Equilibrium
MCQsChemical Equilibrium
Chemical equilibrium is defined as the state at which there is no further change in concentration of reactants and products.
Let us consider a general reversible reaction in a closed vessel-
A + B ⇌ C + D
Initially reaction occurs in forward direction but as the concentration of products increases reaction also starts in backward direction. At a certain stage, rate of forward reaction becomes equal to the rate of backward reaction called equilibrium state.
So, at equilibrium the rate of forward reaction(rf) is equal to the rate of backward reaction(rb).
Chemical equilibrium is defined as the state at which there is no further change in concentration of reactants and products.
Let us consider a general reversible reaction in a closed vessel-
A + B ⇌ C + D
Initially reaction occurs in forward direction but as the concentration of products increases reaction also starts in backward direction. At a certain stage, rate of forward reaction becomes equal to the rate of backward reaction called equilibrium state.
So, at equilibrium the rate of forward reaction(rf) is equal to the rate of backward reaction(rb).
Derivation of Equilibrium Constant
Law of Mass Action
Guldberg and Waage states that the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances.
Let us consider a reversible homogeneous chemical reaction which is in equilibrium state at a particular temperature-
aA + bB ⇌ cC + dD
Assume active masses of A, B, C and D be the [A] [B] [C] and [D] respectively at equilibrium.
According to law of mass action-
rate of forward reaction(rf) ∝ [A]a [B]b
rate of backward reaction(rb) ∝ [C]c [D]d
or, rf = Kf [A]a [B]b
rb = Kb [C]c [D]d
At equilibrium-
rf = rb
or, Kf [A]a [B]b = Kb [C]c [D]d
or, Kf / Kb = [C]c [D]d / [A]a [B]b
or, Kc = [C]c [D]d / [A]a [B]b
Kc is known as equilibrium constant and has a definite value for different chemical reaction at particular temperature.
So, we can say that equilibrium constant(Kc) at a given temperature is the ratio of the rate constants of forward and backward reactions.
Use of Partial Pressures Instead of Concentration
For gaseous reactions, partial pressures are conveniently used since at any fixed temperature partial pressure is directly proportional to concentration.
Now, Kc becomes Kp and [ ] becomes P
Kp = PCc PDd/ PAa PBb
Unit: mole per liter. or atm
Law of Mass Action
Guldberg and Waage states that the rate of a chemical reaction is directly proportional to the product of the active masses of the reacting substances.
Let us consider a reversible homogeneous chemical reaction which is in equilibrium state at a particular temperature-
aA + bB ⇌ cC + dD
Assume active masses of A, B, C and D be the [A] [B] [C] and [D] respectively at equilibrium.
According to law of mass action-
rate of forward reaction(rf) ∝ [A]a [B]b
rate of backward reaction(rb) ∝ [C]c [D]d
or, rf = Kf [A]a [B]b
rb = Kb [C]c [D]d
At equilibrium-
rf = rb
or, Kf [A]a [B]b = Kb [C]c [D]d
or, Kf / Kb = [C]c [D]d / [A]a [B]b
or, Kc = [C]c [D]d / [A]a [B]b
Kc is known as equilibrium constant and has a definite value for different chemical reaction at particular temperature.
So, we can say that equilibrium constant(Kc) at a given temperature is the ratio of the rate constants of forward and backward reactions.
Use of Partial Pressures Instead of Concentration
For gaseous reactions, partial pressures are conveniently used since at any fixed temperature partial pressure is directly proportional to concentration.
Now, Kc becomes Kp and [ ] becomes P
Kp = PCc PDd/ PAa PBb
Unit: mole per liter. or atm
Relation between KP and KC
Let us consider a reversible homogeneous chemical equilibrium reaction-
aA + bB ⇌ cC + dD
According to law f mass action-
Kc = [C]c [D]d / [A]a [B]b -----(equation-1)
and
Kp = (PC)c (PD)d/ (PA)a (PB)b -----(equation-2)
we know that PV = nRT (ideal gas equation)
or, P = (n/V)RT
or, P = (active mass)RT
or, PA = [A]RT
similarly-
PB = [B]RT
PC = [C]RT
PD = [D]RT
Putting these values in equation-2, we get-
Kp = [C]c (RT)c [D]d (RT)c / [A]a (RT)c [B]b (RT)c
Kp = [C]c [D]d / [A]a [B]b X (RT)(c+d)/(RT)(a+b)
or, Kp = KC X (RT)(c+d)-(a+b)
or, Kp = KC X (RT)Δngas
Case -1
Δngas = 0
then KP = KC
Case -2
Δngas = Positive
then KP > KC
Case -3
Δngas = Negative
then KP < KC
Let us consider a reversible homogeneous chemical equilibrium reaction-
aA + bB ⇌ cC + dD
According to law f mass action-
Kc = [C]c [D]d / [A]a [B]b -----(equation-1)
and
Kp = (PC)c (PD)d/ (PA)a (PB)b -----(equation-2)
we know that PV = nRT (ideal gas equation)
or, P = (n/V)RT
or, P = (active mass)RT
or, PA = [A]RT
similarly-
PB = [B]RT
PC = [C]RT
PD = [D]RT
Putting these values in equation-2, we get-
Kp = [C]c (RT)c [D]d (RT)c / [A]a (RT)c [B]b (RT)c
Kp = [C]c [D]d / [A]a [B]b X (RT)(c+d)/(RT)(a+b)
or, Kp = KC X (RT)(c+d)-(a+b)
or, Kp = KC X (RT)Δngas
Case -1
Δngas = 0
then KP = KC
Case -2
Δngas = Positive
then KP > KC
Case -3
Δngas = Negative
then KP < KC
Le-Chatelier's Principle
According to this principle, if a system at equilibrium is subjected to a change of concentration, pressure or temperature then the equilibrium is shifted in such a way as to nullify the effect of change. Le-Chatelier's principle is applicable for both chemical and physical equilibrium.
Chemical Equilibrium
Reactant(R) ⇌ Product(P)
1. Change in concentration
In an equilibrium, on increasing the concentrations of reactants, equilibrium shift in favour of products (i.e. forward reaction) while on increasing the concentrations of the products equilibrium shift in favour of reactants (i.e. backward reaction).
2. Change in Pressure
Pressure has no effect on the reactions of liquids and solids. When the pressure on the gaseous system is increased, the volume decreases i.e. the total number of moles present per unit volume increases.
According to Le-Chatelier's principle, the equilibrium shifts in that direction in which there is decrease in number of moles. If there is no change in number of moles of gases in a reaction then a pressure change does not affect the equilibrium
When ∆n = 0
As per Le Chatelier's principles, there will be no effect on Equilibrium and Product Formation on changing the pressure.
When ∆n = +ve
An increase in pressure or decrease in volume will decrease the product formation. Decrease of pressure or increase of volume will increase the product formation.
When ∆n = -ve
An increase in pressure or decrease in volume will increase the formation of the product.
3. Change in temperature
If the temperature of the system at equilibrium is increased then reaction will proceed in that direction in which heat can be used. Thus for endothermic reaction, increase in temperature will favour the forward reaction. For exothermic reactions, increase in temperature will favour the backward reaction.
4. Effect of Catalyst
A catalyst is a substance that changes the rate of reactions (increase or decrease) without quantitatively taking part in the reaction.
In a reversible reaction, the change in rate of reaction is the same for both forward and backward reactions.
According to Le Chatelier's principle, the presence of the catalyst may speed up or slow down the attainment of equilibrium but will not affect the equilibrium concentration.
5. Effect of Addition of Inert Gas
At constant pressure, if an inert gas is added it will increase the volume of the system. Therefore. the equilibrium will shift in a direction in which there is an increase in the number of moles of gases.
At constant volume, if an inert gas is added the relative molar concentration of the substance will not change. Hence. the equilibrium position of the reaction remains unaffected.
Physical Equilibrium
Those reaction in which change in only and only physical states (solid, liquid and gas) of substance takes place without any chemical change, is called physical reaction.
1. Ice-water system (melting of ice)
Melting of ice is endothermic reaction (absorption of heat) and decrease in volume.
H2O(solid) ⇌ H2O(liquid)
Increase of temperature and pressure will favour the melting of ice into water.
2. Water-water vapour system (Vapourisation of water)
Vapourisation of water is an endothermic and condensation of vapour into water is an exothermic.
Favourable conditions for conversion of water into vapour are high tempeprature and low pressure.
When the temperature is increased, the equilibrium shifts towards right side. So rise in temperature will increase the vapour.
When the pressure is increased, The equilibrium shifts towards left side. So increase in pressure will favour the rate of condensation of vapour into water.
3. Solubility of gases
Gas + H2O ⇌ Aqueous Solution
Solubility of gases increases with increasing pressure which dissolves in a solvent(H2O) with a decrease in volume.
According to this principle, if a system at equilibrium is subjected to a change of concentration, pressure or temperature then the equilibrium is shifted in such a way as to nullify the effect of change. Le-Chatelier's principle is applicable for both chemical and physical equilibrium.
Chemical Equilibrium
Reactant(R) ⇌ Product(P)
1. Change in concentration
In an equilibrium, on increasing the concentrations of reactants, equilibrium shift in favour of products (i.e. forward reaction) while on increasing the concentrations of the products equilibrium shift in favour of reactants (i.e. backward reaction).
2. Change in Pressure
Pressure has no effect on the reactions of liquids and solids. When the pressure on the gaseous system is increased, the volume decreases i.e. the total number of moles present per unit volume increases.
According to Le-Chatelier's principle, the equilibrium shifts in that direction in which there is decrease in number of moles. If there is no change in number of moles of gases in a reaction then a pressure change does not affect the equilibrium
When ∆n = 0
As per Le Chatelier's principles, there will be no effect on Equilibrium and Product Formation on changing the pressure.
When ∆n = +ve
An increase in pressure or decrease in volume will decrease the product formation. Decrease of pressure or increase of volume will increase the product formation.
When ∆n = -ve
An increase in pressure or decrease in volume will increase the formation of the product.
3. Change in temperature
If the temperature of the system at equilibrium is increased then reaction will proceed in that direction in which heat can be used. Thus for endothermic reaction, increase in temperature will favour the forward reaction. For exothermic reactions, increase in temperature will favour the backward reaction.
4. Effect of Catalyst
A catalyst is a substance that changes the rate of reactions (increase or decrease) without quantitatively taking part in the reaction.
In a reversible reaction, the change in rate of reaction is the same for both forward and backward reactions.
According to Le Chatelier's principle, the presence of the catalyst may speed up or slow down the attainment of equilibrium but will not affect the equilibrium concentration.
5. Effect of Addition of Inert Gas
At constant pressure, if an inert gas is added it will increase the volume of the system. Therefore. the equilibrium will shift in a direction in which there is an increase in the number of moles of gases.
At constant volume, if an inert gas is added the relative molar concentration of the substance will not change. Hence. the equilibrium position of the reaction remains unaffected.
Physical Equilibrium
Those reaction in which change in only and only physical states (solid, liquid and gas) of substance takes place without any chemical change, is called physical reaction.
1. Ice-water system (melting of ice)
Melting of ice is endothermic reaction (absorption of heat) and decrease in volume.
H2O(solid) ⇌ H2O(liquid)
Increase of temperature and pressure will favour the melting of ice into water.
2. Water-water vapour system (Vapourisation of water)
Vapourisation of water is an endothermic and condensation of vapour into water is an exothermic.
Favourable conditions for conversion of water into vapour are high tempeprature and low pressure.
When the temperature is increased, the equilibrium shifts towards right side. So rise in temperature will increase the vapour.
When the pressure is increased, The equilibrium shifts towards left side. So increase in pressure will favour the rate of condensation of vapour into water.
3. Solubility of gases
Gas + H2O ⇌ Aqueous Solution
Solubility of gases increases with increasing pressure which dissolves in a solvent(H2O) with a decrease in volume.
Reaction Quotient(Q)
Reaction quotient is a measure of the relative amounts of products and reactants present in a reaction at a particular point of time.
For reversible reaction
aA + bB ⇌ cC + dD
where a, b, c and d are the stoichiometric coefficients for the balanced reaction, we can calculate reaction quotient using the following equation-
Q = [C]c [D]d / [A]a [B]b
Reaction quotient applied at any stage of reaction but Equilibrium constant only at equilibrium.
Reaction quotient(Q) can predict the direction of reaction by comparing with equilibrium constant(K).
1. When Q = K, then reaction is in equilibrium state.
2. When Q < K the concentration of the reactants is higher than it would be at equilibrium. In order to reach equilibrium, the reaction will favor the forward reaction. So, the rate of forward reaction increases.
3. When Q > K the concentration of the products is higher than it would be at equilibrium. In order to reach equilibrium, the reaction will favor the backward reaction. So, the rate of backward reaction increases.
If the value of Q is ≥ 103, Product predominates
If the value of Q is ≤ 10-3, Reactant predominates
If the value of Q is in between 10-3 to 103, significant amounts of both products and reactants
Reaction quotient is a measure of the relative amounts of products and reactants present in a reaction at a particular point of time.
For reversible reaction
aA + bB ⇌ cC + dD
where a, b, c and d are the stoichiometric coefficients for the balanced reaction, we can calculate reaction quotient using the following equation-
Q = [C]c [D]d / [A]a [B]b
Reaction quotient applied at any stage of reaction but Equilibrium constant only at equilibrium.
Reaction quotient(Q) can predict the direction of reaction by comparing with equilibrium constant(K).
1. When Q = K, then reaction is in equilibrium state.
2. When Q < K the concentration of the reactants is higher than it would be at equilibrium. In order to reach equilibrium, the reaction will favor the forward reaction. So, the rate of forward reaction increases.
3. When Q > K the concentration of the products is higher than it would be at equilibrium. In order to reach equilibrium, the reaction will favor the backward reaction. So, the rate of backward reaction increases.
If the value of Q is ≥ 103, Product predominates
If the value of Q is ≤ 10-3, Reactant predominates
If the value of Q is in between 10-3 to 103, significant amounts of both products and reactants
The value of the reaction quotient Q, for the cell
Zn(s)∣Zn2+(0.01 M) ∥ Ag+(1.25 M)∣Ag(s) is
1. 156
2. 125
3. 1.25 x 10-2
4. 6.4 x 10-3 ✓
Relationship among Equilibrium Constant, Reaction Quotient and Gibbs Free Energy
We know that-
∆G = ∆Go + RTlnQ
at equilibrium-
∆G = 0 and Q = Kc
putting these values in the above equation we get-
0 = ∆Go + RTlnKc
or, ∆Go = − RTlnKc
or, lnKc = − ∆Go/RT
or, Kc = e− ∆Go/RT
If ∆Go < 0
then − ∆Go/RT is positive i.e. e− ∆Go/RT > 1
or, Kc > 1
Thus, the reaction is spontaneous or reaction will proceed in forward direction.
If ∆Go > 0
then − ∆Go/RT is negative i.e. e− ∆Go/RT < 1
or, Kc < 1
Thus, the reaction is nonspontaneous or reaction will proceed in forward direction to very little extent.
We know that-
∆G = ∆Go + RTlnQ
at equilibrium-
∆G = 0 and Q = Kc
putting these values in the above equation we get-
0 = ∆Go + RTlnKc
or, ∆Go = − RTlnKc
or, lnKc = − ∆Go/RT
or, Kc = e− ∆Go/RT
If ∆Go < 0
then − ∆Go/RT is positive i.e. e− ∆Go/RT > 1
or, Kc > 1
Thus, the reaction is spontaneous or reaction will proceed in forward direction.
If ∆Go > 0
then − ∆Go/RT is negative i.e. e− ∆Go/RT < 1
or, Kc < 1
Thus, the reaction is nonspontaneous or reaction will proceed in forward direction to very little extent.
Degree of Ionisation or Degree of Dissociation
As we know that the weak acids or bases do not ionize completely so an equilibrium exists between the ionized and unionized species. The degree of ionisation may be defined as the fraction of total amount of a weak acid or a base that exists in the ionized form. It is denoted by a Greek letter 'α'. The equilibrium constant can be used to calculate the degree of ionisation of a weak acid or a base.
Consider a weak acid HA which partially dissociates in its aqueous solutions and the following equilibrium is established-
HA ⇌ H+ + A−
Initial Concn.: C 0 0
Equi. Concn.: C(1-α) Cα Cα
So, The equilibrium constant-
Ka = [H+] [A−]/[HA]
or, Ka = Cα.Cα/C(1-α)
or, Ka = Cα2/(1-α)
Since the acid HA is very weak, α << 1; so, we can neglect α in comparison to 1
so, Ka = Cα2
or, α = √Ka/C
or, α = √KaV (as 1/C = V)
So if we know the value of the dissociation constant of the acid and the concentration or volume of the weak acids we can find its degree of dissociation or ionisation.
Per cent Dissociation or Ionization = Degree of dissociation X 100%
Similarly, we can derive an expression for degree of dissociation of a weak base-
α = √KbV (as 1/C = V)
As we know that the weak acids or bases do not ionize completely so an equilibrium exists between the ionized and unionized species. The degree of ionisation may be defined as the fraction of total amount of a weak acid or a base that exists in the ionized form. It is denoted by a Greek letter 'α'. The equilibrium constant can be used to calculate the degree of ionisation of a weak acid or a base.
Consider a weak acid HA which partially dissociates in its aqueous solutions and the following equilibrium is established-
HA ⇌ H+ + A−
Initial Concn.: C 0 0
Equi. Concn.: C(1-α) Cα Cα
So, The equilibrium constant-
Ka = [H+] [A−]/[HA]
or, Ka = Cα.Cα/C(1-α)
or, Ka = Cα2/(1-α)
Since the acid HA is very weak, α << 1; so, we can neglect α in comparison to 1
so, Ka = Cα2
or, α = √Ka/C
or, α = √KaV (as 1/C = V)
So if we know the value of the dissociation constant of the acid and the concentration or volume of the weak acids we can find its degree of dissociation or ionisation.
Per cent Dissociation or Ionization = Degree of dissociation X 100%
Similarly, we can derive an expression for degree of dissociation of a weak base-
α = √KbV (as 1/C = V)
Ionic Product of Water
H2O ⇌ H+ + OH−
Equilibrium Constant-
K = [H+] [OH−]/[H2O]
or, K . [H2O] = [H+] [OH−]
or, Kw = [H+] [OH−] (as the concentration of H2O is constant)
The constant Kw, is called the dissociation constant or ionic product constant of water.
For pure water-
[H+] = [OH−] = 10−7M at 25oC
so, pH = pOH = 7 at 25oC
and Kw = 10−7 X 10−7 = 10−14
and, pKw = 14
Number of water moles in 1 liter of water = 1000/18 = 55.5 moles
number of water molecules in 1 liter water = 55.5 NA
Concentration of H+ or OH− ions = 10−7mol/liter
H2O ⇌ H+ + OH−
Equilibrium Constant-
K = [H+] [OH−]/[H2O]
or, K . [H2O] = [H+] [OH−]
or, Kw = [H+] [OH−] (as the concentration of H2O is constant)
The constant Kw, is called the dissociation constant or ionic product constant of water.
For pure water-
[H+] = [OH−] = 10−7M at 25oC
so, pH = pOH = 7 at 25oC
and Kw = 10−7 X 10−7 = 10−14
and, pKw = 14
Number of water moles in 1 liter of water = 1000/18 = 55.5 moles
number of water molecules in 1 liter water = 55.5 NA
Concentration of H+ or OH− ions = 10−7mol/liter
Relation in between pH and pOH
Kw = [H+] [OH−]
taking negative log on both sides we get-
−logKw = −log[H+] −log[OH−]
or, pKw = pH + pOH
For neutral water-
[H+] = [OH−]
pKw = pH + pOH
or, pKw = pH + pH
or, 2pH = pKw
or, pH = pKw/2
similarly,
or, pKw = pOH + pOH
or, 2pOH = pKw
or, pOH = pKw/2
OR, pH = pOH = pKw/2
Kw = [H+] [OH−]
taking negative log on both sides we get-
−logKw = −log[H+] −log[OH−]
or, pKw = pH + pOH
For neutral water-
[H+] = [OH−]
pKw = pH + pOH
or, pKw = pH + pH
or, 2pH = pKw
or, pH = pKw/2
similarly,
or, pKw = pOH + pOH
or, 2pOH = pKw
or, pOH = pKw/2
OR, pH = pOH = pKw/2
Degree of ionisation of water (α)
H2O ⇌ H+ + OH−
According to Ostwald's dilution law
[H+] = Cα
or, α = [H+] / C
or, α = 10−7/55.5 = 1.8 X 10−9
or, α = 1.8 X 10−7%
H2O ⇌ H+ + OH−
According to Ostwald's dilution law
[H+] = Cα
or, α = [H+] / C
or, α = 10−7/55.5 = 1.8 X 10−9
or, α = 1.8 X 10−7%
Ionisation Constant of water (K)
H2O ⇌ H+ + OH−
Equilibrium Constant-
K = [H+] [OH−]/[H2O]
K = 10−7 X 10−7/55.5
K = 1.8 X 10−16
H2O ⇌ H+ + OH−
Equilibrium Constant-
K = [H+] [OH−]/[H2O]
K = 10−7 X 10−7/55.5
K = 1.8 X 10−16
Solubility Product (Ksp)
solubility product is defined as- in the saturated solution, the product of concentrations of the ions raised to a power equal to the number of times, the ions occur in the equation representing the dissociation of the electrolyte at a given temperature.
Consider in general, the electrolyte of the type AxBywhich is dissociated as-
AxBy ⇌ xA+y + yB−x
Applying law of mass action-
K = [A+y]x [B−x]y/[AxBy]
or, K [AxBy] = [A+y]x [B−x]y
for saturated solution-
[AxBy] = constant
so, Ksp = [A+y]x [B−x]y
solubility product is defined as- in the saturated solution, the product of concentrations of the ions raised to a power equal to the number of times, the ions occur in the equation representing the dissociation of the electrolyte at a given temperature.
Consider in general, the electrolyte of the type AxBywhich is dissociated as-
AxBy ⇌ xA+y + yB−x
Applying law of mass action-
K = [A+y]x [B−x]y/[AxBy]
or, K [AxBy] = [A+y]x [B−x]y
for saturated solution-
[AxBy] = constant
so, Ksp = [A+y]x [B−x]y
Common Ion Effect
The phenomenon in which the degree of dissociation of a weak electrolyte is suppressed by the addition of a strong electrolyte having an ion common to weak electrolyte is known as common ion effect.
Let us consider dissociation of a weak electrolyte (acetic acid)-
CH3COOH ⇌ CH3COO− + H+
The equilibrium constant-
Ka = [CH3COO−] [H+]/[CH3COOH]
Now, if sodium acetate is added to this solution-
The concentration of CH3COO− ion in the solution increases and thus, in order to have Ka constant [H+] must decrease or the concentration of undissociated acetic acid must increase. In other words, the dissociation of acetic acid is suppressed on addition of CH3COONa to its solution.
pH and Common-Ion Effect
When the conjugate ion of a buffer solution is added to it, the pH of the buffer solution changes due to the common ion effect.
The phenomenon in which the degree of dissociation of a weak electrolyte is suppressed by the addition of a strong electrolyte having an ion common to weak electrolyte is known as common ion effect.
Let us consider dissociation of a weak electrolyte (acetic acid)-
CH3COOH ⇌ CH3COO− + H+
The equilibrium constant-
Ka = [CH3COO−] [H+]/[CH3COOH]
Now, if sodium acetate is added to this solution-
The concentration of CH3COO− ion in the solution increases and thus, in order to have Ka constant [H+] must decrease or the concentration of undissociated acetic acid must increase. In other words, the dissociation of acetic acid is suppressed on addition of CH3COONa to its solution.
pH and Common-Ion Effect
When the conjugate ion of a buffer solution is added to it, the pH of the buffer solution changes due to the common ion effect.
Salt Hydrolysis
Salt hydrolysis is the phenomenon of interaction of cations and anions of a salt with H2O in order to produce an acidic nature or an alkaline nature.
Salt + Water ⇌ Acid + Base ΔH =+ve
The net effect of dissolving a salt (which undergoes hydrolysis) is to break up the water molecules (hydrolysis) to produce a weak acid or weak base or both and thus, phenomenon is always endothermic.
The process of salt hydrolysis is actually reverse the process of neutralization.
1. Hydrolysis of Strong Acid and Strong Base (SA - SB) types of salt-
Hydrolysis of salt of Strong Acid and Strong Base is not possible as both cation and anion are not reactive. Aqueous solution of these type of salt is neutral in nature,so, pH of the solution is 7.
2. Hydrolysis of Strong Acid and Weak Base (SA - WB) types of salt-
NH4Cl + H2O ⇌ NH4OH + HCl
NH4+ H2O ⇌ NH4OH + H+
In this type of salt hydrolysis, cation reacts with H2O so it is called cationic hydrolysis. The cation of the salt which come from weak base is reactive. Solution is acidic in nature as [H+] is increased. So the pH of the solution is less than 7.
Relation between Kh, KW and Kb
NH4+ H2O ⇌ NH4OH + H+
so, Hydrolysis Constant (Kh)
Kh = [NH4OH] [H+]/[NH4+] -----(equation-1)
NH4OH ⇌ NH4+ + OH−
so, Ionization Constant of base (Kb)
Kb = [NH4+] [OH−]/[NH4OH] -----(equation-2)
and
H2O ⇌ H+ + OH−
so, Ionic Product of water (KW)
KW = [H+] [OH−]/[H2O] -----(equation-3)
Now if we multiplying equation-1 and equation-2, we get equation-3
Kh X Kb = KW
so, Kh = KW/Kb
Degree of Hydrolysis
NH4+ + H2O ⇌ NH4OH + H+
Initial Conc.: C 0 0
Con. after
some times: C − Ch Ch Ch
so, Kh = Ch2/(1-h)
h is very less in comparision to 1
so, Kh = Ch2
or, h = √Kh/C
or, h = √KW/Kb.C (as Kh = KW/Kb)
pH of the Solution
[H+] = Ch = C.√KW/Kb.C (as h = √KW/Kb.C )
or, [H+] = √KWC/Kb
Taking negative log on both sides we get-
−log[H+] = −log√KWC/Kb
or, pH = −log(KWC/Kb)1/2
or, pH = −1/2[logKWC + logC − logKb]
or, pH = −1/2logKWC −1/2logC −(−1/2logKb)
or, pH = 1/2pKW −1/2logC −1/2pKb
or, pH = 7 −1/2logC −1/2pKb
3. Hydrolysis of Weak Acid and Strong Base (WA - SB) types of salt
Similar as Hydrolysis of Strong Acid and Weak Base (SA - WB) types of salt.
Salt hydrolysis is the phenomenon of interaction of cations and anions of a salt with H2O in order to produce an acidic nature or an alkaline nature.
Salt + Water ⇌ Acid + Base ΔH =+ve
The net effect of dissolving a salt (which undergoes hydrolysis) is to break up the water molecules (hydrolysis) to produce a weak acid or weak base or both and thus, phenomenon is always endothermic.
The process of salt hydrolysis is actually reverse the process of neutralization.
1. Hydrolysis of Strong Acid and Strong Base (SA - SB) types of salt-
Hydrolysis of salt of Strong Acid and Strong Base is not possible as both cation and anion are not reactive. Aqueous solution of these type of salt is neutral in nature,so, pH of the solution is 7.
2. Hydrolysis of Strong Acid and Weak Base (SA - WB) types of salt-
NH4Cl + H2O ⇌ NH4OH + HCl
NH4+ H2O ⇌ NH4OH + H+
In this type of salt hydrolysis, cation reacts with H2O so it is called cationic hydrolysis. The cation of the salt which come from weak base is reactive. Solution is acidic in nature as [H+] is increased. So the pH of the solution is less than 7.
Relation between Kh, KW and Kb
NH4+ H2O ⇌ NH4OH + H+
so, Hydrolysis Constant (Kh)
Kh = [NH4OH] [H+]/[NH4+] -----(equation-1)
NH4OH ⇌ NH4+ + OH−
so, Ionization Constant of base (Kb)
Kb = [NH4+] [OH−]/[NH4OH] -----(equation-2)
and
H2O ⇌ H+ + OH−
so, Ionic Product of water (KW)
KW = [H+] [OH−]/[H2O] -----(equation-3)
Now if we multiplying equation-1 and equation-2, we get equation-3
Kh X Kb = KW
so, Kh = KW/Kb
Degree of Hydrolysis
NH4+ + H2O ⇌ NH4OH + H+
Initial Conc.: C 0 0
Con. after
some times: C − Ch Ch Ch
so, Kh = Ch2/(1-h)
h is very less in comparision to 1
so, Kh = Ch2
or, h = √Kh/C
or, h = √KW/Kb.C (as Kh = KW/Kb)
pH of the Solution
[H+] = Ch = C.√KW/Kb.C (as h = √KW/Kb.C )
or, [H+] = √KWC/Kb
Taking negative log on both sides we get-
−log[H+] = −log√KWC/Kb
or, pH = −log(KWC/Kb)1/2
or, pH = −1/2[logKWC + logC − logKb]
or, pH = −1/2logKWC −1/2logC −(−1/2logKb)
or, pH = 1/2pKW −1/2logC −1/2pKb
or, pH = 7 −1/2logC −1/2pKb
3. Hydrolysis of Weak Acid and Strong Base (WA - SB) types of salt
Similar as Hydrolysis of Strong Acid and Weak Base (SA - WB) types of salt.
