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Unit 6
How do we control chemical change?
Chemistry XXI
The central goal of this unit is to help you identify
the structural and environmental factors that can
be used to control chemical reactions.
M1. Characterizing Interactions
Recognizing interactions
between reacting molecules.
M2. Changing the Environment
Exploring the. influence of
external factors.
M3. Analyzing the Products
Analyzing the effect of charge
stability.
M4. Selecting the Reactants
Evaluating the impact of
electronic and steric effects.
Unit 6
How do we control
chemical change?
Chemistry XXI
Module 2: Changing the Environment
Central goal:
To analyze the effect
concentration,
temperature, and nature
of the solvent on
reaction extent.
The Challenge
Transformation
How do I change it?
Chemistry XXI
Drugs must travel through
different part of our body before
reaching their final target.
How can we predict the
effect of different
environmental conditions
on their structure and
properties?
How can we take advantage of this knowledge to
control their behavior?
Reaction Control
Chemistry XXI
The extent (Thermodynamics) and rate (Kinetics) to
which a substance, like an drug, reacts with
another, like water, depends on the environmental
conditions.
We can affect and control
chemical reactions by
changing the concentration of
reactants and products, the
temperature and pressure of
the surroundings, or the
nature of the solvent in which
the process takes place.
Fast Processes
Proton (H+) transfer in water is a fast process.
Thus, we are more interested in controlling the
thermodynamics than the kinetics of acid-base
reactions in liquid water.
Chemistry XXI
CONCENTRATION
Thus, we will focus
our attention on how
to control
reaction extent
in this case.
TEMPERATURE
SOLVENT
Concentration Effects
Let us consider an acidic drug HA that undergoes
this type of reaction when dissolved in water:
Chemistry XXI
HA(aq) + H2O(l)
A (aq) + H3O+(aq)
For example,
phenobarbital, the most
widely used
anticonvulsant worldwide.
[ H 3O  ][ A ]
Ka 
[ HA]
pKa = 7.4
What factors will determine the values of [H3O+],
[A-], and [HA] at equilibrium?
Concentration Effects
HA(aq) + H2O(l)
A (aq) + H3O+(aq)
[ H 3O  ][ A ]
Ka 
[ HA]
Chemistry XXI
The actual concentration of each species at
equilibrium depends on the values of Ka and the
initial concentrations [HA]o, [A-]o, and [H3O+]o.
Let us now analyze the case in which the initial
values of [A-]o and [H3O+]o are negligible
compared to the value of [HA]o.
Equilibrium Values (Acids)
If the initial concentration of HA in water is Co(mol/L)
and we assume that x amount reacts with water:
Chemistry XXI
HA(aq) + H2O(l)
Initial
Co
Final
Co- x
0
x
If we assume that
x << Co but x >> 1x10-7,
and we know that:

A (aq) + H3O+(aq)

[ H 3O ][ A ]
Ka 
[ HA]
1 x 10-7
x + 1 x 10-7
[ x][ x]
Ka 
[Co ]
x  Co K a  (Co K a )
1/ 2
Higher Co  Higher x
Let’s Think
pKa = 7.4
Phenobarbital (HA) has poor
solubility in water ~ 1.0 g/L.
M(C12H12N2O3) = 232.2 g/mol
Estimate the pH of a saturated solution
of this drug.
x  Co K a  (Co K a )
Chemistry XXI
1/ 2
3
x  (4.3x10 x10
7.4 1/ 2
)
Co  1.0
g 1 mol
x
 4.3x10 3 M
L 232.2 g
x  1.3 x10 5 M
Are our assumptions (x >> 1.0 x 10-7) valid?
pH = -log (x) = 4.9
Equilibrium Values (Bases)
A similar procedure can be followed to determine
the equilibrium concentrations when a base reacts
with water:
Chemistry XXI
B(aq) + H2O(l)
BH+(aq) + OH-(aq)
Initial
Co
0
Final
Co- x
x
1 x 10-7
x + 1 x 10-7
[ x ][ x ]
Kb 
[Co ]
[ BH  ][OH  ]
Kb 
[ B]
x  Co Kb  (Co K w / K a )
1/ 2
K a Kb  K w
Let’s Think
Due to the poor solubility
of its acid form (HA),
phenobarbital is often
administered as an ionic
salt of its basic form (A-)
(100 times more soluble).
pKa = 7.4
M(NaC12H11N2O3) = 254.2 g/mol
Chemistry XXI
If we know that this reaction will take place in water:
A (aq) + H2O(l)
HA(aq) + OH-(aq)
[ HA][OH  ]
Kb 
[ A ]
Estimate the pH of a a saturated solution of
sodium phenobarbital (100. g/L).
Let’s Think
HA(aq) + OH-(aq)
A (aq) + H2O(l)
Initial
Co
Final
Co- x
0
x
M(NaC12H11N2O3) = 254.2 g/mol
Chemistry XXI
Co  100.
g 1 mol
x
 3.93x101 M
L 254.2 g
1 x 10-7
x + 1 x 10-7
[ HA][OH  ]
Kb 
[ A ]
[ x ][ x ]
Kb 
[Co ]
K a Kb  K w
x  Co Kb  (Co K w / K a )1/ 2  (3.93x101 x1014 / 107.4 )
x  [OH  ]  3.14 x104
pOH = 3.5  pH = 10.5
Higher Co  Higher x
Concentration Effects
Imagine now that you have an acid in equilibrium
in aqueous solution and you decide to add more
acid. What would you expect to happen to the
concentration of the other species?
HA(aq) + H2O(l)
Chemistry XXI
Rate f  k f [ HA][ H 2O]
Rateb k b[ A ][ H 3O  ]
Kinetic Argument
A (aq) + H3O+(aq)
Increasing [HA] increases
Ratef compared to Rateb.
More A- and H3O+ will be
produced until the rates
become equal again.
Thermodynamic Argument


Chemistry XXI
[ H 3O ][ A ]
Actual
Q
Value
[ HA]n
<
[ H 3O  ]eq [ A ]eq Equilibrium
Ka 
Value
[ HA]eq
Let’s Think
CH
CH
HC
Imagine you have a 0.125 M
aqueous solution of
aspirin, an acid drug with
pKa = 3.5, in equilibrium.
CH
C
HO
C
C
CH3
O
C
O
O
a) Estimate the pH of the solution.
Chemistry XXI
b) Predict what would happen to the pH when:
 you add more A-;
 you add more HA;
 you add OH you add more H2O;
Use both, kinetic and thermodynamic arguments.
Let’s Think
Use the simulation at
http://www.chem./arizona.edu/chemt/C21/sim
Chemistry XXI
Acid
to verify your estimates and predictions.
Chemistry XXI
Let’s Think
Concentration Effects
Chemistry XXI
Understanding how the concentration of one
species affects the concentrations of the others is
crucial to predict and control the form that a drug
will take in different parts of our body.
Drugs go through various
parts of our body that
have relatively fixed but
different values of pH.
How do the drugs
change? Where are they
more likely to be
absorbed?
Equilibrium Ratios
The ratio of the conjugate forms of and acid-base
pair is determined by the equilibrium constant:


[ H 3O ][ A ]
Ka 
[ HA]
Ka
[ A ]

[ HA] [ H 3O  ]
Chemistry XXI
By taking logarithms, this relationship can be
transformed into:
[ A ]
log
 log K a  log[ H 3O  ]  pH  pKa
[ HA]
Henderson-Hasselbalch
Equation
[ A ]
pH  pKa  log
[ HA]
Henderson-Hasselbalch
How much of a an acid is
in A- or HA form depends
on the pH of the medium
where we put it.
Chemistry XXI
For example,
[ A ]
log
 pH  pKa
[ HA]
[HA] = [A-]
when pH = pKa
When analyzing drugs, it is useful to calculate the
percentage of the drug that exist in acid or basic form
in different parts of the body:

[
A
]

%A  
x100 
[ A ]  [ HA]
100
100

[ HA] 1  10 pKa  pH
1 
[A ]
% Ionization = % AConsider a drug with a pKa = 4.0.
100
%A 
1  10 pKa  pH
% HA  100  % A

120
Chemistry XXI
[A-] < [HA]
if pH < pKa
% Species
100
80
[HA]
[A-]
60
if pH > pKa
40
[A-] > [HA]
20
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
pH
Let’s Think
100
%A 
1  10 pKa  pH

Determine the dominant form of each of these drugs
in different parts of the body.
Chemistry XXI
Drug
Stomach
(pH = 2.0)
Duodenum
(pH = 6.0)
Jejunum
(pH = 7.5)
Aspirin
(Acid, pKa = 3.5)
3.1% A-
99.7% A-
99.99% A-
Phenobarbital
(Acid, pKa = 7.4)
4x10-4% A-
3.8% A-
55.7% A-
Ephedrine
(Base, pKa = 9.6)
3x10-6% A-
3x10-1% A-
0.8% A-
What % of the drug is ionized at each point?
Where is the drug more likely to be absorbed?
Temperature Effects
The extent of a chemical reaction can also be
controlled by changing the temperature of the system.
Chemistry XXI
As we have seen before,
the higher T the greater
the fraction of molecules
with enough energy to
react.
k  Ae
Ea

RT
T
k
Reaction rate
increases with T.
Temperature Effects
The effect of temperature on reaction rate depends
on the activation energy Ea of the reaction.
k  Ae
Ea

RT
The effect is more pronounced
the higher the value of Ea.
Chemistry XXI
Ep
Eaf
Eab
R
P
Reaction Coordinate
Thus, for a system in
equilibrium, the forward
and backward rates are
not affected in the same
proportion and there is a
shift in the equilibrium.
Let’s Think
Ep
Eaf
Eab
R
P
Use a kinetic argument
to make a prediction
about the effect on the
equilibrium for these two
types of reactions.
Reaction Coordinate
Chemistry XXI
Ep
Eaf
Eab
P
R
Reaction Coordinate
Endothermic/Exothermic
processes
(DHo > 0/DHo < 0) shift
towards
products/reactants at
higher T.
Thermodynamic Argument
Chemistry XXI
The same conclusion can
be derived by analyzing the
effect of T on the
equilibrium constant:
K e
DGorxn = DHorxn–TDSorxn
This approach allows us to
see that it is actually the
sign of DHorxn which
determines the effect of T.
Exothermic
o
DGrxn
(
)
RT
K e
T
K
Endothermic T
K
o
o
DH rxn
DS rxn
(

)
RT
R
Let’s Think
Use the simulation at
http://www.chem./arizona.edu/chemt/C21/sim
Chemistry XXI
Acid
H 2O
molecules
not shown
in the
simulation
to determine whether the reaction is
exothermic or endothermic.
Chemistry XXI
Let’s Think
How does the temperature affect the pH of
this solution?
Solvent Effects
Chemistry XXI
The rate and extent of a reaction can also be
controlled by changing the solvent in which the
process takes place.
A given solvent can
stabilize or destabilize
the reactants or products
of a reaction, or the
transition state.
Sol1
DG
Sol2
R
Thus, they may affect both
reaction rate and extent.
P
Reaction Coordinate
Solvent Effects
In acid-base reactions, the effect of the solvent is
crucial because it is actually one of the reactants:
Chemistry XXI
HA + SH
A + SH2+
Many drugs are insoluble in water.
Thus, to measure their acid-base it
is common to use other solvents,
such as methanol (CH3OH).
The Ka of most carboxylic drugs in
CH3OH decreases by a factor of 105
compared to that in water.
How do you explain it?
Let′s think!
Formation
of ions is
less likely in
less polar
solvents.
Reaction Control
Chemistry XXI
Our analysis reveals the central role that
environmental factors play in the extent and rate of
chemical processes:
 We can control the extent and rate of chemical
reactions by altering the concentration of reactants
and products, modifying the temperature and
pressure of the system, or changing the solvent in
which the reaction takes place.
 The effect of these factors is better understood by
considering both kinetic and thermodynamic
arguments.
Chemistry XXI
Let′s apply!
Assess what you know
Amino Acids and Proteins
As we know, proteins are natural polymers made of
amino acid chains.
Amino Acid
Chemistry XXI
Amine
Carboxyl
Peptide
bond
Predict
Let′s apply!
Chemistry XXI
The average pH inside cells is close to 7.4 (similar to
blood plasma). Calculate the % ionization of these
amino acid residues and predict which of them will
mostly be in their ionized forms.
Base
pKa = 10.8
Acid
pKa = 8.3
Acid
pKa = 3.9
Acid
pKa = 10.1
99.96%
0.2%
Neutral
11.2%
99.97%
Protein Folding
The presence of charged groups helps the
protein to fold due to ion-ion interactions and
ion-dipole interactions between residues.
Chemistry XXI
Protein folding can
be represented as a
chemical process:
Unfold
Fold
This process is affected by temperature.
Let′s apply!
Predict
The unfolded form of
a protein is favored a
higher temperatures.
Chemistry XXI
The unfolded species
does not have
catalytic properties.
How do you explain the effect of
temperature on folding from the kinetic
and the thermodynamic perspectives?
Chemistry XXI
Imagine someone gives you the pKa of
a drug. Work with a partner making a
list of the things you could tell that
person about the properties of the
drug outside and inside your body.
Changing the Environment
Summary
Chemistry XXI
We can control the extent of a reaction by altering
the concentration of reactants and products,
modifying the temperature, or changing the
solvent in which the reactions takes place.
Given the expression and value of the dissociation
constant for and acid or base in water (pKa, pKb), we
can evaluate things such as:
pH of solution;
Degree of dissociation as function of pH;
Effect on pH of changes in C and T.
Chemistry XXI
For next class,
Investigate what structural features of
substances can be used to predict their relative
acid strength.
How can we predict whether one substance will
be a stronger acid than another by analyzing
their molecular structure?