[OasisJuice]

A 10.00 mL solution of 0.2500 M CH3COONa is titrated with 0.1300 M HCl to the equivalence point.
CH3COO− + H+ → CH3COOH

volume of HCl required=19.23 mL

Calculate the pH of the solution at the equivalence point.

Now what I did was first
total volume added = 10mL+19.23mL = 29.23mL
Moles of CH3COO- reacted = Moles of CH3COOH made = 0.25M*0.01L = 0.0025mol
Molarity = 0.0025mol/(0.02923L) = 0.08553 M

x^2 / 0.0855 = Ka of CH3COOH

x^2/0.0855 = 1.8197E-5
x=1.555E-6

-log=pOH = 5.81
14-5.81 = 8.19

Which is wrong.  Please help, because I've tried this question over 16 times,:((((

[Borek]

CH₃COOH dissociation is not producing OH^-...

And x is not 1.555E-6.

[OasisJuice]

CH₃COOH dissociation is not producing OH^-...

And x is not 1.555E-6.

Oh....Well Im not sure what it's producing then...H+ ions?  So then by just solving for x (My calculation's wrong?) and doing the log of that, I would get the answer?

[Borek]

Well Im not sure what it's producing then...H+ ions?

Hardly surprising, acetic acid dissociates giving H⁺.

So then by just solving for x (My calculation's wrong?) and doing the log of that, I would get the answer?

Yes, you just did some mistake in your math.