Acids and Bases in aqueous
solution.
protonation curve for glycine
2.5
O
2
pKa2 = 2.33
-
NH3+
O
1.5
nbar
glycine
1
pKa1 = 9.45
0.5
0
0
5
pH = pKa2
10
pH
pH = pKa1
15
The properties of water:
~ 2.75 Å
H-bonds
Water is held together
by H-bonding, which
gives it its surprisingly
high B.Pt. and M.Pt.
for such a small
molecule. Note that
the heavier H2S molecule
is a gas at room
temperature. Water is also
weakly ionized into H+
and OH- ions, as
discussed next.
The self-ionization of water:
Water ionizes weakly to produce OH- and H+ ions, where Kw is the
ionization constant of water, according to :

H2O(l)
H+ (aq)
+
OH-(aq)
Kw = 10-14
This means that the product of the concentration (molarity) of [H+]
and [OH-] must always be 10-14.
Kw
=
10-14
=
[H+] [OH-]
[1]
Thus, for pairs of [H+] and [OH-] concentrations in solution we have:
[H+] = 1.0 M, then [OH-] = 10-14/1.0
[H+] = 0.01 M, then [OH-] = 10-14/0.01
[H+] = 10-7 M, then [OH-] = 10-14/10-7
[H+] = 10-12 M, then [OH-] = 10-14/10-12
[H+] = 10-15 M, then [OH-] = 10-14/10-15
= 10-14 M
= 10-12 M
= 10-7 M
= 10-2 M
= 10 M
pH and pOH
The concentration of [H+] and [OH-] can be expressed as
the pH and pOH: (p in pH or pOH means ‘potency’, and
means that any constant ‘K’ or concentration with a ‘p’ in
front of it is represented as pK = -log K , or pH = -log [H+])
pH = - log [H+]
pOH = - log [OH-]
We can show that:
pKw =
14.0
= pH
+
pOH
Thus, for any solution where the pH is known,
pH + pOH = 14.0, or pOH = 14 – pH.
[H+]
pH
pOH
[OH-]
10 M
1M
0.01 M
10-5.7 M
10-7 M
10-9.3 M
10-12 M
10-14 M
10-15.09 M
-1.0
0.0
2.0
5.7
7.0
9.3
12.0
14.0
15.09
15.0
14.0
12.0
8.3
7.0
4.7
2.0
0.0
-1.09
10-15 M
10-14 M
10-12 M
10-8.3 M
10-7.0 M
10-4.7 M
10-2 M
1M
12 M
Acid dissociation constants (Ka):
A weak Brønsted acid such as acetic acid (CH3COOH)
will dissociate to give off protons in aqueous solution
according to:
acetic acid
acetate ion
CH3COOH (aq)  CH3COO- (aq) +
H+ (aq)
conjugate acid
proton
conjugate base
The extent of such dissociation is controlled by the acid
dissociation constant, Ka.
Ka
=
[ CH3COO- ] [ H+ ]
[ CH3COOH ]
The acid dissociation constant of
acetic acid:
The value of Ka for acetic acid is 10-4.84, so the pKa is 4.84.
This gives us the expression:
10-4.84 =
[ CH3COO- ] [ H+ ]
[ CH3COOH ]
[2]
This expression can be used to solve problems relating to
the ionization of acetic and other acids. For example, what
is the pH of a 0.1 M solution of acetic acid? Equation 2
must be obeyed at all times, so we have:
10-4.84 = [CH3COO-][H+]/[0.1] = [H+]2/[0.1]
so [H+] = (10-4.84 x 0.1)0.5 = 10-2.92 M, pH = 2.92.
Notice that in solving the above problem, ionization of acetic
acid produces equal concentrations of H+ and CH3COOions. One assumes that the amount of ionization is small, so
that [CH3COOH] is not corrected for the small amount that it
decreases, and we assume that [CH3COOH] is still ≈ 0.1 M.
b) What is the pH of a 1 M solution of ammonia if the pKa of
NH4+ is 9.2?
Note: pKa of an acid plus pKb of its conjugate base = pKw
pKa(NH4+) + pKb(NH3) = pKw = 14.0.
pKb NH3 = 14.0 – 9.2 = 4.8, so Kb = 10-4.8.
10-4.8 = [NH4+] [OH-] / [NH3]
so again we have: [OH-] = (10-4.8 x 1)0.5, = 10-2.4.
If [OH-] = 10-2.4 M, then pOH = 2.4, and pH = 14-2.4 = 11.6
The species distribution diagrams of
acids and bases:
We can calculate the percentage of an acid (e.g.
CH3COOH) or its conjugate base (CH3COO-) that is
present as the acid or the conjugate base at any given pH
value if the pKa is known (4.84). Thus, for acetic acid at
pH 3.7 we have:
10-4.84 =
[CH3COO-] [H+] / [CH3COOH]
=
[CH3COO-] [10-3.7] / [CH3COOH]
[CH3COO-] / [CH3COOH] = 10-4.84 / 10-3.7 = 10-1.14 = 0.072
so percent of [CH3COO-] at pH 3.7 =
100% x (0.072/(1+0.072)) = 7.2 %.
% of [CH3COOH] = 100 – 7.2 = 92.8 %.
percent as species indicated.
species distribution vs pH acetic
acid
110
100
90
80
70
60
50
40
30
20
10
0
CH3COO-
CH3COOH
pH50 = pKa
2
3
4
5
pH
6
7
8
The relationship between pH50 and the pKa
The pH at which the concentrations of CH3COOH and CH3COO- in
solution are equal, i.e. both are 50%, is known as the pH50. On the
previous slide it was indicated that the pKa of the acid equals pH50.
This arises simply as follows:
Ka
10-4.84
or pKa
=
10-4.84
=
=
=
[ H+ ]
pH50
[ CH3COO- ] [ H+ ]
[ CH3COOH ]
[CH3COO-] = [CH3COOH]
at pH50, so these cancel
Thus, we saw that for CH3COOH the crossover point for
[CH3COOH] and [CH3COO-] occurred at pH 4.84, which is the pKa.
Similarly, on the next slide, we see that the crossover point for
[NH4+] and [NH3] occurs at pH 9.22, which is the pKa of NH4+.
species
distribution
vs
pH
of
The species distribution diagram of
ammonia/ammonium ion
percent as species indicated.
ammonia/ammonium ion:
110
100
90
80
70
60
50
40
30
20
10
0
NH4+
NH3
pH50 = pKa
5
7
9
pH
11
13
Multiprotic acids and bases:
Many acids have more than one ionizable proton, and many bases
have more than one proton acceptor site. A familiar example of this
is triprotic phosphoric acid:
H3PO4
H2PO4HPO42-



H2PO4- +
HPO42- +
PO43+
H+
H+
H+
pKa3
pKa2
pKa1
=
=
=
2.15
7.20
12.38
It can be shown that for a monoprotic acid, when the concentration
of the acid and its conjugate base are equal (both 50%), then pKa =
pH. For polyprotic acids the average extent of protonation at any pH
is given by the function nbar. For the phosphate ion:
nbar = average number of protons bound per phosphate
The relationship between nbar, pH, and
the pKa values of multiprotic acids
On a previous slide we showed that for a monoprotic acid such as
CH3COOH, the pKa = pH50. At pH50 for a monoprotic acid, in fact,
nbar = 0.5. This is the point at which half of the acetate ions have a
proton on them, while half do not. For polyprotic acids, similar
considerations apply. At nbar = 0.5 for phosphate, we have [HPO42-]
= [PO43-], and the pH at nbar = 0.5 = pKa1. In fact the half-values of
nbar give us the pKa values from the corresponding pH values:
nbar
pH = pKan
0.5
1.5
2.5
pH = pKa1
pH = pKa2
pH = pKa3 etc.
Thus from a curve of nbar versus pH for phosphoric acid, we can
estimate the pKa values from the pH values corresponding to nbar
= 0.5, 1.5, and 2.5:
H3PO4
protonation
Plot of nbar
versus
pH forcurve
phosphate:
3.5
H3PO4
3
pKa3 = 2.15
nbar
2.5
H2PO4-
2
pKa2 = 7.20
HPO42-
1.5
pKa1
= 12.38
1
PO43-
0.5
0
0
5
pH = pKa3
pH
pH pH
= pKa2
10
15
pH = pKa1
Species distribution diagram for phosphoric acid
H3PO4
H2PO4-
HPO42-
pH
PO43-
protonation curve for glycine
2.5
2
O
O
O
HO
NH3+
-
NH3+
O
-
O
NH2
glycine
1.5
nbar
pKa2 = 2.33
1
pKa1 = 9.45
0.5
0
0
5
pH = pKa2
10
pH pH = pK
a1
pH
15
O-
O-
O
O
-
O
+
NH+
-H
OH
NH+
-
+
O
H
O-
+
N
-
O
O
pKa2 = 6.13
-H+
HN
-
O
-
pKa1 = 9.46
O
-H+
-
O
NH+ +HN
O
-
O
O
-
O
O-
O
O
H+
-
O
HO
pKa3 = 2.69
(nitrilotriacetate)
O
N
O-
N
-
O
O
O
EDTA
O
-H+
NTA
O
O
O
H+
pKa1 = 9.52
O
H+
O
-
O
O-
O-
N
H
O
O
-H
+
pKa2 = 2.52
-
O
O
+
O
O
O
O-
O
O
NH+ +HN
O
-
O
O-
O
(ethylenediamine
tetraacetate)
Lewis acids and bases:
You will recall Brønsted acids and bases,
where a Brønsted acid is a proton donor and
a Brønsted base is a proton acceptor. A
broader definition is that of Lewis Acids and
Bases, where a Lewis Acid is an electron
acceptor, and a Lewis Base is an electron
donor.
F-, a
Lewis Base
BF3, a
Lewis Acid
Gilbert Newton Lewis
(1875-1946)
Lewis Acid accepts
electrons from the
Lewis Base to form
a complex.