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Responses of Chemical Equilibria
Chapter 9 of Atkins: Section 9.5
The Response of Chemical Equilibria to pH
Proton transfer equilibria (acid and base reactions) are established on
timescales as short as nanoseconds.
The Response of Chemical Equilibria to Conditions
These types of reactions (characteristic of solutions of Brønsted acids and
bases) are highly responsive to the activity of the hydronium ion, H3O+
(the best representation we have of the hydrogen ion species in solution).
Acid-base equilibria in water
The pH of acids and bases
Acid-base titrations
The pH curve
Buffers
Indicators
Since the scale of H3O+ concentrations (activities) can range from 10-14 M
and lower to 10 M and beyond, we choose to use a convenient common
logarithm scale.
pH = −log a H O +
3
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Chemistry 451 - Fall 2003
Lecture 22 - 1
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The pH Scale - Alkaline to Acidic
Chemistry 451 - Fall 2003
Lecture 22 - 2
Acid-Base Equilibria
We measure pH using many sophisticated and non-sophisticated means.
Early on we often make approximation that:
pH = −log a H O +
3
a H O + = [H 3O + ]
3
Since many thermodynamic observables depend on pH itself (and not
hydronium ion concentration specifically) there is often no need to make
this approximation.
We need to consider three equilibrium constants to discuss the effect of
pH on equilibria involving the transfer of protons between species.
Ka, Kb, and Kw
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Chemistry 451 - Fall 2003
Lecture 22 - 3
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Chemistry 451 - Fall 2003
Lecture 22 - 4
1
Equilibrium Constants
Equilibrium Constants
2 H2O(l)
HA(aq) + H2O(l)
B(aq) + H2O(l)
2 H2O(l)
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H3O+(aq) + A–(aq)
HB+(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
Ka =
Kb =
a H O + aA −
3
a HA
a HB + aOH −
pK w = pH + pOH
Also, the equilibrium constant of the conjugate acid to B (the
species HB+) is related to that of the base by:
3
Lecture 22 - 5
K w = K aK b
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Chemistry 451 - Fall 2003
Lecture 22 - 6
The pH of Acids and Bases
For a weak acid or base, the extent of the reaction is not very advanced.
For a strong acid or base, the reaction heavily favors the products.
H3O+(aq) + A–(aq)
pOH = −log aOH −
If we also use the notation:
Strong and Weak Acids and Bases
HA(aq) + H2O(l)
Kw = 1.008 x
3
10–14
pKw = 14.00
aB
K w = a H O + aOH −
Chemistry 451 - Fall 2003
At
25oC,
K w = a H O + aOH −
H3O+(aq) + OH–(aq)
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
First, the approximation will be made that the activity of the species in
solution are equivalent to the numerical values of concentration given in
moles per liter (M).
For a weak acid, the extent of proton transfer is so small that it is usually
assumed that the molar concentration of HA is unchanged from its nominal
value. Also, the molar amounts of H3O+ and A– formed from the reaction
are equal (we amount from the autoprotolysis of water is considered
minimal), and we can find:
[H O ][A ] = [H O ]
+
Ka =
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Chemistry 451 - Fall 2003
Lecture 22 - 7
Prof. Mueller
3
[HA]
+ 2
−
3
[HA]
or
Chemistry 451 - Fall 2003
[H O ]=
+
3
K a [HA]
Lecture 22 - 8
2
The pH of Acids and Bases
HA(aq) + H2O(l)
pH =
Acid-Base Titrations
H3O+(aq) + A–(aq)
We can better understand the stages of chemical equilibrium in acid/base
reactions by following the progress of a titration of a weak acid, HA, with a
solution of a strong base, MOH.
1
1
pK a − log[HA]
2
2
Volume of acid = VA
Molar concentration of acid is initially A0
Similarly for a weak base equilibrium:
Molar concentration of base (titrant) is B.
B(aq) + H2O(l)
pH = pK w −
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HB+(aq) + OH–(aq)
Goal: calculate pH at any stage in the titration.
1
1
pK b + log[B]
2
2
Chemistry 451 - Fall 2003
This can be done “exactly” - assuming that
activities are given by concentrations - and
approximately making certain assumptions.
Lecture 22 - 9
Acid-Base Titrations
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Chemistry 451 - Fall 2003
Lecture 22 - 10
Acid-Base Titrations - Initial pH
Stoichiometric point (also called the equivalence point): enough titrant has
been added for the amount of strong base it supplies to match the amount
of weak acid in the analyte.
Upon the addition of VB of titrant, the total volume of the analyte solution
is increased to V = VA + VB.
Assumptions:
- since HA is weak, amount of HA present is much greater than amount of
A -.
- HA provides hydronium ions that far outnumber those arising from
autoprotolysis of water.
At the starting point of the titration:
pH =
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 11
Prof. Mueller
1
1
pK a − log A0
2
2
Chemistry 451 - Fall 2003
Lecture 22 - 12
3
Acid-Base Titrations - Addition of Base
The Henderson-Hasselbelch Equation
As some base is added (but well before stoichiometric point), the
concentration of A- ions comes almost entirely from the following reaction
(which formally is the formation of a salt, MA):
pH = pK a − log
pH = pK a − log
HA(aq) + OH–(aq) → A–(aq) + H2O (l)
Neglecting the small concentration of A– from the remaining weak acid, we
can set [A–] = S, where S is the salt concentration present after reaction of
HA with the titrant.
At this point, nHA remaining is A0VA minus the amount converted to salt,
or:
[HA] =A’= A0VA/V - S
Then:
a H O + aA − a H O + S
Ka = 3
= 3
a HA
A′
or:
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pH = pK a − log
A′
S
[acid]
[base]
When the molar concentration of acid and salt (acting as base) are equal:
pH = pKa
Hence the pKa of an acid can be measured directly from the pH of the
mixture. Practically, this is accomplished by recording the pH during the
titration and then examining the data for the pH halfway to the
stoichiometric point.
A′
S
Chemistry 451 - Fall 2003
Lecture 22 - 13
Acid-Base Titrations
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Chemistry 451 - Fall 2003
Lecture 22 - 14
Acid-Base Titrations - Stoichiometric Point
At the stoichiometric point, the hydronium ion concentration in the solution
comes from the autoprotolysis of water, and the OH- ions are produced by
the reaction below:
A–(aq) + H2O (l) → HA(aq) + OH–(aq)
Since only a small amount of HA is formed in this way, [A–] = S, and [OH–]
= [HA]. Hence
Kb =
a HA aOH −
aA −
[OH ]
− 2
=
S
Using Kw = KaKb, we can rearrange and substitute into the above equation
and conclude that at the stoichiometric point:
pH =
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 15
Prof. Mueller
1
1
1
pK a + pK w + log S
2
2
2
Chemistry 451 - Fall 2003
Lecture 22 - 16
4
Acid-Base Titrations - “Exact” Solution
(
)(
)(
+
+
+
VB K a A0 [H 3O ]+ K w − [H 3O ] [H 3O ]+ K a
=
+
+ 2
+
VA
[H3O ]+ K a [H 3O ] + B0 [H3O ]− K w
(
2
Acid-Base Titrations - “Exact” Solution
)
(
)(
)(
+
+
+
VB K a A0 [H 3O ]+ K w − [H 3O ] [H 3O ]+ K a
=
+
+ 2
+
VA
[H 3O ]+ K a [H 3O ] + B0 [H 3O ]− K w
)
(
2
)
)
This equation is derived using
(a) Electrical neutrality
(b) Conservation of A - containing species
(c) Conservation of M+- containing species
pKa
and the definitions for Ka and Kw.
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 17
Using Acid-Base Equilibria: Buffers
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 18
Using Acid-Base Equilibria: Indicators
Buffer action takes advantage of the slow variation of the pH in the vicinity
of A’ = S in the titration curve.
HIn(aq) + H2O (l)
In–(aq) + H3O+(aq)
Normally, HIn is some large, water-soluble, weakly acidic organic
molecule. Could be di-protic, such as phenolphthalein.
This gives the solution the ability to
oppose changes in pH when small
amounts of strong acids or bases
are added.
Example: What is the pH of an
aqueous buffer solution that
contains 0.200 mol L-1 KH2PO4
and 0.100 mol L-1 K2HPO4 ?
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 19
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 20
5
Using Acid-Base Equilibria: Indicators
HIn(aq) + H2O (l)
In–(aq) + H3O+(aq)
K in =
a H O + aIn −
3
a HIn
Then we can find the ratio of acid and base at a given pH:
log
pH < pKin
pH > pKin
pH = pKin
[HIn] = pK
[In ]
−
in
− pH
indicator in acidic form
indicator in basic form
endpoint
Choose an indicator so that the dramatic passing through the end-point
coincides with the stoichiometric point of the titration.
Prof. Mueller
Chemistry 451 - Fall 2003
Lecture 22 - 21
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