a 25.0 ml aliquot of vinegar was diluted to 250 ml in volumetric flask. Titration of...
a 25.0 ml aliquot of vinegar was diluted to 250 ml in volumetric flask. Titration of 50.0 ml aliquots of diluted solution required an average of 34.88 ml of 0.09600 M NaOH. Express the acidity of the vinegar in terms of percentage (w/v) of acetic acid.
Homework Answers
The balanced reaction
CH3COOH + NaOH --> CH3COONa + H2O
Moles of NaOH = molarity x volume
= 0.09600 Mol/L x 34.88 ml x 1L/1000 mL
= 0.00334848 mol
From the stoichiometry of the reaction
Stoichiometric Molar ratio of CH3COOH : NaOH = 1 : 1
moles of CH3COOH in the 50 ml aliquot = 0.00334848 mol
Moles of CH3COOH in flask
= 0.00334848 mol x 250 mL / 50 mL
= 0.0167424 mol
Mass of CH3COOH = moles x molecular weight
= 0.0167424 mol x 60 g/mol
= 1.0045 g
Acidity of vinegar in 25 mL of aliquot = 1.0045 g / 25 mL
= 0.04018176 x 100%
= 4.02%
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