Question

A 25.00 mL sample of vinegar was titrated with 39.27 mL of 0.4293 M NaOH. Calculate the concentration of acetic acid in the vinegar sample.
HC2H302 + NaOH NaC2H302 + H2O

Answer 1

Answer:

0.6743 M

Explanation:

HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O

First we calculate how many NaOH moles reacted, using the definition of molarity:

• Molarity = moles / volume

• moles = Molarity * volume

• 0.4293 M * 39.27 mL = 16.86 mmol NaOH

One NaOH moles reacts with one acetic acid mole, so the vinegar sample contains 16.86 mmoles of acetic acid as well.

Finally we calculate the concentration (molarity) of acetic acid:

• 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M