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Chapter 3 problems with solutions CHEM 3510-001 Fall 2021 Biological Chemistry I.pdf

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9/11/21, 4:57 PMChapter 3 problems with solutions: CHEM 3510-001 Fall 2021 Biological Chemistry I1/18Chapter 3 problems with solutionsChapter 3 problems2) Relationship between the titration curve and acid base properties of glycine.a. This represents the fully protonated form of Gly, which would only be present if no eq of NaOHhave been added to the solution.b. To answer this, you much think about the equilibria that Gly would undergo. For the averagenet charge to be +1/2, we need to be at the half equivalence point for titration of thecarboxylate. This only happens at (II) or at the pKa of the carboxylate.c. This occurs at the pKa of the amino group. We are not shown pKa, but know that to reach thatpoint, we need to have added 1.5 eq of NaOH. So the answer is IV.d. See c for the reasoning. This only occurs when 0.5 eq of NaOH have been added (II).e.f. The maximal buffering capacity occurs at the pKa ±1. So in the case of this graph, the correctanswers are II and IV.g. The average net charge is zero at the pI. The pI of Gly is 2+10/2=6.0. So, this occurs at pointIII.h. The carboxyl is fully titrated when 1 eq of NaOH has been added (III).i. Gly is fully titrated when 2 eq of NaOH have been added (V)j. The zwitterionic species shown would only occur at the pI (III)k. The average net charge on Gly is -1 only at point V.l. This occurs at the half equivalence point of titration of the carboxylate (pKa 2.34), so at II.m. III represents the isoelectric point.n. Titration is done when 2 equivalents of NaOH have been added (V).o. The worst regions are I, III, and V.3) How much Ala is present as the completely uncharged species?OK, lets think a little aboutthis. Even though we can draw the uncharged structure, which is shown for Ala, the pKa of thecarboylate is far lower than that of ammonia. So, in solution, we will never be able to produce theuncharged form. Therefore, the zwitterionic form will predominate. (b) The pI of Ala is 6 (assumingpKa of 2 and 10 for the carboxylate and amino groups).We need to figure out what fraction will beunprotonated at the carboxyl and deprotonated at the amino group at pH 6. So lets solve the HHequation twice for each group at pH 6.0Fraction of COOH protonated:6=2+log [-COO-]/[ -COOH]10= [-COO-]/[ -COOH]4
9/11/21, 4:57 PMChapter 3 problems with solutions: CHEM 3510-001 Fall 2021 Biological Chemistry I2/18So the probability of having the carboxyl group protonated is 1 in 10000.Now what is the probability of having the amino group deprotonated at pH 6?6=10+log[-NH ]/[-NH]-4 = log [-NH]/[-NH]10= [-NH]/[-NH]So the probability of having the amino group deprotonated is 1 in 10,000.The probability therefore, that at pH 6, the amino group will be deprotonated and the carboxylateprotonated is 1 in 10.

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