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10 20 40 60 80 100 120 140 160 -Titration curve 1st Derivative 0.159 0.106 0.053 0.000 2nd Derivative 0.0061 0.0000 -0.0061 -0.0122 20 40 60 80 100 120 140 160 0.01 M NaOH (µL) |

10 20 40 60 80 100 120 140 160 -Titration curve 1st Derivative 0.159 0.106 0.053 0.000 2nd Derivative 0.0061 0.0000 -0.0061 -0.0122 20 40 60 80 100 120 140 160 0.01 M NaOH (µL) |

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
Chapter15: Acid-base Equilibria
Section: Chapter Questions
Problem 81E: Which of the indicators in Fig. 14-8 could be used for the titrations in Exercises 61 and 63?
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The second end point in the titration curve does not have an easy to define equivalence point and you are worried your estimates will have a lot of error.  One of your podmate’s older sister is in grad school and she suggested using derivatives to improve the precision of the calculation.  She also sent along the plot shown for some of her biochemical research, which suddenly makes you realize that maybe what you are learning in your gen chem lab is useful.   

 

What is the value of using derivatives to  find the equivalence point?

10 20 40 60 80 100 120 140 160 -Titration curve 1st Derivative 0.159 0.106 0.053 0.000 2nd Derivative 0.0061 0.0000 -0.0061 -0.0122 20 40 60 80 100 120 140 160 0.01 M NaOH (µL) |
Transcribed Image Text:10 20 40 60 80 100 120 140 160 -Titration curve 1st Derivative 0.159 0.106 0.053 0.000 2nd Derivative 0.0061 0.0000 -0.0061 -0.0122 20 40 60 80 100 120 140 160 0.01 M NaOH (µL) |
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