A student titrates 25.0 ml of vinegar with 0.317 m naoh. the initial naoh burret reading is 0.15 ml and the final reading is 19.35 ml. what is the molarity of hoac in the vinegar sample?

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