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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: damib on March 30, 2018, 05:19:39 AM

Title: Precipitation of BaCl2, CaCO3 and K2SO4
Post by: damib on March 30, 2018, 05:19:39 AM

Hello,
I am doing precipitation of 5 g of BaCl₂·2H₂O, 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.

Title: Re: Precipitation of BaCl2, CaCO3 and K2SO4
Post by: Borek 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.