Titration Curve - Acetic Acid with NaOH?

Question

Given: [NaOH] = 0.100 M
pKa = 4.60
Ka = 2.50 x 10^ -5
Vol. of NaOH to reach Equivalence = 9.7 mL
Vol. of NaOH to reach 1/2 Equivalence = 4.85 mL

Question:

1. Calculate the concentration (M), of the diluted acetic acid solution?

2. Calculate the concentration of the original acetic acid solution?

Help!

Answer 1

9.7/1000 x 0.1 moles of NaOH reacted with the CH3COOH, but you have not said what volume of acid was used.
Neither have you said by what factor the original acid was diluted.
The Ka does not affect the concentration of the acid.