Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Not sure where to start to calculate the molar concentration (molarity) determined by titration of the HCl solution and the HC2H3O2 solution.  Volume of HCl = 5ml Volume of HC2H3O2 = 5ml NaOH Concentration = 0.100 M  20ml of water  Reaction equations are shown in net ionic form. H+ (aq) + OH– (aq) → H2O(l) HC2H3O2(aq) + OH– (aq) → H2O(l) + C2H3O2– (aq)   Volume of NaOH at Endpoint= 4.79ml  (HCl Trial)  Volume of NaOH at Equivalence Point = 5.083ml (HCl Trial)  Volume of NaOH at Endpoint= 3.67ml (HC2H3O2 Trial)  Volume of NaOH at Equivalence Point =4.025 (HC2H3O2 Trial)

Not sure where to start to calculate the molar concentration (molarity) determined by titration of the HCl solution and the HC2H3O2 solution.  Volume of HCl = 5ml Volume of HC2H3O2 = 5ml NaOH Concentration = 0.100 M  20ml of water  Reaction equations are shown in net ionic form. H+ (aq) + OH– (aq) → H2O(l) HC2H3O2(aq) + OH– (aq) → H2O(l) + C2H3O2– (aq)   Volume of NaOH at Endpoint= 4.79ml  (HCl Trial)  Volume of NaOH at Equivalence Point = 5.083ml (HCl Trial)  Volume of NaOH at Endpoint= 3.67ml (HC2H3O2 Trial)  Volume of NaOH at Equivalence Point =4.025 (HC2H3O2 Trial)

Chemistry
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...
See similar textbooks
icon
Related questions
Question

Not sure where to start to calculate the molar concentration (molarity) determined by titration of the HCl solution and the HC2H3O2 solution. 

Volume of HCl = 5ml

Volume of HC2H3O2 = 5ml

NaOH Concentration = 0.100 M 

20ml of water 

Reaction equations are shown in net ionic form.

H+ (aq) + OH– (aq) → H2O(l)

HC2H3O2(aq) + OH– (aq) → H2O(l) + C2H3O2– (aq)

 

Volume of NaOH at Endpoint= 4.79ml  (HCl Trial) 

Volume of NaOH at Equivalence Point = 5.083ml (HCl Trial) 

Volume of NaOH at Endpoint= 3.67ml (HC2H3O2 Trial) 

Volume of NaOH at Equivalence Point =4.025 (HC2H3O2 Trial) 

 

Expert Solution
Step 1

Here, we need to use the expression

MacidVacid=MbaseVbase

which is valid at the equivalence point.

trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Acid-Base Titrations
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS