1. pH calculations
strong acids
complete dissociation HA  H+
+ A-
What is the pH of a
0.040 M solution
of HClO4?
HClO4  H+
+ ClO4-
[H+] =
[ClO4-] =
0.040 M
pH =
- log [H+]
pH =
1.40
[H+]
from H2O
= 1.0 x 10-7 M
ignore autoionization
if [HA]
> 10-7 M

2. pH calculations
strong bases
complete dissociation
What is the pH of a
0.011 M solution
of Ca(OH)2?
Ca(OH)2
+ H2O 
Ca2+ (aq)
+ 2(OH-) (aq)
[OH-] =
2 [Ca2+] =
0.022 M
pOH =
- log [OH-] pH
+ pOH
= 14.00
pOH =
1.67
pH = 12.33
Na2O
+ H2O 
2 Na+ (aq)
+ 2 OH- (aq)

3. pH calculations
weak acids
incomplete dissociation HA  H+
+ A-
Ka =
[H+]eq
[A-]eq
= x2
[HA]i
- x
[HA]eq
[HA]eq
= [HA]i
- x  x2 Ka
[HA]i Ka
very small
assume x
<<
[HA]i if x
x 100
< 5%
approximation
O.K.
[HA]i x
x 100%
percent ionization =
[HA]
assume H+
from HA
>> H+
from H2O
(1 x 10-7 M)

4. pH calculations
weak acids
incomplete dissociation
What is the Ka of a
0.10 M
CHOOH,
pH = 2.38
CHOOH  H+
+ CHOO-
[H+]eq =
= 4.2 x 10-3
[H+]
[CHOO-]
Ka = = x2
= (4.2 x 10-3)2
=1.8 x 10-4
0.10
[CHOOH] x
[HA] (M)
[H+] (M)
[A-] (M)
4.2x10-3
x 100
Initial
0.10
0.00
0.00
0.10 -x +x +x
Change
4.2 %
Equil. x x x
% ionization

5. successive iterations x = 7.91 x 10-3 pH = 2.10
Calculate the pH of a
0.100 M HF solution
Ka = 6.8 x 10-4
= [H+]
[F-] HF  H+
+ F-
[HF]
[HF]
[H+]
[F-]
6.8 x 10-4 = x2 I
0.100
0.00
0.00 x C -x +x +x
x =
[H+] =
8.24 x 10-3 E x x x
pH =
2.08
x 100
= 8.25 %
5% assumption not good
0.100
solve quadratic or
successive iterations
.00790
.00791
x = 7.91 x 10-3
pH = 2.10
.00791

6. Calculate the pH of a
0.100 M HF solution
pH = 2.10
% ionization =
[H+]
x 100
= 7.91 x 10-3
x 100 =
7.91 %
[HF]
0.10
Calculate the pH of a
0.010 M HF solution
pH = 2.64
[H+] =
2.3 x 10-3
% ionization =
2.3 x 10-3
x 100
= 23%
0.010
HF (aq)
H+ (aq)
+ F- (aq) 
1 mol solute
2 mol solute
increase concentration

7. Find the pH of a
M solution
of H2CO3
H2CO3 
H+ +
HCO3-
Ka1 =
4.3 x 10-7
HCO3- 
H+ +
CO32-
Ka2 =
5.6 x 10-11
4.3 x 10-7 = x2
x = 4.0 x 10-5
x 100 =
1.10 %
3.7 x 10-3
3.7x10-3
pH =
4.40
[H2CO3]
[H+]
[HCO3-] I
3.7 x 10-3
0.00
0.00
[CO32-] = C -x +x +x E
3.7x10-3-x x x
3.7 x 10-3
4.0 x 10-5
4.0 x 10-5

8. Find the pH of a
M solution
of H2CO3
H2CO3 
H+ +
HCO3-
Ka1 =
4.3 x 10-7
HCO3- 
H+ +
CO32-
Ka2 =
5.6 x 10-11
5.6 x 10-11=
(4.0 x y)
(y)
y = 5.6 x 10-11
= [CO32-]
4.0 x y
pH determined by Ka1
[HCO3-]
[H+]
[CO32-]
% ionization I
4.0 x 10-5
4.0 x 10-5
0.00
1.4 x 10-4 % C -y +y +y E
4.0 x 10-5-y
4.0 x y y

9. pH calculations
weak bases
incomplete dissociation
What is the [OH-] of a
0.15 M
solution of NH3
Kb =
1.8 x 10-5
NH3 + H2O 
NH4+
+ OH-
[NH4+]
[OH-]
Kb = = x2
= 1.8 x 10-5
[NH3] x
x = 1.64 x 10-3
= [OH-]
pOH =
2.79
pH =
11.21
[NH3] (M)
[NH4+] (M)
[OH-] (M)
Initial
0.15
0.00
0.00
1.64x10-3 -x +x +x
Change
0.15
Equil. x x x
1.1 %