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

Chemistry

Ronan Kelley

30 January, 09:20

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?

+1

Answers (1)

Know the Answer?

Teddy · 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