A researchers titrated a 100 mL solution of 0.1 M glycine at pH 1.72 with 2 M NaOH solution. She then monitored the pH and plotted the results in the graph shown here. The key points in the titration are designated I through V. 12 10 9.60 8 pH 6 4 11.30 2 5.97 2.34 (1) (II) 0.5 › (III) (IV) I 1.0 1.5 OH (equivalents) (V) 2.0
A researchers titrated a 100 mL solution of 0.1 M glycine at pH 1.72 with 2 M NaOH solution. She then monitored the pH and plotted the results in the graph shown here. The key points in the titration are designated I through V. 12 10 9.60 8 pH 6 4 11.30 2 5.97 2.34 (1) (II) 0.5 › (III) (IV) I 1.0 1.5 OH (equivalents) (V) 2.0
Chemical Principles in the Laboratory
11th Edition
ISBN:9781305264434
Author:Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Publisher:Emil Slowinski, Wayne C. Wolsey, Robert Rossi
Chapter24: The Standardization Of A Basic Solution And The Determination Of The Molar Mass Of An Acid
Section: Chapter Questions
Problem 2ASA: In an acid-base titration, 21.16 mL of an NaOH solution are needed to neutralize 20.04 mL of a...
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For each of the statements, identify the appropriate key point(s) (I, II, III, IV, V) in the titration.
![A
researchers titrated a 100 mL solution of 0.1 M glycine
at pH 1.72 with 2 M NaOH solution. She then monitored
the pH and plotted the results in the graph shown here. The
key points in the titration are designated I through V.
12
10
8
pH 6
4
2
11.30
9.60
5.97
2.34
(II)
I
0.5
(III)
(IV)
1.0
1.5
OH- (equivalents)
(V)
2.0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81f8cdc7-827f-46be-9170-c83342326c3d%2Ff5e24c79-3eff-486b-abde-39d55f7c1f55%2Fbcenxbs_processed.png&w=3840&q=75)
Transcribed Image Text:A
researchers titrated a 100 mL solution of 0.1 M glycine
at pH 1.72 with 2 M NaOH solution. She then monitored
the pH and plotted the results in the graph shown here. The
key points in the titration are designated I through V.
12
10
8
pH 6
4
2
11.30
9.60
5.97
2.34
(II)
I
0.5
(III)
(IV)
1.0
1.5
OH- (equivalents)
(V)
2.0
![For each of the statements, identify the appropriate key
point(s) in the titration.
The average net charge of glycine is +1.
Half of the amino groups are ionized.
The average net charge of glycine is 0.
The average net charge of glycine is -1.
This is the isoelectric point for glycine.
Glycine has its maximum buffering capacity at
these points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81f8cdc7-827f-46be-9170-c83342326c3d%2Ff5e24c79-3eff-486b-abde-39d55f7c1f55%2Fygiwhq_processed.png&w=3840&q=75)
Transcribed Image Text:For each of the statements, identify the appropriate key
point(s) in the titration.
The average net charge of glycine is +1.
Half of the amino groups are ionized.
The average net charge of glycine is 0.
The average net charge of glycine is -1.
This is the isoelectric point for glycine.
Glycine has its maximum buffering capacity at
these points.
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