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Topic: Acid Dissociation Constant help  (Read 3436 times)

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Offline jdmbo1

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Acid Dissociation Constant help
« on: April 04, 2007, 05:00:05 PM »
This is from a practice problem that I can't seem to figure out.

Acetic Acid M = .096
pH of Acetic Acid = 2.86
NaOH M = .101
Volume of NaOH = 8.11 ml
Volume of Acetic Acid and NaOH = 19.96 ml
pH for this buffer = 4.36

So far I figured out the [H+] for the volume of the acetic acid and NaOH as 4.47x10-5. My next step is to figure out the [A-] and [HA] of this solution. More information. I figured out the [H+] for acetic acid as .0014 and the [HA] for acetic acid as .0941. In the answer section, they give the values for [A-] of acetic acid after addition of NaOH as 0.0292 M and [HA] of acetic acid after addition of NaOH as .0387. I can't seem to replicate those answers. Can someone show me the basic concept on how to find solve this.

Offline Borek

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Re: Acid Dissociation Constant help
« Reply #1 on: April 04, 2007, 05:21:05 PM »
Treat neutralization as limiting reagent question and don't forget about dilution.
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Offline jdmbo1

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Re: Acid Dissociation Constant help
« Reply #2 on: April 04, 2007, 09:16:51 PM »
ok, so I figured out the [A-], .101 M NaOH x .00811 L = .000819
19.96+8.11 = 28.07 ml --> .002807 L
(.000819)/(.002807) = .0292. Are those calculations correct?

How would I find the [HA] now?

Offline Borek

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Re: Acid Dissociation Constant help
« Reply #3 on: April 05, 2007, 02:48:45 AM »
Rest of the acid.
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