Transcript Lecture 1

 This week in lab:
 Tuesday/Wednesday: Check-in and Pipette calibration
 Thursday/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 acids
 Hint: You may think of the unknown amino acid containing
both HA+/- and H2A+ forms of the amino acid
 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
strong
1.92
Sulfuric acid
pKa3
12.35
 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.
H H H
N
S
O
N
O
CH3
CH3
COO-K+
 Diprotic acids undergo the following equilibria:
 H2L+
 HL
HL + H+
L- + H+
Ka1
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.
 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
 Leucine
pKa1=2.33
pKa2=9.75
 Since Ka1 >>> Ka2, only the first equilibrium has to be
considered at low pH-values
 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).
 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 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
 Example: Titration of 10 mL of 0.050 M H2L+ with
0.050 M NaOH
 Two reactions have to be considered
 H2L+ + OHHL + H2O
(1)
 HL
+ OHL- + 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)
pK a1  pK a 2 2.33  9.75
pH 

2
2
 pH=6.04
 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
HL + OH-
LInitial
Change
Equilibrium
5.0*10-4 moles
(=0.010 L *0.050 M)
-x
(5.0 * 10 4  x )
0.030L
HL
OH-
~0
~0
+x
+x
x
y
0.030L
x
y
0.030L
 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
 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 = 2.5*10-4 moles
 Find concentration of base
 c = 2.5*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
 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
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 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
 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.