Titraion problem?

Question

consider the titration of 25.0 ml of 0.100M acetic acid with 0.100M NAOH.

Calculate the pH when 11.50 mL of 0.10M NAOH has been added.

Answer 1

find moles:

0.0250 litres @ 0.100mol/litre =  0.00250 moles acid

0.0115 litres @ 0.100 mol/litre = 0.00115 moles of NaOH

they react to produce 0.00115 moles of sodium acetate, and in so doing... decrease the acid to 0.00135 moles

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now we will find the new molarities:

0.036986

aectic acid: 0.00135 moles / 0.0365 litres = 0.0370 molar

acetate : 0.00115 moles / 0.0365 litres = 0.0315 molar

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HC2H3O2 --> C2H3O2-  &  H+

Ka = [C2H3O2-] [ H+] / [HC2H3O2]

1.8e-5 = [0.0315] [ H+] / [0.0370]

H+ = 2.11 e-5

pH = 4.675

your answer (2 sig figs in the K) :  pH = 4.68