Vinegar Titration Lab

Topics: Sodium hydroxide, Base, Acid Pages: 3 (574 words) Published: October 8, 2014
Vinegar Titration Lab
Procedure:
1.) Obtain ~100 mL of NaOH. Record the molarity on the data table. 2.) Set up a ring stand with a buret.
3.) Place 5mL of vinegar in a 125 mL Erlenmeyer flask. Dilute the vinegar with 25 mL of water and add two drops of phenolphthalein. 4.) Fill the buret with NaOH. Record the initial volume of the buret in your data table. 5.) Titrate the vinegar sample until the first faint pink color does not disappear. 6.) Record the final volume of the buret in your data table. 7.) Repeat steps 3 through 6 until you have three trials that agree within five percent. Equipment:

Ring stand, goggles, buret, 125mL Erlenmeyer flask, 150mL beaker, excess vinegar and NaOH. Data Table:

Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Trial 6
Trial 7
Vinitial
2.6
3.9
4.9
2.85
0
2.9
1.1
Vfinal
21.3
21.3
23.3
21.89
19.3
21.4
20.0
ΔV
18.7
17.4
18.4
19.04
19.3
18.5
18.9
(data chosen from best trials out of 13)
Average of the trials: = 18.6057
Percent agreement from the average of each trial:
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Trial 6
Trial 7
100.507%
93.5197%
98.8944%
102.334%
103.732%
99.4319%
101.582%

Trials 1, 6 and 7 agree the closest to 100%, so I averaged them and got an average of 18.7 mL NaOH added to the vinegar. To find the molarity of the acetic acid in the vinegar and the percent of vinegar that is acetic acid, I used the mole ratio of the neutralization reaction that occurs between sodium hydroxide and acetic acid.

For every mole of sodium hydroxide added to the vinegar, one mole of acetic acid reacts with it. Since the mole ration of sodium hydroxide to acetic acid in this reaction is 1:1, then if I find the moles of sodium hydroxide I can find the grams of acetic acid in 5mL of vinegar. To do this, I do:

.0042075 moles NaOH means .0042075 moles of HC2H3O2 reacted as well. To find the mass, I do:

The density of vinegar is 1.01 g/mL, and so the mass of the 5mL of vinegar in the...
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