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Chm 217 post lab 8 titration of vinegar, Exercises of Chemistry

post lab questions for lab 8: titration of vinegar

Typology: Exercises

2023/2024

Uploaded on 05/21/2024

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Post-Lab Questions:
1. The main sources of error in this experiment were in the process of using the burette.
There could be too much NaOH added due to the vinegar due to not closing the
stopcock of the burette in time. There could also be errors in the calculations of the
mass percentage.
2. Sally’s lab book:
a. Given the mass of the vinegar of 25.20 g and the volume of the vinegar as 25.0
mL, the density of the vinegar can be found using 25.20 g/25.0 mL = 1.008 g/mL.
b. The mass of acetic acid is 3.5%, which means that the vinegar contains 3.5 grams
of acetic acid. The moles of acetic acid can be found using the equation 3.5g x
25.02 g/100g acetic acid = 0.876 g acetic acid. The mass of acetic acid needs to
be divided by 60 g in the equation 0.876 /60 = 0.015 mol acetic acid.
c. The volume of NaOH can be calculated since 1 mL of acetic acid is equivalent to 1
mL of NaOH. The molarity of NaOH is 0.201 M, which means the volume can be
calculated by 1000 mL x 0.0145 M NaOH/0.201 M = 72.14 mL of NaOH.
d. The molarity of acetic acid can be calculated by 73.55 mL x 0.201 M/100 mL =
0.148 M. since there are only 25 mL of acetic acid, 0.148 M x 4 = 0.5912 M is the
molarity of acetic acid.
e. The mass % of 1 mole acetic acid is 60 g, which can be multiplied by 0.5912 to
equal 35.47 g acetic acid. 100 g of vinegar contained 35.47 g x 100 g/1000 g =
3.55 g acetic acid. The mass % of acetic acid is 3.55%, which is equivalent to the
given 3.5%.
f. Errors:
i. If the burette was not dry when the NaOH was added, it would increase
the amount of NaOH needed to complete the experiment.
ii. The bubble in the tip of the burette would affect the volume of the NaOH
solution as it was being drained out but would otherwise not affect the
experiment.
iii. If there is a bubble at the beginning of the experiment, the volume of
NaOH would be measured as more than the actual amount released,
which could skew the measurement.
iv. The excess H2O would possibly mean that more NaOH would need to be
added as well as more vinegar, since the volume is increased.
v. The solute would be titrated too strongly with the solvent, which would
have a strong impact on the results.

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Post-Lab Questions:

  1. The main sources of error in this experiment were in the process of using the burette. There could be too much NaOH added due to the vinegar due to not closing the stopcock of the burette in time. There could also be errors in the calculations of the mass percentage.
  2. Sally’s lab book: a. Given the mass of the vinegar of 25.20 g and the volume of the vinegar as 25. mL, the density of the vinegar can be found using 25.20 g/25.0 mL = 1.008 g/mL. b. The mass of acetic acid is 3.5%, which means that the vinegar contains 3.5 grams of acetic acid. The moles of acetic acid can be found using the equation 3.5g x 25.02 g/100g acetic acid = 0.876 g acetic acid. The mass of acetic acid needs to be divided by 60 g in the equation 0.876 /60 = 0.015 mol acetic acid. c. The volume of NaOH can be calculated since 1 mL of acetic acid is equivalent to 1 mL of NaOH. The molarity of NaOH is 0.201 M, which means the volume can be calculated by 1000 mL x 0.0145 M NaOH/0.201 M = 72.14 mL of NaOH. d. The molarity of acetic acid can be calculated by 73.55 mL x 0.201 M/100 mL = 0.148 M. since there are only 25 mL of acetic acid, 0.148 M x 4 = 0.5912 M is the molarity of acetic acid. e. The mass % of 1 mole acetic acid is 60 g, which can be multiplied by 0.5912 to equal 35.47 g acetic acid. 100 g of vinegar contained 35.47 g x 100 g/1000 g = 3.55 g acetic acid. The mass % of acetic acid is 3.55%, which is equivalent to the given 3.5%. f. Errors: i. If the burette was not dry when the NaOH was added, it would increase the amount of NaOH needed to complete the experiment. ii. The bubble in the tip of the burette would affect the volume of the NaOH solution as it was being drained out but would otherwise not affect the experiment. iii. If there is a bubble at the beginning of the experiment, the volume of NaOH would be measured as more than the actual amount released, which could skew the measurement. iv. The excess H2O would possibly mean that more NaOH would need to be added as well as more vinegar, since the volume is increased. v. The solute would be titrated too strongly with the solvent, which would have a strong impact on the results.