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A 3.455 g sample of vinegar is titrated. It requires 16.83 mL of 0.2145 M NaOH...

A 3.455 g sample of vinegar is titrated. It requires 16.83 mL of 0.2145 M NaOH to get to the phenolphthalein end point. Calculate the weight percent of acetic acid in this sample of vinegar.

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Answer #1

NaOH= 16.83 mL of 0.2145M

number of moles of NaOH= 0.2145M x 0.01683L= 0.00361 moles

CH3COOH + NaOH ------------- CH3COONa   + H2O

1 mole            1mole

According to equation

1 mole of NaOH = 1 mole of CH3COOH

0.00361 mole of NaOH = 0.00361 mole of CH3COOH

number of moles of acetic acid = 0.00361 moles

molar massof acetic acid = 60 gram/mole

mass of acetic acid present in 0.00361 moles of acetic acd = 0.00361 x60 = 0.2166 grams

mass of acetic acid = 0.2166 grams

Mass of vinegar = 3.455 grams

The percent concentration by mass of acetic acid in sample of vinegar will be

%m/m = mass of acetic acid/mass of vinegar x100

           = 0.2166/3.455 x100 = 6.629 %

%m/m = 6.629 %.

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