1. Lecture 1
Determination of the Concentration and the Acid Dissociation Constants of an Unknown Amino Acid

2. 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 (3/31-4/6)
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.

3. 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
pKa3
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
12.35
Sulfuric acid
-10
1.92

4. 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

5. 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.

6. Diprotic Acids II Relevance: Example: Bicarbonate buffer system
pH<6.37: [H2CO3] > [HCO3-] >>> [CO32-]
pH=pKa1=6.37: [H2CO3]=[HCO3-]
6.37<pH<8.35: [HCO3-] > [H2CO3] >> [CO32-]
8.35<pH<10.32: [HCO3-] > [CO32-] >> [H2CO3]
pH==pKa2=10.32: [HCO3-]=[CO32-]
pH>10.32: [CO32-] > [HCO3-] >>> [H2CO3]
Relevance:
pH-value of blood:
pH=7.4: 91.5% HCO3-/8.5 % H2CO3
H2CO3
H2CO3

7. Example I
Leucine
Since Ka1 >>> Ka2, only the first equilibrium has to be considered at low pH-values.
pKa1=2.33
pKa2=9.75

8. Example II What is the pH-value of a 0.05 M H2L+ solution?
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).
Ka1 =

9. Example III
However, using the number above we can find the true concentration.
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.
Ka2 =
L- =
= x10-10

10. 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= Veq/ Veq Veq 2 Veq Veq

11. Titration II
Example: Titration of 10 mL of M H2L+ with M NaOH
Two reactions have to be considered
H2L OH HL + H2O (1)
HL OH L- + H2O (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)
 pH=6.04

12. 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). Note that the total volume of the solution is 30.0 mL L- + H2O HL OH- L- HL
OH-
Initial
5.0*10-4 moles (= L *0.050 M) ~0
Change -x +x
Equilibrium

13. Titration IV Step 5 (continued): Determine pKb of L- Determine [OH-]
Using the quadratic equation, one obtains y = [OH-] = 9.38*10-4 M (= 5.5 % of M) pOH =  pH=10.97

14. Titration V
Step 6: After 25.0 mL of base have been added, all H2L+ has been converted to L-. This required mL of base to accomplish.
There is an excess of 5.0 mL of base in the solution:
Find number of moles of base excess
n = L * M = 2.50*10-4 moles
Find concentration of base
c = 2.50*10-4 moles/ L = 7.14*10-3 M
Find pOH and pH
pOH = -log([OH-]) = 2.15
pH = 14 – pOH = 11.85

15. Summary for Leucine
The six points of interest in the titration of M leucine with 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

16. Summary for Leucine
Note that the pH-change around the first equivalence point is much larger than around the second equivalence point.
How could the pH-values of the other points shown in the graph be calculated?

17. 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.

18. 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 12, 2016 or April 13, 2016 at the beginning of the lab section.