DETERMINATION OF THE CONCENTRATION AND THE ACID DISSOCIATION CONSTANTS OF AN UNKNOWN AMINO ACID

Lecture 1

This week in lab: Tuesday/Wednesday: Check-in and Pipette calibrationThursday/Friday: Start of experiment 8: “The determination

of the concentration and the acid dissociation constants of an amino acid”

Experiment 8 takes a total of two lab periods (4/3-4/9)Today’s lecture: Titration of an unknown amino acidsHint: You may think of the unknown amino acid containing

both HA+/- and H2A+ forms of the amino acid

Scheduling

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)

Polyprotic Acids

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 strong 1.92

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 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.

Amino Acids

HOOC (S) NH

(S)

NH2

O

COOCH3

H2N (S) COOH

R H

+H3N (S) COO-

R H

Zwitterion

N

HN

O

H

O

SH

CH3

CH3

COO-K+

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 I

Example: Bicarbonate buffer system pH<6.37: [H2CO3] > [HCO3

-] >>> [CO32-]

pH=6.37: [H2CO3]=[HCO3-]

6.37<pH<8.35: [HCO3-] > [H2CO3] >> [CO3

2-] 8.35<pH<10.32: [HCO3

-] > [CO32-] >> [H2CO3]

pH=10.32: [HCO3-]=[CO3

2-] pH>10.32: [CO3

2-] > [HCO3-] >>> [H2CO3]

Relevance: pH-value of blood: 7.35-7.45

pH=7.4: 91.5% HCO3-/8.5 % H2CO3

Diprotic Acids II

H2CO3

H2CO3

Leucine

Since Ka1 >>> Ka2, only the first equilibrium has to be considered at low pH-values

Example I+H3N COOH +H3N COO-

+ H+

H2L+ HL

Ka1= 4.69*10-3

+H3N COO-

HL

H2N COO-

L-

+ H+ Ka2= 1.79*10-10

pKa1=2.33

pKa2=9.75

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+])=1.88For the calculation above, we assumed that the second

equilibrium was unimportant (L- ≈ 0).

Example II

x2

[0.05 x]Ka1 =

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.

Example III

[H][L ][HL]Ka2 =

L- = Ka2[HL][H]

= 1.79x10-10

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

Titration I

V= 0 Veq/2 Veq 1.5 Veq 2 Veq 2.5 Veq

Example: Titration of 10 mL of 0.050 M H2L+ with 0.050 M NaOHTwo 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=2.33Step 3: After 10.0 mL of base were added, the first equivalence

point is reached (=isoelectric point)

pH=6.04

Titration II

2759332

2pKpKpH 2a1a ..

Step 4: After 15.0 mL of base have been added, [HL]=[L-] pH=pKa2=

9.75Step 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 HL + OH-

Titration III

L- HL OH-

Initial 5.0*10-4 moles (=0.010 L *0.050 M)

~0 ~0

Change -x +x +xEquilibrium

L0300x1005 4

.)*.(

yL0300

x

.y

L0300x

.

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 0.0167 M) pOH = 3.03 pH=10.97

Titration IV

510

1410*59.5

10*79.110*0.1

a

wb K

KK

yLyKb

][

2

Step 6: After 25.0 mL of base have been added, all H2L+ has been converted to L-. This required 20.0 mL of baseto accomplish.

There is an excess of 5.0 mL of base in the solutionFind number of moles of base

n = 0.0050 L * 0.050 M = 2.5*10-4 moles Find concentration of base

c = 2.5*10-4 moles/0.0350 L = 7.14*10-3 MFind pOH and pH

pOH = -log([OH-]) = 2.15pH = 14 – pOH = 11.85

Titration V

The six points of interest in the titration of 0.050 M leucine with 0.050 M NaOH are

Summary for Leucine

Point Base added Equivalence pH-value CommentsP1 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

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

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 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 torinse the burette into the labeled waste container. Do not pour un-neutralized NaOHsolutions 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

Individual Work

Use Excel for plotting titration curves and first-derivative graphs. The pKa’s 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’s 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’s.

The report is due on April 15, 2014 or April 16, 2014 at the beginning of the lab section.

Report