3. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equa
ID: 277260 • Letter: 3
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3. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equal to the isoelectric point of alanine, the net charge on alanine is zero. Two structures can be drawn that have a net charge of zero, but the pre- dominant form of alanine at its pI is zwitterionic.
(a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pI?
(b) What fraction of alanine is in the completely un- charged form at its pI? Justify your assumptions.
12 11.30 09.60__._.-IV pH 65.97 -2.34 0.5 1.0 1.5 2.0 OH (equivalents) 3. How Much Alanine Is Present as the Completely Uncharged Species? At a pH equal to the isoelectric point of alanine, the net charge on alanine is zero. Two structures can be drawn that have a net charge of zero, but the pre- dominant form of alanine at its pl is zwitterionic CH CH H&N; ??? Uncharged ?? Zwitterionic (a) Why is alanine predominantly zwitterionic rather than completely uncharged at its pl? (b) What fraction of alanine is in the completely un charged form at its pl? Justify your assumptionsExplanation / Answer
(a) Key point: I
The given glycine is positively charged. Positively charged species occurs mostly in the lowest pH. In this case, the more protonation occurs at pH 0.
(b) Key point: II
At pKa, half of the protons are removed from ?-carboxyl group. For glycine pKa =2.34. At pKa or pK1 of 2.34, half of the protons are rmoved from glycine (charge = -1/2). Glycine already has a proton in the amino group. So the total charge = (-1/2)+1 = +1/2.
(c) Key point: IV
At pK2, half of the protons are removed from the ?-amino group.
(d) Key point: II
In carboxyl group,
[A-] = [HA]
=>[A-]/[HA] = 1
Henderson-Hasselbalch equation,
pH = pKa + log ([A-]/[HA]) = pKa + log(1) = pKa + 0
pH = pKa
(e) Key point: IV
In amino group,
[A-] = [HA]
=>[A-]/[HA] = 1
Henderson-Hasselbalch equation,
pH = pKa + log ([A-]/[HA]) = pKa + log(1) = pKa + 0
pH = pKa
(f) Key point: II AND IV
In the pKa regions, glycine abstract protons from acid or base abstract proton from glycine with minimum pH change. So glycine has the maximum buffering capacity.
(g) Key point: III
Average charge is zero when pH = pI = 5.97.
(h) Key point: III
At point II, the carboxyl group loses half of the protons. When reaching pI (Point III), the carboxyl group loses the remaining protons, leading to first equivalent point.
(i) Key point: V
When reaching pK2, half of the protons from the amino group is lost. The remaining is lost when the pH is 11.3.
(j) Key point: III
Amino group is protonated and the carboxyl group is deprotonated, resulting in a net charge of zero. The net charge is zero at pI (key point III).
(K) Key point: V
At pK2, the amino group is deprotonated, resulting in zero charge at amino group. The carboxyl group is also deprotonated, redulting in negative charge (-1) at carboxyl group. Net charge = 0+ (-1) = -1.
(l) Key point: II
At pK1, the carboxyl group is half deprotonated.
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