Homework Help is Here – Start Your Trial Now!
Science
Chemistry
Part 2: Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted. Use stoichiometry to determine moles of the acid reacted. Divide moles acid by volume of the sample used to get molarity (mol/L). Average the molarity for all trials. 0.1137 Average Molaity Trial 1 Trial 2 ● ● Data Volume of HCI Initial Burette Reading Final Burette Reading Net Volume NaOH Moles NaOH Used Moles HCI Used Molarity of HCI Average Molarity of HCI 10AL 0.15 Me 8.58 8.43 0.00058 BUFE 00009 34 $80.000958 40.74 Males Hel Net Volung 0.000 9/8.43 0.01 0.958 1.284 0.0084 01137 lome 8.5PML of 15.88 ML 7.3m2 0.000 0830 Door 0.00083. 0.083 Na OH Trial 3 10 ML 15.88ML 23.452 7.344 7.7712 23. 0.000983 0488 00009 0.00972 0.0883 10ML = 0.01 L Trial 4 10ML 23.65ML 7.1072 0-000883 30.75ML 37.28m2 0.000807 8.000f 0.000807 Trial 5 0807 10 ML 30:2512 4.53m2 0.000742 0.000742 0.0742

Part 2: Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted. Use stoichiometry to determine moles of the acid reacted. Divide moles acid by volume of the sample used to get molarity (mol/L). Average the molarity for all trials. 0.1137 Average Molaity Trial 1 Trial 2 ● ● Data Volume of HCI Initial Burette Reading Final Burette Reading Net Volume NaOH Moles NaOH Used Moles HCI Used Molarity of HCI Average Molarity of HCI 10AL 0.15 Me 8.58 8.43 0.00058 BUFE 00009 34 $80.000958 40.74 Males Hel Net Volung 0.000 9/8.43 0.01 0.958 1.284 0.0084 01137 lome 8.5PML of 15.88 ML 7.3m2 0.000 0830 Door 0.00083. 0.083 Na OH Trial 3 10 ML 15.88ML 23.452 7.344 7.7712 23. 0.000983 0488 00009 0.00972 0.0883 10ML = 0.01 L Trial 4 10ML 23.65ML 7.1072 0-000883 30.75ML 37.28m2 0.000807 8.000f 0.000807 Trial 5 0807 10 ML 30:2512 4.53m2 0.000742 0.000742 0.0742

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

Can you confidently determine if the molarities are accurate? Explain why or why not.

 

Part 2: Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted Use stoichiometry to determine moles of the acid reacted. ● ● Divide moles acid by volume of the sample used to get molarity (mol/L), Average the molarity for all trials. 0.1137 Data Volume of HCI Initial Burette Reading Final Burette Reading Net Volume NaOH Moles NaOH Used Moles HCI Used Molarity of HCI Average Molarity of HCI 0.0084. Average Molaity Trial 1 Trial 2 10AL 0.15 Me 8.58 8.43 B5 0.000958 BULA 1200009% 34 $80.000958 60.874 Males Hel Net volun 0.000 9/8.43 0.01 0.958 1.284 01137 lome 8.5PML 15.88 ML 7.3m2 0.000830 2007 0.00083. 0.083 Na OH Trial 3 10 ML 15.88ML 23.4512 7.3ML 7.7712 23. 0.000983 468 0.02973 0.0883 10ML = 0.01L Trial 4 10ML 23.65ML 30.75ML 7.1072 0-000883 0.000807 0.0001 0.000807 0807 Trial 5 10 ML 30:35A2 37.28m2 4.53m2 0.000742 0.000742 0.0742
Transcribed Image Text:Part 2: Use the volume of NaOH used and the molarity of NaOH to calculate moles of NaOH reacted Use stoichiometry to determine moles of the acid reacted. ● ● Divide moles acid by volume of the sample used to get molarity (mol/L), Average the molarity for all trials. 0.1137 Data Volume of HCI Initial Burette Reading Final Burette Reading Net Volume NaOH Moles NaOH Used Moles HCI Used Molarity of HCI Average Molarity of HCI 0.0084. Average Molaity Trial 1 Trial 2 10AL 0.15 Me 8.58 8.43 B5 0.000958 BULA 1200009% 34 $80.000958 60.874 Males Hel Net volun 0.000 9/8.43 0.01 0.958 1.284 01137 lome 8.5PML 15.88 ML 7.3m2 0.000830 2007 0.00083. 0.083 Na OH Trial 3 10 ML 15.88ML 23.4512 7.3ML 7.7712 23. 0.000983 468 0.02973 0.0883 10ML = 0.01L Trial 4 10ML 23.65ML 30.75ML 7.1072 0-000883 0.000807 0.0001 0.000807 0807 Trial 5 10 ML 30:35A2 37.28m2 4.53m2 0.000742 0.000742 0.0742
Objective: In this simple titration, you will learn the procedure and calculations involved in titrations. You will react a solution of NaOH with a standard acid KHP. You will then use the calculated (standardized) concentration of NaOH to calculate the concentration of an HCI solution. Concepts: Titrations Data Table: Titration Data Part 1: Use the molar mass of KHP to calculate moles of KHP reacted. Use stoichiometry to determine moles of NaOH reacted. Divide moles NaOH by volume used to get molarity (mol/L). Average the molarity for all trials. Mass of KHP Initial Burette Reading Final Burette Reading Moles KHP Used Moles NaOH Used Molarity NaOH Average Molarity of NaOH Trial 1 0.49 0.88 21.25 Introduction to Titrations Data Sheet Net Volume 1.25 -0.88 16.25-0.92 NaOH 15.33 =20.37 Mass of kw Moles of Text P 0.0024 = 0.00 24 Males NAOld Net volume 0.1178 Trial 2 + 0.37 0.92 16.25 حممه 0.0018 to Titrations 0.0018 0.1171 0.1137 mol/L Trial 3 0.45 14.25 34.38 34.38 -16.45 = 18.13 20024 0.0022 0.0022 0.1213 + Trial 4 0.31 19.72 35.7 35.7-19.72 - 15.98 00024 0.0015 0.0015 0.09387 Trial 5 0.44 1.15 19.72 19.72-1-15 -18.57 0-0024 0.0022 douzz 0-1185
Transcribed Image Text:Objective: In this simple titration, you will learn the procedure and calculations involved in titrations. You will react a solution of NaOH with a standard acid KHP. You will then use the calculated (standardized) concentration of NaOH to calculate the concentration of an HCI solution. Concepts: Titrations Data Table: Titration Data Part 1: Use the molar mass of KHP to calculate moles of KHP reacted. Use stoichiometry to determine moles of NaOH reacted. Divide moles NaOH by volume used to get molarity (mol/L). Average the molarity for all trials. Mass of KHP Initial Burette Reading Final Burette Reading Moles KHP Used Moles NaOH Used Molarity NaOH Average Molarity of NaOH Trial 1 0.49 0.88 21.25 Introduction to Titrations Data Sheet Net Volume 1.25 -0.88 16.25-0.92 NaOH 15.33 =20.37 Mass of kw Moles of Text P 0.0024 = 0.00 24 Males NAOld Net volume 0.1178 Trial 2 + 0.37 0.92 16.25 حممه 0.0018 to Titrations 0.0018 0.1171 0.1137 mol/L Trial 3 0.45 14.25 34.38 34.38 -16.45 = 18.13 20024 0.0022 0.0022 0.1213 + Trial 4 0.31 19.72 35.7 35.7-19.72 - 15.98 00024 0.0015 0.0015 0.09387 Trial 5 0.44 1.15 19.72 19.72-1-15 -18.57 0-0024 0.0022 douzz 0-1185
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Concentration Terms
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