, because was one reaction
that deals with K-Na exchange and also is very familiar to me.. So I posted it..
So one correct example would be NaCl and KCl found in "Na light" salt. Solubilities are (from wiki):
NaCl: 35.6 g (0 °C)/ 39.1 g (100 °C) (per 100mL pure water)
KCl: 28.1 g (0 °C)/ 56.7 g (100 °C) (per 100mL pure water)
Bacause of both having Cl- ions, the solubilites of these two salts would be quite different (will be much less) from the listed ones.. So in a very
concentrated hot solution of NaCl and KCl, NaCl is much more prone to ppt out first.. But I dont know in fact since I dont tried this.. Purifying KCl
out of agricultural grade KCl is much easier and cheaper..
Ah, thanks to correct me in steps in reactions too. Dont knew KHCO3 being used on fire extinguishers.
Other day I was tempted to try solvay synthesis with KCl to obtain KHCO3 but I just realised that KHCO3/KCl solubilies are similair.. And separating
the two would be messy..
"The secret of freedom lies in educating people, whereas the secret of tyranny is in keeping them ignorant."
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