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Topic: Help balancing this chem equation  (Read 10316 times)

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Offline revolve

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Help balancing this chem equation
« on: February 08, 2008, 02:29:15 PM »
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

Offline Borek

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Re: Help balancing this chem equation
« Reply #1 on: February 08, 2008, 03:16:25 PM »
Start with Ba. Treat CH3COO- as one thing. Don't worry about fractions at first stage - you may remove them later.

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Offline revolve

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Re: Help balancing this chem equation
« Reply #2 on: February 08, 2008, 03:47:17 PM »
Start with Ba? Isn't Ba already balanced?

Offline Arkcon

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Re: Help balancing this chem equation
« Reply #3 on: February 08, 2008, 04:21:13 PM »
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).
« Last Edit: February 08, 2008, 06:25:07 PM by Arkcon »
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Offline Borek

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Re: Help balancing this chem equation
« Reply #4 on: February 08, 2008, 04:26:03 PM »
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...
« Last Edit: February 08, 2008, 04:48:12 PM by Borek »
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Offline revolve

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Re: Help balancing this chem equation
« Reply #5 on: February 09, 2008, 09:05:22 PM »
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.
« Last Edit: February 09, 2008, 09:20:46 PM by revolve »

Offline Yggdrasil

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Re: Help balancing this chem equation
« Reply #6 on: February 09, 2008, 09:11:24 PM »
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?

Offline revolve

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Re: Help balancing this chem equation
« Reply #7 on: February 09, 2008, 09:26:08 PM »
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.




Offline enahs

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Re: Help balancing this chem equation
« Reply #8 on: February 09, 2008, 09:49:32 PM »
It is only wrong because you are writing Ba2+; when it is in fact Ba2+. "One" Ba atom with a positive 2 charge.


Offline Arkcon

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Re: Help balancing this chem equation
« Reply #9 on: February 09, 2008, 09:58:07 PM »
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?
« Last Edit: February 09, 2008, 10:26:10 PM by Arkcon »
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline revolve

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Re: Help balancing this chem equation
« Reply #10 on: February 09, 2008, 10:24:43 PM »
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
« Last Edit: February 09, 2008, 11:23:05 PM by revolve »

Offline Arkcon

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Re: Help balancing this chem equation
« Reply #11 on: February 09, 2008, 10:33:09 PM »
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
« Last Edit: February 09, 2008, 11:06:49 PM by Arkcon »
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

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