Morality and Mass percent of Vinegar

Question

A 10.00 -mL sample of vinegar, an aqueous solution of acetic acid (HC2H3O2), is titrated with .5073M NaOH, and 17.64 mL is required to reach the equivalence point.

What is the molarity of the acetic acid?

If the density of the vinegar is 1.006g/cm^3, what is the mass percent of acetic acid in the vinegar?

Answer 1

HC2H3O2 + NaOH ==> C2H3O2Na + H2O ... balanced equation (note 1:1 ratio between acid and base)

moles of NaOH used = 0.5073 moles/L x 0.01764 L = 0.008949 moles

moles HC2H3O2 present = 0.008949 moles NaOH x 1 mole HC2H3O2/mole NaOH = 0.008949 moles

Molarity of acetic acid = 0.008949 moles/0.0100 L = 0.8949 M (to 4 significant figures)

10.00 ml vinegar x 1.006 g/ml = 10.06 g vinegar

0.8949 moles acetic acid/liter x 0.0100 liters = 0.008949 moles acetic acid in 10.00 ml of vinegar

0.008949 moles acetic acid x 60.052 g /mole = 0.5374 g acetic acid

Mass percent acetic acid = 0.5374 g/10.06 g (x100%) = 5.342%