Question:
A 20.00 mL sample of vinegar was titrated with 1.50 M NaOH
Last updated: 7/25/2022
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A 20.00 mL sample of vinegar was titrated with 1.50 M NaOH. If it took 18.50 mL of NaOH to reach the equivalence or stoichiometric point, what is the mass percent of acetic acid (CH₃CO₂H) in the vinegar? Assume the density of vinegar is 1.00 g/mL. CH₃CO₂H + NaOH --> H₂O + CH₃CO₂Na