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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: AL on October 22, 2010, 12:46:12 PM

Title: Barium and sulfuric acid
Post by: AL 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!

Title: Re: Barium and sulfuric acid
Post by: Borek on October 22, 2010, 01:31:01 PM

Is it present as a metal, not in ionic form?

Not that it changes final result.

Title: Re: Barium and sulfuric acid
Post by: AL 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?

Title: Re: Barium and sulfuric acid
Post by: Borek on October 22, 2010, 06:25:16 PM

1:1 stoichiometry, that's right.