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Chemistry 130
Acid and Base equilibria
Dr. John F. C. Turner
409 Buehler Hall
[email protected]
Chemistry 130
Acids and bases
The Brønsted-Lowry definition of an acid states that any material that
produces the hydronium ion is an acid. A Brønsted-Lowry acid is a
proton donor:
HA
H2 O l
H3 O+aq
A-aq
H 3 O+aq or H+aq
The hydronium ion,
has the same structure as ammonia. It
is pyramidal and has one lone pair on O.
Chemistry 130
Acids and bases
The Brønsted-Lowry definition of a base states that any material that
accepts a proton is a base. A Brønsted-Lowry base is a proton acceptor:
B-aq
H2 O l
HO-aq
BH+aq
In aqueous solution, a base forms the hydroxide ion:
Chemistry 130
Conjugate acids and bases
For any acid-base reaction, the original acid and base are complemented
by the conjugate acid and conjugate base:
NH3 aq
H2O l
HO-aq
NH+4 aq
On the RHS,
On the LHS,
Water is the proton donor
Hydroxide ion is the proton acceptor
Ammonium ion is the proton donor
Ammonia is the proton acceptor
Water and hydroxide ion are conjugate acid and base
Ammonia and ammonium ion are also conjugate acid and base
Chemistry 130
Conjugate acids and bases
The anion of every acid is the conjugate base of that acid
The cation of every base is the conjugate acid of that base
Acid-base conjugates exist due to the dynamic equilibrium that exists in
solution.
Chemistry 130
NH3 aq
H2O l
HO-aq
NH+4 aq
NH3 aq
H2O l
HO-aq
NH+4 aq
Acid and base constants
We use equilibrium constants to define the position of equilibrium and to
reflect the dynamic nature of the system.
For an acid, we define
HCl aq
KA =
H2 O l
+
-
H 3 O aq
[Products]
=
[Reactants]
Cl aq
+
-
[H 3 O aq ][Cl aq ]
[HCl aq ]
Water is not included as the change in concentration of water is negligible
(water is ~55 M when pure) for a dilute solution.
Chemistry 130
Acid and base constants
We use equilibrium constants to define the position of equilibrium and to
reflect the dynamic nature of the system.
For a base, we define
NaOH aq
KB =
H2 O l
[Products]
=
[Reactants]
+
Na aq
-
OH aq
[Na+aq ][OH -aq ]
[NaOH aq ]
Again, water is not included as the change in concentration of water is
negligible (water is ~55 M when pure) for a dilute solution.
Chemistry 130
Acid and base constants
We also use the logarithm of the acid or base constant as an indicator of
acid or base strength:
+
KA =
-
[H 3 O aq ][Cl aq ]
[HCl aq ]
pK A = − lg K A
+
KB =
-
[Na aq ][OH aq ]
[NaOH aq ]
pK B = − lg K B
Note that the logarithm used here is to base 10, not the natural log
lg = log10
ln = loge
Chemistry 130
Conjugate acid and base strengths
The dynamic nature of the equilibrium between the conjugate acid-base pair
means that a strong acid will have a weak conjugate base
A strong base will have a weak conjugate acid.
In both these cases, 'strong' and 'weak' are defined qualitatively the position
of the acid or base equilibrium.
A strong acid forms almost exclusively the hydronium ion and the
concentration of the undissociated acid is negligible; the conjugate base
must be weak.
Chemistry 130
Conjugate acid and base strengths
There are a variety of strong acids – any material that
has a larger pKA than the hydronium ion will form the
hydronium ion in solution.
Any material with an acid constant smaller than the
hydronium ion will establish a measurable
equilibrium between the acid and the dissociated
hydronium ion and associated conjugate base – the
anion.
Water therefore acts as a leveling solvent and
restricts the degree of acidity possible in aqueous
solution.
Chemistry 130
Acid
pK A
HI
HBr
HCl
H 2 SO4
−9
−8
−6
−3
+
H 3 O aq
− 1.7
HNO3
− 1.3
Autoprotolysis of water
Water can act as both an acid and a base – it can form the hydronium ion
as well as the hydroxide ion in solution.
Pure water also undergoes a 'self-equilibrium':
H2 O l
KW =
H2 O l
+
H 3 O aq
[ Products]
=
[Reactants]
+
-
-
OH aq
[H 3 O+aq ][OH -aq ]
2
[H 2 O l ]
− 14
K W = [H 3 O aq ][OH aq ] = 1× 10
pK W = 14
This is autoprotolysis or autionization.
Chemistry 130
pKW , pH and pOH
The autoprotolysis of water relates the hydronium ion concentration and the
hydroxide ion concentration to the autoprotolysis constant of water:
H2 O l
+
H2 O l
-
H 3 O aq
OH aq
K W = [H 3 O+aq ][OH -aq ] = 1× 10− 14
+
-
+
-
pK w = − lg10 [H 3 O aq ][OH aq ] = − lg 10 [H 3 O aq ] − lg 10 [OH aq ] = 14
+
-
pK w = − lg10 [H 3 O aq ] − lg10 [OH aq ] = pH
pK w = pH
pOH = 14
pOH = 14
This relationship dictates the concentrations of hydroxide and hydronium
ion in all aqueous solutions
Chemistry 130
Simple pH and pOH calculations
Q:
What is the pH of a solution of 0.0002 M solution of HI?
−1
Molarity of solution = 0.0002 mol L
The ionization equation of HI in water is
HI aq
H2 O l
+
H 3 O aq
The K A of HI is very large and we know that HI
+
Therefore, [ H 3 O aq ] = 0.0002 mol L
lg10 [ H 3 O+aq ] = − 3.699
pH = − lg10 [H 3 O+aq ] = 3.699
Chemistry 130
−1
-
I aq
aq
is a strong acid
Simple pH and pOH calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.0002 M solution of HI?
We have already calculated the pH of the solution
+
pH = − lg10 [ H 3 O aq ] = 3.699
The relationship between pH, pOH and pK W is
pK w = pH
pOH = 14
3.699
pOH = 14
pOH = 14 − 3.699 = 10.301
-
pOH = − lg 10 [OH aq ] = 10.301
[OH- aq ] = 10− 10.301 = 5 × 10− 11 mol L− 1
Chemistry 130
Simple pH and pOH calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.00075 M solution of KOH?
pOH calculation
−1
[OH aq ] = 0.00075 mol L
-
lg [OH aq ] = − 3.125
pOH = − lg 10 [OH -aq ] = 3.125
As
pK w =
pH
pH =
As
pH =
pH
pOH = 14
3.125 = 14
14 − 3.125 = 10.875
+
− lg 10 [ H 3 O aq ]
+
lg 10 [ H 3 O aq ] = − 10.875
[ H 3 O+aq ] = 10− 10.875 = 1.33 × 10− 11 mol L− 1
Chemistry 130
Weak acids and bases
These calculations are straightforward as HI is a strong acid and KOH is a
strong base.
For weak acids and bases, which do not fully ionize in solution, we have to
use the rules for equilibria with which we are already familiar.
We also use
pK A = − lg K A
pK B = − lg K B
which we have defined for the generic acid reaction and base reaction in
water.
Chemistry 130
Weak acid/base calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.5 M solution of acetic acid?
CH 3 CO2 H aq
Acetic acid
H2 O l
Initial concentrations
Change
Equilibrium concentrations
+
H 3 O aq
−5
-
CH 3 CO2 aq
Acetate ion
K A = 1.8 × 10
+
-
CH 3 CO2 H aq
H 3 O aq
CH 3 CO2 aq
0.5
−x
0.5− x
≈0
x
x
0
x
x
At equilibrium, the new concentrations are:
[CH 3 CO2 H aq ] = 0.5− x
-
[CH 3 CO2
+
aq
] = x
[H 3 O aq ] = x
Chemistry 130
Weak acid/base calculations
−5
The acid constant for acetic acid isK A = 1.8 × 10
whereas the autoprotolysis constant for water isK W = 1 × 10− 14
The contribution of the self-ionization of water is of the order of 10-7 to the
concentration of hydronium ion.
In general, if the acid constant for the weak acid is sufficiently large in
comparison to the autoprotolysis constant for water, we can ignore the
autoprotolysis of water with respect to the hydrogen ion concentration
Chemistry 130
Weak acid/base calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.5 M solution of acetic acid?
CH 3 CO2 H aq
Acetic acid
H2 O l
+
H 3 O aq
-
CH 3 CO2 aq
Acetate ion
We now set up the equilibrium constant for aqueous acetic acid:
+
[ CH 3 CO2 aq ][ H 3 O aq ]
KA =
[CH 3 CO2 H aq ]
and use the concentrations that we have already calculated:
[CH 3 CO2 H aq ] = 0.5− x
+
[CH
CO
][
H
O
3
2 aq
3
aq ]
[CH 3 CO-2 aq ] = x
KA =
=
[CH 3 CO2 H aq ]
+
[H 3 O aq ] = x
Chemistry 130
−5
K A = 1.8 × 10
x x
0.5− x
Weak acid/base calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.5 M solution of acetic acid?
CH 3 CO2 H aq
Acetic acid
+
H2 O l
H 3 O aq
-
CH 3 CO2 aq
Acetate ion
−5
K A = 1.8 × 10
We now solve the equilibrium equation that we have set up:
KA =
x x
−5
= 1.8 × 10
0.5− x
2
x
0.5− x
−5
= 1.8 × 10
There are two ways of solving this equation
1. use the quadratic formula
Chemistry 130
2.
make the assumption
that
x≪
0.5
Weak acid/base calculations
Q:
Calculate pOH, the pH and the concentration of hydroxide ion in a
solution of 0.5 M solution of acetic acid?
CH3 CO2 H aq
Acetic acid
H2 O l
+
H 3 O aq
-
CH 3 CO2 aq
Acetate ion
−5
K A = 1.8 × 10
Using assumption 2, that x≪ 0.5
x2
x2
−5
KA =
≈
= 1.8 × 10
0.5− x
0.5
x 2 ≈ 0.5 × 1.8 × 10− 5 ≈ 9 × 10− 6
1
−6 2
x ≈ 9 × 10
≈ 0.003
+
−1
[H 3 O aq ] = 0.003 mol L
pOH = 14− pH = 11.48
Chemistry 130
+
pH = − lg[H 3 O aq ] = 2.52
[OH -aq ] = 3.33 × 10− 12 mol L− 1
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
NH 2 OH aq
H2O l
Hydroxylamine
-
OH aq
Initial concentrations
Change
Equilibrium concentrations
−9
+
NH 3 OH aq
KB = 9.1 × 10
Hydroxylammonium
-
+
NH 2 OH aq
OH aq
NH 3 OH aq
1.5
−x
1.5− x
≈0
x
x
0
x
x
At equilibrium, the new concentrations are:
[NH 2 OH aq ] = 1.5− x
+
[NH 3 OH aq ] = x
-
[OH aq ] = x
Chemistry 130
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
NH 2 OH aq
H2O l
Hydroxylamine
-
OH aq
+
We now set up the base equilibrium constant for aqueous hydroxylamine:
+
[ NH3 OH aq ][ OH aq ]
KA =
[ NH 2 OH aq ]
and use the concentrations that we have already calculated:
[ NH 2 OH aq ] = 1.5− x
+
2
[
NH
OH
][
OH
+
x
3
aq
aq ]
[ NH 3 OH aq ] = x
KB =
=
[ NH 2 OH aq ]
1.5− x
[OH aq ] = x
Chemistry 130
−9
NH 3 OH aq
K B = 9.1 × 10
Hydroxylammonium
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
NH 2 OH aq
H2O l
Hydroxylamine
-
OH aq
+
The equation that we need to solve is:
+
2
[NH 3 OH aq ][OH aq ]
x
−9
KB =
=
= 9.1 × 10
[ NH 2 OH aq ]
1.5− x
Using the quadratic method,
x2
−9
= 9.1 × 10
1.5− x
2
−9
−8
−9
so x = 1.5− x ⋅ 9.1 × 10 = 1.365 × 10 − 9.1 × 10 x
2
−9
−8
x
9.1 × 10 x − 1.365 × 10 = 0
Chemistry 130
−9
NH 3 OH aq
KB = 9.1 × 10
Hydroxylammonium
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
NH 2 OH aq
H2 O l
Hydroxylamine
-
−9
+
OH aq
NH 3 OH aq
K B = 9.1 × 10
Hydroxylammonium
2
−
b±
b
− 4ac
2
For a quadratic of the form ax
bx
c = 0, x =
2a
−9
−8
In this case, a = 1 b = 9.1 × 10 c = − 1.365 × 10 and so
−9
x =
−9 2
− 9.1 × 10 ± 9.1 × 10
−8
− 4⋅ 1⋅ − 1.365 × 10
2
−9
1
−8 2
− 9.1 × 10 ± 5.460000008 × 10
x =
2
Chemistry 130
1
2
−4
= 1.168 × 10
−1
mol L
1
2
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
NH 2 OH aq
H2 O l
Hydroxylamine
-
OH aq
+
The equation that we need to solve is:
+
2
[ NH3 OH aq ][ OH aq ]
x
−9
KB =
=
= 9.1 × 10
[ NH 2 OH aq ]
1.5− x
Using the assumption that x≪ 1.5
2
2
x
x
≈
= 9.1 × 10− 9
1.5− x
1.5
2
−9
−8
so x = 1.5 ⋅ 9.1 × 10 = 1.365 × 10
x ≈
1
−8 2
1.365 × 10
Chemistry 130
−9
NH 3 OH aq
K B = 9.1 × 10
Hydroxylammonium
≈ 1.168 × 10− 4 mol L− 1
Weak acid/base calculations
Q:
Calculate pOH, pH and the concentrations of both hydroxide ion and
hydronium in a solution of 1.5 M solution of hydroxylamine?
-
NH 2 OH aq
H2 O l
Hydroxylamine
OH aq
−9
+
NH 3 OH aq
K B = 9.1 × 10
Hydroxylammonium
In this case there is no difference between the methods to the third place of
−4
−1
decimal and so x = [OH aq ] = 1.168 × 10 mol L
-
−4
pOH = − lg10 [OH aq ] = − lg10 1.168 × 10
As pOH
pH = 14
pH = 14 − pOH = 14 − 3.93 = 10.07
+
and so as pH = − lg10 [H 3 O aq ]
[H 3 O+aq ] = 10− 10.07 = 8.56 × 10− 11 mol L− 1
Chemistry 130
= 3.93
Weak acid/base calculations
Given that pH and pOH are related via the relationship
+
-
− 14
K W = [H3 O aq ][OH aq ] = 1× 10
pK w = pH
pOH = 14
then in aqueous solution, only one of these quantities is required for all to
be calculable.
Similarly, if KA is known for an acid or KB is known for a base, then using the
change on the establishment of equilibrium will give the hydronium ion
concentration or the hydroxide concentration and so all is known.
If we know the pH or pOH, we can also calculate KA or KB for the system,
using similar methods.
Chemistry 130
Weak acid/base calculations
Q:
Calculate pKB for dimethylamine given that a 0.164 M solution has a
pH of 11.98.
Me2 NH aq
H2 O l
Dimethylamine
-
+
OH aq
Me2 NH 2 aq
Dimethylammonium
−1
When [Me2 NH aq ] = 0.164 mol L , pH = 11.98
pH
pOH = 14
so
11.98
pOH = 14 so pOH = 14 − 11.98 = 2.02
Given that pOH = − lg10 [OH-aq ]
-
− 2.02
[OH aq ] = 10
Chemistry 130
−3
= 9.55 × 10
−1
mol L
Weak acid/base calculations
Q:
Calculate pKB for dimethylamine given that a 0.164 M solution has a
pH of 11.98.
Me2 NH aq
H2 O l
Dimethylamine
-
+
OH aq
Me2 NH 2 aq
Dimethylammonium
−3
-
From the pH calculation, [OH aq ] = 9.55 × 10
Me2 NH aq
−1
mol L
-
OH aq
+
Me2 NH 2 aq
Initial concentrations
0.164
≈0
0
−3
−3
−3
Change
− 9.55 × 10
9.55 × 10
9.55 × 10
−3
−3
−3
Equilibrium concentrations 0.164− 9.55 × 10
9.55 × 10
9.55 × 10
Chemistry 130
Weak acid/base calculations
Q:
Calculate pKB for dimethylamine given that a 0.164 M solution has a
pH of 11.98.
-
Me2 NH aq
H2 O l
Dimethylamine
+
OH aq
Me2 NH 2 aq
Dimethylammonium
We now know all the concentrations requred to calculate K B
−3
[Me2 NH aq ] = 0.164− 9.55 × 10
−1
= 1.5445 × 10
−1
mol L
[OH- aq ] = 9.55 × 10− 3 mol L− 1
−3
+
[Me2 NH 2 aq ] = 9.55 × 10
+
KB =
-
[Me2 NH 2 aq ][OH aq ]
[ Me2 NH aq ]
pK B = − lg10 K B = 3.23
Chemistry 130
−1
mol L
−3
=
−3
9.55 × 10
9.55 × 10
1.5445 × 10− 1
= 5.9 × 10− 4
Polyprotic acids
A polyprotic acid is one that can ionize more than once. Common examples
include
Sulfuric acid
H 2 SO4
Phosphoric acid
H 3 PO4
Carbonic acid
H 2 CO3
Note that acids such as acetic acid are not polyprotic
CH 3 CO2 H aq
-
CH 3 CO2 aq
Chemistry 130
H2 O l
H2 O l
+
H 3 O aq
+
H 3 O aq
-
CH 3 CO2 aq
2-
CH 2 CO2 aq
Polyprotic acids
Each ionization of a polyprotic acid has an associated acid constant.
For phosphoric acid H 3 PO4
H 3 PO4
aq
H 3 O aq
aq
H2 O l
H 3 O aq
-
H 2 PO4
2-
HPO4
+
H2 O l
aq
H2 O l
+
+
H 3 O aq
−3
-
H 2 PO4
aq
−8
2-
HPO4
aq
K A 2 = 6.3 × 10
− 13
3-
PO4
K A 1 = 7.1 × 10
K A 3 = 4.3 × 10
aq
Note that each acid constant differs from the one before and, although the
first ionization, in this case, is strong, the others are not.
−3
H 3 PO4 aq
K A 1 = 7.1 × 10
Strong acid
H 2 PO-4
aq
− 13
2-
HPO4
Chemistry 130
K A 2 = 6.3 × 10− 8
aq
K A 3 = 4.3 × 10
Weak acid
Weak acid
Polyprotic acids
If the first ionization of a polyprotic acid is described by a large acid
constant, then equal concentrations of the hydronium ion and
dihydrogenphosphate ons are produced.
The equilibrium for the second ionization is
H 2 PO-4 aq
H3 O+aq
H2 O l
K A2 = 6.3 × 10− 8
HPO24 aq
and the equilibrium constant is
2-
K A2 =
[HPO4
+
aq ][H 3 O aq ]
-
[H 2 PO4
2-
K A 2 = [HPO4
aq ]
2-
=
-
[HPO4 aq ][H2 PO4
-
[H2 PO4
aq
]
aq
]
= 6.3 × 10− 8
−8
aq ] = 6.3 × 10
and so the acidity of the second ionization is independent of the initial
concentration of phosphoric acid
Chemistry 130
Chemistry 130
Acid and Base equilibria
Dr. John F. C. Turner
409 Buehler Hall
[email protected]
Chemistry 130
Salts of strong and weak acids
When a salt is dissolved, the equilibria for the conjugate acid and base are
established.
Dissolving the salt introduces the conjugate acid or the conjugate base into
the solution and the normal equilibria occur
Salts of strong acids and strong bases form neutral solutions
Salts of strong acids and weak bases form acidic solutions
Salts of weak acids and strong bases form basic solutions
The pH of solutions of salts of weak acids and weak bases depend on the
acid constant of the acid and the base constant of the base
Chemistry 130
Salts of strong and weak acids
The reaction that occurs when an anion associated with a weak acid is
dissolved in water changes the pH of the solution.
This happens because the anion is the conjugate base of the associated
acid and the acid-base equilibrium for that acid is established.
For nitrous acid, the associated anion is nitrite,NO-2
-
NO-2 aq
H2 O l
OH-aq
HNO2 aq
+
KB =
[OH aq ][ HNO2 aq ]
-
NO2 aq
-
[ H 3 O aq ]
+
[OH aq ][ HNO2 aq ] [ H 3 O aq ]
We can multiply K B by
to give K B =
⋅
+
+
[ H 3 O aq ]
NO2 aq
[ H 3 O aq ]
-
+
[OH aq ][ HNO2 aq ] [ H 3 O aq ]
KB =
⋅
NO2 aq
[ H 3 O+aq ]
−4
For nitrous acid, K A,HNO = 7.2 × 10
2
Chemistry 130
Salts of strong and weak acids
For nitrous acid, the associated anion is nitrite,
-
NO-2 aq
OH- aq
H2 O l
HNO2
KB =
aq
+
[OH aq ][ HNO2 aq ]
-
NO2 aq
-
[ H 3 O aq ]
+
[OH aq ][ HNO2 aq ] [H 3 O aq ]
We can multiply K B by
to give K B =
⋅
+
+
[ H 3 O aq ]
NO2 aq
[H 3 O aq ]
[OH -aq ][ HNO2 aq ] [ H 3 O+aq ]
KW
KB =
⋅
=
+
K A,HNO
NO2 aq
[ H 3 O aq ]
2
−4
For nitrous acid, K A,HNO = 7.2 × 10
2
KB =
1 × 10− 14
−4
7.2 × 10
Chemistry 130
− 11
= 1.39 × 10
and so
Salts of strong and weak acids
For nitrous acid, the associated anion is nitrite,
-
NO-2 aq
H2 O l
OH-aq
HNO2 aq
KB =
+
[OH aq ][ HNO2 aq ]
-
NO2 aq
-
[ H 3 O aq ]
+
[OH aq ][ HNO2 aq ] [ H 3 O aq ]
We can multiply K B by
to give K B =
⋅
+
+
[ H 3 O aq ]
NO2 aq
[ H 3 O aq ]
[OH- aq ][ HNO2 aq ] [ H 3 O+aq ]
KW
KB =
⋅
=
+
K A,HNO
NO2 aq
[ H 3 O aq ]
2
−4
For nitrous acid, K A,HNO = 7.2 × 10
2
− 14
1 × 10
− 11
KB =
=
1.39
×
10
−4
7.2 × 10
Chemistry 130
and so
Strength of conjugate acids and bases
The acid and base strength of a conjugate acid-base pair, such as
nitrite-nitrous acid
NO-2 − HNO2
acetic acid-acetate
CH3 CO2 H − CH3 CO-2
ammonia-ammonium
NH 3 − NH+4
are related by the relationship
K A KB = K W
pK A
Chemistry 130
pK B = pK W = 14
Strength of conjugate acids and bases
This situation occurs because the conjugate base of a weak acid and the
conjugate acid of a weak base are both appreciably strong.
The acid or base strength of a conjugate acid or base of a strong base or
acid is extremely weak and is negligible in most applications.
The appreciable strength of a conjugate acid or base and the presence of
an equilibrium, because the base or acid is weak means that additions of
acid or base to a solution that contains the acid-base conjugate pair will not
effect the pH of the solution greatly.
These solutions are termed 'buffers'.
Chemistry 130
Buffer solutions
A buffer solution is one that contains a conjugate acid-base pair and is used
to provide a relatively constant pH in chemical reactions, biological and
medical systems and in industrial settings.
Because of the presence of the conjugate acid-base, we can write
HA aq
Weak acid
-
KA =
A- aq
H2 O l
Weak acid
+
[ A aq ][ H 3 O aq ]
[ HA aq ]
H 3 O+aq
and so
KA
[HA aq ]
-
[ A aq ]
= [H 3 O+aq ]
from which we can calculate the pH if we know the concentrations of the
acid and conjugate base and the acid constant for the acid.
Chemistry 130
Buffer solutions
Given that
-
A aq
H2 O l
HA aq
Weak acid
Weak acid
+
aq
[ H3 O
] = KA
[HA aq ]
-
[ A aq ]
+
aq
pH = − lg10 [ H 3 O
pH = − lg10 K A
Chemistry 130
+
H 3 O aq
] = − lg10 K A
lg10
[ A -aq ]
[ HA aq ]
[ HA aq ]
-
[ A aq ]
= − lg10 K A − lg 10
[ HA aq ]
-
[ A aq ]
Buffer solutions
For a buffer, the pH is given by
-
pH = p K A
lg10
[A aq ]
[HA aq ]
which is the Henderson-Hasselbalch equation.
Chemistry 130
Chemistry 130