General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter15: Acids And Bases
Section: Chapter Questions
Problem 15.100QP: A 2.500-g sample of a mixture of sodium carbonate and sodium chloride is dissolved in 25.00 mL of...
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Question
Calculate the percent difference in the two values for the molarity of the NaOH solution by:
%Difference =[M1-M2]\Mavg x100%
Given: 1st value for the molarity of NaOH = 0.0647 M
2nd value for the molarity of NaOH= 0.2627 M
NaOH vs H2SO4
Burette solution is NaOH and the pipette solution is 10.0 mL of 0.205 M H2SO4
Titration | Initial burette reading | Final burette reading | Volume of NaOH consumed | Average volume of NaOH |
Approximate | 0.0 mL | 31.8 mL | 31.8 mL | 31.7 mL |
Titration 1 | 0.0 mL | 31.5 mL | 31.5 mL | |
Titration 2 | 0.0 mL | 31.7 mL | 31.7 mL |
To find the average volume
Average volume = 31.8 mL + 31.5 mL + 31.7 mL331.8 mL + 31.5 mL + 31.7 mL3
= 31.7 mL
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
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