Question

45 THE TITRATION OF VIR EXPERIMENT 8 REPORT SHEET Name Section Date Attach your completed Post-Lab Questions to this Report Sheet to hand in next week. INTRODUCTION By titrating vinegar with a standardized solution of sodium hydroxide, the experimental % acetic acid in the vinegar was determined and compared to the manufacturer's reported value. DATA 5% acetic acid on vinegar Trial 1 Trial 2 Trial 3 Volume of Vinegar Used (mL) Mass S Vinegar Reported on Bottle Concentration of NaOH (M) Initial Burette Reading (mL) Final Burette Reading ImL) Volume of NaOH used (ml) 0.250 0170 17.87 0.250 11.87 34.59 0.250 26:19 43.08 CALCULATIONS Show one example calculation on the next page. Trial 1 Trial 2 Trial 3 Average Moles of NaOH Used (mol) Moles of Acetic Acid Reacted (mol) Mass of Acetic Acid Reacted (g) Molar Concentration of Acetic Acid (M) Mass % of Acetic Acid % error for calculated versus manufacturer's value for % acetic acid in vinegar

Answer 1

Commercially sold vinegar is a 5% solution of acetic acid. The concentration of a given sample of vinegar can be verified by titrating with a solution of a base such as NaOH whose molarity is known.
The reaction between vinegar and NaOH is a neutralization reaction.

Here, volume of vinegar used for each titration should be specified.
Let us assume, the volume of vinegar used in each titration = 5 mL
Molarity of NaOH = 0.250 M in each sample.
Volume of NaOH used in trial 1 = Final reading - Initial reading = 17.87 mL - 0.07 mL =17.17 mL
Number of moles NaOH reacted = Molarity x Volume in L = 0.250 M x 0.01717 L = 0.0042925
According to the balanced equation, the moles of CH₃COOH reacted = 0.0042925
Molarity of the acetic acid solution = Number of moles/Volume in L =  0.0042925/0.005 L = 0.8585 M
Molar mass of CH₃COOH = 60.052 g/mol
Mass of CH₃COOH reacted = number of moles x Molar mass = 0.0042925 x 60.052 = 0.25777321 g
Assuming the density of vinegar = density of water = 1 g/mL
Mass of the samle of vinegar = density x Volume = 1 g/mL x 5 mL = 5g
Mass percentage of CH₃COOH in the sample of vinegar = (mass of CH₃COOH / Mass of solution) x 100
Mass percentage of CH₃COOH in the sample of vinegar = (0.25777321 g/5 g) x 100 = 5.1554 %

Using the above steps we can complete the given table:

CH3COOH(aq) + NaOH(aq) + CH3COONa(aq) + H2O

	Trial 1	Trial 2	Trial 3	Average
Volume of NaOH used(mL)	17.87-0.07 = 17.17 mL	34.59 -17.87 = 16.72 mL	43.08 -26.19 = 16.89 mL
Moles of NaOH used	0.0042925	Molarity x V in L = 0.250 x 0.01672 L = 0.00418	0.250 x 0.01689 = 0.0042225
Moles of acetic acid reacted	0.0042925	0.00418	0.0042225
Mass of acetic acid reacted(g)	0.25777321 g	Number of moles x Molar mass = 0.00418 x 60.052 = 0.25101736 g	0.0042225 x 60.052 = 0.25356957 g
Molar conc. of acetic acid (M)	0.8585 M	0.00418/0.005 = 0.836 M	0.0042225/0.005 = 0.8445 M
Mass % of acetic acid	5.1554 %	(0.25101736/5) x100 = 5.020 %	(0.25356957/5) x100 = 5.071 %	(5.1554 + 5.020 + 5.071)/3 = 5.0821 %

% Error = {(5.0821 - 5)/5} x 100 = (0.0821/5) x 100 = 1.642 %