pH
147
Thus, a 0.1 M solution of HCl in water has
[H 3 O
+ ] = 0.1, and pH = − log 0.1 = 1.
Similarly, a 0.01 M solution has [H 3 O
+ ] = 0.01 and
pH = − log 0.01 = 2, and a 0.001 M solution has
[H 3 O
+ ] = 0.001 and pH = − log 0.001 = 3.
It follows from this that, because we are using a
logarithmic scale, a pH difference of 1 corresponds
to a factor of 10 in hydronium ion concentration.
If the pH is known, then we can calculate the
hydronium ion concentration. Since
pH = − log[H 3 O
+ ]
the hydronium ion concentration is given by
[H 3 O
+ ] = 10
−pH
For example, if the pH = 4, [H 3 O
+ ] = 10
−4
=
0.0001 M.
When we have a strong base, our calculations
need to invoke the ionization constant for water
K w = [H 3 O
+ ][HO
− ] = 10
−14
Thus, the pH of a 0.1 M solution of NaOH in
water is calculated from [HO
− ] = 0.1, and since
[H 3 O
+ ][HO
− ] = 10
−14 , [H 3 O
+ ] must be 10
−13 .
Hence, the pH of a 0.1 M solution of NaOH in
water will be − log 10
−13
= 13. A 0.01 M
solution of NaOH will have [HO
− ] = 10
−2 and
pH = − log 10
−12
= 12, and a 0.001 M solution has
[HO
− ] = 10
−3 and pH = − log 10
−11
= 11.
Weak acids are not completely ionized in aqueous
solutions, and the amount of ionization, and thus
hydronium ion concentration, is governed by the
equilibrium
H 2 O
HA +
A
H 3 O
+
and the equilibrium constant K a we defined above:
K a =
[A
− ][H 3 O
+ ]
[HA]
However, since [H 3 O
+ ] must be the same as [A
− ],
we can write
K a =
[H 3 O
+ ]
2
[HA]
and therefore
[H 3 O
+ ] =
K a [HA]
If we take negative logarithms of both sides, we get
− log[H 3 O
+ ] = −
1
2
log K a −
1
2
log[HA]
which becomes
pH =
1
2
pK a −
1
2
log[HA]
Note: this is simply a variant of the Henderson–Hasselbalch equation below, when [A
− ] =
[H 3 O
+ ].
The calculation of the pH of a weak base may
be approached in the same way. The equilibrium we
need to consider is
H 2 O
B +
HO
BH
+
and the equilibrium constant K b will be defined as
K b =
[HO
− ][BH
+ ]
[B]
However, since [HO
− ] must be the same as [BH
+ ],
we can write
K b =
[HO
− ]
2
[B]
and therefore
[HO
− ] =
K b [B]
Now we need to remember that
K w = [HO
− ][H 3 O
+ ]
so that we can replace [HO
− ] with K w /[H 3 O
+ ]; this
leads to
K w
[H 3 O
+ ]
=
K b [B]
and hence
[H 3 O
+ ] =
K w
√
K b [B]
If we now take negative logarithms of both sides, we
get
− log[H 3 O
+ ] = − log K w +
1
2
log K b +
1
2
log[B]
which becomes
pH = pK w −
1
2
pK b +
1
2
log[B]
147
Thus, a 0.1 M solution of HCl in water has
[H 3 O
+ ] = 0.1, and pH = − log 0.1 = 1.
Similarly, a 0.01 M solution has [H 3 O
+ ] = 0.01 and
pH = − log 0.01 = 2, and a 0.001 M solution has
[H 3 O
+ ] = 0.001 and pH = − log 0.001 = 3.
It follows from this that, because we are using a
logarithmic scale, a pH difference of 1 corresponds
to a factor of 10 in hydronium ion concentration.
If the pH is known, then we can calculate the
hydronium ion concentration. Since
pH = − log[H 3 O
+ ]
the hydronium ion concentration is given by
[H 3 O
+ ] = 10
−pH
For example, if the pH = 4, [H 3 O
+ ] = 10
−4
=
0.0001 M.
When we have a strong base, our calculations
need to invoke the ionization constant for water
K w = [H 3 O
+ ][HO
− ] = 10
−14
Thus, the pH of a 0.1 M solution of NaOH in
water is calculated from [HO
− ] = 0.1, and since
[H 3 O
+ ][HO
− ] = 10
−14 , [H 3 O
+ ] must be 10
−13 .
Hence, the pH of a 0.1 M solution of NaOH in
water will be − log 10
−13
= 13. A 0.01 M
solution of NaOH will have [HO
− ] = 10
−2 and
pH = − log 10
−12
= 12, and a 0.001 M solution has
[HO
− ] = 10
−3 and pH = − log 10
−11
= 11.
Weak acids are not completely ionized in aqueous
solutions, and the amount of ionization, and thus
hydronium ion concentration, is governed by the
equilibrium
H 2 O
HA +
A
H 3 O
+
and the equilibrium constant K a we defined above:
K a =
[A
− ][H 3 O
+ ]
[HA]
However, since [H 3 O
+ ] must be the same as [A
− ],
we can write
K a =
[H 3 O
+ ]
2
[HA]
and therefore
[H 3 O
+ ] =
K a [HA]
If we take negative logarithms of both sides, we get
− log[H 3 O
+ ] = −
1
2
log K a −
1
2
log[HA]
which becomes
pH =
1
2
pK a −
1
2
log[HA]
Note: this is simply a variant of the Henderson–Hasselbalch equation below, when [A
− ] =
[H 3 O
+ ].
The calculation of the pH of a weak base may
be approached in the same way. The equilibrium we
need to consider is
H 2 O
B +
HO
BH
+
and the equilibrium constant K b will be defined as
K b =
[HO
− ][BH
+ ]
[B]
However, since [HO
− ] must be the same as [BH
+ ],
we can write
K b =
[HO
− ]
2
[B]
and therefore
[HO
− ] =
K b [B]
Now we need to remember that
K w = [HO
− ][H 3 O
+ ]
so that we can replace [HO
− ] with K w /[H 3 O
+ ]; this
leads to
K w
[H 3 O
+ ]
=
K b [B]
and hence
[H 3 O
+ ] =
K w
√
K b [B]
If we now take negative logarithms of both sides, we
get
− log[H 3 O
+ ] = − log K w +
1
2
log K b +
1
2
log[B]
which becomes
pH = pK w −
1
2
pK b +
1
2
log[B]