Arginine1214.0VI12.0F12.010.0FIV10.08.08.0pH6.06.04.0-4.02.02.00.00.51.01.52.02.53.0Eguiyslonts of OuBelow you can find the titration curve for arginine. The pH was monitored, and the resultswere plotted as shown in the graph. The key points in the titration are designated I to VI. Foreach of the statements (a) to (f), identify the appropriate key point in the titration. (pKa(СООН): 1.82, рКa (NH2): 8.99, pKa (R): 12.48)Average net charge of +2 predominates:The predominant species is +H3N-CH-((CH2)3-NH-C-NH2-NH2+)-COO- :Average net charge of arginine is 0:d.а.b.C.The pH is equal to the pka of the carboxyl group:Half of the backbone amino groups are protonated:Average net charge of arginine is +1/2:The carboxyl group has been completely titrated:е.f.g.

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Answered: Average net charge of +2 predominates:…

Average net charge of +2 predominates: а. b. The predominant species is +H3N-CH-((CH2)3-NH-C-NH2-NH2+)-COO- : Average net charge of arginine is 0: C.

Curren'S Math For Meds: Dosages & Sol

11th Edition

ISBN:9781305143531

Author:CURREN

Publisher:CURREN

Chapter10: Reconstitution Of Powdered Drugs

Section: Chapter Questions

Problem 7.5P

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Concept explainers

Osmosis

In simple words, osmosis is the slow, gradual movement of a liquid through a membrane. But there are nuances to this process. For instance, there has to be an imbalance in the distribution of solute in the liquid on either side of the membrane, which has to be semipermeable. The liquid in the region with a lower solute concentration moves to the region with a higher concentration, thereby displacing the solution in the process. The objective is to equalize the concentration of molecules on both sides of the membrane; the displacement of liquid can be viewed as a byproduct of this equalization. Since there is no energy required to carry osmosis, it is considered a passive process.

Question

Arginine
12
14.0
VI
12.0F
12.0
10.0F
IV
10.0
8.0
8.0
pH
6.0
6.0
4.0-
4.0
2.0
2.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Eguiyslonts of Ou
Below you can find the titration curve for arginine. The pH was monitored, and the results
were plotted as shown in the graph. The key points in the titration are designated I to VI. For
each of the statements (a) to (f), identify the appropriate key point in the titration. (pKa
(СООН): 1.82, рКa (NH2): 8.99, pKa (R): 12.48)
Average net charge of +2 predominates:
The predominant species is +H3N-CH-((CH2)3-NH-C-NH2-NH2+)-COO- :
Average net charge of arginine is 0:
d.
а.
b.
C.
The pH is equal to the pka of the carboxyl group:
Half of the backbone amino groups are protonated:
Average net charge of arginine is +1/2:
The carboxyl group has been completely titrated:
е.
f.
g.

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