IMPULSE PU SCIENCE COLLEGE, HUBLI
Unit 7
EQUILIBRIUM
Equilibrium represents a state in a process when the observable properties such as temperature,
pressure and concentration do not change with time.
The equilibrium may be established in physical changes and chemical changes. Therefore equilibrium
may be classified as physical equilibrium and chemical equilibrium.
Physical equilibrium: equilibrium in physical process
The types of equilibria exist in three states of matter:
1. Solid ⇌ Liquid
2. Liquid ⇌ Gas
3. Solid ⇌ Gas
1. Solid ⇌ Liquid equilibrium:
When ice and water kept in insulated closed vessel at constant temperature. (273K) and atmospheric pressure.
There is no change of mass of ice and water because rate of conversion of ice into water and from water to ice are
equal at 273 K and atmospheric pressure. This stage is known as solid-liquid equilibrium.
It is a stage in a reversible process at which rate of melting of ice (solid) is equal to rate of
freezing of water (liquid).
Example: H2O(s)
⇌
H2O(l)
(273 K and 1 atm)
(rate of melting of ice = rate of freezing of water)
2. Liquid ⇌ Gas equilibrium:
It is a stage in a reversible process at which rate of evaporation of liquid is equal to rate of
condensation of vapours.
Example: H2O(l)
⇌ H2O(g)
(373 K and 1 atm)
( rate of evaporation = rate of condensation)
a. For any liquid at one atmospheric pressure (1.013 bar ), the temperature at which the liquid and
vapours are at equilibrium is called normal boiling point of the liquid. The normal boiling point of
water is 1000 C at 1atm (1.013 bar).
3. Solid ⇌ Gas equilibrium: It is a stage in a reversible process at which rate of sublimation of solid is
equal to rate of condensation of vapours.
Example:
I2(solid)
⇌
I2(vapour)
(rate of sublimation of solid = rate of condensation of vapours)
NH4Cl(s) ⇌
NH4Cl(vapour)
Camphor(s) ⇌ Camphor (vapour)
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Equilibrium involving Dissolution of Solid or Gases in Liquid
1. Solids in Liquids: It is a stage in a reversible process at which rate of dissolution of solid is equal to
rate of crystallization of solid in a saturated solution at given temperature.
Example:
Sugar (solution) ⇌ Sugar(solid)
(rate of dissolution of sugar = rate of crystallization of sugar)
NaCl (solution) ⇌ NaCl(solid)
2. Gases in Liquids: It is a state of equilibrium between gas molecules dissolved in solution and
vapours of gas above liquid surface at given temperature.
Example : CO2(gas) ⇌ CO2(in solution)
This equilibrium is governed by Henry’s law, which states that
“at any given temperature, the mass of a gas dissolved in a given mass of a solvent is proportional
to the pressure of the gas above the solvent”.
mαp
Characteristics of equilibrium in physical process:
1.
2.
3.
4.
It is possible only in a closed system at a given temperature
It is dynamic ie opposing processes occurs at same rate.
It is characterized by constant value of one of its parameters at a given temperature.
The magnitude of quantities at any stage indicates the extent to which process proceeded before
reaching equilibrium.
There are two types of reactions. They are
1) Reversible reactions: A reaction in which products of the reaction react with each other to form
back the original reactants OR a reaction which can occur in both forward and backward directions is
called reversible process. It is represented by ⇌.
Example : A + B ⇌
C+D
Dissociation of hydrogen iodide: 2HI ⇌ H2 + I2
Synthesis of ammonia by Haber’s process: N2+ 3 H2 ⇌ 2NH3
2) Irreversible reaction: A reaction which can occurs only in forward direction but not in backward
direction is called irreversible reaction.
Ex:
A+B
C+D
NaOH + HCl
NaCl(aq) + AgNO3(aq)
NaCl + H2O
AgCl(s) + NaNO3(aq)
Equilibrium in chemical process
Consider a reversible reaction A + B ⇌ C + D is conducted in a closed vessel at constant temperature.
In the beginning of reaction, the concentration of A and B are maximum and concentration of C
and D are zero. And also (rate) velocity of forward reaction (v) is maximum and velocity (rate)
of backward reaction is zero. As reaction proceeds, the concentration of A and B decrease,
where as the concentration of C and D increases leading to decrease in the rate of forward
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reaction and an increase in the (rate) velocity of backward reaction. After some time, velocity of
forward reaction is equal to velocity of backward reaction. i.e, V1 = V2. The system reaches a
state of chemical equilibrium where there is no change in composition.
Chemical Equilibrium: It is a state of a reversible reaction, in which the rate of forward reaction
is equal to the rate of backward reaction.
Justify chemical equilibrium is dynamic:
Any reversible reaction at equilibrium, the reaction seems to be stopped, because the concentration of
reactant and product do not change with time in this state. But actually the forward and backward
reaction occurs at same rate. So there is a molecular activity.
Hence equilibrium is dynamic.
Molar concentration or active mass:
It is defined as “number of moles of substances present in 1 dm3 of a solution”.
The unit of a molar concentration is mol/dm3. The symbol is [ ].
Law of mass action: This was formulated by Guldberg and Waage
It states that “at constant temperature rate of a reaction is directly proportional to the product of
molar concentration of reactants raised to the power of respective number of moles”.
Consider a reaction A + B ⇌ Products.
Applying law of mass action
Rate of reaction 𝛼 [A] [B]
𝜈 = k[A] [B]
Where k is rate or velocity constant
Note: Applying law of mass action for the following reactions
a) N2 + 3H2 ⇌
2NH3
3
𝜈 =k [N2] [H2]
b) 2HI ⇌ H2 + I2
𝜈 =k [HI]2
Derive equilibrium constant expression (Kc) for a reaction.
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Consider a reversible reaction aA + bB ⇌
cC + dD is conducted in a closed vessel at constant
temperature. Let [A], [B], [C] and [D] be the molar concentration of reactant and product A, B, C & D
respectively and let Rf and Rb be the rate of forward and backward reaction respectively.
Applying law of mass action to the forward reaction.
Rate of forward reaction (Rf ) 𝛼 [A]a [B]b
∴ Rf = k1 [A]a [B]b
Where k1 is rate constant of forward reaction.
Applying law of mass action to the backward reaction.
Rate of backward reaction (Rb) 𝛼 [C]c [D]d
Rb = k2[C]c [D]d
where k2 is rate constant of a backward reaction.
At equilibrium. Rf = Rb
k1 [A]a [B]b = k2 [C]c [D]d
∴
[𝐶]𝑐
𝑘
Kc = 𝑘1 = [𝐴]𝑎
2
[𝐷]𝑑
[𝐵]𝑏
where Kc is equilibrium constant expressed in terms of concentration
This is known as equilibrium law or law of chemical equilibrium.
“ At a given temperature, the ratio of the product of the molar concentration of the product to
that of the reactants, each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation is constant
“Equilibrium constant is the ratio of velocity (rate) constant of a forward reaction to the velocity (rate)
constant of a backward reaction” or Kc = Rf /Rb
“It is the ratio of product of molar concentration of products is to product of molar concentration of
reactants raised to power of respective number of moles”.
Note: Write Kc expression for the following reactions.
⇌
a) N2 + 3H2
2NH3
[NH3 ]2
Kc = [N
b) PCl5
2 ] [H2 ]3
⇌ PCl3 + Cl2
Kc =
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[PCl3 ][Cl2 ]
[PCl5 ]
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⇌
c) H2 + I2
Kc =
2HI
[HI]2
[H2 ] I2 ]
Derive Kp expression for gaseous reversible homogeneous reaction
⇌
aA + bB
cC + dD
Consider a gaseous reversible reaction aA + bB ⇌
cC + dD is conducted in a closed vessel at
constant temperature, Let PA, PB, Pc and PD be the partial pressure of reactants and products and V1 and
V2 be the velocity of forward reaction and backward reactions.
Applying law of mass action to the forward reaction.
V1 𝛼 PaA × PbB
∴ V1 = k1PaA × PbB
Where k1 is velocity constant of a forward reaction.
Applying law of mass action to the backward reaction.
V2 𝛼 PcC × PdD
V2 = k2 × PcC × PdD
Where k2 is velocity constant of a backward reaction.
At equilibrium,
V1 = V2
k1 × PAa × PbB = k2 × PcC × PdD
k
Kp =k1 =
2
Pc c × PD d
PA a × PB b
∴ Equilibrium constant is the ratio of products of partial pressure of products to the product of partial
pressure of reactants raised to the power of respective number of moles.
Note : Write Kp expression for the following reactions:
⇌ 2NH3(g)
a) N2(g) + 3H2(g)
PNH32
Kp =
⇌
b) PCl5
PN2 × PH23
PCl3 + C12
Kp =
Ppcl3 × Pcl2
Ppcl5
Derive the equation which relates Kp and Kc
Kp expression for a reversible reaction aA + bB
Kp =
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𝑃𝐶 𝑐 × 𝑃𝐷 𝑑
𝑃𝐴 𝑎 × 𝑃𝐵 𝑏
⇌
cC + dD is
- - - - (1)
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Kc =
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
Ideal gas equation for ‘n’ mole of a gas is PV = nRT
𝑛
n
∴ P = CRT
P = 𝑣 RT
∴ PA = [ A ]a RTa
∵v=C
PB = [B]b RTb
Pc = [C]c RTc
PD = [D]d RTd
&
Substituting these in equation (1)
[𝐶]𝑐 𝑅𝑇 𝑐 × [𝐷]𝑑 𝑅𝑇 𝑑
Kp = [𝐴]𝑎
∴ Kp =
𝑅𝑇 𝑎 × [𝐵]𝑏 𝑅𝑇 𝑏
[𝐶]𝑐 [𝐷]𝑑 ×
𝑅𝑇 𝑐+𝑑
[𝐴]𝑎 [𝐵]𝑏
𝑅𝑇 𝑎+𝑏
∴ Kp = Kc × RT (c+d) - (a+b)
∴
Kp = Kc × 𝑅𝑇 Δn
Write the relationship between Kp and Kc for the following reactions:
a) N2(g) + 3H2(g)
⇌ 2NH3(g)
∆ng = 2 – 4 = - 2
Kp = Kc × 𝑅𝑇 ∆𝑛𝑔
Kp = Kc × 𝑅𝑇 −2
∴
∴ Kp < Kc
Kc
Kp = RT2
b) PCl5(g)
⇌
∆𝑛 = 2 – 1 = 1
Kp = Kc × 𝑅𝑇 ∆𝑛
Kp = Kc × 𝑅𝑇 1
∴ Kp = Kc × RT
∴ Kp > Kc
PCl3(g) + Cl2(g)
c) 2HI(g)
⇌ H2(g) + I2(g)
∆𝑛 = 2 – 2 = 0
Kp = Kc × 𝑅𝑇 0
Kp = Kc × 𝑅𝑇 1
∴ Kp = Kc
Relationship between equilibrium constant of a reaction and its multiples:
Chemical equation
equilibrium constant
aA
+ bB
⇌
cC
+
dD
Kc
cC
+ dD
⇌
aA
+
bB
K/c = K
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c
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naA
𝑎𝐴
𝑛
𝑐𝐶
𝑛
+
+
⇌ ncC
+ nbB
𝑏𝐵
𝑛
𝑑𝐷
𝑛
⇌
⇌
𝑐𝐶
𝑛
𝑎𝐴
𝑛
+
ndD
𝑑𝐷
+
+
K//c = Knc
KIIIc = 𝑛√𝐾𝑐
𝑛
𝑏𝐵
KIVc =
𝑛
1
𝑛
√𝐾𝑐
Homogeneous Equilibrium: The equilibrium in which all reactants and products are in same phase.
1. N2(g) + 3H2(g)
⇌
2NH3(g)
3+
2. Fe (aq) + SCN (aq) ⇌ Fe(SCN)2+(aq)
3. 2HI(g) ⇌ H2(g) + I2(g)
Example:
Heterogeneous equilibrium:
The equilibrium in which reactants and products are in different phase is called heterogeneous
equilibrium.
Example: Let us consider dissociation of CaCO3
⇌
CaCO3(s)
CaO(s) + CO2(g)
Kc =
[CaO(s)] [CO2(g)]
[CaCO3(s)]
(We can simplify chemical equilibrium in heterogeneous system, the molar concentration of pure solids and
liquids is constant. Because molar concentration of pure solids and liquid is independent of the amount of the
substances and it is considered as constant.
Simplified equilibrium constant is
⇌
2) Ni(s) + 4CO(g)
Kc =
Kc = [CO2(g)] or Kp = Pco2
Ni (CO)4(g)
[𝑁𝑖(𝐶𝑂)4(𝑔)]
[𝐶𝑂(𝑔)]4
3. Ag2 O(s) + 2HNO3(aq)
⇌ 2AgNO3(aq) + H2O(l)
[AgNO3(aq)]2
Kc = [HNO
3
(aq)]2
Characteristics of equilibrium constants Kc / Kp




The value of equilibrium constant is independent of initial concentrations of reactants and
products.
Equilibrium constant depends on temperature. It is constant for a given reaction at constant
temperature.
Equilibrium constant is applicable only when concentration of reactants and products have
attained equilibrium.
The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium
constant for the forward reaction. (ie Kc = x then K = 1/x)
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Application of equilibrium constant:
1) It is used to predict the extent of reactions.
2) It is used to predict the direction of the reaction.
3) It is used to calculate equilibrium concentrations.
Predicting the extent of a reaction:
Equilibrium constant is directly proportional to the concentration of product and inversely proportional
to the concentration of reactants. It shows that higher the value of K, faster is the reaction and it nearly
to completion. But smaller the value of K, slower is the reaction and the reaction proceeds rarely.
If Kc >103, ie if Kc is very large, the reaction proceeds nearly to completion
If Kc < 10 -3, ie if Kc is very small, the reaction proceeds rarely.
If Kc is in range of 10-3 to 103, appreciable concentration of both reactants and products are present.
Predicting the direction of the reaction:
It helps to predict the direction of the reaction at any stage by using reaction quotient (Q). It is same as
the equilibrium constant except the concentrations are not necessarily equilibrium values.
For a general equation, aA
+ bB
Qc =
Qc =



⇌
cC + dD
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
product of concentration of products
product of concentration of reactants
If Qc > Kc, the reaction will proceed in the backward direction until the equilibrium is achieved
(reactant side)
If Qc < Kc, the reaction will proceed in the forward direction until the equilibrium is achieved
(product side)
If Qc = Kc, the reaction is at equilibrium
Relationship between equilibrium constant, reaction quotient and Gibbs energy:
The relationship between ∆G, = ∆𝐺 0 , and Kp is ∆G=∆G0 +RT ln Kp.
At equilibrium, ∆G=0, and Kp = Qc
∴ 0 =∆G0 + RT In Qc
∆G0 = - RT ln Qc
∆G0 = - 2.303 RT log10 Qc
Factors affecting position of equilibria :Important factors which affect equilibria are
1)
2)
3)
4)
5)
Concentration
Pressure
Temperature
Catalyst
Addition of inert gas etc.
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The effect of change in conditions on equilibrium can be predicted by using Le Chatelier’s principle.
It states that, “if a stress is applied to a system at equilibrium, the equilibrium is shifted in such a
way so as to cancel the applied stress. OR
If a system at equilibrium is subjected to a change in temperature, pressure or concentration, the
equilibrium shifts in the direction that tends to nullify the effect of the change.
1) Effect of concentration change:
When reactants are added to a system at equilibrium, the system tends to decrease the concentration of
reactant to convert it into product. So forward reaction is favorable and equilibrium shifts to right side.
The increase in concentration of any of product shifts equilibrium towards backward direction.
2) Effect of temperature change:
 When temperature of the system is increased at equilibrium, the system tends to decreases
the temperature by the absorption of heat so endothermic reaction is favorable and
equilibrium shifts to such side.
(the increase in temperature shifts equilibrium in the direction of the endothermic reaction.
 When temperature of the system is decreased at equilibrium, the system tends to increase
the temperature by the liberation of heat. So exothermic reaction is favorable and
equilibrium shifts to such side.
(the decrease in temperature shifts equilibrium in the direction of the exothermic reaction.
Example: N2(g) + 3H2(g)
⇌
2NH3(g)
ΔH = −93.38 kJ/mol
Low temperature is favorable for high yield of ammonia, but practically very low temperature
slow down the reaction and thus a catalyst is used to increase rate of the reaction.
3) Effect of pressure change:
 If pressure of a system is increased at equilibrium, the system tends to shift the equilibrium
in the direction of decreasing gaseous moles.
 If pressure of a system is decreased at equilibrium, the system tends to shift the equilibrium
in the direction of increasing gaseous moles.
 Pressure has no effect if the gaseous reactants and products have equal moles
4) Catalyst: catalyst has no effect on the position of equilibrium because it alters both forward and
backward reactions to the same extent, but equilibrium attains quickly.
5) Rare gas: Addition of inert gas at constant volume does not change the partial pressure or molar
concentrations of the substance involved in a reaction. So the added gas does not take part in a
reaction. ∴ addition of inert gas has no effect on position of equilibrium at constant volume.
Note: But at constant pressure, addition of inert gas increases the number of moles. So equilibrium
shifts to that side (if ∆n(g) = 0, inert gas has no effect].
Ionic equilibrium in solution: Chemical compounds are of two types.
1. Electrolytes: These are substances which conduct electricity both in molten state and in
solution.
Ex: NaCl, H2SO4, CH3COOH etc
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2. Non electrolytes: These are substances which do not conduct electricity both in solid and in
solution.
Ex: C6H6, C2H5OH, urea, glucose etc
Electrolytes are of two types. They are
1. Strong electrolytes: There are electrolytes, which undergoes complete ionization in aqueous
solution.
Ex: Strong acid: HCl, HNO3, H2SO4, HClO4 etc
Strong base: KOH, NaOH, LiOH, Ba(OH)2 etc
Soluble salt: NaCl, Na2SO4 CH3COONa, KBr etc
Ex: NaCl → Na + + Cl2. Weak electrolytes: These are electrolytes which undergoes partial ionization in aqueous
solution.
Ex : Weak acid: CH3COOH, H3PO4, H2S etc
Weak base: NH4OH Ca(OH)2, Al(OH)3 etc
Ex: NH4OH ⇌
NH+4 + OH −
Ionic equilibrium : It is the equilibrium established between ions and unionized molecules in an
aqueous solution of the weak electrolyte.
Ex: CH3COOH(aq) ⇌ CH3COO-(aq) + H+ (aq)
Acids, Bases and Salts:
Acids are substances, which have sour taste, turns blue litmus paper into red, and they reacts with bases
gives salt and water. They can also react with carbonates and bicarbonates liberating CO2 gas.
Bases are substances, which have bitter taste, turns red litmus paper into blue, and they reacts with
acids forms salt and water.
Salts: These are substances, which are formed by the combination of acids and bases.
There are four types of salts.
1.
2.
3.
4.
Salt of strong acid and strong base ex : NaCl, K2SO4, NaNO3 NaBr, KBr
Salt of strong acid and weak base ex: NH4Cl, FeCl3, CuCl2, CuSO4
Salt of weak acid and strong base ex: CH3COONa, Na2CO3, NaHCO3
Salt of weak acid and weak base ex: CH3COONH4, HCOONH4, (NH4)2CO3, NH4HCO3
Theories of acids and bases: There are three important theories of acids and bases. They are :
1. Arrhenius theory of acids and bases:
According to this theory “Acids are substances, which can donate proton (or Hydrogen
ion)”
Ex: HCl, H2SO4, HNO3 etc.
HCl(aq)
H+(aq) + Cl-(aq)
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Basicity of an acid: It is equal to number of replaceable hydrogen atom present in one
molecule of an acid.
So HCl, HNO3 are monobasic. H2SO4, H2C2O4 are dibasic, H3PO4 is tribasic acid.
2.
Bases are substances, which can donate hydroxyl ions.
Ex: NaOH, KOH, etc.
NaOH(aq)
Na+(aq) + OH-(aq)
Acidity of a base: “It is equal to number of replaceable hydroxyl groups present in one
molecule of base”.
So NaOH, KOH are monoacidic, Ca(OH)2, Ba(OH)2, are diacidic bases.
Bronsted - Lowry –theory of acids and bases:
According to this theory “Acids are substances which denote proton and bases are substances
which accept proton”.
H3 O+ (aq)
Ex: HCl(aq) + H2O(l) ⇌
Acid
base
Cl− (aq)
+
conjugate acid
conjugate base
In this example, HCl donate proton, so it is an acid and H2O accept proton, so it is base.
NH4 + (aq)
NH3(aq) + H2O(l) ⇌
base
acid
conjugate acid
+
OH − (aq)
conjugate base
Conjugated acid base pairs: The acid-base pair which differs only by one proton.
Ex : HCl & Cl− and H2O and H3O + etc
Note: If Bronsted acid is a strong acid, then its conjugate base is a weak base and vice versa.
Each conjugate acid has one extra proton and each conjugate base has one less proton.
Conjugated bases for acids:
Conjugated acids
Conjugated bases
Conjugated acids
Conjugated bases
HCl
Cl-
H2S
HS-
H2O
OH-
HNO3
NO3-
H3O+
H2O
HClO4
ClO4-
H2SO4
HSO4-
NH3
NH2-
NH4
NH3
Amphoteric substances: These are substances, which can act both as an acid as well as a base.
Ex: H2O, NH3, etc.
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Lewis theory of acids and bases:
According to this theory, acids are substances which accept pair of electrons and bases are substances
which donate pair of electrons.
Ex: NH3 + BF3
⟶ H3N:
BF3
In above example, NH3 donate pair of electron, so it is base and BF3 accept pair of electron, so it is an
acid.
Example: electron deficient species like AlCl3, Co3+, Mg2+ etc acts as Lewis acids and H2O, NH3, OHwhich can donate a pair of electrons, can act as Lewis bases.
Relative Strength of acids and bases: It depends on ability to donate proton or ability to donate
hydroxyl ion. Acids and bases that dissociate completely to give large number of H+ and OH-ions are
strong acids and bases.
But in case of weak acids and weak bases, they undergoes partial ionization to give low concentration
of H+ and OH-ions.
The relative strength of weak acid and weak bases can be compared in terms of dissociation constant or
ionization constant.
Dissociation constant (Ionisation constant) of weak acids (Ka)
Consider an aqueous solution of weak acid HA, ionizes partially in H2O, can be represented by a
reaction.
HA
⇌
H3O+ + A Applying law of mass action
K=
[ H3 O+ ] [A− ]
[HA]
K =
[H3 O+ ] [A− ]
HA
+
∴ Ka = [H3 O ] [A
HA
−]
Ka is called ionization constant or dissociation constant of acid Ka is a measure of acid strength.
It shows that, larger the value of Ka, stronger is the acid and smaller the value of Ka, weaker is the
acid.
Ex: Ka values of A and B are 1.8 × 10-5 & 1.8 × 10-6 respectively. Between these A is strong and B is
weak.
Ionization constant for weak bases: (Kb )
Consider an aqueous solution of weak base BOH, ionizes partially in H2O, can be represented as
BOH
⇌
B+ + 𝑂𝐻 −
Applying law of mass action
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K=
K=
[B+ ] [OH− ]
[BOH]
[𝐵+ ] [𝑂𝐻 − ]
Kb =
𝐵𝑂𝐻
[𝐵+ ] [𝑂𝐻 − ]
[𝐵𝑂𝐻]
Kb is the ionization constant or dissociation constant of weal base. Kb is a measure of base strength
It shows that, larger the value of Kb, stronger is the base and smaller the value of Kb, weaker is the
base.
Ex: Kb values of P and Q are 1.8 × 10 – 5 & 1.85 × 10 – 5 respectively. Between these, Q is strong and P
is weak.
Derive an expression for Ostwald dilution laws:
[Ionization constant for weak electrolytes]
Consider a weak electrolyte HA, ionizes partially in water can be represented as
⇌ H+ + A –
HA
Let ‘c’ mol / dm3 be the initial concentration of HA and let ‘𝛼’ be the degree of dissociation. i.e out of
‘c’ mole dissociate to give ‘cα’ mole of H+ and ‘cα’ mole of A – . ∴ the remaining concentration of HA
at equilibrium is (c – cα ) mole.
⇌ H+ + A –
HA
Initial conc.
C
0
0
Conc at eqm
C–Cα
cα
cα
Applying law of mass action.
i.e, K =
K=
[H+ ] [A− ]
HA
𝑐𝛼 ×𝑐𝛼
(𝑐−𝑐𝛼)
𝛼2 𝑐 2
K = 𝑐(1−𝛼)
𝑐𝛼2
K =1−𝛼
It is the mathematical expression of Ostwald dilution law.
For weak electrolyte ≪< 1 ∴ 𝛼 = 1 ;
Ka
Ka = C𝛼 2 or α = √ C
Note: 1) Ostwald dilution law states that, degree of dissociation of an electrolyte increases with
dilution and it approaches unity at infinite dilution. It is called limiting value,
2. Degree of dissociation or ionization (𝛼) : It is defined as the fraction of the total quantity of
electrolyte ionized at equilibrium or the amount of electrolyte dissociates per mole.
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3. The equation used to calculate [H+] = √𝑲𝒂 × 𝜶 or c 𝜶 for weak acids and [OH
√𝑲𝒃 × 𝒄 or c 𝜶 for weak bases.
–
] =
Ionic product of water: K 𝝎
Water is a weak electrolyte & it undergoes partial ionization can be represented as
⇌
H2O
H+ + OH –
Applying law of mass action,
K=
[H+ ][ OH− ]
H2 O
K[H2O] = [H+] [OH – ]
K𝜔 = [ H+ ] [OH – ]
“Ionic product of water is the product of molar concentration of hydrogen ion and hydroxyl ion
in water or in any aqueous solution”.
At 298K, K𝜔 = 10 – 14 M2
∴ at 298K [H+ ] [OH – ] = 10 – 14
In acidic solutions, [ H+ ] > [ OH – ] > 10-7 in basic solutions, [ OH+ ] > [ H+ ] < 10-7 and in neutral
solutions or in water [ H+ ] = [OH – ] = 10 – 7 M
Common ion effect: Suppression of degree of dissociation of weak electrolyte by the addition of
strong electrolyte containing common ion is called common ion effect.
Ex : Consider partial ionization of CH3COOH,
CH3COOH ⇌ CH3COO – + H+
Applying law of mass action,
K=
[CH3 COO− ] [ H+ ]
CH3 COOH
To this small quantity of CH3COONa is added, so [CH3COO –] ion increases. In order to keep ‘K’
constant, [H+] should decreases and [CH3COOH] must increases. ∴ degree of dissociation of
CH3COOH is suppressed. It can also suppressed by adding HCl or any other strong electrolyte
containing common ion.
Ex : 2 The degree of dissociation of NH4OH can suppressed by adding NH4Cl
Relation between Ka and Kb:
Considering the conjugate acid- base dissociation equilibrium reaction
B(aq) + H2O(aq) ⇌
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BH+(aq) + OH – (aq)
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Applying law of mass action
Kb =
[𝐵𝐻 + ] [𝑂𝐻 − ]
[𝐵]
Multiplying and dividing the above expression by [H+]
Kb=
[𝐵𝐻 + ] [𝑂𝐻 − ][𝐻 + ]
[𝐵][𝐻 + ]
=[OH – ] [H+] × [BH+ ] / [B] [H+]
∵ Ka =
[𝐵][𝐻 + ]
[𝐵𝐻 + ]
Ka × Kb = K𝜔
Kb = Kw / Ka
Equilibrium constant for a net reaction:
It is obtained after adding equilibrium constants of a individual reactions.
i.e KNET = K1 × K2 × - - - ex: In case of a conjugate acid base pair NH4 and NH3
we see
NH4(aq) + H2O(1) ⇌
H3 O(aq) + NH3(aq)
Ka = [H3O+] [NH3] / [NH4] - - - - - - (1)
NH3(aq) + H2O(1) ⇌
NH4(aq) + OH – (aq)
Kb = [ NH4 ] [OH – ] /[NH3] - - - - -(2)
Ka x Kb = [H+ ] [OH – ] = 10 – 14 =Kw
Polybasic acids and polyacidic bases:
Polybasic acids are acids, which can donate more than one H+ ion.
Poly acidic bases are bases, which can donate more than one OH – ion. In poly basic acids, Ka1, Ka2,
Ka3 are first ionization constant, second ionization constant and third ionization constant respectively
but Ka1 > Ka2 > Ka3 because more difficult to remove proton from a negative ion due to electrostatic
forces.
Ex: H2A is stronger than HA – and it is stronger than A –
Factors affecting acid strength:
Acidic nature depends on
a) The Bond strength of H-A (or size)
b) Polarity of H-A bond (electro negativity)
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a) The Bond strength of H-A : The strength of H-A bond is weak in case strong acid. Hence
cleavage of bond becomes easier, where as the strength of H-A bond is high is case of weak
acids.
As the size of elements increases down a group, bond strength decreases, so acid strength
increases.
Size increases.
HF < HC1 < HBr < HI
Acid strength increases
b) Polarity of H-A bond: In case of strong acids the electro negativity difference between H and A is
large, hence H-A bond become more polar therefore cleavage of bond becomes easier, whereas
polarity of H-A bond is less in case of weak acids.
As the electro negativity increases in the same row, polarity of the substance increases. ∴ acidic
strength increases.
Electro negativity
CH4 < NH3 < H2O < HF
Acid strength increases
PH
of a solution:
It was proposed by Sorensen. Hydronium ion concentration in molarity expressed on logarithmic scale
known as pH scale.
PH of a solution: It is the negative logarithm to the base ten of molar concentration of H+ ion in
mol/dm3.
1
∴ PH = – log10 [H+] or PH = log10 [𝐻 +]
On pH Scale, acidic solution has pH< 7, basic solution has pH> 7 and neutral solution has pH=7
Note: PH of blood is 7.35, so it is slight basic.
POH of a solution: It is the negative logarithm to the base ten of molar concentration of hydroxyl ion in
a solution.
POH = – log10 [OH – ]
Note : At 298K, PH + POH = 14
In neutral compounds or neutral solution: PH = POH = 7
Show that PH + POH = 14 at 298 K
We know that, [H+] [OH – ] = 10 – 14
Taking – log10 on both sides
–log10 [H+] [OH – ] = – log10 10 – 14
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–
–
[log10 [H+ ] + log10 [OH – ]] = –( – 14) × log10 10
log10 [H+] + – log10 [OH –] = 14 × log10 10
PH + POH = 14
Pka of a solution: It is the negative logarithm to the base ten of ionization constant of weak acids.
i.e. PKa = – log10Ka
larger the value of Ka, smaller the value of Pka, stronger is the acid and vice versa.
Pkb of a solution: It is the negative logarithm to the base ten of ionization constant of weak bases.
i.e Pkb = – log10Kb
larger the value of Kb, smaller the value of Pkb, stronger is the base and vice versa
Note: Pkb value of P and Q are 2.8 & 6.2, which is strong. P is strong.
𝝎
Show that pKa + pKb = 14 = 𝑷𝒌
We know that, at 298K, Ka × Kb = K𝜔
Taking –log10 on both sides
∴ –log10 ka × kb = –log10 k𝜔 = –log10 – 14
∴ –log10 ka +(–) log10 kb = –log10 k𝜔 = –(–14) × log10 10
By definition ∴
Pka + Pkb = 𝑃𝑘𝑤 = 14
Hydrolysis of a salt and PH of a solution:
The process of interaction between water and cations or anions or both of salts to produce either acidic
or alkaline solution is known as hydrolysis of salts
Hydrolysis of salts of following types:
Salts of strong acid and strong base: They do not undergo hydrolysis, because their solution is
neutral due to hydration of ions. ∴ PH of a solution of salt of strong acid and strong base is neutral ie
pH= 7. Ex: NaCl
Salt of strong acid and weak base: These salts undergo hydrolysis to produce acidic solution. In this
category of salt, cation undergoes hydrolysis.
Example:NH4Cl(aq) ⇌
NH4+(aq) + Cl-(aq)
NH4+ ions undergo hydrolysis with water to form NH4OH and H+ ions.
NH4+(aq) + H2O ⇌
NH4OH(aq) + H+(aq)
NH4OH being a weak base remains almost unionized, but H+ ion concentration increase. Thus pH of a
solution is less than 7.
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Salt of strong base and weak acid: These salts undergo hydrolysis to produce basic solution. In this
category of salt, anion undergoes hydrolysis.
Example: CH3COONa(aq) ⇌ CH3COO-(aq) + Na+(aq)
Acetate ions undergo hydrolysis with water to form acetic acid and OH- ions.
CH3COO-(aq) + H2O ⇌
CH3COOH(aq) + OH-(aq)
Acetic acid being a weak acid remains almost unionized, but OH- ion concentration increase. Thus pH
of a solution is more than 7.
Salt of weak acid and weak base : Salts of this undergoes hydrolysis to form neutral solution.
When ammonium acetate is dissolved in water, it undergoes hydrolysis as
CH3COONH4(aq) + H2O ⇌
CH3COOH(aq) +NH4OH(aq)
Acetic acid and ammonium hydroxide are weak electrolytes and ionize partially. They undergo
ionization to the same extent. Therefore [H+] = [OH-]. Hence the solution is almost neutral.
CH3COOH ⇌
NH4OH ⇌
CH3COO- + H+
NH4+ + OH-
pH is determined by their pK values
∴ PH of such solution = 7 + ½ (Pka – Pkb )
Buffer solutions: These are solutions, which can resist change in their pH value by adding small
quantity of an acid and small quantity of a base.
There are three types of buffer solutions. They are
1. Acidic buffer. Ex: CH3COOH + CH3COONa
2. Basic buffer ex : NH4OH + NH4C1
3. Neutral buffer Ex : CH3COONH4
The equation used to calculate pH of acidic buffer is pH = pka + log10
POH of a basic buffer is POH = Pkb + log10
[𝑠𝑎𝑙𝑡]
[𝑏𝑎𝑠𝑒]
[𝑠𝑎𝑙𝑡]
𝑎𝑐𝑖𝑑
and
. These are Hendersons equations.
PH of many body fluids are controlled by buffer action and any deviation indicates malfunctioning of
the body. Buffer solutions are used to determine pH of unknown solutions.
Solubility (S): it is defined as the number of moles of the solute dissolved in 1dm3 of the saturated
solution,
Solubility equilibrium of sparingly soluble salts: Solubility depends on lattice enthalpy of salts and
the salvation enthalpy of ions in solution. Generally a salt dissolve in a particular solvent, if its
salvation enthalpy is greater than its lattice enthalpy.
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Every salt has its characteristic solubility which depends on temperature. On the basis of solubility,
salts are three types.
1.
2.
3.
4.
Salt having solubility more than 0.1 mol/L are considered as soluble in water.
Salt having solubility between 0.01 - 0.1 mol/L are considered as slightly soluble in water.
Salt having solubility more than > 0.1 mol/L are considered as soluble in water.
Salt having solubility less than 0.01 mol/L are considered as sparingly soluble water.
In a saturated solution of sparingly soluble salt solution contains both ions and undissolved salt. ∴ an
equilibrium is established between both ions and undissolved salt. It is called solubility equilibria.
AB(s)
A+ + B –
⇌
Unionized electrolyte
ions
Consider a saturated solution of sparingly soluble electrolyte AB, it contains both ions and undissolved
solute. Therefore an equilibrium is established between ions and solid.
i.e, AB(s) ⇌ A+ + B –
Applying law of mass action
K=
[A+ ] [ B− ]
[AB(s)
K[AB(s)] = [A+] [B –]
+
–
∴ Ksp = [A ] [B ]
for a pure solid substances, the concentration remains constant.
Ksp is called Solubility product
Solubility product (Ksp) or solubility product constant:
“Solubility product is defined as the product of molar concentration of constituent ions raised to their
appropriate powers in a saturated solution of sparingly soluble electrolyte at a given temperature”.
Note: Solubility product equation for different electrolytes
1.
2.
3.
4.
5.
6.
AB2 type : Ksp = [A++] [B –]2
A2B type : Ksp = [A+]2 [B – – ]
AB3 type : Ksp = [A+++] [B – ]3
AB4 type : Ksp = [A4+] [B – ]4
A3B4 type : Ksp = [A4+]3 [B3 – ]4
MxXy type : Ksp = [Ap+]x [Bq –]y
Relationship between solubility and solubility product:
Consider a saturated solution of sparingly soluble electrolyte AB. An equilibrium is established
between both ions and undissolved solute.
∴ AB(s)
⇌
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A+ + B –
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Let”S” mol/dm3 be the solubility of AB, so it gives ‘S’ mole of A+ and ‘S’ mol of B –
∴ Ksp = [A+] [B – ]
Ksp = s × s
∴ Ksp = s2
It is the solubility product expression of AB type electrolyte.
“It is an expression, which relates solubility and solubility product of an electrolyte”.
Note: Write solubility product expression for different electrolytes.
1. AB2 type: Ksp = [A 2+] [B – ]2
= s × (2s)2 = 4s3
2. A2B type : Ksp = [A+]2 [B 2 – ]
= (2s)2 × s
Ksp = 4s3
3. AB3 type: Ksp = [A3+] [B – ]3
= s × (3s)3
Ksp = 27S4
4. AB4type: Ksp = [A4+] [B –]4
Ksp = s × (4s)4
Ksp = 256 s5
Ionic product: IP or Qsp
It is the product of molar concentration of constituent ions raised to their appropriate power in any
solution of an electrolyte.
Solubility product principles:
1. If IP < Ksp, it is unsaturated solution ∴ precipitation does not takes place.
2. If IP = Ksp, it is just saturated solution ∴ precipitation does not take place
3. If IP > Ksp, it is super saturated solution ∴ precipitation takes place.
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Distinction between Ionic product and solubility product
Common ion effect on solubility of ionic salts.
According to Le Chateliers principle, if we increase the concentration of any one of ions, it combines
with ions of opposite charge, so some of the salt will be precipitated till Ksp = Qsp.
If we decrease the concentration of any one of the ions, more salt will dissolve to increase the
concentration of both the ions till once Ksp = Qsp.
Ex: When HCl is passed through saturated solution of NaCl, Cl– ion increases. So it combines with
Na+ of NaCl and forms precipitate of NaCl.
Thus, due to common ion effect solubility may decrease or increase till once again Ksp = Qsp.
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