Lecture 1
Scheduling
• This week in lab:
• Tuesday/Wednesday: Check-in and Pipette calibration
• Thursday/Friday: Start of experiment 8:
“Determination of the concentration and the acid
dissociation constants of an unknown amino acid”
• Experiment 8 takes a total of two lab periods (4/3-4/8)
• Today’s lecture: Titration of an unknown amino acids
• Hint: You may think of the unknown amino acid
containing both HA+/- and H2A+ forms of the amino
acid
Polyprotic Acids
• Definition: Polyprotic acids, also known as polybasic acids, are able to
donate more than one proton per acid molecule, in contrast to monoprotic
acids that only donate one proton per molecule. The protons are usually
released one at a time.
• Examples: sulfuric acid (H2SO4=O2S(OH)2), phosphoric acid (H3PO4=
OP(OH)3), carbonic acid (H2CO3=OC(OH)2), oxalic acid ((COOH)2),
all amino acids (H2N-CHR-COOH)
Acid
pKa1
pKa2
Ascorbic acid
4.10
11.60
Carbonic acid
6.37
10.32
Malic acid
3.40
5.20
Oxalic acid
1.27
4.27
Phthalic acid
2.98
5.28
Phosphoric acid
2.15
7.20
Sulfuric acid
-10
1.92
pKa3
12.35
Amino Acids
•
Amino acids are the building blocks of proteins and
enzymes
• Amino acids have the form H2N-CHR-COOH where R is
a side chain
• Proteins dominantly contain the (S)-enantiomer
(exception: (R)-cysteine, glycine (achiral))
• NutraSweet (aspartame, artificial sweetener) is a famous
dipeptide composed of phenylalanine and aspartic acid
• Penicillins are tripeptides (L-Cysteine, D-Valine,
L-Aminoadipic acid)
• The isoelectric point (pI) is the pH value at which the
molecule carries no net electrical charge (HL).
• At this point, the amino acid displays its lowest solubility in polar
solvents (i.e., water, salt solutions) and does not migrate in the
electrical field either.
• Important for the chromatographic separation of peptides and
proteins
H H H
N
S
O
N
O
CH3
CH3
COO-K+
Diprotic Acids I
• Diprotic acids undergo the following equilibria:
•
• H2L+
HL + H+
Ka1
• HL
L- + H+
Ka2
• Three possible forms in solution: H2L+, HL, L• The solution contains all three species at any given
time. The individual concentration depends on the
pH-value.
Diprotic Acids II
•
Example: Bicarbonate buffer system
• pH<6.37: [H2CO3] > [HCO3-] >>> [CO32-]
• pH=6.37: [H2CO3]=[HCO3-]
• 6.37<pH<8.35: [HCO3-] > [H2CO3] >> [CO32-]
• 8.35<pH<10.32: [HCO3-] > [CO32-] >> [H2CO3]
• pH=10.32: [HCO3-]=[CO32-]
• pH>10.32: [CO32-] > [HCO3-] >>> [H2CO3]
• Relevance:
• pH-value of blood: 7.35-7.45
• pH=7.4: 91.5% HCO3-/8.5 % H2CO3
H2CO3
H2CO3
Example I
• Leucine
pKa1=2.33
pKa2=9.75
• Since Ka1 >>> Ka2, only the first equilibrium has to be
considered at low pH-values
Example II
• What is the pH-value of a 0.05 M H2L+ solution?
x2
Ka1 =
[0.05  x]
• The 5 % rule fails in this case. Thus, the quadratic
formula has to be used here.
•
x =1.31 * 10-2 M = [H+] (=26.2 %>>5 %)
•
pH= -log([H+])=1.88
• For the calculation above, we assumed that the
second equilibrium was unimportant (L- ≈ 0).
Example III
• However, using the number above we can find the
true concentration.
[H  ][L ]
Ka2 =
[HL]
Ka2[HL]
-10
=
1.79x10
L- =
[H ]
• With [HL] = [H+] =1.31 * 10-2 M.
• The calculation shows that the concentration
of L- is indeed very low compared to the other
concentrations.
Titration I
• Titration of diprotic acid has six points of interest
•
•
•
•
•
•
P1: Initial pH-value
P2: pH-value at halfway to first equivalence point (pH=pKa1)
P3: pH-value at first equivalence point
P4: pH-value at halfway to second equivalence point (pH=pKa2)
P5: pH-value at first equivalence point
P6: pH-value after adding excess of base
V= 0
Veq/2
Veq
1.5 Veq 2 Veq 2.5 Veq
Titration II
• Example: Titration of 10 mL of 0.050 M H2L+ with
0.050 M NaOH
• Two reactions have to be considered
• H2L+ + OH• HL + OH-
HL + H2O
L- + H2O
(1)
(2)
• Step 1: Initial pH-value (see previous calculation)
• Step 2: After 5.0 mL of base have been added,
[H2L+]=[HL]
 pH=pKa1=2.33
• Step 3: After 10.0 mL of base were added, the first equivalence point
is reached (=isoelectric point)
•
pK a1  pK a 2 2.33  9.75
pH 

2
2
 pH=6.04
Titration III
•
•
Step 4: After 15.0 mL of base have been added,
[HL]=[L-]
 pH=pKa2= 9.75
Step 5: After 20.0 mL of base were added, the second
equivalence point is reached. Since all of HL is converted
to L-, the hydrolysis of L- has to be considered (ICE).
L- + H2O
Initial
Change
Equilibrium
HL + OHL-
HL
OH-
5.0*10-4 moles
(=0.0100 L *0.050 M)
~0
~0
-x
+x
+x
(5.0 *10 4  x )
0.0300L
x
y
0.0300L
x
y
0.0300L
Titration IV
• Step 5 (continued):
• Determine pKb of L• Determine [OH-]
K w 1.0 *10 14
5
Kb 


5
.
59
*
10
K a 1.79 *10 10
Kb 
y2
[ L ]  y
• Using the quadratic equation, one obtains
y = [OH-] = 9.38*10-4 M (= 5.5 % of 0.0167 M)
pOH = 3.03
 pH=10.97
Titration V
• Step 6: After 25.0 mL of base have been added, all H2L+
has been converted to L-. This required 20.0 mL of base
to accomplish.
• There is an excess of 5.0 mL of base in the solution
• Find number of moles of base
• n = 0.0050 L * 0.050 M = 2.50*10-4 moles
• Find concentration of base
• c = 2.50*10-4 moles/0.0350 L = 7.14*10-3 M
• Find pOH and pH
• pOH = -log([OH-]) = 2.15
• pH = 14 – pOH = 11.85
Summary for Leucine
• The six points of interest in the titration of 0.050 M
leucine with 0.050 M NaOH are
Point
Base added
Equivalence
pH-value
Comments
P1
0.0 mL
0.0
1.88
P2
5.0 mL
0.5
2.33
=pKa1
P3
10.0 mL
1.0
6.04
=(pKa1+pKa2)/2=pI
P4
15.0 mL
1.5
9.75
=pKa2
P5
20.0 mL
2.0
10.97
P6
25.0 mL
2.5
11.85
• Question: How could the pH-value at V=4.0 mL or
V=14.0 mL be calculated?
Individual Work
•
•
•
•
•
•
In lab, this week on Thursday and Friday
The student obtains a standardized NaOH solution.
The students have used pH meters before (Chem 14BL) so the calibration should
go rather smoothly. If you do not remember how to do it anymore, please review
it in the lab manual (page 12).
Make sure to keep the standardized sodium hydroxide and the unknown amino acid
solution. DO NOT store your standard solution (NaOH) in volumetric flasks.
Use other glassware (i.e., Erlenmeyer flask, jar) to store the solutions (ask your TA).
The student has to perform three titrations of the unknown amino acid solution
(until pH=12)
Clean-up
•
•
Neutralize all titrant solutions with citric acid until the pH paper turns light green or
orange before discarding in the drain. Pour the small amount of waste NaOH used to
rinse the burette into the labeled waste container. Do not pour un-neutralized NaOH
solutions down the drain.
At the end of the assignment, place the capped bottles of unused NaOH and amino
acid on the lab cart for return to the lab support
Report
•
•
•
•
•
Use Excel for plotting titration curves and first-derivative graphs.
The pKa-values of the amino acid are determined from the full titration graph
To determine pKa1 and pKa2, locate the volume on the graphs half way between the two
equivalence point volumes determined from the expanded derivative curves. The pH at
this point is in the titration is equal to pKa2.
Next, measure an equal distance on the graph to the left of Vep1. The pH at this point is
equal to pKa1.
Error Analysis:
•
•
•
•
Calculate the relative average deviation in the concentrations of your amino acid.
Compare the relative average deviation with the inherent error calculated by propagating
the errors in measurements of the pipet, the volumes determined from the graphs, and the
standard base solution.
Estimate the absolute error in your pKa-values by considering the variability you had in
the pH’s of the solutions at the |DVep/2| points in the three titrations. Report the range for
each of the pKa-values.
The report is due on April 14, 2014 or April 15, 2015 at the beginning of the lab
section.