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Topic: Precipitation of BaCl2, CaCO3 and K2SO4  (Read 2502 times)

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Offline damib

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Precipitation of BaCl2, CaCO3 and K2SO4
« on: March 30, 2018, 05:19:39 AM »
Hello,
I am doing precipitation of 5 g of BaCl2ยท2H2O, 5 g of CaCO3 and 3,7 g of K2SO4 in 100 ml of H2O in laboratory next week and I just need confirmation that I assume correct reactions.

CaCO3 isn't soluble at all, so firstly happens BaCl2+K2SO4 :rarrow: BaSO4 :spindown:+2KCl.
CaCO3 and BaSO4 is then vacuum filtrated and KCl solution is left for crystalization.
Then I put 100 ml of dilluted HCl in the beaker with filter cake, so this happens,
CaCO3+2HCl :rarrow: H2CO3(boiling) :spinup:+CaCl2, so another filter cake from vacuum filtration will consist of BaSO4.
Now I put solution with CaCl2 together with 5,5 g solution of Na2CO3 in 40 ml of water together, so what happens is
CaCl2+Na2CO3 :rarrow: CaCO3 :spindown:+2NaCl.
Calcium carbonate is then vacuum filtrated.

Is it how it works, or am I missing something crucial?
Thanks for help.

Offline Borek

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Re: Precipitation of BaCl2, CaCO3 and K2SO4
« Reply #1 on: March 30, 2018, 07:14:15 AM »
General logic looks OK, there are some fine points to consider. Be sure to use a reasonable excess of acid to dissolve the carbonate and excess of sodium carbonate later (beware: sodium carbonate is almost always hydrated to some extent, so what you will be weighing will not follow Na2CO3 formula).

Plus, there are no such things "as completely insoluble", but that probably doesn't matter too much for the experiment.
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