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

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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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)

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

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 % = ___________

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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