Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: revolve on February 08, 2008, 02:29:15 PM
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CH3COOH + Ba(OH)2 --> Ba + CH3COO + H2O
How do you approach this? I keep getting fractions when trying to balance O, or H that lead me no where. Could someone show me a step by step for this one? Help would be greatly appreciated. Thank you
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Start with Ba. Treat CH3COO- as one thing. Don't worry about fractions at first stage - you may remove them later.
http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-inspection
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Start with Ba? Isn't Ba already balanced?
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Start with Ba? Isn't Ba already balanced?
No. Consider this, if it were: HCl + NaOH, you'd know exactly what to do, right? (You might not, in that case, whoops, my bad, ask again).
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Ba is balanced only if you assume that you start with exactly one molceule of Ba(OH)2. Not accidentally it is a very good assumption in this case. Remember: CH3COO- is a complete ion, its oxygen, carbon nor hydrogen don't need to be balanced separately as long as CH3COO- is balanced on the whole.
I am assuming you are aware of the fact that it should be Ba2+ and CH3COO- on the right...
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CH3COOH + Ba(OH)2 --> Ba2+ + CH3COO- + H2O
Ok, So I have 1 Ba on both sides. I balance CH3COO- as a whole. So CH3COO- is missing an H on the left so I've got 4 H's on the left and 3 H's on the right. So if I balance these as a whole then I've got to use fractions correct?
CH3COOH --> 4/3CH3COO-
Then multiply through by 3
3CH3COOH --> 4CH3COO-
???
This is really confusing me. Dunno why this one is giving me a hard time, it seems others ones I have done have been much easier. I'm just overlooking something really really simple.
Thanks for your help by the way.
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CH3COO- can be treated like a polyatomic ion (the acetate ion). For simplicity, let's represent this polyatomic ion by the symbol Ac.
HAc + Ba(OH)2 --> Ba2+ + Ac- + H2O
can you balance this equation now?
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HAc + Ba(OH)2 --> Ba2+ + Ac- + H2O
Ok there are 3 H on the left and 2 on the right so
HAc + Ba(OH)2 --> Ba2+ + Ac- + 3/2H2O
then multiply through by 2?
2HAc + 2Ba(OH)2 --> 2Ba2+ + 2Ac- + 3H2O
I think this is wrong already.
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It is only wrong because you are writing Ba2+; when it is in fact Ba2+. "One" Ba atom with a positive 2 charge.
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Try it this way:
H[Acetate] + Ba(OH)2 -->
You know 3 things:
1). Acetate has a -1 charge, H has a +1 charge
2). Barium has a +2 charge (second column of peroiodic table) (OH) has a -1 charge, so you'll need two in this molecule
3). Acid and bases neutralize each other into water, that's one H+ and one OH-
So ...
H[Acetate] + Ba(OH)2 --> Ba([Acetate])2 + H2O
Now, of course, you know that the barium and acetate exist in solutions as ions, like you did in your equation. But putting them together in a molecule insures that you'll balance properly. Of course, now it's not balanced, two acetates on the right one on the left, two (OH)'s on the left and only one water on the right -- how will you fix it?
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Arkcon thank you very much and the others who helped.
Arkcon, yes it does make it easier when you combine the ions in the product.
2 CH3COOH + Ba(OH)2 = Ba2+ + 2 CH3COO- + 2 H2O
Thanks again
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See there you go, like I said in the beginning, you know:
HCl + NaOH --> NaCl + H2O
There are only two changes:
Instead of Cl- we have an organic anion
And every mole of barium (or calcium, even radium if you're crazy) hydroxide needs 2 equalvilents of acid to neutralize it.
If you're bored, have some fun with H2SO4 or H3PO4