ACID-BASE TITRATIONPRELA BOUESTIONSNAMENaOH according to the following equationHC,H O2NAOHNACH O2НОНacid.WELBL02. Calculate the grams of acetic acid in one liter of the vinegar. 52.5gS. The 4.00 mL sample of vinegar had a mass of 4.056 g. Calculate the grams of vinegar inone liter.3.4. Calculate the % acetic acid in the vinegar.g acetic acid x 100g vinegar% acetic acid||95

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A 4.00 mL sample af vinegar (active ingredient, noetic acnd) was titratedd yith 0 125OM JESTIONS NAME NaOH according to the following equation HC H,O2 NACH3O +HOH HOPN acid mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 0-7872M

A 4.00 mL sample af vinegar (active ingredient, noetic acnd) was titratedd yith 0 125OM JESTIONS NAME NaOH according to the following equation HC H,O2 NACH3O +HOH HOPN acid mL of base was needed to reach the end point. Calculate the molarity of the acetic acid. 0-7872M

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.28QAP

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Average the two percentages and calculate the percentage difference. given : Concentration of acetic acid in the 100 mL sample of titration 1 = 3.238 % Concentration of acetic acid in the 100 mL sample of titration 2 = 3.099 % NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of…