A vinegar titration was completed using a pH meter as an indicator of the changi

Question

A vinegar titration was completed using a pH meter as an indicator of the changing pH. A 25.00 mL sample of diluted vinegar (diluted by a factor of 5) was placed in a beaker and subsequently titrated with 0.1098 M NaOH. A derivative curve of the titration suggested that the equivalence point occurred at 40.90 mL. Calculate the mass percent of CH_3COOH. (The density of vinegar is 1.008 g/mL.) Moles of base at the equivalence point = Moles of acid at the equivalence point = Mass of acid = Volume of concentrated vinegar = Mass of vinegar solution = Mass percent of acetic acid =

Explanation / Answer

Titration of vinegar

moles of base used = 0.1098 M x 40.90 ml = 4.49082 mmol

moles of acid present in solution = 4.49082 mmol

mass of acid present = 4.49082 mmol x 60.05 g/mol/1000 = 0.270 g

Volume of concentrated vinegar = 5 ml

mass of vinegar solution = 5 ml x 1.008 g/ml = 5.040 g

Mass percent of acetic acid = 0.270 x 100/5.040 = 5.36%

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