PH after titration of Weak Acid with a Strong Base

1. Homework Statement
Titrating .1006M CH3COOH with .2012M NaOH. pKa=4.74, Ka=1.8*10^-5
What is the pH at the equivalence point of the titration?

2. Homework Equations
pKa = -log(Ka) pH = -log[H+]

3. The Attempt at a Solution

Alright, after setting up an ICE table (in order to find [H+] at the equivalence point) I realized that when you work it out Ka = [H+], however when I plugged that in I got a pH in th 4-5 range when it should be above 7.
Here is my Ka reasoning
Ka = [A-][H+]/[HA]
Ka = [Co+x][x]/[Co-x] Co = Initial Concentration
x is very small (Weak acid, no/little disassociation)
Ka = x
Ka = [H+]

Am I missing something or is my logic just wrong? Anything that could point me in the correct direction would be greatly appreciated.

Answers and Replies

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symbolipoint
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Education Advisor
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The pH at equivalence is the same as if you dissolved the sodium salt in water, without any acid or base. Now, you calculation should be based on just the effect of the sodium salt alone in the water.

Late answer but i came across this question when i was looking for the same answer.

Don't forget the answer your getting is pOH!
Therefore pH = 14 - pOH (where your pOH will be between 5-6 for this type of question)
pH = > 7