Question

Lab Report: Titration of Vinegar Experimental Data Trial 1 Trial 2 Initial Buret Reading Trial 3 OML OML OML Final Buret Reading 40 6 ML 40-3 ML 41-7 ML Volume of NaOH used 40.0ML 4 7 40.3 ML Molarity of NaOH used Volume of vinegar used 5ML SML 5ML Data Analysis 1. Write the balanced equation for the neutralization reaction between aqueous sodium hydroxide and acetic acid. The Molarity of Acetic Acid in Vinegar Trial 1 Trial 2 Trial 3 Moles of NaOH used in titration Moles of acetic acid neutralized in vinegar sample Molarity of acetic acid in vinegar Average molarity Determine the molarity of acetic acid in vinegar. Show all calculations for one trial 2. below.

Answer 1

1)

CH3COOH (aq) + NaOH(aq) ---> CH3COONa (aq) + H2O(l)

2) for trial 1
given
molarity of NaOH M1 = 0.1 M
volume of NaOH V1 = 40.6 ml
moles of NaOH titrated ,M1V1= 0.1M *40.6 ml=4.06 milli mols
volume of vinegar V2 = 5ml

molarity of vinegar M2 = ?

moles of NaOH titrated = moles of vinegar present
which is equal to 4.06 milli mols

So
molarity of acetic acid = number of moles/volume

molarity of vinegar M2=4.06 m.mols/5ml=0.812 M

	trial 1	trial2	trial3
moles of NaOH in millimoles	4.06	4.03	4.17
moles of acetic acid in millimoles	4.06	4.03	4.17
molarity of acetic acid in M	0.812	0.806	0.834
Average molarity

Average molarity =[0.812 + 0.806 +0.834]/3 = 0.817 M

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