Answer :
NaOH = 19.35 - 0.15
= 19.2 mL
19.2 mL x 0.317 M = 6.086 millimoles NaOH
6.086 mmol NaOH neutralized 6.086 mmol acetic acid
6.086 mmol acetic acid / 25.0 mL = 0.2435 mmol/mL
= 0.2435 mol/L
Acetic acid was 0.2435 M