The exact definition of pH is:
pH  - log(aH )

aH is called the activity of the hydrogen ion.

The activity of a species X can be written as
aX

γX [X]
where γX is called the activity coefficient.
781
The exact definition of pH is:
pH  - log(aH )

aH is called the activity of the hydrogen ion.

The activity of a species X can be written as
aX

γX [X]
where γX is called the activity coefficient. For fairly
dilute solutions, γX  1 , so that
aX

[X]
782
So the definition pH = - log[H+] will only apply
under dilute conditions.
783
So the definition pH = - log[H+] will only apply
under dilute conditions.
At 25 oC:
If pH < 7, the solution is acidic
784
So the definition pH = - log[H+] will only apply
under dilute conditions.
At 25 oC:
If pH < 7, the solution is acidic
If pH > 7, the solution is basic
785
So the definition pH = - log[H+] will only apply
under dilute conditions.
At 25 oC:
If pH < 7, the solution is acidic
If pH > 7, the solution is basic
If pH = 7, the solution is neutral
786
So the definition pH = - log[H+] will only apply
under dilute conditions.
At 25 oC:
If pH < 7, the solution is acidic
If pH > 7, the solution is basic
If pH = 7, the solution is neutral
Typical pH scale is:
acidic
basic
1
7
14
neutral
787
NEUTRAL
788
789
The p notation:
790
The p notation:
In general:
pX = - log(X)
791
The p notation:
In general:
pX = - log(X)
examples:
pH = -log[H+]
792
The p notation:
In general:
examples:
pX = - log(X)
pH = -log[H+]
pOH = -log[OH-]
793
The p notation:
In general:
examples:
pX = - log(X)
pH = -log[H+]
pOH = -log[OH-]
pK = -log K
794
Acid dissociation constants
795
Acid dissociation constants
Acid dissociation constant: The equilibrium
constant for the dissociation of an acid.
796
Acid dissociation constants
Acid dissociation constant: The equilibrium
constant for the dissociation of an acid.
Consider the dissociation of the monoprotic generic
acid HA,
797
Acid dissociation constants
Acid dissociation constant: The equilibrium
constant for the dissociation of an acid.
Consider the dissociation of the monoprotic generic
acid HA,
HA(aq)

+ H2O  H3O+(aq) + A-(aq)
798
Acid dissociation constants
Acid dissociation constant: The equilibrium
constant for the dissociation of an acid.
Consider the dissociation of the monoprotic generic
acid HA,
HA(aq)

+ H2O  H3O+(aq) + A-(aq)
[H3O ][A - ]
Kc 
[HA][H2O]
799
For weak acids the % of dissociation is low, therefore
the molar concentration of H2O is essentially
constant.
800
For weak acids the % of dissociation is low, therefore
the molar concentration of H2O is essentially
constant. Therefore we write
[H3O ][A - ]
Kc[H2O] 
[HA]
801
For weak acids the % of dissociation is low, therefore
the molar concentration of H2O is essentially
constant. Therefore we write
[H3O ][A - ]
Kc[H2O] 
[HA]
Now set Ka  Kc[H2O] , so that
802
For weak acids the % of dissociation is low, therefore
the molar concentration of H2O is essentially
constant. Therefore we write
[H3O ][A - ]
Kc[H2O] 
[HA]
Now set Ka  Kc[H2O] , so that
[H3O ][A - ]
Ka 
[HA]
803
For weak acids the % of dissociation is low, therefore
the molar concentration of H2O is essentially
constant. Therefore we write
[H3O ][A - ]
Kc[H2O] 
[HA]
Now set Ka  Kc[H2O] , so that
[H3O ][A - ]
Ka 
[HA]
The subscript a refers to acid, so Ka is an aciddissociation equilibrium constant.
804
If we write the acid dissociation in simplified
fashion:
+
HA(aq) 
H
+
A
(aq)
(aq)

805
If we write the acid dissociation in simplified
fashion:
+
HA(aq) 
H
+
A
(aq)
(aq)

Then we can write directly:
 ][A - ]
[H
Ka 
[HA]
806
If we write the acid dissociation in simplified
fashion:
+
HA(aq) 
H
+
A
(aq)
(aq)

Then we can write directly:
 ][A - ]
[H
Ka 
[HA]
The strength of the acid HA is quantitatively
measured by the value of Ka.
807
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
808
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
809
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
Need to set up an ICE table.
HCO2H
H+(aq)
HCO2-(aq)
810
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
Need to set up an ICE table.
HCO2H
H+(aq)
HCO2-(aq)
0.10 M
0M
0M
811
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
Need to set up an ICE table.
HCO2H
H+(aq)
HCO2-(aq)
0.10 M
0M
0M
-y
+y
+y
812
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
Need to set up an ICE table.
HCO2H
H+(aq)
HCO2-(aq)
0.10 M
0M
0M
-y
+y
+y
0.10 – y
y
y
813
Example problem: Calculate the pH of a 0.10 M
formic acid (methanoic acid) solution.
Ka = 1.7 x 10-4 for methanoic acid.
+
The reaction is: HCO2H 
 H (aq) + HCO2 (aq)
Need to set up an ICE table.
HCO2H
H+(aq)
HCO2-(aq)
0.10 M
0M
0M
-y
+y
+y
0.10 – y
y
y
[H ][HCO2- ]
Ka 
[HCO2H]
814
Therefore
2
y
Ka 
(0.10 - y)
815
Therefore
2
y
Ka 
(0.10 - y)
We will look at two different approaches to solve
this.
816
Therefore
2
y
Ka 
(0.10 - y)
We will look at two different approaches to solve
this.
Approach 1 (exact): The preceding result can be
written as
817
Therefore
2
y
Ka 
(0.10 - y)
We will look at two different approaches to solve
this.
Approach 1 (exact): The preceding result can be
written as
y2  1.7 x 104 y  1.7 x 105  0
818
Therefore
2
y
Ka 
(0.10 - y)
We will look at two different approaches to solve
this.
Approach 1 (exact): The preceding result can be
written as
y2  1.7 x 104 y  1.7 x 105  0
Now use
2
b

b
 4ac
y 
2a
819
The solutions are:
-4
-4 2
-5
1.7x10

(1.7x10
)

4(

1.7x10
)
y
2
820
The solutions are:
-4
-4 2
-5
1.7x10

(1.7x10
)

4(

1.7x10
)
y
2
that is y = 4.0 x 10-3 M or y = -4.2 x 10-3 M
821
The solutions are:
-4
-4 2
-5
1.7x10

(1.7x10
)

4(

1.7x10
)
y
2
that is y = 4.0 x 10-3 M or y = -4.2 x 10-3 M
The latter solution is unphysical.
822
The solutions are:
-4
-4 2
-5
1.7x10

(1.7x10
)

4(

1.7x10
)
y
2
that is y = 4.0 x 10-3 M or y = -4.2 x 10-3 M
The latter solution is unphysical.
Therefore [H+] = y = 4.0 x 10-3 M
823
The solutions are:
-4
-4 2
-5
1.7x10

(1.7x10
)

4(

1.7x10
)
y
2
that is y = 4.0 x 10-3 M or y = -4.2 x 10-3 M
The latter solution is unphysical.
Therefore [H+] = y = 4.0 x 10-3 M
Hence pH = -log(4.0 x 10-3) = 2.4
824
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
825
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
826
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
Therefore
2
y
Ka 
(0.10 - y)
827
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
Therefore
2
y
Ka 
(0.10 - y)
simplifies to y2  0.10 Ka , that is
828
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
Therefore
2
y
Ka 
(0.10 - y)
simplifies to y2  0.10 Ka , that is
y   0.10 Ka   1.7x105
829
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
Therefore
2
y
Ka 
(0.10 - y)
simplifies to y2  0.10 Ka , that is
y   0.10 Ka   1.7x105
therefore y = 4.1 x 10-3 M, and the pH = 2.4
830
Approach 2 (approximate): Because Ka is small we
expect y to be small, so try the approximation
0.10 - y  0.10
Therefore
2
y
Ka 
(0.10 - y)
simplifies to y2  0.10 Ka , that is
y   0.10 Ka   1.7x105
therefore y = 4.1 x 10-3 M, and the pH = 2.4
Clearly this approach is quicker!
831
When does the approximate approach
work?
832
When does the approximate approach
work?
For an initial concentration of dissociating acid –
call it [C], and an equilibrium constant Ka, then:
833
When does the approximate approach
work?
For an initial concentration of dissociating acid –
call it [C], and an equilibrium constant Ka, then:
4
[C]
 4x10 the error will be approx. 0.5% or less.
If
Ka
834
When does the approximate approach
work?
For an initial concentration of dissociating acid –
call it [C], and an equilibrium constant Ka, then:
4
[C]
 4x10 the error will be approx. 0.5% or less.
If
Ka
4
[C]
If
 1x10 the error will be approx. 1% or less.
Ka
835
When does the approximate approach
work?
For an initial concentration of dissociating acid –
call it [C], and an equilibrium constant Ka, then:
4
[C]
 4x10 the error will be approx. 0.5% or less.
If
Ka
4
[C]
If
 1x10 the error will be approx. 1% or less.
Ka
2
[C]
 4x10 the error will be approx. 5% or less.
If
Ka
836
Summary: The approximate approach will work
well if Ka is small and the initial concentration of
acid is not small.
837
Summary: The approximate approach will work
well if Ka is small and the initial concentration of
acid is not small.
For the formic acid problem
[C]  0.10  5.9x102
Ka 1.7x10- 4
So we would expect a bit less than a 5% error.
838
Summary: The approximate approach will work
well if Ka is small and the initial concentration of
acid is not small.
For the formic acid problem
[C]  0.10  5.9x102
Ka 1.7x10- 4
So we would expect a bit less than a 5% error.
For this problem, 5% of the [H+] concentration is
about 0.2x10-3 and so the pH would lie between
2.37 and 2.41, which when rounded to two sig
figures, would be 2.4, in agreement with the exact
approach.
839
Percent Dissociation
840