Ionic equilibrium chapter 3(12th HSC Maharashtra state board)

z
Ionic equilibrium
Presentedby:FreyaCardozo
1

z
Electrolyte
 The substances which give rise to ions when dissolved in water are electrolytes.
 The non electrolytes are those which do not ionize and exist as molecules in
aqueous solutions.
 Electrolytes can be classified as
1. Strong electrolyte : The electrolytes Ionizing completely or almost completely
are strong electrolytes. For example : strong acids, strong bases and salts.Eg.HCl
2. Weak electrolytes :The electrolytes which dissociate to a smaller extent in
aqueous solution are weak electrolytes. Weak acids and weak bases belong to
this class. Eg. CH3COOH
2

z
Representation
We represent strong electrolytes using 
Since weak electrolytes do not completely ionize we represent it by double
headed arrow showing equilibrium
3

z
Degree of dissociation (∝)
 The degree of dissociation of an electrolyte is defined as a
fraction of total number of moles of the electrolyte that
dissociates into its ions when the equilibrium is attained.
 If 'c' is the molar concentration of an electrolyte the equilibrium
concentration of cation or anion is (∝ × c) mol dm-3.
4

z
Acid base theories

 Arrehnius theory
 Bronsted-Lowry theory.
 Lewis theory
5

z
Arrhenius theory
 Acid- Donates H+ ion
 Base- Donates OH- ion
6

z
Drawback
It does not account for the basicity of NH3
and Na2CO3 which do not have OH group.
7

z
Bronsted-Lowry theory -proton transfer
theory.
 Acid -Acid is a substance that donates a proton (H⊕) to another
substance.
 Base-Base is a substance that accepts a proton (H⊕) from
another substance.
8

z
Conjugate acid- bases
The base produced by accepting the proton from an acid is the conjugate base
of that acid. Likewise the acid produced when a base accepts a proton is called
the conjugate acid of that base. A pair of an acid and a base differing by a proton
is said to be a conjugate acid-base pair.
9

z
Lewis – electron pair Theory
 Acid : Any species that accepts an electron pair is called Lewis acid.
 Base : Any species that donates an electron pair is called Lewis base.
10

z
Amphoteric nature of water
 Water has the ability to act as an acid as well as a base. Such
behaviour is known as amphoteric nature
 Reaction
11

z
Ionisation of weak and strong acid and
bases
12

z
Dissociation constant of weak acids
and weak bases
13

z Ostwald’s dilution law
 For weak acids
Consider an equilibrium of weak acid HA that exists in solution partly as the undissociated species HA
and partly H⊕ and A- ions.
Suppose, 1 mol of acid HA is initially present in volume V dm3 of the solution
Then at equilibrium fraction dissociated would be ∝, where ∝ is degree of dissociation of the acid.
The fraction of an acid that remains undissociated would be (1 - ∝).
15

z
c=n/V
16

z
degree of dissociation of a weak acid is
inversely proportional to the square root of its
concentration or directly proportional to
the square root of volume of the solution
containing 1 mol of the weak acid.
18

z
For weak base
19

z
Type 1 numericals- Based on Ostwalds
law

K= Dissociation constant
C= concentration*( in M)
Alpha= degree of dissociation
22

z
Solve
A weak monobasic acid is 0.05% dissociated in 0.02 M solution.
Calculate dissociation constant of the acid
Given : c= 0.02M
%alpha= 0.05 alpha=0.05/100
To find : Ka=?
Formula:
Solution:
23

z
Solve
The dissociation constant of
NH4OH is 1.8 × 10-5. Calculate its degree
of dissociation in 0.01 M solution
Given :Kb = 1.8 × 10-5
c=0.01M
To find: alpha=?
TIP: The value for for alpha should always be less than 1
24

z
Questions
A weak monobasic acid is
12% dissociated in 0.05 M solution. What
is percent dissociation in 0.15 M solution.
26

z
Calculate [H3O⊕] in 0.1 mol dm3 solution of acetic acid.
Given : Ka [CH3COOH] = 1.8 × 10-5
27

z
Concept of pH and
pOH
28

z
Autoionization of water
Pure water ionizes to a very small extent. The ionization equilibrium
of water is represented as,
29

z
A majority of H2O molecules are undissociated, consequently concentration of water
[H2O] can be treated as constant
Kw = KK' is called ionic product of water. The product of molar concentrations of
hydronium (or hydrogen) ions and hydroxyl ions at equilibrium in pure water at the
given temperature is called ionic product of water.
30

z
pH scale
 Instead of expressing acidity in terms of hydronium ion conc we use
this scale
 Given by Sorenson
 Definition:
pH of a solution as the negative logarithm to the base 10, of the
concentration of H⊕ ions in solution in mol dm-3
 Mathematical expression
32

z
 pOH of a solution can be defined as the negative logarithm to the
base 10, of the molar concentration of OH ions in solution.
Thus, pOH = -log10[OH- ]
33

z
Relationship between pH and pOH
34

z
Types of solutions based on pH
Neutral solutions
For pure water or any aqueous neutral solution at 298 K
[H3O⊕] = [OH -] = 1.0 × 10-7 M
Hence, pH = -log10[H⊕] = -log10[1 × 10-7] = 7
Acidic solutions
In acidic solution, there is
Excess of H3O⊕ ions, or [H3O⊕] > [OH ] Hence, [H3O⊕] > 1 × 10-7 and pH < 7
Basic solutions
In basic solution, the excess of OH ions are present that is [H3O⊕] < [OH ] or
[H3O⊕] < 1.0 × 10-7 with pH > 7.
35

z
Type2 numericals-
Based on pH and pOH
36

z
Questions
Calculate pH and pOH of 0.01 M HCl solution.
37

z
pH of a solution is 3.12.
Calculate the concentration of H3O⊕ ion.
pH= -log[H+]
-3.12= log[H+]
-3-0.12+1-1= log[H+]
-4+0.88= log[H+]
-4.88= log[H+]
[H+]=7.586 x 10-4
38

z
A weak monobasic acid is
0.04 % dissociated in 0.025M solution.
What is pH of the solution ?
Given : alpha= 0.04/100
c=0.025M
To find: pH=?
[H+]= alpha x c= 0.0004x 0.025= 10-5
pH=5
39

z
The pH of monoacidic weak base is 11.2. Calculate its percent
dissociation in 0.02 M solution.
40

z
Mcq
 The pH of 10-8 M of HCl is
1. 8
2. 7
3. less than 7
4. greater than 7
 For pH > 7 the hydronium
ion concentration would be
1. 10-7
2. < 10-7 M
3. > 10-7 M
4. ≥ 10-7 M
41

z
Concept of Hydrolysis
42

z
Hydrolysis of water
 When a salt is dissociated in water, it dissociates completely into its constituent
ions.
 The solvent water dissociates slightly as,
Pure water is neutral and [H3O⊕] = [OH -].
 If the ions of the salt do not interact with water, the hydronium and hydroxyl ion
concentrations remain equal and the solution is neutral
43

z
 But when anions or cations of the salt dissociate and interact with the
ions of water either the hydronium ions or hydroxyl ions concentration
increases , making the solution acidic or basic
 Such a reaction between the ions of salt and the ions of water is called
hydrolysis of salt.
 Hydrolysis of salt is defined as the reaction in which cations or anions
or both ions of a salt react with ions of water to produce acidity or
alkalinity (or sometimes even neutrality).
44

z
Acidic and bases
Strong acids Strong bases Weak acids Weak bases
HCl NaOH CH3COOH NH4OH
H2SO4 KOH HF HCN
HNO3 Cu(OH)2
45

z
4 different types of Salts
 Strong acid and strong base
 Strong acid and weak base
 weak acids and strong base
 weak acids and weak base
46

z
Salt of strong acid and base - NaCl
 salt of strong acid HCl and a strong base NaOH. Both are electrolytes and
completely dissociate
 NaCl  Na+ + Cl-
 The ions Na⊕ and Cl- have no tendency to react with water because the product
formed after reactn are strong Electrolytes And they completely dissociate to ions
47

z
 reactants and the products are the same.
 This implies that neither the cation nor anion of the salt reacts with water or
there is no hydrolysis.
 Equality H3O⊕ = OH- produced by ionization of water is not disturbed and
solution is neutral
48

z
Strong acid weak base CuSO4
 From acid H2SO4 and base Cu(OH)2
 It dissociate to give following ions
 sulphate ions would react with water to form H2SO4 that completely dissociate so they won’t
make any difference in the pH
 But the Cu2+ ions will react with the water as follows
 Due to the presence of excess of H3O⊕ions, the resulting solution of CuSO4 become acidic and
turns blue litmus red.
49

z
 Give reason why To get clear solution of CuSO4, the addition of H2SO4
would be required.
1. Formation of sparingly soluble Cu(OH)2by hydrolysis makes the aqueous
solution of CuSO4 turbid.
2. If H2SO4, that is H3O⊕ ions are added, the hydrolytic equilibrium shifts to
the left.
3. A turbidity of Cu(OH)2dissolves to give a clear solution. To get clear solution
of CuSO4, the addition of H2SO4 would be required.
50

z
Question
Why an aqueous solution of NH4Cl is acidic while that of HCOOK basic ?
51

z
Salts of weak acids and strong bases-
CH3COONa
 Acid is CH3COOH and base is NaOH
 It dissociate to form Na+ ions which donot react with water beacuse strong
Electrolyte NaOH will form
 However the CH3COO- ions react
 As a result of excess OH ions produced the solution becomes basic. The solution
of CH3COONa is therefore basic.
52

z
Salts of weak acids and weak base
 When salt BA of weak acid HA and weak base BOH is dissolved in water, it
dissociates completely as
BA(aq)  B⊕(aq) + A (aq)
 The hydrolysis reaction involves the interaction of both the ions of the salt with
water,
53

z
Salt of weak acid and weak base for which Ka > Kb
Eg.NH4F
 weak acid HF and weak base NH4OH
 Ka =7.2 × 10-4 Kb=1.8× 10-5
 NH4 hydrolyses to a higher extent and thus Conc of H+ is more making soltn acidic
NH4⊕ ions are slightly stronger as acid than F ions as base.
 The solution of NH4F is thus only slightly acidic and turns blue litmus red
54

z
Salt of weak acid and weak base for which
Ka< Kb Eg.NH4CN
 Acid- HCN and base – NH4OH
 Ka = 4.0 × 10-10 Kb=1.8×10-5
 When NH4CN is dissolved in water, it hydrolyses as
 The CN ions hydrolyse to a greater extent than NH4⊕ ions, The reaction produces more
OH ions than the H3O⊕ ions produced in reaction. The solution of NH4CN is, basic and
turns red litmus blue.
55

z
Salt of weak acid and weak base for
which Ka = Kb CH3COONH4.
 Acid- CH3COOH and base- NH4OH
 Ka= 1.8 × 10-5 and Kb= 1.8× 10-5
As Ka = Kb, the relative strength of
acid and
base produced in hydrolysis is the same.
Therefore, the solution is neutral.
56

z
Overview
Type Example pH
Strong acid and strong
base
NaCl Neutral
Strong acid and weak
base
CuSO4 Acidic
Weak acid and strong
base
CH3COONa Basic
Weak acid and weak
base
Ka>Kb NH4F Acidic
Ka<Kb NH4CN Basic
Ka=Kb CH3COONH4 Neutral
57

z
Buffers
58

z
What are buffers
 Buffer solution is defined as a solution which resists drastic changes in pH when a
small amount of strong acid or strong base or water is added to it.
 Buffers are composed of 2 substance a weak acid/base + corresponding salt of
weak acid/base with strong acid/base
 types of buffers
1. Acidic buffer
A solution containing a weak acid and its salts with strong base is called an acidic
buffer solution. Eg. CH3COOH and CH3COONa
2. Basic buffer
A solution containing a weak base and its salt with strong acid is the basic buffer
solution. Eg.NH4OH and NH4Cl
59

z
Handerson Hasselbalch equation
60

z
Buffer action*:Sodium acetate acetic
acid buffer
 Weak acid -CH3COOH
 salt- CH3COONa (made of weak acid CH3COOH and strong base NaOH)
 CH3COONa is a strong Electrolyte thus dissociates completely
 Due to dissociation there is large concentration of CH3COO- as it dissociates
completely
 acetic acid is a weak acid, the concentration of undissociated CH3COOH
molecules is usually high
62

z
 If a small amount of strong acid is added to this solution the added H⊕ ions will be
consumed by the conjugate base CH3COO- present in large concentration.
 if small amount of base is added, the added OH ions will be neutralized by the large
concentration of acetic acid
 The acid or base added thus can not change the [H⊕] or [OH ] concentrations and, pH
of the buffer remains unchanged.
63

z
Properties of buffer
The pH of a buffer solution does not change appereciably
i. by addition of small amount of either strong acid or strong base,
ii. on dilution ( because the Conc terms in the H.H eqn do not
involve water)
iii. When it is kept for long time.
64

z
Application
 In biochemical system:
pH of blood in our body is maintained at 7.36 - 7.42 due to (HCO3 + H2CO3) buffer. A mere
change of 0.2 pH units can cause death.
 Agriculture
Soil contains carbonates, phosphate etc that buffer soil pH .choice of fertilizers depends
upon pH of soil.
 Industry
Buffers play an important role in paper, dye, ink, paint and drug industries.
 Medicine
Penicillin preparations are stabilized by addition of sodium citrate as buffer. When citric acid
is added to milk of magnesia (Mg(OH)2), magnesium citrate is formed, which is a buffer.
65

z
 In qualitative analysis
a pH of 8 to 10 is required for precipitation of cations IIIA group. It
is maintained with the use of (NH4OH + NH4Cl) buffer.
66

z
Type 3 numericals
Based on Henderson-Hasselbach equation
 Ka/Kb Dissociation constant for acid / base
 pKa= -logKa
 [salt] concentration of salt
 [acid]/[base]: Tip- To know if it is salt or acid see which has H+
67

z
Type 3 Numericals
Calculate the pH of buffer solution containing 0.05 mol NaF per litre
and 0.015 mol HF per litre. [Ka = 7.2 × 10-4for HF]
68

z
Calculate the pH of buffer solution composed of 0.1 M weak base BOH
and 0.2 M of its salt BA. [Kb = 1.8× 10-5 for the weak base]
69

z
Which of the following solution will have pH value equal to 1.0 ?
a. 50 mL of 0.1M HCl + 50mL of 0.1M NaOH
b. 60 mL of 0.1M HCl + 40mL of 0.1M NaOH
c. 20 mL of 0.1M HCl + 80mL of 0.1M NaOH
d. 75 mL of 0.2M HCl + 25mLof 0.2M NaOH
70

z
Blood in human body is highly buffered at pH of
a. 7.4 b. 7.0
c. 6.9 d. 8.1
Which of the following is a buffer solution ?
a. CH3COONa + NaCl in water
b. CH3COOH + HCl in water
c. CH3COOH+CH3COONa in water
d. HCl + NH4Cl in water
71

z
Solubility
product
72

z
Solubility
 The solubility of a compound is the amount in grams that
dissolves per unit volume (which may be 100 mL or 1L of its
saturated solution).
 Depending on solubility we can classify salts as
1. soluble
2. insoluble
3. Sparingly soluble
73

z
Solubility equilibria
 Suppose some powdered sparingly soluble salt such as AgCl is put into water
and stirred vigorously. A very small amount of AgCl dissolves in water to form
its saturated solution. Most of the salt remains undissolved. Thus, solid AgCl is
in contact with its saturated solution.
 AgCl is a strong electrolyte. Hence the quantity of AgCl that dissolves in water
dissociates completely into its constituent ions, Ag⊕ and Cl- .
 A dynamic equilibrium exists between undissolved solid AgCl and the
dissolved ions, Ag⊕ and Cl- , in the saturated Solution.
 This equilibrium, called solubility equilibrium
74

z
Defn-Solubility product
In the saturated solution of sparingly soluble salt the product of equilibrium
concentrations of the constituent ions raised to the power equal to their respective
coefficients in the balanced equilibrium expression at a given temperature is called
solubility product.
77

z
Relationship between solubility and
solubility product
 Molar solubility : The number of moles of a compound that dissolve to
give one litre of saturated solution is called its molar solubility.
80

z
Type 4 numericals
Based on S and Ksp
82

z
Calculating ksp
83

z
Questions
What is the relationship between molar solubility and solubility
product for salts given below
i. Ag2CrO4
ii. Ca3(PO4)2
iii. Cr(OH)3
85

z
The solubility product of AgBr is 5.2 × 10-13. Calculate its solubility in mol
dm-3 and g dm-3(Molar mass of AgBr = 187.8 g mol-1)
86

z
The solubility product of a sparingly
soluble salt AX is 5.2×10-13. Its
solubility in mol dm-3 is
a. 7.2 × 10-7 b. 1.35 × 10-4
c. 7.2 × 10-8. d. 13.5 × 10-8
87

z
Conditions for precipitation
 Defn- solubility product Ksp : Thus, in the saturated solution of sparingly soluble salt the
product of equilibrium concentrations of the constituent ions raised to the power equal to
their respective coefficients in the balanced equilibrium expression at a given temperature is
called solubility product
 Defn- ionic product IP : Similar to Ksp just difference is, it is product of concentration Of ion
under any conditions
 Ksp specifically says conc of ions at equilibrium
 condition
1. IP = Ksp; the solution is saturated and solubility equilibrium exists.
2. IP > Ksp; the solution is supersaturated and hence precipitation of the compound will occur.
3. If IP < Ksp, the solution is unsaturated and precipitation will not occur.
88

z
Type 5 numericals
Based on I.P and Ksp
Precipitation will occur or no?
 If I.P>Ksp then precipitation will occur
 I.P product of concentration of ions at any time whereas Ksp is at
equilibrium
89

z
A solution is prepared by mixing equal volumes of 0.1M MgCl2 and 0.3M
Na2C2O4 at 293 K. Would MgC2O4 precipitate out ? Ksp of MgC2O4 at 293
K is 8.56 × 10-5.
90

z
If 20.0 cm3 of 0.050 M Ba(NO3)2 are mixed with 20.0 cm3
of 0.020 M NaF, will BaF2 precipitate ? Ksp of BaF2 is 1.7 × 10-6 at
298 K.
91

z
Common ion
effect
92

z
Common ion effect
 Consider a solution of weak acid CH3COOH and its soluble ionic salt
CH3COONa.
 CH3COOH is weak electrolyte so dissociates partially
 CH3COONa is strong Electrolyte thus completely dissociates
93

z
 Both the acid and the salt produce CH3COO- ions in solution.
 CH3COONa Dissociates completely. Therefore it provides high
concentration of CH3COO- ions.
 According to Le-Chatelier principle, the addition of CH3COO- from
CH3COONa to the solution of CH3COOH, shifts equilibrium of dissociation
of CH3COOH to left.
 Thus reverse reaction is favoured in which CH3COO- combines with H⊕ to
form unionised CH3COOH.
 Hence dissociation of CH3COOH is supressed due to presence of
CH3COONa containing a common CH3COO- ion.
94

z
Defn – common ion effect
The common ion effect states that the ionisation of a weak electrolyte is
suppressed in presence of a strong electrolyte containing an ion common to
the weak electrolyte.
95

z
Solubility and common ion effect
The presence of a common ion also affects the solubility of a sparingly soluble
salt. Consider, the solubility equilibrium of AgCl,
96

z
 Suppose AgNO3 is added to the saturated solution of AgCl. The salt AgNO3
being a strong electrolyte dissociates completely in the solution.
The dissociation of AgCl and AgNO3 produce a common Ag⊕ ion. The
concentration of Ag⊕ ion in the solution increases due to complete dissociation
of AgNO3
97

z
 According to to Le-chatelier's principle the addition of Ag⊕ ions
from AgNO3 to the solution of AgCl shifts the solubility
equilibrium of AgCl from right to left.
 Thus AgCl precipitates
 Therefore presence of common ion decrease solubility Of
sparingly soluble salt
 Ksp remains the same
98

z
Questions
 Why cations are Lewis acids ?
 Why is KCl solution neutral to litmus?
 The dissociation of H2S is suppressed in the presence of HCl.
Name the phenomenon.
 Acetic acid is 5% ionised in its decimolar solution. Calculate the
dissociation constant of acid
99

Ionic equilibrium chapter 3(12th HSC Maharashtra state board)

More Related Content

What's hot

Similar to Ionic equilibrium chapter 3(12th HSC Maharashtra state board)

More from Freya Cardozo

Recently uploaded

Ionic equilibrium chapter 3(12th HSC Maharashtra state board)

Editor's Notes