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Topic: Barium and sulfuric acid  (Read 12601 times)

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Offline AL

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Barium and sulfuric acid
« on: October 22, 2010, 12:46:12 PM »
Hi,
I am having trouble getting started with this problem. I hope my equation is correct at least. If I have a 20,000 mg/L of barium in solution, what is the amount of sulfuric acid to pull out all the barium as barium sulfate? 

Ba + H2SO4 -->  BaSO4 + 2H

I appreciate any help to get me started. Thank you!
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Offline Borek

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Re: Barium and sulfuric acid
« Reply #1 on: October 22, 2010, 01:31:01 PM »
Is it present as a metal, not in ionic form?

Not that it changes final result.

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Offline AL

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Re: Barium and sulfuric acid
« Reply #2 on: October 22, 2010, 01:45:25 PM »
Right, as a metal.

So since 20,000 mg/L Ba2+ = 0.1456 mol Ba2+ and it reacts in a 1:1 ratio with sulfuric acid, does that mean I would also need 0.1456 mol/L sulfuric acid in order to get all the barium to precipitate as barium sulfate?
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Offline Borek

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Re: Barium and sulfuric acid
« Reply #3 on: October 22, 2010, 06:25:16 PM »
1:1 stoichiometry, that's right.
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