Acetic Acid in Vinegar

Question

DATA

Vinegar                    Trial 1      Trial 2    Trial 3        

Final Volume, mL             5.16        11.98    17.98

Initial Volume in mL            0          5.16    11.98

Volume used in titration, mL      5.16        6.82    6.00

Standardized NaOH           Trial 1      Trial 2    Trial 3        

Final Volume, mL             24.13       37.35    35.19

Initial Volume in mL            0         13.39    8.69

Volume used in titration, mL     24.13       23.96    26.50

Concentration of NaOH, M      0.04M

Calculations 

Molarity of HC₂H₃O₂ in vinegar ________________  ______________ ___________________

Average molarity                            M

Mass of one liter of vinegar                    g

Mass of HC₂H₃O₂ in one liter of vinegar          g

Percent of HC₂H₃O₂ in your sample             %

mass percent of acetic acid in vinegar = (mass of acetic acid/ mass of vinegar solution) x 100

Answer 1

Trial molarities: 0.187, 0.141, 0.177 M  |  Average 0.168 M  |  10.1 g acetic acid per litre  |  1.01% by mass

Step 1 — molarity from each trial

At the equivalence point, moles of NaOH equal moles of acetic acid (1:1, monoprotic acid and monobasic base). Since both are volumes in mL, use M_aV_a = M_bV_b and the mL cancel:

M_acid = (M_NaOH × V_NaOH) / V_vinegar

Trial 1: (0.04)(24.13) / 5.16 = 0.187 M
Trial 2: (0.04)(23.96) / 6.82 = 0.141 M
Trial 3: (0.04)(26.50) / 6.00 = 0.177 M

Average = 0.168 M

Step 2 — mass of one litre of vinegar

This line on your data sheet is asking for a measurement — weigh a known volume and compute the density. If you did not record it, use 1.00 g/mL (vinegar is dilute aqueous), giving 1000 g per litre. State the assumption in your report.

Step 3 — mass of acetic acid in one litre

M(HC₂H₃O₂) = 60.05 g/mol

m = (0.168 mol)(60.05 g/mol) = 10.1 g

Step 4 — percent by mass

% = (10.1 / 1000) × 100 = 1.01%

Two things worth raising in your discussion

1. Your trials do not agree well. The spread runs from 0.141 to 0.187 M — about 25% of the mean, where a careful titration should repeat within 1–2%. Trial 2 is the outlier. Running Dixon's Q-test: Q = (0.177 − 0.141) / (0.187 − 0.141) = 0.78, against Q_crit = 0.97 for n = 3. So you cannot statistically reject trial 2 — with only three points the test has almost no power — and you should keep it in the average. But do say the precision was poor. Likely causes: overshooting the endpoint, an air bubble in the burette tip, or inconsistent judgement of the phenolphthalein colour change.

2. Your result is well below commercial vinegar. Store vinegar is about 5% acetic acid (roughly 0.83 M), and you obtained 1.01%. That is a factor of five. Two likely explanations: the vinegar was diluted before the titration — common in teaching labs, and 5× dilution would explain this exactly — or the NaOH concentration is not really 0.04 M. Check your procedure for a dilution step, and if there was one, multiply your result by the dilution factor before comparing to the label.

Do not treat that gap as an error to hide. Identifying it, and naming the most probable cause, is the part of a lab report that actually earns marks.