Biochemistry I
Lecture 3
January 18, 2013
Solvation – Continued:
ii) Hydrophilic (polar) compounds
(e.g. methanol):
H3C
O
H
iii) Hydrophobic (apolar) compounds
(e.g. methane).
http://chem.ps.uci.edu/~kcjanda/Group/Re
search_hydrates.html
O
iv)Amphipathic (or amphiphilic)
compounds are both polar (usually
charged) and have a substantial nonpolar
section (e.g. fatty acids). These can form
micelles if the nonpolar part is sufficiently
large. Micelles are aggregates of
amphipathic molecules that sequester the
nonpolar part on the inside, much like the
inside of an orange.
O
Lecture 3: Acid-Base Equilibria
OLI Quiz: Q1
Readings: Horton: 2.7-2.9. Nelson & Cox: Chapter 2.2.
Key Terms:
 Equilibrium constant
 Acid dissociation constant
 pH = pKa + log[A-]/[HA]
 Calculation of degree of protonation.
 Acid strength & chemical structure.
Why pH is important in Biochemistry.
i) Molecular interactions can be sensitive to pH.
Changing the pH can change the charge on molecules.
ii) Biological activity can be sensitive to pH.
e.g. ionized groups may be required for function.
1
Biochemistry I
Lecture 3
January 18, 2013
3A. Acids and Bases:
pH: pH is measured as the -log[H+], smaller pH, more acidic the
solution. Neutral pH is 7.0. At this pH there are an equal
number of H+ and OH- ions in solution. [H+]=10-7 M.
 Acid: can donate protons to water, forming its conjugate
base and an hydronium ion.
 Base: can accept protons
The following describes ionization or dissociation of the
proton from the acid.
HA  H 2O  A  H 3O 
acid + water conjugate + hydronium
base
Monoprotic acid: releases one proton (Acetic acid)
Diprotic acid: releases two protons (Malonic acid)
O
Triprotic acid: releases three protons (Phosphoric acid)
3B. Characterization of Acid Strength Using pKa.
When a proton dissociates from an acid, it becomes bound to a
water molecule, generating a hydronium ion, in addition to the
deprotonated acid:
pKa
OH
O
OH
HO
Since the concentration of water is essentially constant, it can
be ignored and we can write a modified equilibrium reaction
that just focuses on the species of interest:
pKa1
OH
OH
pKa1
HA → A- + H+
and write the equilibrium constant for the dissociation:
This is given a special name, the 'k-a', or 'k-acidity'. The acidity
constant, Ka is a fundamental property of the acid, it does not depend on
the pH of the solution. However, it does depend on the chemical
structure and environment of the acidic group.
3C. pKa: Since the pH scale is used to characterize [H+], it is useful to
express the acidity constant in the same way, by taking its negative log,
giving the “p-K-a”:
pKa = - log Ka
Note: When the pH = pKa, 50% of the acid is protonated.
3D. Acid Strength and pKa
The stronger the acid the more the reaction:
HA → A- + H+
is to the right.
How does the KA & pKa vary as the acid strength increases? 𝐾𝑎 =
2
O
OH
[𝐴− ][𝐻 +]
[𝐻𝐴]
O
O
pKa2
O
HO
O
HO
P
O
P
O
O
O
HO
HA +H2O → A- + H3O+
O
O
O
HO
O
O
P
pKa2
O
O
O
O
pKa3
P
O
O
Biochemistry I
Lecture 3
January 18, 2013
3E. Prediction of Protonation State at any pH:
In many cases only one of the two species
(protonated or deprotonated) may be biologically
active.
Given the pKa of the ionizable group, and the pH of
the solution, we would like to calculate the
following:
 The fraction that is protonated: fHA.
 The fraction that is deprotonated: fABeginning with the equilibrium constant for ionization:
[ H  ][ A  ]
[ HA]
Take –log of both sides:
f HA 
Ka 



[ H ][ A ] 

 log K a   log

[
HA
]





[ A ] 

 log K a   log[H  ]  log

[ HA] 





[ A ] 

pK a  pH  log

[
HA
]




Defining
[ HA]
[ HA]  [ A ]
1

[ A ]
1
[ HA]
R
f HA

[A ]
[ HA]
1

(1  R )
f A 
[ A ]
[ HA]  [ A ]
[ A ]

f A 
[ HA]
1 [A ]
[ HA]

R
(1  R )
R=[A-]/[HA]
pH  pK a  log([ A ] /[ HA])
pH  pK a  log( R)
Example: Calculate the fraction protonated of the side chain of
Histidine, an amino acid found in proteins. pKa = 6.0.
pH  pK a  log( R)
10( pH  pK a )  R
Fraction Protonated
H
N
+
N
H
N
+
+
N
H
H
pH
R=10(pH-pKa)
FHA=1/(1+R)
4
R = 10 (4 - 6) = 10-2
FHA = 1/(1 + 0.01) = 0.99
5
R = 10 (5 - 6) = 10-1
FHA = 1/(1 + 0.10) = 0.91
6
R = 10 (6 - 6) = 100
FHA = 1/(1 + 1) = 0.5
7
R = 10 (7 - 6) = 10+1
FHA = 1/(1 + 10) = 0.091
8
R = 10 (8 - 6) = 10+2
FHA = 1/(1 + 100) = 0.01
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
pH
3
Biochemistry I
Lecture 3
January 18, 2013
3F. Chemical Structure and Acidity:
Key Concepts:
 The strength of an acid depends on the ability to break the A-H bond, e.g. NH versus OH.
 For a given type of acid, the strength depends on the relative stability of HA and A-, which can be
affected by the chemical groups nearby and the environment.
H
N
+
H
N
H
Amine, pKa ~10.7
H
H
Easier to break an N-H bond versus an O-H
bond, therefore an amine is a stronger acid
than an alcohol.
Ethanol
H
O
pKa ~14
O
(Ser,Thr sidechain)
H
H
H
O
H
H
H
pKa ~ 4.0
O
(Glu,Asp sidechain)
F
O
F
H
O
F
F
Acetic Acid
O
F
H
O
H +
H N
H
O H
F
pKa ~0
O
H +
H N
H
O
Trifluoroacetic acid
O
Negative charge delocalized over C=O, lower
in energy, therefore a carboxylate is a
stronger acid than an alcohol.
Electronegative fluorine atoms withdraw
negative charge from carboxyl group, a
stronger acid than acetic.
O
O
Glycine (pKa =2.0)
(an amino acid)
Practice:
1. Assume that the histidine containing protein also contains a group that ionizes with a pKa of 4, draw the
curve of fraction protonated versus pH for that group on the same graph. The curve for the histidine is
already plotted for you.
Fraction Protonated
2. Which region of the plot corresponds to both groups being fully deprotonated.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
pH
4