Dissociation of water Understanding 18.1 Lewis acids and bases Understanding 18.2 Calculations involving acids and bases

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