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Topic: Equivalence point of a "titration"  (Read 1408 times)

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Offline professordad

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Equivalence point of a "titration"
« on: February 24, 2013, 12:27:19 PM »
"Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH."

("Sample exercise 17.8" in "Chemistry: The Central Science" by "Brown Lemay Bursten".)

(The solution implied that "50.0 mL" NaOH was necessary for titration)
So there are 0.00500 mol acetic acid and 0.00500 mol NaOH.  The reaction of acetic acid and NaOH is
HC2H3O2 + NaOH  :rarrow:  H2O + NaC2H3O2
After the reaction, there should be 0.00500 mol NaC2H3O2, implying 0.00500 mol C2H3O2-

Now I would use HC2H3O2  ::equil::  H+ + C2H3O2-.  But HC2H3O2 has a concentration of 0 M.  So I can't use this equation..

The book said to use the reaction
"C2H3O2- (aq) + H2O (l)  ::equil::  HC2H3O2 (aq) + OH- (aq)"

Later in the solution the author(s) wrote
"Kb = [HC2H3O2-][OH-] / [C2H3O2-] = (x)(x) / (0.0500 - x)"

First, where did they get that reaction and why do they use this one over the one I mentioned?
Also, is the minus sign attached to the acetic acid concentration expression not supposed to be there?  Is it a typo?
Finally, why do you let the concentration of acetic acid be x when you already found that it was 0 M?

Thanks.

Offline Borek

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Re: Equivalence point of a "titration"
« Reply #1 on: February 24, 2013, 02:41:07 PM »
You are calculating pOH of the solution of conjugate base.
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