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Titration Revisted

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Submitted By yayun
Words 1616
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Titrations Revisited

By Drew Rutherford
Concordia College

Introduction

The first experiment today will be the titration of acetic acid in vinegar. Vinegar is a solution of acetic acid, an organic acid of formula CH3COOH (MW = 60.0526 g/mole). In order to be sold as vinegar, it needs to meet the FDA’s guideline of 5.00% acetic acid by mass. Knowing that the density of vinegar is 1.04 g/mL and analyzing this solution by titration, chemists can determine the mass percentage of acetic acid in a sample of vinegar. The titration reaction is given below:

CH3COOH + NaOH ( CH3COONa + H2O Reaction 1

A customer has purchased a vinegar solution at a local thrift store at a 20% discount and he believes that the sample of vinegar he has purchased does not meet FDA regulations. He is suing the thrift store for the $0.36 he feels that he has been cheated out of. The court has asked you to analyze the sample and render your verdict. Does this sample conform to the FDA guideline?

If 2.00 mL of the thrift store vinegar required 9.73 mL of 0.150 M NaOH to reach the endpoint, then 0.00877 g of acetic acid was present.

9.73 mL NaOH x 1 liter NaOH x 0.150 mole NaOH x 1 mole CH3COOH x 60.0526 g CH3COOH 1000 mL 1 liter 1 moles NaOH 1 mole CH3COOH

= 0.0876 grams CH3COOH in sample

2.00 mL vinegar x 1.04 g vinegar = 2.08 g vinegar 1 mL

0.0876 g CH3COOH x 100 = 4.21 % CH3COOH by mass
2.08 g vinegar sample

You can find the molarity of acetic acid in vinegar by dividing the number of moles of acetic acid in vinegar by the volume of vinegar used.

Molarity of CH3COOH = 0.00146 moles CH3COOH 0.00200 L

Molarity of CH3COOH = 0.73 M

The thrift store vinegar does not meet the FDA guidelines.
The second experiment today will be the determination of the mass…...

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