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 answer:
<span>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</span>
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