Molarity of titrant (NaOH):  0.4550 M HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l) Trial # First Second Third Fourth Initial buret reading 0.15 mL 2.43 mL 1.32 mL 0.58 mL Final buret reading 18.62 mL 20.87 mL 20.03 mL 19.14 mL Volume of titrant used 18.47 mL 18.44 mL 18.71 mL 18.56 mL   4) Calculate the molarity of the acetic acid in the vinegar solution (Show your work).  use FW for moles-->grams acetic acid. Molarity acetic acid = _____________ M   5) Calculate the weight % of acetic acid in the vinegar. How does this compare with the % listed on the label (5.00%)?  (For this calculation assume that density of vinegar is 1.03 g/mL and of course, show your work). Weight % = ___________   6) If you didn’t get the same weight % of acetic acid as listed on the vinegar label (5.00 %), what are two things (be specific) that could’ve happened during the experiment that could explain the variation from the expected weight %? To do this you need the mass of the acetic acid and the mass of the vinegar you are provided the density of the vinegar. You can use the density for volume-->grams vinegar

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...
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Molarity of titrant (NaOH):  0.4550 M

HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l)

Trial #

First

Second

Third

Fourth

Initial buret reading

0.15 mL

2.43 mL

1.32 mL

0.58 mL

Final buret reading

18.62 mL

20.87 mL

20.03 mL

19.14 mL

Volume of titrant used

18.47 mL

18.44 mL

18.71 mL

18.56 mL

 

4) Calculate the molarity of the acetic acid in the vinegar solution (Show your work).

 use FW for moles-->grams acetic acid.

Molarity acetic acid = _____________ M

 

5) Calculate the weight % of acetic acid in the vinegar. How does this compare with the % listed on the label (5.00%)?  (For this calculation assume that density of vinegar is 1.03 g/mL and of course, show your work).

Weight % = ___________

 

6) If you didn’t get the same weight % of acetic acid as listed on the vinegar label (5.00 %), what are two things (be specific) that could’ve happened during the experiment that could explain the variation from the expected weight %?

To do this you need the mass of the acetic acid and the mass of the vinegar

you are provided the density of the vinegar. You can use the density for volume-->grams vinegar

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