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Topic: Percent Dissociation of HA  (Read 7235 times)

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Offline skibum143

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Percent Dissociation of HA
« on: September 02, 2010, 11:03:48 PM »
Hi...I'm confused about the % Dissociation on the attached problem:

What is the percent dissociation of HX? (image attached)

Our book defines percent dissociation as HA dissociated / HA initial * 100 is the % HA dissociated. But when I use that formula for the HX reaction, my answer is wrong. (4 dissociated / 10 initial = 40%) Please *delete me* I don't know the molarity in the problem so I was just using the number of molecules.

Offline ContinuousProcess

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Re: Percent Dissociation of HA
« Reply #1 on: September 03, 2010, 01:03:34 AM »
Looking at the HX box I see six HA and four particles coming from two dissociated HA's.

Offline Borek

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Re: Percent Dissociation of HA
« Reply #2 on: September 03, 2010, 03:36:12 AM »
(4 dissociated / 10 initial = 40%)

Why 4, why 10?
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Offline skibum143

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Re: Percent Dissociation of HA
« Reply #3 on: September 03, 2010, 10:32:56 AM »
I guess I'm confused on what the "Initial" HA count would be. I thought since HA + H20 -> H30+ + A-, that there would have been 10 HA originally, to end up with 6HA, 2H30+ and 2A-. But maybe since each HA creates 1H30+ and 1A-, there were only 8 HA originally?

So would that mean 2 / 8 = 25%?

I'll wait to try that until I get some feedback since I'm on my last try on my HW. Thanks so much for your *delete me*!

Offline opti384

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Re: Percent Dissociation of HA
« Reply #4 on: September 03, 2010, 11:16:25 AM »
Quote
Ithere were only 8 HA originally?

Yes.

Offline skibum143

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Re: Percent Dissociation of HA
« Reply #5 on: September 03, 2010, 12:04:08 PM »
I see. Thanks again for your *delete me*

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