1. Mol NaOH = L HCl * Molarity of HCl * Stoichiometry (mol NaOH / mol HCl)Molarity of HCL= 0.116Amount of NaOH, mL = Final NaOH (ml) – Initial NaOH (mL) : 30.2-12= 18.2 mL = 0.0182 L L HCL= 0.02 L2. Mol HC2H3O2 = L NaOH * Molarity of NaOH (from # I) * Stoichiometry (mol HC2H3O2/ mol NaOH)Amount of HC2H3O2, mL = Final HC2H3O2 (mL) – Initial HC2H3O2 (mL) 22.95-0.15= 22.8 mL = 0.0228 LAmount of NaOH, mL = Final NaOH (ml) – Initial NaOH (mL) 23.09-0.01= 23.08 mL = 0.02308 L

Answered: 1. Mol NaOH = L HCl * Molarity of HCl *…

Introductory Chemistry: A Foundation

9th Edition

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 34QAP: 34. For each of the following solutions, the number of moles of solute is given, followed by the...

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1. Mol NaOH = L HCl * Molarity of HCl * Stoichiometry (mol NaOH / mol HCl)

Molarity of HCL= 0.116
Amount of NaOH, mL = Final NaOH (ml) – Initial NaOH (mL) : 30.2-12= 18.2 mL = 0.0182 L
L HCL= 0.02 L

2. Mol HC₂H₃O₂ = L NaOH * Molarity of NaOH (from # I) * Stoichiometry (mol HC₂H₃O₂/ mol NaOH)

Amount of HC₂H₃O₂, mL = Final HC₂H₃O₂ (mL) – Initial HC₂H₃O₂ (mL) 22.95-0.15= 22.8 mL = 0.0228 L
Amount of NaOH, mL = Final NaOH (ml) – Initial NaOH (mL) 23.09-0.01= 23.08 mL = 0.02308 L

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