This document discusses chemical equilibrium, including definitions, concepts, and factors that affect equilibrium. It defines equilibrium as a state where the forward and reverse reaction rates are equal, resulting in constant concentrations. The equilibrium constant, K, relates concentrations or pressures of products and reactants. A system at equilibrium adjusts in response to changes in concentration, pressure, volume, or temperature to partially counteract the change according to Le Chatelier's principle. Temperature particularly affects equilibrium based on whether the reaction is endothermic or exothermic.

1. Equilibrium is a state in which there are no observable changes
as time goes by.
 It is reversible rxn in which rate forwards equals with
reverse reactions.
 Equilibrium is achieved when:
1. When the rate of the forward reaction is equal to the rate
of the backward reaction.
 Rate of production of product is the same as the rate of
the reformation of the reactants.
2. The concentrations of the reactants and products remain
constant.
6.1 Equilibrium Concepts and Acid-base Equilibrium
CHAPTER-6
1

2. Law of Mass Action
• Law of Mass Action- For a reversible reaction
at equilibrium and constant temperature, a
certain ratio of reactant and product
concentrations has a constant value
equilibrium concentration (K).
• The Equilibrium Constant (K)- A number
equal to the ratio of the equilibrium
concentrations of products to the equilibrium
concentrations of reactants each raised to the
power of its stoichiometric coefficient.
2

3. • Reaction Quotient: Symbol Q, a mathematical
expression relating the concentrations of
reactants and products at any stage of a reaction.
For the general reaction:
At equilibrium:
At any stage
K =
[C]c[D]d
[A]a[B]b
aA (g) + bB (g) c C (g) + d D (g)
Q =
[C]c[D]d
[A]a[B]b
3

4. 4

5. 6.2 Chemical Equilibrium
• Chemical equilibrium is defined by K.
• The magnitude of K will tell us if the
equilibrium reaction favors the reactants or
the products.
• If K » 1……..favors products
• If K « 1……..favors reactants
• If K = 0…….. At equilibrium
If Q < K, the reaction goes forward | to increase [product]
if Q = K, the system is at equilibrium |the law of mass action
if Q > K, the reaction goes backward | to decrease [product] 5

6. 6

7. 7

8. 8

9. 9

10. 10

11. Equilibrium Constant Expressions
• Equilibrium constants can be expressed
using Kc or Kp.
• Kc uses the concentration of reactants and
products to calculate the equilibrium
constant.
• Kp uses the pressure of the gaseous
reactants and products to calculate the
equilibrium constant..
11

12. The relation between Kc and Kp
• Equilibrium Constant Equations
• But from equations of ideal gases we have ,
PV= nRT, P = n/V RT
P = cRT
aA (g) + bB (g) cC (g) + dD (g)
Kc =
[C]c[D]d
[A]a[B]b
Kp =
PcCPd
PaAPbB
D
12

13. Equilibrium Constant Expressions
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
Dn = (c + d) – (a + b)
Kp =
[RTC]c [RTD]d
[RTA]a[RTB]b
Kp =KC
[RT]c [RT]d
[RT]a[RT]b
13

14. Equilibrium Constant Calculations
The equilibrium concentrations for the reaction between carbon
monoxide and molecular chlorine to form COCl2 (g) at 740C
are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M.
Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
Kc = [COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7
14

15. Equilibrium Constant Calculations
• The equilibrium constant Kp for the reaction is 158 at
1000K. What is the equilibrium pressure of O2 if the
PNO = 0.400 atm and PNO = 0.270 atm?
Kp =
2
PNO PO
2
PNO
2
2
PO2 = Kp
PNO
2
2
PNO
2
PO2 = 158 x (0.400)2/(0.270)2 = 347 atm
15

16. Equilibrium Constant Calculations
Consider the following equilibrium at 295K:
The partial pressure of each gas is 0.265 atm. Calculate
Kp and Kc for the reaction.
NH4HS (s) NH3 (g) + H2S (g)
Kp = P
NH3 H2S
P = 0.265 x 0.265 = 0.0702
Kp = Kc(RT)Dn
Kc = Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K
Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4
16

17. 6.3 TYPES OF EQUILIBRIA
• A homogeneous equilibrium is one in which all reactants
and products (and any catalysts, if applicable) are present in
the same phase.
• These solutions are most commonly either liquid or gaseous
phases, as shown by the examples below:
Homogeneous Equilibria
17

18. Heterogeneous Equilibrium
• A heterogeneous equilibrium involves reactants and
products in two or more different phases.
• Examples
18

19. Writing Equilibrium Constant Expressions
• The concentrations of pure solids, pure liquids and
solvents do not appear in the equilibrium constant
expressions.
• The equilibrium constant is a dimensionless quantity.
• In quoting a value for the equilibrium constant, you
must specify the balanced equation and the temperature.
• If a reaction can be expressed as a sum of two or more
reactions, the equilibrium constant for the overall
reaction is given by the product of the equilibrium
constants of the individual ..
19

20. Predicting the Direction of a Reaction
• The Kc for hydrogen iodide in the following
equation is 53.4 at 430ºC. Suppose we add
0.243 mol H2, 0.146 mol I2 and 1.98 mol HI to
a 1.00L container at 430ºC. Will there be a net
reaction to form more H2 and I2 or HI?
H2 (g) + I2 (g) → 2HI (g)
[HI]0
2
[H2]0 [I2]0
Kc =
[1.98]2
[0.243][0.146]
Kc = Kc = 111
20

21. Calculating Equilibrium Concentrations
• At 1280ºC the equilibrium constant (Kc) for
the reaction is 1.1 x 10-3. If the initial
concentrations are [Br2] = 0.063 M and [Br] =
0.012 M, calculate the concentrations of these
species at equilibrium.
Br2 (g) 2Br (g)
Let x be the change in concentration of Br2
Br2 (g) 2Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063
-x
0.063 - x
0.012
+2x
0.012 + 2x
21

22. Calculating Equilibrium Concentrations
[Br]2
[Br2]
Kc =
Kc =
(0.012 + 2x)2
0.063 - x
= 1.1 x 10-3
4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x2 + 0.0491x + 0.0000747 = 0
ax2 + bx + c = 0 -b ± b2 – 4ac

2a
x =
x = -0.00178
x = -0.0105
22

23. Br2 (g) 2Br (g)
Initial (M)
Change (M)
Equilibrium (M)
0.063 0.012
-x +2x
0.063 - x 0.012 + 2x
At equilibrium, [Br] = 0.012 + 2x = -0.009 M
At equilibrium, [Br2] = 0.063 – x = 0.0648 M
or 0.00844 M
Calculating Equilibrium Concentrations
23

24. Factors that Affect Chemical Equilibrium
• Le Châtlier’s Principle- if an external stress is
applied to a system at equilibrium, the system
adjusts in such a way that the stress is partially
offset as the system reaches a new equilibrium
position.
24

25. Factors that Affect Chemical Equilibrium
• Chemical Equilibrium represents a balance
between forward and reverse reactions.
• Changes in the following will alter the
direction of a reaction:
–Concentration
–Pressure
–Volume
–Temperature
25

26. Changes in Concentration
Change Shift in Equilibrium
Increase in [Products] left
Decrease in [Products] right
Increase in [Reactants] right
Decrease in [Reactants] left
26

27. Changes in Volume and Pressure
• Changes in pressure primarily only concern gases.
• Concentration of gases are greatly affected by pressure
changes and volume changes according to the ideal gas
law.
PV = nRT
P = (n/V)RT
P = cRT
27

28. Changes in Pressure and Volume
Change Shift in Equilibrium
Increase in Pressure Side with fewest moles
Decrease in Pressure Side with most moles
Increase in Volume Side with most moles
Decrease in Volume Side with fewest moles
28

29. Reactions that shift right when pressure increases
and shift left when pressure decreases
Consider the reaction:
2SO2(g) + O2(g) ⇄ 2SO3(g),
1. The total moles of gas decreases as reaction
proceeds in the forward direction.
2. If pressure is increased by decreasing the volume
(compression), a forward reaction occurs to reduce
the stress.
3. Reactions that result in fewer moles of gas favor
high pressure conditions.
29

30. Reaction that shifts left when pressure increases,
but shifts right when pressure decreases
Consider the reaction: PCl5(g) ⇄ PCl3(g) + Cl2(g);
1. Forward reaction results in more gas molecules.
2. Pressure increases as reaction proceeds towards
equilibrium.
3. If mixture is compressed, pressure increases, and
reverse reaction occurs to reduce pressure;
4. If volume expands and pressure drops, forward
reaction occurs to compensate.
5. This type of reactions favors low pressure
condition
30

31. Reactions not affected by pressure changes
Consider the following reactions:
1. CO(g) + H2O(g) ⇄ CO2(g) + H2(g);
2. H2(g) + Cl2(g) ⇄ 2HCl(g);
1. Reactions have same number of gas molecules
in reactants and products.
2. Reducing or increasing the volume will cause
equal effect on both sides – no net reaction will
occur.
3. Equilibrium is not affected by change in
pressure.
31

32. Changes in Temperature
• Equilibrium position vs. Equilibrium constant
• A temperature increase favors an endothermic
reaction and a temperature decrease favors and
exothermic reaction.
Change Endo. Rx Exo. Rx
Increase T K decreases K increases
Decrease T K increases K decreases
32

33. Changes in Temperature
Consider: N2O4(g) ↔ 2NO2(g)
The forward reaction absorbs heat; endothermic
heat + N2O4(g) ↔ 2NO2(g)
So the reverse reaction releases heat; exothermic
2NO2(g) ↔ N2O4(g) + heat
Changes in temperature??
33

34. Consider the following exothermic reaction:
N2(g) + 3H2(g) ⇄ 2NH3(g); ∆Ho
= -92 kJ,
The forward reaction produces heat => heat is
a product.
When heat is added to increase temperature,
reverse reaction will take place to absorb the heat;
If heat is removed to reduce temperature, a net
forward reaction will occur to produce heat.
Exothermic reactions favor low temperature
conditions.
34

35. The Effect Temperature on Equilibrium
Consider the following endothermic reaction:
CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g), DHo
= 205 kJ
1. Endothermic reaction absorbs heat  heat is a
reactant;
2. If heat is added to increasing the temperature, it will
cause a net forward reaction.
3. If heat is removed to reduce the temperature, it will
cause a net reverse reaction.
4. Endothermic reactions favor high temperature
condition.
35

36. For spontaneous reactions, ∆G decreases (becomes less negative) as the reaction
proceeds towards equilibrium.
Δ G = Δ G
o
+ RT ln Q
G = free energy at any moment
G = standard-state free energy
R = ideal gas constant = 8.314 J/mol-K
T = temperature (Kelvin)
lnQ = natural log of the reaction quotient
At equilibrium, ∆G = 0; Q = K
∆G = ∆Go
+ RTlnK = 0
∆Go
= -RTlnK
lnK = - ∆Go
/RT (∆Go
calculated at temperature T)
Equilibrium constant, K = e-(∆Go/RT)
∆G◦ < 0, K > 1; reaction favors products formation
∆G◦ > 0, K < 1; reaction favors reactants formation
∆G◦ = 0, K = 1; reaction favors neither reactants
nor products
Free Energy and Equilibrium Constant
36

37. Consider the reaction: N2(g) + 3H2(g)  2NH3(g),
At 25o
C, ∆Go
= -33 kJ
lnK = -(-33 x 103
J/(298 K x 8.314 J/K.mol)) = 13
K = e13 = 4.4 x 105 (reaction goes to completion)
At 250o
C, ∆Go
= 12 kJ;
lnK = -(12 x 103
J/(523 K x 8.314 J/K.mol)) = -2.8
K = e-2.8 = 0.061 (very little product is formed)
Calculating K from ∆Go
37

38. 1) The equilibrium mixture is found to contain 0.07,
0.11, 0.03 and 0.03 moles of CO, H2, CH4 and H2O
respectively for the reaction:
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
What is the equilibrium constant, K?
Solution
CO (g) + 3 H2 (g)  CH4 (g) + H2O (g)
start 0.07 0.11 0.03 0.03
(0.03) (0.03)
K = —————— = 9.7
(0.07) (0.11)3
Equilibrium calculations
38

39. 2) 1.000 mole of H2 gas and 1.000 mole of I2 vapor
are introduced into a 5.00-liter sealed flask. The
mixture is heated to a certain temperature and the
following reaction occurs until equilibrium is
established.
H2(g) + I2(g) ⇄ 2HI(g)
At equilibrium, the mixture is found to contain
1.580 mole of HI.
(a) What are the concentrations
of H2, I2 and HI at equilibrium?
(b) Calculate the equilibrium constant Kc.
39

40. Solution:
——————————————————————————
H2(g) + I2(g) ⇄ 2 HI(g)
——————————————————————————
Initial [ ], M: 0.200 0.200 0.000
Change in [ ], M: -0.158 -0.158 + 0.316
Equilibrium [ ], M 0.042 0.042 0.316
——————————————————————————
]
][I
[H
[HI]
2
2
2
2
2
(0.316)
0.042
( )
=
Kc = = 57
40

41. 3) The equilibrium concentrations for the reaction
between carbon monoxide and molecular chlorine
to form COCl2 (g) at 74◦C are [CO] = 0.012 M,
[Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate
the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) ⇄ COCl2 (g)
Kc =
[COCl2]
[CO][Cl2]
=
0.14
0.012 x 0.054
= 220
Kp = Kc(RT) ∆n
Dn = 1 – 2 = -1 R = 0.0821 L-atm mol-1 K-1
T = 273 + 74 = 347 K
Kp = 220 x (0.0821 x 347)-1 = 7.7 41

42. 4) At 1123K, Kc = 0.153 for the reaction:
C(s) + CO2 (g) ⇄ 2 CO (g) DHo = 172.5 kJ
What is Kp?
If initially the partial pressure of CO2 was 1.50 atm, what is the total
pressure at equilibrium?
How will the equilibrium shift if temperature is decreased to 900K?
Solution: R in L-atm mol-1 K-1
Kp = 0.153 (R T)1 = 0.153(0.08206*1123) = 14.1
C(s) + CO2 (g) ⇄ 2 CO (g)
1.50-x 2x equilibrium pressure
(2x)2 / (1.50 – x) = 14.1; 4x2 + 14.1x –21.2 = 0
x = {- 14.1 + (14.12 + 4*4*21.1)} / (2*4) = 1.13 atm
total pressure = 1.50-x+2x = 1.50+x = 1.50 + 1.13 = 2.63 atm
P(CO2) = 1.50 – 1.13 = 0.37 atm; P(CO) = 2*1.13 = 2.26
P(CO) decreases at 900K (<1123K) in the endothermic reaction (+
DHo).  shift left
42

43. 5) At 600K, the partial pressure of NO2 in a equilibrium
mixture is 0.10 atm,
N2O4 ⇄ 2 NO2 Kp = 3.06 L atm mol 1 K 1
What is the partial pressure of N2O4?
What is the percent of dissociation of N2O4?
What percentage of molecules are NO2?
Solution:N2O4 (g) ⇄ 2 NO2(g) Kp = 3.06 L atm mol1 K1
x 0.10 let x be the unknown
0.102 / x = 3.06; x = 0.12/3.06 = 0.0033 atm P(N2O4)
percent dissociated N2O4 = 0.05 / (0.0033+0.05)
= 0.94 = 94%
% of NO2 molecule = 0.10/(0.10+0.0033) = 0.97 = 97% 43

44. 6.4. Concepts of acid-base
Arrhenius concept
Acids: dissociates in water to form hydrogen ions
 increases the concentration of hydrogen (H+) ions in an
aqueous solution
 Protonation of water yields the hydronium ion (H3O+)
H+ + H2O H3O+
 bare proton (H+) does not exist as a free species in
aqueous solution.
Bases: dissociates in water to form hydroxide ions
 increases the concentration of hydroxide (OH–) ions in an
aqueous solution.
acid–base reactions is a neutralization reaction

45. Limitations of the Arrhenius Definition
 Restricted to aqueous solutions and refer to the concentration of the solvated
ions
 Pure H2SO4 or HCl dissolved in toluene are not acidic, but could donate a
proton to toluene.
 A sodium amide (NaNH2) in liquid ammonia is not alkaline, but amide ion
(NH2
−) will readily deprotonate ammonia.
Br∅nsted-Lowery concept- conjugate acid-base pairs
 Acid: A compound that donates a proton to another compound
 Base: Compound that accepts a proton .
 An acid-base reaction is, thus, the transfer of a proton from a donor
(acid) to an acceptor (base).

46. 1
2
Examples
pyridine molecules (C5NH5) undergo base ionization when dissolved in
water, yielding hydroxide and pyridinium ions.
The reaction between water and ammonia

47. 3
4
5
the two possible acid-base reactions for two amphiprotic (Species
capable of either donating or accepting protons), bicarbonate ion
and water
Liquid water undergoes autoionization (like molecules react to yield ions)
the ion-product constant for w autoionization water, Kw

48. Lewis concept
Lewis Acid: a species that accepts an electron pair (i.e., an
electrophile) and will have vacant orbitals, or the ability to form a
vacant orbital.
Examples of Lewis Acids: H+, K+, Mg2+, Fe3+, BF3, , AlCl3,
Br2.
Lewis Base: a species that donates an electron pair (i.e., a
nucleophile) and will have lone pair electrons
Examples of Lewis Bases: OH-, F-, H2O, ROH, NH3, SO4
2-, H-,
CO, C6H6.
Bond formed between Lewis Acid and Base when both electrons
come from the Lewis base is a coordinate covalent bond (or dative
covalent bond).

49. three general types of Lewis Acids
 Molecules with an incomplete octet
 Molecules with double bonds
 metal ions with vacant d-orbitals
Incomplete Octets-Boron Trifluoride (BF3)
Figure 6.7: General reaction scheme where a molecule with an incomplete
octet functions as a Lewis Acid

50. Metal Cations: have vacant d-orbitals that can accept electrons and thus behave as
Lewis Acids.
Double Bonds: has 4 electrons shared between 2 atoms in a pi and sigma orbital.
Fig. 6.8 General reaction scheme where a molecule with a double bond
functions as a Lewis Acid
Figure 6.9: General reaction scheme where a molecule with a metal cation
functions as a Lewis Acid

51. 6.5 pH and pOH
● The Danish biochemist Søren Sørensen proposed the term pH in
1909
● H3O+ and OH- ions are present both in pure water and in all aqueous
solutions.
● Concentrations of these ions in a solution determines the solution’s
properties and the chemical behaviors of its other solutes
● Expressing acidity or basicity in terms of the concentration of H3O+
or OH- can be cumbersome because the values tend to be very small.
● Very small concentrations of H3O+ is conveniently expressed in terms
of pH and that of OH- is expressed in terms of pOH (in terms of
logarithmic scale)

52. …Cont’d
● The pH of a solution is defined as
pH = -log[H3O+] ⇨ [H3O+] = 10-pH
● Likewise, the hydroxide ion molarity may be expressed as a p-function, or
pOH:
pOH = -log[OH-] ⇨ [OH-] = 10-pOH
● The relationship between ionic product of water (KW) and, [H3O+] and [OH-]
is
KW = [H3O+][OH-] ; KW = 1..0 x 10-14 at 25 oC
-log KW = -log([H3O+][OH-]) = -log[H3O+] + -(log[OH-]
pKW = pH + pOH ⇨ 14.00 = pH + pOH at 25 oC
● The sum of a solution’s pH and pOH is not always 14. It depends on
temperature.

53. …Cont’d
● As the temperature changes, the exact values will change because
the value of Kw changes. However, the relationship
pH + pOH = pKw will remain the same.
● The ionization of water increases as temperature increases.
Therefore, the ion product, Kw, also increases as temperature
increases.
Figure 1: The pH Scale

54. …Cont’d
● Summary of relations for acidic, basic and neutral solutions
Acidic: [H3O+] > [OH-], [H3O+] < 1.0x10-7 M, pH < 7
Neutral: [H3O+] = [OH−], [OH−] = 1.0x10-7 M, pH = 7
Basic: [H3O+] < [OH-], [OH−] > 1.0x10-7 M, pH > 7
Table 2: Approximate pH range of some common materials at 25 oC
Material pH Material pH Material pH
1 M HCl 0 Cherries 3.2–4.7 Saliva 6.5–7.5
Gastric juice 1.0–3.0 Tomatoes 4.0–4.4 Pure water 7.0
Lemons 1.8–2.4 Beer 4-4.5 Blood 7.3–7.5
Vinegar 2.4–3.4 Bananas 4.5–5.7 Eggs 7.6–8.0
Soft drinks 2.0–4.0 Bread 5.0–6.0 Urine 5-7
Apples 2.9–3.3 Rainwater 5.5–5.8 Sea water 8.0–8.5
Grapefruit 2.9–3.4 Potatoes 5.6–6.0 Milk of magnesia 10.5
Oranges 3.0–4.0 Milk 6.3–6.6 Household ammonia 11.9

55. …Cont’d
Example 1: What is the pH of a 1.0 x 10-3 M NaOH solution?
Solution: NaOH is a strong base that dissociates completely. Thus,
[OH−] = 1.0 x 10–3 M ⇨ pOH = -log[OH-] = -log(1.0 x 10–3 ) = 3
pH + pOH = 14.0 at 25 oC ⇨ pH = 14.0-pOH = 14.0-3 = 11
Example 2: The pH of household ammonia was measured to
be 11.28. Determine its [H3O+]?
Solution: pOH = 14.00 - pH = 14.00 - 11.28 = 2.72
pOH = -log[OH-]. ⇨ log[OH-] = 10-pOH
[OH-] = 10-2.72 = 1.9 x 10-3 M

56. …Cont’d
Exercises: 1) The pH of an aqueous solution is 3.67. Determine its
[H3O+].
2) Determine the pH, [H3O+], [NO3
-], and [OH−] at 25 oC in a 0.60 M
aqueous solution of HNO3(aq), a strong acid.
Answer: [H3O+] = [NO3
-] = 0.60 M; [OH-] = 1.7 x 10-14 M, pH = 778
3) Arrange the following solutions in order from lowest to highest pH:
0.10 M HCl, 0.10 M H2SO4, and 0.10 M HF.
4) Calculate [Ba2+], [OH−], and [H3O+] at 25 oC in a 0.25 M aqueous
solution of barium hydroxide, a strong base.
Answer: [Ba2+] = 0.25 M; [OH−] = 0.50 M; [H3O+] = 2.0 x 10−14 M

57. 6.6 Relative strength of acids and base
● Different acids dissociate to different extents in aqueous solution.
● A strong acid is one that ionizes amost completely (to a large extent) in
aqueous solution. A strong acid is a strong electrolyte.
● Acids that dissociate to only a small extent (partially) are weak acids.
A weak acid is a weak electrolyte.
● Weak acids: Hydrofluoric acid (HF), sulfurous acid (H2SO3)
(diprotic), acetic acid (HC2H3O2), carbonic acid (H2CO3), (diprotic),
formic acid (HCHO2), phosphoric acid (H3PO4) (triprotic),
hydrocyanic acid (HCN), nitrous acid (HNO2), hypochlorous acid
(HOCl)
● Only a small fraction of the weak acid molecules transfer a proton to
water, and the solution, therefore, contains mainly undissociated HA
molecules along with small amounts of H3O+ and the conjugate base,
A-.
CH3COOH + H2O ⇋ CH3COO- + H3O+
HF + H O ⇋ H O+ + F-

58. …Cont’d
● Strong acid: Perchloric acid (HClO4), hydroiodic acid (HI),
hydrobromic acid (HBr), hydrochloric acid (HCl), sulphuric acid
(H2SO4) and nitric acid (HNO3)
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
● A weak acid shows smaller values for colligative properties than a
strong acid
● A base that dissociate almost completely in water are called strong
bases.
● A base that dissociates only slightly in aqueous solution is a weak
base.
● Examples of strong bases: Lithium hydroxide (LiOH), sodium
hydroxide (NaOH), potassium hydroxide (KOH), rubdium hydroxide
(RbOH), cesium hydroxide (CsOH), calcium hydroxide (Ca(OH)2),
strontium hydroxide (Sr(OH)2), barium hydroxide (Ba(OH)2),
thallium(I) hydroxide (TlOH),

59. …Cont’d
● Examples of weak bases: Ammonia (NH3), methylamine
(CH3NH2), ethylamine (C2H5NH2), pyridine (C5H5N),
aniline (C6H5NH2)
● In solutions of the same concentration, stronger acids ionize
to a greater extent, and yield higher concentrations of
hydronium ions than do weaker acids.
● In solutions of the same concentration, stronger bases ionize
to a greater extent, and yield higher concentrations of
hydroxide ions than do weaker bases.

60. 6.6.1 Ionization of weak acids
● The strength of a weak acid, that is, extent of dissociation of
a weak acid is quantified by the acid dissociation constant
(Ka) which is the equilibrium constant for the ionization of
the weak acid.
● For a weak acid HA, HA + H2O ⇋ H3O+ + A-
● Note that water has been omitted from the equilibrium
equation because its concentration in dilute solutions is
essentially the same as that in pure water (55.5 M) and pure
liquids are always omitted from equilibrium equations.
+ -
3
a
[H O ][A ]
K =
[HA]

61. …Cont’d
● The larger the Ka, the stronger the acid.
● Strong acids can be assumed, for practical purposes, to
ionize completely in water, and [HA] is zero and the acid
ionization constant is immeasurably large (Ka ≈ ∞).
● For weak acid, Ka is small and they mainly exist in the form
of undissociated molecule.
● For example, about 1.3% of the solute molecules in a 0.1 M
acetic acid solution are ionized at room temperature. The
remaining 98.7% of the acetic acid molecules, CH3COOH,
remain nonionized.

62. Table 2: Ionization constants of some weak acid at 25 oC
●The small value of Ka for hydrocyanic acid means that only a
small fraction of the hydrocyanic acid molecules dissociate.
Most of them remain in the undissociated form as HCN(aq).
Exercise: Write ionization equation and Ka expression for aqueous
solution of the following weak acids.
a) C6H5COOH b) Cl2CHCOOH c) HOBr d) HCOOH
Acid Ka Acid Ka
HF (hydrofluoric acid) 7.1 x 10-4 ClCH2COOH (chloroacetic acid) 1.36 x 10-3
HCN (hydrocyanic acid) 4.9 x 10-10 CH3CH(OH)COOH (lactic acid) 1.39 x10-4
HNO2 (nitrous acid) 4.5 x 10-4 C6H5COOH (benzoic acid) 6.5 x 10-5
CH3COOH (acetic acid) 1.75 x 10-5 C6H5OH (phenol) 1.3 x 10-10
HCOOH (formic acid) 1.77 x 10-4 C9H8O4 (acetylsalicylic acid; aspirin) 3.0 x 10-4

63. 6.6.1.1 Percent ionization of weak acids
● The magnitude of Ka indicates the strength of an acid (as Ka increases,
so does the strength of the acid). Another measure of the strength of
an acid is its percent ionization (%α), which is defined as
● The stronger the acid, the greater the percent ionization. For a
monoprotic acid HA, the concentration of the acid that undergoes
ionization is equal to the concentration of the H+ ions or the
concentration of the A- ions at equilibrium. Therefore, we can write
the percent ionization as
ionized acid concentration at equilibrium
Percent ionization = x 100
initial concentration of acid
+ -
3 eq eq
o o
[H O ] [A ]
Percent ionization (%α) = x 100 = x 100
[HA] [HA]

64. …Cont’d
● Unlike the Ka value, the percent ionization of a weak
acid varies with the initial concentration of acid,
typically decreasing as concentration increases. The
more dilute the solution, the greater the percent
ionization.
Figure 2: Dependence of percent ionization on the initial concentration
of acid. Note that at very low concentrations, all acids (weak and strong)
are almost completely ionized.

65. …Cont’d
Example: Calculate the pH and the concentrations of all the
species in a 0.250 M aqueous solution of benzoic acid,
C6H5COOH(aq). The value of Ka for benzoic acid is 6.3 x 10-5
M.
Solution: C6H5COOH + H2O ⇋ H3O+ + C6H5COO-
Ka= [C6H5COO-][H3O+]/[C6H5COOH] = [H3O+]2/[C6H5COOH]
[H3O+] = [C6H5COO-] = 3.968x10-3 M
[C6H5COOH]eq = (0.250 - 3.968x10-3) M = 0.246 M
pH = -log [H3O+] = -log (3.968x10-3) = 2.40
+ -5 -3
3 a 6 5
[H O ] = K [C H COOH] = 6.3 x 10 x 0.250 = 3.968x10 M

66. …Cont’d
Percent ionization = 1.59%
Exercises: 1) The pH of a 0.40 M solution of formic acid,
HCOOH(aq), is 2.08 and that of a 0.40 M solution of
hydrocyanic acid, HCN(aq), is 4.80. Compare the percentages
of acid molecules that are dissociated in the two solutions.
Answer: HCOOH, 2.1%; HCN, 0.0040%
2) Calculate the percent ionization of acetic acid in the
following solutions. a) 1.0 M acetic acid solution with a pH
of 2.40 b) 0.10 M acetic acid solution with a pH of 2.90
c) 0.010 M acetic acid solution with a pH of 3.40
- + -3
6 5 eq 3 eq
6 5 o o
[C H COO ] [H O ] 3.968x10 M
Percent ionization (%α) = x 100 = x 100 = x 100
[C H COOH] [HA] 0.250 M

67. 6.6.2 Ionization of weak bases
● Weak bases accept a proton from water to give the conjugate acid of
the base and OH- ions.
● The reaction for ionization of a weak base (B) in aqueous solution can
be written as
B(aq) + H2O(l) ⇋ BH+(aq) + OH-(aq)
● The expression for the equilibrium constant is written as
Kb = [BH+][OH-]/[B]
● The higher the Kb the stronger the base is.
● Some examples of reactions of weak bases are as follows:
NH3(aq) + H2O(l) ⇋ NH4
+(aq) + OH-(aq)
HOCH2CH2NH2(aq) + H2O(l) ⇋ HOCH2CH2NH3
+(aq) + OH-(aq)
CN-(aq) + H2O(l) ⇋ HCN(aq) + OH-(aq)
IO3
-(aq) + H2O(l) ⇋ HIO3(aq) + OH-(aq)
Exercise: Write ionization equation and Kb expression for aqueous
solution of the following weak bases
a) CH3NH2 b) NH3 c) C5H5N

68. …Cont’d
Example: Find the concentration of hydroxide ion, in a 0.25 M solution
of trimethylamine, a weak base. Calculate the pH and the percent
ionization (the fraction ionized) of trimethylamine,. Kb = 7.4 x 10−5
Solution: (CH3)3N(aq) + H2O(l) ⇋ (CH3)3NH+(aq) + OH−(aq)
Kb = [CH3)3NH+][OH-]/[CH3)3N] = [OH-]2/[CH3)3N]
[CH3)3NH+] = [OH−] =
This is less than 5% of the initial concentration (0.25), so the assumption
is justified/acceptable.
pOH = -log[OH−] = -log(4.3x10-3) = 2.37
pH + pOH = 14 ⇨ pH = 14-pOH = 14-2.37 = 11.63
Percent ionization = ([OH−]eq/[CH3)3N]o) x 100
Percent ionization = (4.3x10-3/0.25) x 100 = 1.72%
+ -
eq eq
o o
[BH ] [OH ]
Percent ionization of base = x 100 = x 100
[B] [B]
-5 -3
b 3
K [(CH ) N] = 7.4 x 10 x 0.25 = 4.3x10 M


69. ….Cont’d
Exercises: 1) Aniline, C6H5NH2(l), is used in the
manufacture of dyes and various pharmaceuticals. The
solubility of aniline in water at 25 °C is 3.50 grams per
100.0 mL of solution. Calculate the pH and the
concentrations of all the species in a saturated aqueous
solution of aniline at 25 oC.
Answer: [OH−] = [C6H5NH3
+] = 1.7 x 10-5 M
[C6H5NH2] = 0.376 M
[H3O+] = 5.9 x 10−10 M; pH = 9.23
2) CH3NH2 has a pKb value of 3.34. What is the pH of an
aqueous solution of CH3NH2 of concentration 0.10
mol.dm-3?

70. 6.6.3 Relative strengths of conjugate acid-base pairs
● An acid transfers a proton to a base to form the conjugate base of the original
acid and the conjugate acid of the original base.
● Conjugate acid-base pairs differ by a proton, which is present in the acid form
and missing in the base form.
● Consider a conjugate acid-base pair HA/A- whose ionization equilibrium
equations and ionization constant expressions are:
HA(aq) + H2O(l) ⇋A-(aq) + H3O+(aq) Ka = [A-][H3O+]/[HA]
A-(aq) + H2O(l) ⇋ OH-(aq) + HA(aq) Kb = [OH-][HA]/[A-]
2H2O(l) ⇋ OH-(aq) + H3O+(aq)
● Thus, adding the above two chemical equations yields the equation for the
autoionization for water. The equilibrium constant for a summed reaction is
equal to the product of the equilibrium constants for the added reactions,
and so
Ka x Kb = ([A-][H3O+]/[HA]) x ([OH-][HA]/[A-])
Ka x Kb = [H3O+][OH-] = KW
● For any conjugate acid–base pair, the product of the acid-dissociation
constant for the acid and the base-dissociation constant for the base always
equals the ion-product constant for water:
Ka x Kb = Kw

71. …Cont’d
● The stronger the acid, the weaker the conjugate base. The
stronger the base, the weaker the conjugate acid.
Exercises: 1) Determine the conjugate base for each of the
following acids:
a) HClO(aq) b) NH4
+(aq) c) HSO4
-(aq) d) OH-
2) What are the conjugate acids of
a) CN−(aq) b) HSO3
–(aq) c) SO3
2−(aq)? d) OH-
3) Given that Ka = 6.3 x 10-4 M at 25 oC for the acid HF(aq),
calculate Kb and pKb for F-(aq).
4) The values of Kb for ammonia (NH3) and methylamine
(CH3NH2) are 1.8 x 10-5 and 4.38 x 10-4, respectively.
Calculate the Ka values for the corresponding conjugate
acids.

72. Table 3: Relative strengths of conjugate acid–base Pairs
●

73. Table 4: Relative strengths of select conjugate acid-aase Pairs

74. 6.6.4 Predicting the direction of acid–base reactions
● When beginning with equal concentrations of reactants and
products, the proton is always transferred to the stronger base.
● This means that the direction of reaction to reach equilibrium is
proton transfer from the stronger acid to the stronger base to give
the weaker acid and the weaker base:
Stronger acid + Stronger base → Weaker acid + Weaker base
Example: Decide which species (reactants or products) are favored
at the completion of the reaction.
SO4
2-(aq) + HCN(aq) ⇋ HSO4
-(aq) + CN-(aq)
Solution: If you compare the relative strengths of the two acids
HCN and HSO4
-, you see that HCN is weaker. Or, comparing the
bases SO4
2- and CN-, you see that SO4
2- is weaker. Hence, the
reaction would normally go from right to left.

75. …Cont’d
The reactants are favored. Check that the stronger acid has a
weaker conjugate base, and vice versa. The reaction goes from
stronger acid and base to weaker acid and base.
Exercises: 1) Determine whether each of the following reactions
favors the reactants, products, or neither.
a) NaOH(aq) + KCl(aq) ⇋ NaCl(aq) + KOH(aq)
b) HF(aq) + NaOH(aq) ⇋ NaF(aq) + H2O(l)
c) CH3COONa(aq) + H2O(l) ⇋ CH3COOH(aq) + NaOH(aq)
d) NH4
+(aq) + H2O(l) ⇋ NH3(aq) + H3O+(aq)
Answer: a) neither b) products c) reactants d) reactants
2) Rank the following species in order of increasing acidity:
a) NH4
+, H2O, HF, HSO4
- b) CH3COOH, H2O, HCOOH, F-
Answer: a) H2O < NH4
+ < HF < HSO4
-
b) F- < H2O < CH3COOH < HCOOH
-2 - -
4 4
weaker acid stronger base
weaker base stronger acid
SO (aq) + HCN(aq) HSO (aq) + CN (aq)


76. 6.6.5 Leveling effect of solvents
● Leveling effect is the effect that makes strong acids (or bases)
appear equal in strength in a given solvent because they ionize
or dissociate completely.
● Acids that are stronger than H3O+ quantitatively transfer
their protons to the water to form H3O+. We cannot
measure any differences in the acidities of the strong
acids (HCl, HBr, HI, HNO3, H2SO4, HClO4) in water,
because they all ionize completely.
● This phenomenon is called the leveling effect—the
solvent makes the strong acids appear equal, or level,
in acidity.
● Strong acids will yield identical H3O+ concentrations in
water, independent of their individual Ka values.

77. …Cont’d
● Strong bases are leveled in a similar manner. All
strong bases react stoichiometrically to form
hydroxide ion, which is the strongest base that can
exist in water. The strongest base that can exist in
aqueous solution is hydroxyl ion (OH-)
● Acting as a base, water levels the strength of all
strong acids by making them appear equally strong,
and acting as an acid, it levels the strength of all
strong bases as well.

78. …Cont’d
● Acids that are equally strong in water show
differences in strength when dissolved in a solvent
that is a weaker base than water, such as acetic acid
and liquid HF.
● To measure the differences in acid strength for
“strong” acids, the acids must be dissolved in a
solvent that is less basic than water. In such solvents,
the acids will be “weak,” and so any differences in
the extent of their ionization can be determined.
● For example, the binary hydrogen halides HCl, HBr,
and HI are strong acids in water but weak acids in
ethanol (strength increasing
HCl < HBr < HI).

79. Factors affecting acid strength
● Three factors affect the acidity of the X-H bond in an acid: 1) the
polarity of the bond, 2) the size of the atom X, and 3) the charge
on the ion or molecule. A fourth factor has to be considered to
understand the acidity of the X-OH group in a family of
compounds known as “oxyacids.”
● Within any group in the periodic table, the acidity of the hydrides
(HX) increases from top to bottom with an increase in atomic
size of X.
Example: a) HF <HCl < HBr < HI b) H2O < H2S < H2Se < H2Te
● Across a period in the periodic table, acids become stronger as
the electronegativity of X increases and the H¬X bond becomes
more polar
Example: CH4 < NH3 < H2O < HF

80. …Cont’d
● Acids in which the acidic hydrogen is bonded directly to oxygen
in an H-O- bond are called oxoacids. Three of the strong acids-
nitric, HNO3; perchloric, HClO4; and sulfuric, H2SO4-are
oxoacids.
● The strengths of the oxoacids HOX (H–O–X) increase with
increasing electronegativity of X. The strengths of the oxoacids
(HO)mXOn increase with increasing n, the number of oxygen
atoms that are not bound to hydrogen atoms.
Example: a) HOI < HOBr < HOCl
b) HClO <HClO2 < HClO3 < HClO4
c) H2SO4 > H2SO3 d) HNO3 > HNO2
● The charge on a molecule or an ion can influence its ability to act as an
acid or a base. Compounds become less acidic and more basic as the
negative charge increases.
Basicity: H2PO4
- < HPO4
2- < PO4
3-
Acidity: H3PO4 > H2PO4
- > HPO4
-2

81. 6.7 Buffer solution
● A buffer solution is a solution that contains a mixture of a weak acid
its salt or a mixture of a weak base and its salt.
Example: a) CH3COOH/CH3COONa, HCN/NaCN, HF/KF
b) NH3/NH4Cl
● The solution has the ability to resist changes in pH upon the addition of
small amounts of either acid or base.
● Buffers are very important to chemical and biological systems.
● The pH of blood is about 7.4, whereas the gastric juice in our stomachs
has a pH of about 1.5.
● These pH values, which are crucial for proper enzyme function and the
balance of osmotic pressure, are maintained by buffers in most cases.

82. 6.7.1 How buffers work: The common-ion effect
● Let us consider a buffer made from a weak acid, HA, and its conjugate base,
A-. If hydrogen ions are added to our buffer, they react with the base:
H3O+ + A- → H2O + HA
● The equilibrium constant for this reaction is quite large; a large equilibrium
constant means that the reaction will proceed nearly to completion. The
reaction between the added hydrogen ions and the weak base effectively
removes the added hydrogen ions from the solution, thus adding hydrogen
ions to a buffer solution generally does not cause a large change in the pH of
the solution.
● Similarly, the weak acid in the buffer reacts with any added hydroxide ions:
OH- + HA → H2O + A-
● This reaction also goes to completion. The weak acid in the buffer removes
excess hydroxide ions from the solution, thus adding hydroxyl ions to a
buffer solution will not cause a large change in the pH of the solution.
● The equilibrium constants for the neutralizations of both added hydrogen ions
and added hydroxide ions are large, so these reactions go to completion and
the system tends to keep the pH nearly constant; thus, it is a buffer system.

83. ...Cont’d
● Buffer capacity of a buffer is the amount of strong acid or base that
can be added before the pH of the solution changes significantly.
● The buffer capacity is a measure of the amount of acid or base that can
be effectively neutralized by the buffer.
● The maximum buffering capacity is obtained when the concentrations
of the weak acid and its salt (or the weak base and its salt) are equal.

84. 6.7.2 Henderson–Hasselbalch equation
● This equation is used for calculating pH of a buffer solution.
● For a solution consisting of a weak acid and a salt of the weak acid:
pH = pKa + log([salt]/[acid]
where pKa refers to the weak acid.
● For a solution consisting of a weak base and a salt of the weak acid:
pOH = pKb + log([salt]/[base]
where pKb refers to the weak base.
● In general, pH of a buffer solution is given by
pH = pKa + log([base]/[acid]
● In general, if Ka for the acid is larger than 1x10-7, the buffer will be
acidic. If Kb is larger than 1x10-7, the buffer is basic.

85. 6.7.3 Preparing a buffer solution of specified pH
● To prepare a buffer solution of a given pH, the procedure is as follows:
i) Choose a weak acid with a pKa value close to the required pH of the
buffer; the weak acid may be the salt of a polybasic acid, e.g.
NaH2PO4.
ii) Choose an appropriate salt of the weak acid.
iii) Use the Henderson–Hasselbalch equation to determine the
[base]/[acid] ratio needed to attain the correct pH.
iv) Remember that for maximum buffering capacity, [base]/[acid] = 1.
Exercise: A buffer solution of pH 7.23 is required. Choose a suitable
weak acid and calculate the ratio of [base]/[acid] required. Suggest what
acid and base combination might be appropriate.
Answer: Possible components for the buffer solution are NaH2PO4
(acid) and Na2HPO4 (base) with [base]/[acid] ratio of 1.05.

86. ...Cont’d
Example: Consider 100 mL of an acetate buffer that is 0.120 M acetic
acid and 0.120 M sodium acetate. a) Calculate the pH of the buffer
b) Calculate the change in pH observed when 5.0 mL of 0.050 M HCl is
added to 100.0 mL of the buffer
c) Calculate the change in pH observed when 5.0 mL of 0.050 M NaOH
is added to 100.0 mL of the buffer. Ka of HAc = 1.8x10-5
Solution: a) pH = pKa + log([salt]/[acid]
pH = pKa + log([CH3COO-]/[CH3COOH]
pH = -log(1.8x10-5) + log(0.120/0.120) = 4.74
b) H3O+ + CH3COO- → H2O + CH3COOH
mmol of acid added = 0.050 M x 5.0 mL = 0.25 mmol
mmol of CH3COO- after adding acid = 0.120 Mx100 mL- 0.050 M x 5.0 mL
mmol of CH3COO- after adding acid = 11.75 mmol
[CH3COO-] after adding acid = 11.75 mmol/(100 mL + 5 mL) = 0.1119 M

87. ...Cont’d
mmol of CH3COOH after adding acid = 0.120 Mx100 mL + 0.050 M x 5.0 mL
mmol of CH3COOH after adding acid = 12.25 mmol
[CH3COOH] after adding acid = 12.25 mmol/(100 mL + 5 mL) = 0.1167 M
pH = pKa + log([CH3COO-]/[CH3COOH]
pH = -log(1.8x10-5) + log(0.1119/0.1167) = 4.72
c) OH- + CH3COOH → H2O + CH3COO-
mmol of base added = 0.050 M x 5.0 mL = 0.25 mmol
mmol of CH3COOH after adding base = 0.120 Mx100 mL - 0.050 M x 5.0 mL
mmol of CH3COOH after adding base = 11.75 mmol
[CH3COOH] after adding base = 11.75 mmol/(100 mL + 5 mL) = 0.1119 M
mmol of CH3COO- after adding base = 0.120 Mx100 mL+ 0.050 M x 5.0 mL
mmol of CH3COO- after adding base = 12.25 mmol
[CH3COO-] after adding base = 12.25 mmol/(100 mL + 5 mL) = 0.1167 M
pH = pKa + log([CH3COO-]/[CH3COOH]
pH = -log(1.8x10-5) + log(0.1167/0.1119) = 4.76

88. Exercises
1) Calculate the pH of solutions that is 0.0250 mol sodium nitrite,
NaNO2, in 250.0 mL of 0.0410 M nitrous acid, HNO2.
2) Calculate the pH that results when the following solutions are mixed.
a) 25.0 mL of 0.250 M HF and 20.0 mL of 0.360 M NaF
b) 20.0 mL of 0.144 M NH3 and 10.0 mL of 0.152 M NH4Cl
c) 350.0 mL of 0.150 M pyridinium chloride and 650.0 mL of 0.450
M pyridine
3) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system.
What is the pH after the addition of 20.0 mL of 0.050 M HCl to 80.0
mL of the buffer solution? What is the pH after the addition of 20.0
mL of 0.050 M NaOH to 80.0 mL of the buffer solution?
4) Which of the following solutions can act as buffer systems:
a) KH2PO4/H3PO4 b) NaClO4/HClO4, c) C5H5N/C5H5NHCl
(C5H5N is pyridine d) KHSO4/H2SO4, e) Na2HPO4/NaH2PO4
f) KNO2/HNO2 g) KHSO3/K2SO4 h) Na2SO4/NaHSO4
i) NH3/NH4NO3 j) NaI/HI?