I need to separate a combination of Na2CO3 and NaHCO3 through a detailed labratory procedure for ms final project in AP Chemistry. Does anyone have any ideas? We thought about boiling points...
I need to separate a combination of Na2CO3 and NaHCO3 through a detailed labratory procedure for ms final project in AP Chemistry. Does anyone have any ideas? We thought about boiling points ability to dissolve in water but didn't know how we would obtain the mass of the separated substances. Thanks in advance!
We are looking at separating a mixture of NaHCO3 (sodium bicarbonate) and Na2CO3 (sodium carbonate). Both are white solid powders when dry at room temperature and a mixture of the two would be essentially visually undetectable. You cannot do any real chemical reactions here because the two chemicals can be converted to one another use acid and base chemistry, thus destroying the original ratio of the mixture. As far as I can see, there are only two possibilities here. NaHCO3 has a melting point of 50 degrees C which is actually a pretty low melting point for a solid. Na2CO3 has a much higher melting point at 851 degrees C. You could heat the solid mixture up to a bit above 50 degrees C, thus melting the bicarbonate but leaving the carbonate as a solid. The sodium carbonate is more dense so it should sink to the bottom of the flask. You can then decant off the liquid NaHCO3. Decant means to slowly pour off the liquid while leaving the solid sitting in the bottom of the flask.
The other, and probably easier, possibility is selective dissolving. Both chemicals are highly soluble in water, so that isn't useful here. But sodium bicarbonate is somewhat soluble in acetone while sodium carbonate is insoluble in acetone. You could put the mixture into a flask of warm acetone and stir vigorously to get the NaHCO3 to dissolve. The Na2CO3 will remain as a solid. Then pour the mixture through a paper filter. The Na2CO3 will stay in the filter. Then heat the acetone to boil it away, thus leaving the NaHCO3 in the bottom of the flask.