The document discusses the concepts of pH, pOH, and the dissociation of acids and bases. It includes calculations for pH and ion concentrations from various strong and weak acids and bases, along with examples and exercises. The text emphasizes the relationship between the concentrations of H3O+ and OH- ions in solutions and the use of equilibrium laws to determine dissociation constants.

2. H2O + H2O ⇌ H3O+ + OH-
Acid Base
Kw = [H3O+] [OH-]
Where,
Kw: water dissociation
constant for pure
water at 25 ˚C

3. • [H3O+] = [OH-] = 1.0 × 10−7
mol dm-3
• Thus,
Kw = [H3O+] [OH-]
= (1.0 × 10−7
) (1.0 × 10−7
) mol2 dm-6
= 1.0 × 10−14
mol2 dm-6
• log [H3O+] + log[OH-] = log (1.0 × 10−14
)
log [H3O+] + log[OH-] = -14.0
-log [H3O+] + log[OH-]= 14.0
• Thus, pH + pOH = 14.0
or pH + pOH = pKw (-log = p)

4. THE pH of AQUEOUS SOLUTION
• Calculate the pH of an aqueous solution of
Ba(OH)2 0.25 mol dm-3

5. • Ba(OH)2 (aq)  Ba2+
(aq) + 2OH-
(aq)
• pOH = -log [OH-]
= -log (0.50)
= 0.3
Initial
molarity (mol
dm-3)
0.25 0 0
Final molarity
(mol dm-3)
0 0.25 0.50
pH = 14.0 – 0.3
= 13.7

6. What is the molarity of OH- in an aqueous
solution of NaOH if its pH is 13.5?
• pOH = 14.0 – pH
= 14.0 – 13.5
= 0.5
• pOH = -log [OH-]
0.5 = -log [OH-]
log [OH-] = -0.5
[OH-] = 0.3 mol dm-3
SOLUTION :

7. • Calculate the pH value of 1.0 × 10−8
mol dm-3
of HNO3 acid. Explain your answer.
EXERCISE :
(STRONG ACID)

8. • HNO3  H+ + NO2
-
1.0 × 10−8
1.0 × 10−8
1.0 × 10−8
• pH = -log [H+]
= -log (1.0 × 10−8
)
= 8
• For acid, pH <7, [H3O+]> 1.0 × 10−7
consider
[H3O+] from dissociation of water.
• ∴ [H3O+] = [H3O+]w + [H3O+]acid = 1.0 × 10−7
+
1.0 × 10−8

9. • Anions derived from strong acids (such as Cl-
from HCl) and cations from strong bases (such
as Na+ from NaOH) do not react with water to
affect the pH.

10. • Weak acid :
CH3COOH (aq) ⇌ H+
(aq) + CH3COO-
(aq)
Acid
Base

11. * CH3COO-
(aq) + H2O ⇌ CH3COOH (aq) + OH-

12. • Ka . Kb =
𝐶𝐻3 𝐶𝑂𝑂𝐻 [𝑂𝐻−]
[𝐶𝐻3 𝐶𝑂𝑂−]

13. To obtain the value of Kb for the anion
derived from a weak acid
•
• ,
• For the hydrolysis of acetate ion (*),
• Kb =
1.0 × 10−14 𝐾𝑤
1.8 ×10−5 𝐾𝑄
= 5.6 × 10−10

15. • The degree of dissociation of weak acids and
bases, ∝ less than 1 (or less 100%).
• The molarity of H3O+ ions is less than the
molarity of the acid (HA).
•

16. The molarity of H3O+ for weak acids (HA) can be
calculated using equilibrium law.
• HA (aq) + H2O (l) ⇌ H3O+
(aq) + A-
(aq)
Initial molarity
(mol dm-3 ) :
C 0 0
Reacted
concentration :
-X +X +X
Equilibrium
molarity (mol dm-3)
( C – X ) X X
For a weak
acid, X is
small.

17. Acid dissociation constant, Ka
• Each weak acid has an acid dissociation
constant (Ka).
• The relative strength of weak acids are
deduced from the Ka values.
• Strong acids have higher Ka (lower pKa) values
than weaker acids.
Example ;
Methanoic acid (HCOOH); Ka = 1.6 × 10−4
mol 𝑑𝑚−3
pKa = 3.8
Is a stronger acid than ethanoic acid (C𝐻3COOH); Ka = 1.7 × 10−5
mol𝑑𝑚−3
pKa = 4.8

18. Example;
Calculate (a) pKa, (b) pH, and (c) degree of
dissociation , of 0.01 mol dm-3 nitrous acid,
HNO2 [Ka = 5.1 × 10-4 mol dm-3]

19. Answers ;
(a) pKa = - log Ka
= - log (5.1 × 10-4)
= 3.3
(b) HNO2 (aq) + H2O ⇌ H3O+
(aq) + NO2
-
(aq)
Initial : 0.01 mol dm-3
Reacted : - X X X
At equilibrium : (0.01 – X) mol dm-3 X X

20. Ka =
𝐻3
𝑂
+
[𝑁𝑂2
−
]
[𝐻𝑁𝑂2
]
5.1 × 10-4 M =
𝑋2
0.01 −𝑋 𝑀
, X is small
∴ (0.01 – X) ≈ 0.01
=
𝑋2
0.01 𝑀
X2 = 5.1 × 10−6
M2
X = 2.26 × 10−3
M
∴ [H3O+] = 2.26 × 10−3
M
pH = -log [H+]
= -log (2.26 × 10−3
)
= 2.6
X = 𝑲𝒂 . 𝑪
= 𝟓. 𝟏 × 𝟏𝟎−𝟒(𝟎. 𝟎𝟏)
= 𝟓. 𝟏 × 𝟏𝟎−𝟔
= 2.26 × 𝟏𝟎−𝟑 M
Or :

21. (c) degree of dissociation, ∝
∝ =
𝑋
𝐶
× 100
𝐻3 𝑂+
𝐻𝑁𝑂2
=
2.26 ×10−3 𝑀
0.01 𝑀
× 100
= 23 %

22. • The degree of dissociation of weak bases ∝ <
1 < 100% .
• The molarity of OH- ion is less than the
molarity of undissociated base.
• The molarity of OH- from weak bases (B) can
be calculated using equilibrium law.

23. • B (aq) + H2O ⇋ BH+
(aq) + OH-
(aq)
Initial molarity ,
moldm-3
C O O
Reacted -X X X
Equilibrium
molarity , moldm-3
C – X X X

24. • Weak bases has a base dissociation constant,
Kb.
• The relative strength of weak bases can be
deduced by comparing the Kb values.
• Strong bases have higher Kb (lower pKb) values
than weaker bases.

25. Example ;
Ammonia, (NH3 ; Kb = 1.7 × 10−5
𝑚𝑜𝑙𝑑𝑚−3
pKb = 4.8 )
Is a weaker base than methanamine
( CH3NH2; Kb = 4.2 × 10−4
moldm-3
pKb = 3.4)

26. Example ;
Calculate (a) pKb, (b) pH and (c) degree of
dissociation of 0.1 moldm-3 NH2OH.
[Kb = 9.1 × 10−9
mold𝑚−3
]

27. (a) pKb = -log Kb
= -log ( 9.1 × 10−9
)
= 8.0
(b) NH2OH + H2O ⇋ NH3OH+
(aq) + OH-
(aq)
Initial : 0.1 0 0
Reacted : -X +X +X
At equilibrium : (0.1 – X) +X +X

28. Kb =
𝑁𝐻3 𝑂𝐻+ 𝑂𝐻−
𝐵
9.1 × 10−9
M =
𝑋 2
0.1 −𝑋 𝑀
, X is small, (0.1 – X) ≈ 0.1
=
𝑋2
0.1 𝑀
𝑋2
= 9.1 × 10−10
M2
X = 9.1 × 10−10 M2
X = 3.02 × 10−5
M
∴ [OH-] = 3.02 × 10−5
M

29. POH = -log [OH-]
= -log (3.02 × 10−5
)
= 4.5
pH = 14.0 – 4.5
= 9.5

30. (c) degree of dissociation, ∝
∝ =
𝑋
𝐶
× 100%
=
[𝑂𝐻
−
]
[𝑁𝐻2
𝑂𝐻]
× 100%
=
3.02 ×10−5 𝑀
0.1 𝑀
× 100%
= 0.03 %
NH2OH is a very
weak base.
!
NOTE :
Both in (a) and (b)-
neglected the
contribution of the
autoionization of
water to [H+]and[OH-]
because 1.0 × 10−7M is
so small compared with
1.0 × 10−3
M and
0.040M.

31. Worked example :

32. Question 1
a) Calculate the pH of a 1.8× 10−2
M Ba(OH)2
solution.
b) Calculate the pH of HNO3 0.001 moldm-3.

33. Answer
a) Ba(OH)2 is a strong base. It ionise completely to
form 1.8 x 10-2 moldm-3 OH-
(aq).
pOH = -log[OH-]
= -log(1.8x10-2)
= 1.74
pH + pOH = 14
pH = 14 – pOH
= 14 – 1.74
= 12.3

34. (b)
HNO3 is a strong acid.
It ionise completely to form 0.001 moldm-3 H+
(aq).
pH = -log[H+]
= -log(0.001)
= 3

35. Question 2
The pH of rainwater collected in a certain region
of north Malaysia on a particular day was 4.82.
calculate the H+ ion concentration of the
rainwater.

36. Answer
pH = 4.82
pH = -log[H+]
log[H+] = -pH
log10[H+] = -pH
[H+] = 10-pH
[H+] = 10-4.82
= 0.000015
= 1.5 x 10-5 mol dm-3
logab = c
b= ac

37. Question 3
The pH of a certain fruit juice is 3.33. calculate
the H+ ion concentration.

38. Answer
pH = 3.33
[H+] = ?
pH = -log[H+]
3.33 = -log[H+]
log[H+] = -3.33
log10[H+] = -3.33
[H+] = 10-3.33
= 0.000468
= 4.7 x 10-4 M

39. Question 4
In a NaOH solution, [OH-] is 2.9× 10−4
M.
calculate the pH of the solution.

40. Answer
pOH = -log[OH-]
= -log(2.9 x 10-4)
= 3.5
pOH + pH = 14
pH = 14 – pOH
= 14 – 3.5
= 10.5

41. Question 5
The OH- ion concentration of a certain ammonia
solution is 0.88 M. What is the pH of the
solution?

42. Answer
pOH = -log [OH-]
= -log(0.88)
= 0.055517
= 0.06
pH + pOH = 14
pH = 14 – pOH
= 14 – 0.06
= 13.94

44. A. Strong acid and strong bases
Example ;
Calculate the pH of
a) 1.0 × 10−3
M of HCl
b) 0.020 M Ba(OH)2 solution

45. Answer
a) HCl is a strong acid, completely ionized in
solution :
HCl(aq)  H+
(aq) + Cl-
(aq)
HCl(aq)  H+
(aq) + Cl-
(aq)
Initial (M) : 1.0 × 𝟏𝟎−𝟑 M 0.0 0.0
Change (M) : -1.0 × 𝟏𝟎−𝟑
M +1.0 × 𝟏𝟎−𝟑
+1.0 × 𝟏𝟎−𝟑
Final (M) : 0.0 1.0 × 𝟏𝟎−𝟑
1.0 × 𝟏𝟎−𝟑

46. ∴ [ H+] = 1.0 × 10−3
M
pH = -log (1.0 × 10−3
)
= 3.00

47. (b) Ba(OH)2 is a strong base. Each Ba(OH)2 unit
produces two OH- ions:
Ba(OH)2 (aq)  Ba2+
(aq) + 2OH-
(aq)
∴ [OH-] = 0.040 M
pOH = -log (0.040)
= 1.40
pH = 14.00 – 1.40
= 12.60
Initial (M) : 0.020 0.00 0.00
Change (M) : - 0.020 + 0.020 + 2(0.020)
Final (M) : 0.00 0.020 0.040

48. Exercise!!!

49. 1. What are the concentration of all species
present in 1.00 M acetic acid at 25℃? For
HC2H3O2, Ka is 1.8 × 10−5
.
X = 4.2 × 10−3

50. 2. What are the concentration of all species
present in a 0.10 M solution of HNO2 at 25℃.
For HNO2 Ka is 4.5 × 10−4
.
X = 6.5 × 𝟏𝟎−𝟑

51. 3. Give the definition of pH.
pH is a measure
of hydrogenion concentration; a
measure of
the acidity or alkalinity of asolution

52. 4.
a) What are [H+] and [OH-] in a 0.020 M solution
of HCl.
b) What are [H+] and [OH-] in a 0.0500 M
solution of NaOH.

53. 5.
a) What is the pH of a solution that is 0.050 M
in H+?
b) What is the pH of a solution for which [OH-] =
0.030 M ?

54. 6. What is [H+] of a solution with a pH of 10.60 ?
7. The pH of a 0.10 M solution of a weak acid HX
is 3.30. what is the ionization constant of HX?
(Ka = 2.5 × 10−6
)

55. 8.
a) Propanoic acid, HC3H5O2 a weak monoprotic
is 0.72% ionized in 0.25 M solution. What is
the ionization constant for this acid?
b) In 0.25 M solution of benzylamine, C7H7NH2,
the concentration of OH-
(aq) is 2.4 × 10−3
M.
C7H7NH2 + H2O  C7H7NH3
+ + OH-
what is the value of Kb for the aqueous
ionization of benzylamine?

56. • Answer :
a) Ka = 1.3 × 10−5
b) Kb = 2.3 × 10−5