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Biochemistry I
PS2 Recitation
January 18, 2016
1. Go to the following URL and perform a pH titration. Collect data for the
pH versus mls of NaOH added. Use Excel to plot your titration data.
https://andrew.cmu.edu/user/rule/bc_oli/Pset/PS02/titration.html
pH Titration
9
2. You are required to make a 250 ml solution of 0.05 M Bis-Tris to use as a
buffer to perform a biochemical experiment at pH 6.0. You only have the
fully deprotonated form of this acid in the lab, plus 1M solutions of HCl
and NaOH.
a) Sketch the titration curve for the buffer on the right.
This is the answer to question 1.
b) How many moles of deprotonated buffer do you need to add to
make your buffer solution?
moles buffer = 0.05 moles/L x 0.25L = 0.0125 moles
0.76 eq HCl
8
a) A total of 10 ml of NaOH are required to completely titrate the
buffer solution. This would be equal to 1 equivalent. Therefore 5ml is
equal to 0.5 equivalents. The pH at 5ml is pH = 6.5, so the buffer is
Bis-Tris because its pKa is 6.5.
pH
7
0.24 eq OH
6
5
4
3
0 1 2 3 4 5 6 7 8 9 10
ml NaOH
R  [ A ] /[HA]  10 ( pH  pKa)
f HA  1 /(1  R)  [ HA] /[ AT ]
f A  R /(1  R)  [ A ] /[ AT ]
c) Does your answer to b) depend on the pH of the solution?
No, the only thing that changes with pH is the ratio of the protonated (HA) and the deprotonated (A),
they will always sum to AT.
c) Calculate fHA and fA. Do your calculated values agree with what you would estimate from the titration curve?
R = 10(pH – pKa) = 106-6.5= 10-0.5 = 0.316
fHA = 1/(1+R) = 1/(1.316) = 0.76, fA = 0.24
Yes, this is consistent since the pH< pKa so more should be protonated. At pH 6.0 the fraction
deprotonated is ~.24 from the above graph.
d) How many equivalents of HCL or NaOH do you have to add (specify what would be added to the solution).
You have to add HCl since you are starting from the right of the titration curve:
eq HCl = fHA
e) How many moles of HCl or NaOH would you have to add?
moles HCl = eq HCl x [AT] x V = 0.76 x 0.05 x 0.25 = 0.0095
3. The protein shown on the right contains one histidine residue
and one aspartic acid.
a) What would you predict for the pKa of the histidine residue?
Higher or lower than the normal pKa of 6.0? Why? [Hint:
how will the negative charge affect the acidity of the
histidine?]
O
O
O
O
H
N
+
N
H
1.
N
N
H
+
Electrostatic interactions are the most significant
in shifting pKas.
Free amino
Amino acid
2. Only the protonated form of histidine can participate
acid
in protein
in electrostatic interactions, therefore any effects Energy
HA
A
A
HA
on the deprotonated form can be ignored.
3. The negative charge on the aspartic acid residue will
stabilize the + charge on the histidine side chain, therefore lowering the energy of the
protonated (HA) form.
This will reduce the energy
difference between the HA and A form, making it less
100%
likely to ionize, making it a weaker acid with a higher pKa.
b) Assuming the protein is only biologically active when the histidine is
deprotonated, sketch the activity as a function of pH. [Hint: Is the
activity proportional to fHA or fA- ?] This would be proportional to the
50%
pH
0%
fraction deprotonated, i.e. Activity (%) = 100 x R/(1+R). The
activity would be 50% when the pH = pKa since ½ of the group will be deprotonated.
pKa
+
H
PS2 Recitation
4. A tripeptide (Ala-Ala-His) is shown below, fully
protonated at low pH (left) and fully
deprotonated at high pH (right). This diagram
indicates the charge on each ionizable group
when protonated and deprotonated.
The
sidechain of histidine is circled in the right
diagram.
You will use graphical methods to determine the
net charge on the peptide. The overall charge is
just the weighted sum of the charges on each
group:
Fraction Protonated
Biochemistry I
n
i
i
qTOTAL   ( f HA
q HA
 f Ai q iA )
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
January 18, 2016
HIS
COOH
0
1
2
3
4
5
6
7
Amino
8
9 10 11 12
i 1
pH
a) Sketch the fraction protonated versus pH for
each group (amino, His sidechain, carboxyl) on
the graph to the right. When pH=pKa of acid, fHA=0.5, when the pH is one unit lower, fHA=0.9,
50% protonated
etc. (see lecture 3)
50% deprotonated
q= +1.0
q= 0
b) Calculate the charge at pH = 6. Indicate the
contribution of each ionizable group to the
overall charge on the diagram below. Overall
charge = + 0.5, 1:1 mixture of the two
species.
fully protonated
q= + 1
fully deprotonated
q= -1
√ Have someone check your work.
5. Sketch the titration curve for the amino acid
glutamic acid, whose structure is shown
below.
a) Label the axes first, and add the
appropriate scale, in equivalents, on the
bottom. (One equivalent of NaOH is equal
to the number of moles of the weak acid).
b) Utilize the titration curve from question 4
to guide your sketch of this titration curve.


pKa=4
pKa=2
pKa=9.0
Since there are three ionizable groups, the x-axis ranges
from 0 to 3.
There are three pKa values, so there are three inflection
points, each at a ½ equivalent (0.5, 1.5, 2.5 equivalents).
The inflection points have been sketched as linear
segments with a small slope, indicating a buffer region.
6. View the Jmol page that shows methane and butane dissolved in water.
a) What do the dotted purple lines represent?
b) Which of the two compounds would cause a larger decrease in the entropy of water?
a) The dotted lines are hydrogen bonds between the water molecules that surround, or cage, the
dissolved non-polar compound.
b) The water molecules that are shown around the methane and butane are fixed in position, or
ordered. Since butane is bigger than methane, there are more ordered water molecules, thus
butane decreases the entropy of the water to a greater extent.
√ Have someone check your work.