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: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter16: Reactions Between Acids And Bases

Section: Chapter Questions

Problem 16.99QE: A scientist has synthesized a diprotic organic acid, H2A, with a molar mass of 124.0 g/mol. The acid...

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