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 =

Answer 1

Mass of NaoH in 43 g ml of 0.1098 M NOOH = moles times molecular weight times volume of NaoH in lines = 0.1098 times 40 times 40.90/1000 = .0179 g Moles of base of equivalence point = mass/mole. Ut = 0.179/40 = 0.00449 M Moles at out of equivalent point CH_3COOH + NaoH rightarrow CH_3 COONa + H_2 O therefore moles of acid of equivalence point = 0.00449 M mass of acid = moles times molecular weight = 0.00449 times 60.05 = 0.269 g volume of concentrated vinegar = mass/density = 0.269/1.008 = 0.267 mL mass of vinegar solution = 25 g (since mass of vinegar is negligible density of loiter is 1 so 25 ml solution = 25 g) Mass % of acetic acid = .0269/25 times 100 = 1.076 %