148
ACIDS AND BASES
Box 4.4
Calculation of pH: weak acids and bases
Consider a 0.1 M solution of the weak acid acetic
acid (K a = 1.76 × 10
−5 ; pK a = 4.75). Since the
degree of ionization is small, the concentration
of undissociated acid may be considered to be
approximately the same as the original concentration,
i.e. 0.1. The pH can be calculated using the equation
pH =
1
2
pK a −
1
2
log[HA]
Thus
pH = 2.38 − 0.5 × log 0.1
= 2.38 − (−0.5)
= 2.88
The calculation of the pH of a weak base can be
achieved in a similar way; but again, since we have
a base, our calculations need to invoke the ionization
constant for water
K w = [H 3 O
+ ][HO
− ] = 10
−14
and pK a + pK b = 14. Thus, for a 0.1 M solution of
ammonia (conjugate acid pK a = 9.24)
pH = pK w −
1
2 pK b +
1
2 log[B]
and pK b is thus 14 − 9.24 = 4.76. This leads to
pH = 14 − 2.38 + 0.5 × log 0.1
= 14 − 2.38 + 0.5(−1)
= 11.12
Alternatively, we could use
pH =
1
2
pK w +
1
2
pK a +
1
2
log[B]
to get the same result:
pH = 7 + 4.62 + 0.5 × log 0.1
= 11.12
These calculations are for the pH of weak acids and
weak bases. It is well worth comparing the figures
we calculated above for strong acids and bases. Thus,
a 0.1 M solution of the strong acid HCl had pH 1,
and a 0.1 M solution of the strong base NaOH had
pH 13.
Although this produces a similar type of expression
to that for the pH of a weak acid above, it does
employ pK b rather than pK a . To keep to a ‘pK a only’
concept, we need to incorporate the pK a + pK b =
pK w expression. Then we get the alternative formula
pH = pK w −
1
2
(pK w − pK a ) +
1
2
log[B]
or
pH =
1
2
pK w +
1
2
pK a +
1
2
log[B]
Box 4.5
The pH of salt solutions
It should be self-evident that solutions comprised of
equimolar amounts of a strong acid, e.g. HCl, and a
strong base, e.g. NaOH, will be neutral, i.e. pH 7.0
at 25
◦ C. We can thus deduce that a solution of the
salt NaCl in water will also have pH 7.0.
However, salts of a weak acid and strong base or
of a strong acid and weak base dissolved in water
will be alkaline or acidic respectively. Thus, aqueous
sodium acetate is basic, whereas aqueous ammonium
chloride is acidic. pH values may be calculated from
pK a as follows.
Consider the ionization of sodium acetate in water;
this leads to an equilibrium in which AcO
− acts as
base
HOAc
H 2 O
+
+
OAc
HO
We can treat this equilibrium in exactly the same way
as the ionization of a weak base, where we deduced
the pH to be
pH =
1
2 pK w +
1
2 pK a +
1
2 log[B]
Thus, for a 0.1 M solution of sodium acetate in water,
where pK a for the conjugate acid HOAc is 4.75
pH =
1
2 pK w +
1
2 pK a +
1
2 log[B]
= 7 + 2.38 + 0.5 × log 0.1
= 7 + 2.38 + 0.5 × (−1)
= 8.88
If we now consider a 0.1 M solution of ammonium
chloride in water, where pK a for the conjugate acid
NH 4
+ is 9.24, we have the equilibrium
ACIDS AND BASES
Box 4.4
Calculation of pH: weak acids and bases
Consider a 0.1 M solution of the weak acid acetic
acid (K a = 1.76 × 10
−5 ; pK a = 4.75). Since the
degree of ionization is small, the concentration
of undissociated acid may be considered to be
approximately the same as the original concentration,
i.e. 0.1. The pH can be calculated using the equation
pH =
1
2
pK a −
1
2
log[HA]
Thus
pH = 2.38 − 0.5 × log 0.1
= 2.38 − (−0.5)
= 2.88
The calculation of the pH of a weak base can be
achieved in a similar way; but again, since we have
a base, our calculations need to invoke the ionization
constant for water
K w = [H 3 O
+ ][HO
− ] = 10
−14
and pK a + pK b = 14. Thus, for a 0.1 M solution of
ammonia (conjugate acid pK a = 9.24)
pH = pK w −
1
2 pK b +
1
2 log[B]
and pK b is thus 14 − 9.24 = 4.76. This leads to
pH = 14 − 2.38 + 0.5 × log 0.1
= 14 − 2.38 + 0.5(−1)
= 11.12
Alternatively, we could use
pH =
1
2
pK w +
1
2
pK a +
1
2
log[B]
to get the same result:
pH = 7 + 4.62 + 0.5 × log 0.1
= 11.12
These calculations are for the pH of weak acids and
weak bases. It is well worth comparing the figures
we calculated above for strong acids and bases. Thus,
a 0.1 M solution of the strong acid HCl had pH 1,
and a 0.1 M solution of the strong base NaOH had
pH 13.
Although this produces a similar type of expression
to that for the pH of a weak acid above, it does
employ pK b rather than pK a . To keep to a ‘pK a only’
concept, we need to incorporate the pK a + pK b =
pK w expression. Then we get the alternative formula
pH = pK w −
1
2
(pK w − pK a ) +
1
2
log[B]
or
pH =
1
2
pK w +
1
2
pK a +
1
2
log[B]
Box 4.5
The pH of salt solutions
It should be self-evident that solutions comprised of
equimolar amounts of a strong acid, e.g. HCl, and a
strong base, e.g. NaOH, will be neutral, i.e. pH 7.0
at 25
◦ C. We can thus deduce that a solution of the
salt NaCl in water will also have pH 7.0.
However, salts of a weak acid and strong base or
of a strong acid and weak base dissolved in water
will be alkaline or acidic respectively. Thus, aqueous
sodium acetate is basic, whereas aqueous ammonium
chloride is acidic. pH values may be calculated from
pK a as follows.
Consider the ionization of sodium acetate in water;
this leads to an equilibrium in which AcO
− acts as
base
HOAc
H 2 O
+
+
OAc
HO
We can treat this equilibrium in exactly the same way
as the ionization of a weak base, where we deduced
the pH to be
pH =
1
2 pK w +
1
2 pK a +
1
2 log[B]
Thus, for a 0.1 M solution of sodium acetate in water,
where pK a for the conjugate acid HOAc is 4.75
pH =
1
2 pK w +
1
2 pK a +
1
2 log[B]
= 7 + 2.38 + 0.5 × log 0.1
= 7 + 2.38 + 0.5 × (−1)
= 8.88
If we now consider a 0.1 M solution of ammonium
chloride in water, where pK a for the conjugate acid
NH 4
+ is 9.24, we have the equilibrium