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?

Question

Evan Fox

Chemistry

16 March, 11:34

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?

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Kash Hess · Answer 1

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