Molarity of titrant (NaOH):  0.4550 M   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               Write the balanced neutralization reaction that occurs when you titrate the acetic acid (HC2H3O2)  with sodium hydroxide (NaOH). Hint: It’s in the background. HC2H3O2 (aq) + NaOH (aq) → NaC2H3O2 (aq) + H2O (l)   Using data from all 4 titrations, calculate the average volume of standard NaOH solution required to titrate 10.00 mL of the vinegar solution (Show your work).     Average volume to titrate 18.47+18.44+18.71+18.56/4=   Average Volume = 18.545  mL   Calculate the average number of moles of NaOH used in the titrations (show your work).     Average Moles NaOH =              moles   Calculate or determine (give reasoning) the number of moles of acetic acid in the 10.00 mL sample of vinegar. NaOH     Average Moles acid = ____________ moles   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

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 51P
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Molarity of titrant (NaOH):  0.4550 M

 

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

             

Write the balanced neutralization reaction that occurs when you titrate the acetic acid (HC2H3O2)  with sodium hydroxide (NaOH). Hint: It’s in the background.

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

 

  1. Using data from all 4 titrations, calculate the average volume of standard NaOH solution required to titrate 10.00 mL of the vinegar solution (Show your work).

 

 

Average volume to titrate 18.47+18.44+18.71+18.56/4=

 

Average Volume = 18.545  mL

 

  1. Calculate the average number of moles of NaOH used in the titrations (show your work).

 

 

Average Moles NaOH =              moles

 

  1. Calculate or determine (give reasoning) the number of moles of acetic acid in the 10.00 mL sample of vinegar.

NaOH

 

 

Average Moles acid = ____________ moles

 

  1. 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

 

  1. 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 % = ___________

 

  1. 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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