Question

Lab Report: Titration of Vinegar Experimental Data Trial 1 Initial Buret Reading OML 1 Trial 2 ONL Trial 3 OML Final Buret Reading 40.6 ML 40.3 ML 41.7 NL 417 Volume of NaOH used 40.3ML 70.6ML 0.1 Molarity of NaOH used 0.1 Volume of vinegar used 5ML SML 5M2

Answer 1

MaVa = MbVb (Where a = acid, b = base)

The moles of acetic acid in trial 1 = 0.1 mol/L * (40.6/1000) L = 0.00406 mol

i.e. The mass of acetic acid in trial 1 = 0.00406 mol * 60 g/mol = 0.2436 g

Mass of vinegar sample in trial 1 = (40.6 + 5) mL * 1 g/mL = 45.6 g

Mass percent of acetic acid in trial 1 = (0.2436 g/45.6 g)*100 = 0.534%

The moles of acetic acid in trial 2 = 0.1 mol/L * (40.3/1000) L = 0.00403 mol

i.e. The mass of acetic acid in trial 2 = 0.00403 mol * 60 g/mol = 0.2418 g

Mass of vinegar sample in trial 2 = (40.3 + 5) mL * 1 g/mL = 45.3 g

Mass percent of acetic acid in trial 2 = (0.2418 g/45.3 g)*100 = 0.534%

he moles of acetic acid in trial 3 = 0.1 mol/L * (41.7/1000) L = 0.00417 mol

i.e. The mass of acetic acid in sample 3 = 0.00417 mol * 60 g/mol = 0.2502 g

Mass of vinegar sample in trial 3 = (41.7 + 5) mL * 1 g/mL = 46.7 g

Mass percent of acetic acid in trial 3 = (0.2502 g/46.7 g)*100 = 0.536%

The average mass percent of acetic acid = (0.534 + 0.534 + 0.536)/3 = 0.535%