Class 11 Chemistry – Chapter 6: Equilibrium

Part 1: Physical & Chemical Equilibrium

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Chapter Overview

Equilibrium is a state where two opposite processes occur at the same rate, so there is no overall change in the system.

There are two types of equilibrium:

1. Physical Equilibrium
2. Chemical Equilibrium

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1. What is Equilibrium?

Definition

Equilibrium is the state of a reversible process in which the forward and backward processes occur at equal rates, causing all measurable properties of the system to remain constant.

Example

Water kept in a closed container:

Water (liquid) ⇌ Water vapour

• Water continuously evaporates.
• Water vapour continuously condenses.
• After some time,

Rate of evaporation = Rate of condensation

This is equilibrium.

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2. Dynamic Nature of Equilibrium

Equilibrium is dynamic, not static.

Dynamic means

The reactions continue even at equilibrium.

Example:

Imagine two people exchanging balls.

• One throws 5 balls every minute.
• The other also throws back 5 balls every minute.

Although balls keep moving, the number of balls with each person remains unchanged.

Similarly,

Forward reaction continues.

Backward reaction also continues.

Both occur at equal rates.

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Characteristics of Dynamic Equilibrium

✔ Forward reaction continues.

✔ Reverse reaction continues.

✔ Rates are equal.

✔ Concentrations remain constant.

✔ Appears unchanged from outside.

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Physical Equilibrium

Physical equilibrium involves change in physical state only.

No new substance is formed.

Examples

• Ice ⇌ Water
• Water ⇌ Water vapour
• Iodine solid ⇌ Iodine vapour

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Types of Physical Equilibrium

1. Solid ⇌ Liquid Equilibrium

Example

Ice ⇌ Water

At 273 K (0°C) and 1 atm pressure,

Ice melts into water.

Water freezes into ice.

At equilibrium,

Rate of melting = Rate of freezing

Hence,

Mass of ice remains constant.

Mass of water remains constant.

Temperature remains constant.

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Important Point

Normal melting point

The temperature at which solid and liquid coexist in equilibrium at 1 atm pressure.

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2. Liquid ⇌ Vapour Equilibrium

Example

Water ⇌ Water vapour

Inside a closed vessel,

Water molecules evaporate.

Water vapour molecules condense.

Initially,

Evaporation > Condensation

After some time,

Evaporation = Condensation

Equilibrium is reached.

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Vapour Pressure

Definition

The pressure exerted by vapour when liquid and vapour are in equilibrium at a fixed temperature.

Important Facts

✔ Vapour pressure increases with temperature.

✔ More volatile liquids have higher vapour pressure.

✔ Higher vapour pressure means lower boiling point.

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Open vs Closed System

Closed Container

Equilibrium is possible.

Open Container

Water vapour escapes into air.

Condensation becomes very small.

Hence,

Equilibrium cannot be established.

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3. Solid ⇌ Vapour Equilibrium

Some solids change directly into vapour.

This process is called sublimation.

Example

I₂(s) ⇌ I₂(g)

Other examples

• Camphor
• Ammonium chloride (NH₄Cl)

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4. Solid Dissolved in Liquid

Example

Sugar + Water

Initially,

Sugar dissolves.

After saturation,

Some sugar crystallises back.

At equilibrium,

Rate of dissolution = Rate of crystallisation

This occurs only in a saturated solution.

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5. Gas Dissolved in Liquid

Example

CO₂ in soda water

Bottle closed

↓

High pressure

↓

More CO₂ dissolves.

Bottle opened

↓

Pressure decreases.

↓

CO₂ escapes.

↓

New equilibrium is established.

This follows Henry’s Law:

The amount of gas dissolved in a liquid is directly proportional to the pressure of the gas above the liquid.

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General Characteristics of Physical Equilibrium

1. Occurs only in a closed system.
2. Opposite processes continue simultaneously.
3. Rates become equal.
4. Measurable properties remain constant.
5. Equilibrium is dynamic.
6. Temperature must remain constant.

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Chemical Equilibrium

Chemical equilibrium occurs in reversible chemical reactions.

General reaction

A + B ⇌ C + D

Initially,

Forward reaction is faster.

As products are formed,

Reverse reaction starts.

Eventually,

Forward rate = Reverse rate

Chemical equilibrium is established.

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Characteristics of Chemical Equilibrium

✔ Reaction is reversible.

✔ Forward and reverse reactions continue.

✔ Rates become equal.

✔ Concentration remains constant.

✔ No visible change occurs.

✔ Dynamic in nature.

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Example: Haber Process

N₂ + 3H₂ ⇌ 2NH₃

Initially,

Nitrogen and hydrogen react to form ammonia.

As ammonia accumulates,

It starts decomposing back.

Finally,

Rate of NH₃ formation

=

Rate of NH₃ decomposition

Hence,

Chemical equilibrium is established.

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Physical vs Chemical Equilibrium

Physical Equilibrium	Chemical Equilibrium
Only physical change	Chemical change occurs
No new substance formed	New substances formed
Example: Ice ⇌ Water	Example: N₂ + 3H₂ ⇌ 2NH₃
Involves change of state	Involves reversible reaction
Dynamic	Dynamic

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Important Definitions (Exam Ready)

Equilibrium

The state in which forward and reverse processes occur at equal rates and macroscopic properties remain constant.

Dynamic Equilibrium

An equilibrium in which both forward and reverse processes continue at equal rates.

Closed System

A system that does not exchange matter with the surroundings.

Vapour Pressure

The pressure exerted by vapour in equilibrium with its liquid at a fixed temperature.

Saturated Solution

A solution that cannot dissolve more solute at a given temperature.

Sublimation

Direct conversion of a solid into vapour without passing through the liquid state.

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One-Minute Revision

• Equilibrium is dynamic, not static.
• Forward rate = Reverse rate.
• Concentrations remain constant at equilibrium.
• Physical equilibrium involves only a change of state.
• Chemical equilibrium occurs only in reversible reactions.
• Equilibrium is possible only in a closed system.
• Vapour pressure increases with temperature.
• Henry’s Law explains the dissolution of gases in liquids.

Part 2: Law of Chemical Equilibrium & Equilibrium Constant

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1. Law of Chemical Equilibrium

When a reversible reaction reaches equilibrium, the ratio of the concentrations of products to reactants (each raised to the power of their stoichiometric coefficients) remains constant at a fixed temperature.

This constant is called the equilibrium constant.

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General Reaction

$$aA+bB \rightleftharpoons cC+dD$$aA+bB⇌cC+dD

The equilibrium constant is:$$K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$Kc​=[A]a[B]b[C]c[D]d​

Where:

• [A], [B] = Equilibrium concentrations of reactants
• [C], [D] = Equilibrium concentrations of products
• a, b, c, d = Stoichiometric coefficients

Remember: Always use equilibrium concentrations, not initial concentrations.

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2. Equilibrium Constant (K)

Definition

The equilibrium constant is a numerical value that tells us how far a reaction proceeds before reaching equilibrium.

It depends only on temperature for a given balanced reaction.

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Types of Equilibrium Constants

(A) $K_c$Kc​

Used when concentrations are expressed in mol L⁻¹ (M).

(B) $K_p$Kp​

Used when gases are expressed in terms of partial pressure.

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3. Writing the Expression for $K_c$Kc​

Example 1

$$H_2+I_2 \rightleftharpoons 2HI$$H2​+I2​⇌2HI $$K_c=\frac{[HI]^2}{[H_2][I_2]}$$Kc​=[H2​][I2​][HI]2​

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Example 2

$$N_2+3H_2 \rightleftharpoons 2NH_3$$N2​+3H2​⇌2NH3​ $$K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}$$Kc​=[N2​][H2​]3[NH3​]2​

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Example 3

$$4NH_3+5O_2 \rightleftharpoons 4NO+6H_2O$$4NH3​+5O2​⇌4NO+6H2​O $$K_c=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}$$Kc​=[NH3​]4[O2​]5[NO]4[H2​O]6​

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Shortcut Rules

✔ Products are written in the numerator.

✔ Reactants are written in the denominator.

✔ Stoichiometric coefficients become powers.

✔ Only equilibrium concentrations are used.

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4. Reverse Reaction

If the reaction is reversed,

Original reaction$$A+B \rightleftharpoons C+D$$A+B⇌C+D $$K_c=x$$Kc​=x

Reverse reaction$$C+D \rightleftharpoons A+B$$C+D⇌A+B

Then$$K’_c=\frac1{K_c}$$Kc′​=Kc​1​

Formula

$$K_{reverse}=\frac1{K_{forward}}$$Kreverse​=Kforward​1​

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5. Changing the Equation

Suppose$$H_2+I_2 \rightleftharpoons 2HI$$H2​+I2​⇌2HI

has equilibrium constant K.

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Case 1

Reaction reversed

↓

New constant$$\frac1K$$K1​

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Case 2

Multiply whole equation by n

↓

New constant$$K^n$$Kn

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Case 3

Divide equation by 2

↓

New constant$$\sqrt K$$K​

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Memory Trick

Change in equation	New equilibrium constant
Reverse reaction	$1/K$1/K
Multiply by n	$K^n$Kn
Divide by n	$K^{1/n}$K1/n

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6. Homogeneous Equilibrium

A homogeneous equilibrium is one in which all reactants and products are in the same physical state (same phase).

Examples

$$N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$$N2​(g)+3H2​(g)⇌2NH3​(g)

All are gases.

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$$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$$Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)

All are in aqueous solution.

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Characteristics

• Only one phase is present.
• $K_c$Kc​ or $K_p$Kp​ includes all species because all are variable.

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7. Equilibrium Constant in Gaseous Systems

For gaseous reactions,

Instead of concentration,

Partial pressure may be used.

This constant is called KpK_pKp​.

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Example$$H_2+I_2\rightleftharpoons2HI$$H2​+I2​⇌2HI $$K_p=\frac{(P_{HI})^2}{P_{H_2}P_{I_2}}$$Kp​=PH2​​PI2​​(PHI​)2​

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8. Relationship Between $K_p$Kp​ and $K_c$Kc​

The most important formula of this chapter is$$\boxed{K_p=K_c(RT)^{\Delta n}}$$Kp​=Kc​(RT)Δn​

Where

• R = Gas constant
• T = Temperature in Kelvin
• Δn = (Moles of gaseous products) − (Moles of gaseous reactants)

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How to Find Δn

Example$$N_2+3H_2\rightleftharpoons2NH_3$$N2​+3H2​⇌2NH3​

Products = 2

Reactants = 4

Therefore$$\Delta n=2-4=-2$$Δn=2−4=−2

Hence$$K_p=K_c(RT)^{-2}$$Kp​=Kc​(RT)−2

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When is $K_p = K_c$Kp​=Kc​?

If$$\Delta n=0$$Δn=0

Then$$K_p=K_c$$Kp​=Kc​

Example$$H_2+I_2\rightleftharpoons2HI$$H2​+I2​⇌2HI

Products = 2

Reactants = 2

Therefore$$\Delta n=0$$Δn=0

Hence$$K_p=K_c$$Kp​=Kc​

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9. Heterogeneous Equilibrium

A heterogeneous equilibrium is one in which reactants and products are present in different physical states.

Example

$$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$$CaCO3​(s)⇌CaO(s)+CO2​(g)

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Writing the Expression

General expression$$K_c=\frac{[CaO][CO_2]}{[CaCO_3]}$$Kc​=[CaCO3​][CaO][CO2​]​

Since

Pure solids have constant concentration,

they are omitted.

Therefore$$\boxed{K_c=[CO_2]}$$Kc​=[CO2​]​

Similarly,$$K_p=P_{CO_2}$$Kp​=PCO2​​

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Rule

Pure solids and pure liquids are never included in equilibrium constant expressions because their concentrations remain constant.

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10. Units of Equilibrium Constant

The unit depends on the reaction.

Example

$$H_2+I_2\rightleftharpoons2HI$$H2​+I2​⇌2HI

Units cancel.

So,

$K_c$Kc​ has no unit.

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If total powers differ

Units remain.

Example$$N_2O_4\rightleftharpoons2NO_2$$N2​O4​⇌2NO2​

Here $K_c$Kc​ has units of concentration.

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11. Important Properties of Equilibrium Constant

• Valid only at equilibrium.
• Depends only on temperature.
• Independent of the initial concentrations.
• Changes if the balanced equation is changed.
• Reverse reaction has reciprocal equilibrium constant.

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Formula Sheet

General Formula

$$K_c=\frac{\text{Products}^{\text{coefficients}}}{\text{Reactants}^{\text{coefficients}}}$$Kc​=ReactantscoefficientsProductscoefficients​

Gas Phase

$$K_p=\frac{\text{Partial pressure of products}}{\text{Partial pressure of reactants}}$$Kp​=Partial pressure of reactantsPartial pressure of products​

Relation

$$\boxed{K_p=K_c(RT)^{\Delta n}}$$Kp​=Kc​(RT)Δn​

Reverse Reaction

$$K=\frac1K$$K=K1​

Multiply Equation

$$K_{new}=K^n$$Knew​=Kn

Divide Equation

$$K_{new}=K^{1/n}$$Knew​=K1/n

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Exam Tips

• Never use initial concentrations in the $K_c$Kc​ expression.
• Do not include solids or pure liquids in heterogeneous equilibrium expressions.
• Always calculate Δn using only gaseous species when using $K_p = K_c(RT)^{\Delta n}$Kp​=Kc​(RT)Δn.
• Stoichiometric coefficients become powers in the equilibrium constant expression.

Part 3: Applications of Equilibrium Constant & Le Chatelier’s Principle

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1. Applications of Equilibrium Constant

The value of equilibrium constant (K) helps us to:

1. Predict the extent of a reaction.
2. Predict the direction of a reaction.
3. Calculate equilibrium concentrations.

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2. Predicting the Extent of Reaction

The magnitude of K tells how much a reaction proceeds before reaching equilibrium.

Case 1: Large Value of K

If:$$K_c > 10^3$$Kc​>103

Products are present in much larger amounts than reactants.

Meaning:

• Reaction almost completes.
• Equilibrium lies towards products.

Example:$$H_2+Cl_2 \rightleftharpoons 2HCl$$H2​+Cl2​⇌2HCl

Large $K_c$Kc​ value indicates formation of a large amount of HCl.

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Case 2: Small Value of K

If:$$K_c < 10^{-3}$$Kc​<10−3

Reactants dominate.

Meaning:

• Very little product is formed.
• Reaction hardly proceeds.

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Case 3: Intermediate Value of K

If:$$10^{-3}<K_c<10^3$$10−3<Kc​<103

Both reactants and products exist in considerable amounts.

Example:$$H_2+I_2 \rightleftharpoons 2HI$$H2​+I2​⇌2HI

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Summary Table

Value of K	Meaning
Very large	Products dominate
Very small	Reactants dominate
Intermediate	Both present significantly

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3. Reaction Quotient (Q)

Sometimes a reaction mixture is not at equilibrium.

To know the direction in which it will proceed, we use reaction quotient (Q).

It has the same expression as equilibrium constant but uses concentrations at any given time.

For:$$aA+bB\rightleftharpoons cC+dD$$aA+bB⇌cC+dD $$Q_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}$$Qc​=[A]a[B]b[C]c[D]d​

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Comparing Q and K

Case 1: Q = K

The reaction is already at equilibrium.

No change occurs.

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Case 2: Q < K

The amount of products is less than required.

Reaction moves in the forward direction.

More products are formed.

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Case 3: Q > K

The amount of products is more than required.

Reaction moves in the backward direction.

More reactants are formed.

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Quick Trick

Remember:

Q < K → Go forward

Q > K → Go backward

Q = K → Equilibrium

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4. Le Chatelier’s Principle

Statement

When a system at equilibrium is disturbed by changing:

• concentration
• pressure
• temperature

the equilibrium shifts in such a way that it reduces the effect of the disturbance.

In simple words:

A system tries to oppose any change made to it.

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Factors Affecting Chemical Equilibrium

1. Effect of Concentration

Consider:$$H_2+I_2\rightleftharpoons2HI$$H2​+I2​⇌2HI

Increasing Reactant Concentration

Adding more $H_2$H2​ or $I_2$I2​:

• System consumes extra reactants.
• Equilibrium shifts right.
• More HI is formed.

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Increasing Product Concentration

Adding HI:

• System removes extra product.
• Equilibrium shifts left.
• More $H_2$H2​ and $I_2$I2​ form.

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Rule:

Increase concentration:

• Reactant ↑ → Shift towards products
• Product ↑ → Shift towards reactants

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2. Effect of Pressure

Pressure affects only gaseous equilibria.

Consider:$$N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$$N2​(g)+3H2​(g)⇌2NH3​(g)

Number of gas molecules:

Left side:$$1+3=4$$1+3=4

Right side:$$2$$2

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Increasing Pressure

System tries to reduce pressure.

It moves towards fewer gas molecules.

Therefore:$$\text{Equilibrium shifts right}$$Equilibrium shifts right

More ammonia forms.

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Decreasing Pressure

System moves towards more gas molecules.

Equilibrium shifts left.

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Pressure Rule

Higher pressure:

→ Side with fewer gas molecules

Lower pressure:

→ Side with more gas molecules

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Important Note

Pressure has no effect if:$$\Delta n =0$$Δn=0

Example:$$H_2+I_2\rightleftharpoons2HI$$H2​+I2​⇌2HI

Gas molecules:

Reactants = 2

Products = 2

No shift occurs.

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3. Effect of Temperature

Temperature changes the value of equilibrium constant.

Heat is treated as a reactant or product.

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Exothermic Reaction

Heat is released.

Example:$$N_2+3H_2\rightleftharpoons2NH_3+\text{Heat}$$N2​+3H2​⇌2NH3​+Heat

Increasing temperature adds heat.

System removes extra heat.

Equilibrium shifts left.

Therefore:

• Product formation decreases.
• $K$K decreases.

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Endothermic Reaction

Heat is absorbed.

Example:$$CaCO_3+\text{Heat}\rightleftharpoons CaO+CO_2$$CaCO3​+Heat⇌CaO+CO2​

Increasing temperature:

• More products form.
• Equilibrium shifts right.
• $K$K increases.

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Temperature Rule

Change	Equilibrium Shift
Increase temperature (exothermic)	Backward
Increase temperature (endothermic)	Forward
Decrease temperature	Opposite direction

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4. Effect of Catalyst

A catalyst:

• Increases the rate of forward reaction.
• Increases the rate of reverse reaction.
• Helps equilibrium reach faster.

But:

It does not change the position of equilibrium.

It does not change:

• $K_c$Kc​
• $K_p$Kp​
• Equilibrium composition

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5. Effect of Inert Gas

An inert gas does not react.

Example:

Helium (He)

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At Constant Volume

No effect on equilibrium.

Because:

• Concentrations remain unchanged.
• Partial pressures remain unchanged.

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At Constant Pressure

Volume increases.

Equilibrium may shift towards the side having more gas molecules.

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Haber Process and Equilibrium

Industrial preparation of ammonia:$$N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)+Heat$$N2​(g)+3H2​(g)⇌2NH3​(g)+Heat

Conditions:

High Pressure

Favors ammonia formation because products have fewer gas molecules.

Moderate Temperature

Low temperature favors ammonia but slows reaction.

Catalyst

Iron catalyst is used to increase rate.

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Important Points for Exams

⭐ Equilibrium can be disturbed by concentration, pressure, and temperature.

⭐ Catalyst does not change equilibrium position.

⭐ Only temperature changes equilibrium constant.

⭐ Pressure affects only gases.

⭐ Increasing pressure favors the side with fewer gas molecules.

⭐ $Q$Q predicts reaction direction.

Part 4: Ionic Equilibrium – Acids, Bases and pH

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1. Ionic Equilibrium

When electrolytes dissolve in water, they produce ions.

The equilibrium established between ions and undissociated molecules is called ionic equilibrium.

Example:$$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$$CH3​COOH⇌CH3​COO−+H+

Here:

• CH₃COOH = undissociated acid
• CH₃COO⁻ and H⁺ = ions

Both forward and backward ionisation occur continuously.

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2. Electrolytes

Substances that produce ions in aqueous solution are called electrolytes.

They conduct electricity because of the presence of free ions.

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Types of Electrolytes

(A) Strong Electrolytes

They completely ionise in water.

Examples:

• HCl
• HNO₃
• NaOH
• KCl

Example:$$HCl \rightarrow H^+ + Cl^-$$HCl→H++Cl−

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(B) Weak Electrolytes

They partially ionise in water.

Examples:

• CH₃COOH
• NH₄OH

Example:$$CH_3COOH \rightleftharpoons CH_3COO^-+H^+$$CH3​COOH⇌CH3​COO−+H+

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3. Acids and Bases

There are three important theories:

1. Arrhenius theory
2. Brønsted-Lowry theory
3. Lewis theory

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4. Arrhenius Concept

Arrhenius Acid

A substance that produces hydrogen ions (H⁺) in water is called an acid.

Example:$$HCl \rightarrow H^+ + Cl^-$$HCl→H++Cl−

Other examples:

• H₂SO₄
• HNO₃

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Arrhenius Base

A substance that produces hydroxide ions (OH⁻) in water is called a base.

Example:$$NaOH\rightarrow Na^+ + OH^-$$NaOH→Na++OH−

Examples:

• KOH
• Ca(OH)₂

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Limitations of Arrhenius Theory

• Applicable only to aqueous solutions.
• Cannot explain substances that act as acids without producing H⁺ directly.
• Cannot explain basic nature of NH₃.

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5. Brønsted-Lowry Concept

According to this theory:

Acid

A substance that donates a proton (H⁺).

Base

A substance that accepts a proton (H⁺).

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Example:$$HCl+H_2O\rightleftharpoons H_3O^+ + Cl^-$$HCl+H2​O⇌H3​O++Cl−

Here:

• HCl donates H⁺ → Acid
• H₂O accepts H⁺ → Base

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Conjugate Acid-Base Pair

When an acid loses H⁺, it forms its conjugate base.

Example:$$HA \rightleftharpoons H^+ + A^-$$HA⇌H++A−

HA = Acid

A⁻ = Conjugate base

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When a base accepts H⁺:$$B+H^+\rightarrow BH^+$$B+H+→BH+

B = Base

BH⁺ = Conjugate acid

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Important Rule

Strong acid → Weak conjugate base

Strong base → Weak conjugate acid

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6. Lewis Concept

According to Lewis:

Lewis Acid

A substance that accepts an electron pair.

Examples:

• BF₃
• AlCl₃
• H⁺

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Lewis Base

A substance that donates an electron pair.

Examples:

• NH₃
• H₂O
• OH⁻

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Example:$$BF_3+NH_3\rightarrow F_3B\leftarrow NH_3$$BF3​+NH3​→F3​B←NH3​

• BF₃ accepts electrons → Lewis acid
• NH₃ donates electrons → Lewis base

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Comparison of Acid-Base Theories

Theory	Acid	Base
Arrhenius	Gives H⁺	Gives OH⁻
Brønsted-Lowry	Proton donor	Proton acceptor
Lewis	Electron pair acceptor	Electron pair donor

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7. Ionisation Constant

Weak electrolytes establish equilibrium in solution.

The equilibrium constant for ionisation is called the ionisation constant.

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Acid Ionisation Constant ($K_a$Ka​)

For weak acid:$$HA\rightleftharpoons H^+ + A^-$$HA⇌H++A− $$K_a=\frac{[H^+][A^-]}{[HA]}$$Ka​=[HA][H+][A−]​

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Meaning of $K_a$Ka​

Large $K_a$Ka​:

• More ionisation
• Stronger acid

Small $K_a$Ka​:

• Less ionisation
• Weaker acid

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8. Base Ionisation Constant ($K_b$Kb​)

For weak base:$$BOH\rightleftharpoons B^+ + OH^-$$BOH⇌B++OH− $$K_b=\frac{[B^+][OH^-]}{[BOH]}$$Kb​=[BOH][B+][OH−]​

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Large $K_b$Kb​:

• Stronger base

Small $K_b$Kb​:

• Weaker base

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9. Relationship Between $K_a$Ka​ and $K_b$Kb​

For a conjugate acid-base pair:$$K_a \times K_b=K_w$$Ka​×Kb​=Kw​

Where:$$K_w=1.0\times10^{-14}$$Kw​=1.0×10−14

(at 25°C)

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10. Degree of Ionisation

Degree of ionisation tells the fraction of electrolyte molecules that form ions.

It is represented by:$$\alpha$$α

Formula:$$\alpha=\frac{\text{Number of ionised molecules}}{\text{Total number of molecules}}$$α=Total number of moleculesNumber of ionised molecules​

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Factors Affecting Ionisation

1. Nature of electrolyte

Strong electrolytes ionise more.

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2. Concentration

Dilution increases ionisation.

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3. Common ion effect

Presence of a common ion decreases ionisation.

Example:$$CH_3COOH\rightleftharpoons H^+ + CH_3COO^-$$CH3​COOH⇌H++CH3​COO−

Adding sodium acetate:$$CH_3COONa\rightarrow Na^+ + CH_3COO^-$$CH3​COONa→Na++CH3​COO−

Extra acetate ions shift equilibrium backward.

Hence ionisation of acetic acid decreases.

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Important Formula Sheet

Acid constant

$$K_a=\frac{[H^+][A^-]}{[HA]}$$Ka​=[HA][H+][A−]​

Base constant

$$K_b=\frac{[BH^+][OH^-]}{[B]}$$Kb​=[B][BH+][OH−]​

Water ionisation

$$K_w=[H^+][OH^-]$$Kw​=[H+][OH−]

Relationship

$$K_aK_b=K_w$$Ka​Kb​=Kw​

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Quick Revision

✅ Acids donate H⁺.
✅ Bases accept H⁺.
✅ Strong electrolytes completely ionise.
✅ Weak electrolytes partially ionise.
✅ $K_a$Ka​ measures acid strength.
✅ $K_b$Kb​ measures base strength.
✅ Higher $K_a$Ka​ = stronger acid.
✅ Common ion decreases ionisation.

Part 5: pH Scale, Water Ionisation, Buffer Solutions & Solubility Product

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1. Ionisation of Water

Water behaves as both an acid and a base. Therefore, it is called amphoteric.

A small number of water molecules ionise:$$H_2O \rightleftharpoons H^+ + OH^-$$H2​O⇌H++OH−

More accurately:$$H_2O+H_2O\rightleftharpoons H_3O^+ + OH^-$$H2​O+H2​O⇌H3​O++OH−

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Ionic Product of Water ($K_w$Kw​)

The equilibrium constant for ionisation of water is called the ionic product of water.$$K_w=[H^+][OH^-]$$Kw​=[H+][OH−]

At 298 K (25°C):$$\boxed{K_w=1.0\times10^{-14}}$$Kw​=1.0×10−14​

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In Pure Water

Water produces equal amounts of hydrogen and hydroxide ions.

Therefore:$$[H^+]=[OH^-]$$[H+]=[OH−]

So,$$[H^+]^2=10^{-14}$$[H+]2=10−14 $$[H^+]=10^{-7}M$$[H+]=10−7M

and$$[OH^-]=10^{-7}M$$[OH−]=10−7M

Hence pure water is neutral.

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2. pH Scale

The concentration of hydrogen ions is represented by pH.

Definition

$$\boxed{pH=-\log[H^+]}$$pH=−log[H+]​

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Similarly,$$\boxed{pOH=-\log[OH^-]}$$pOH=−log[OH−]​

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Relationship Between pH and pOH

Since:$$K_w=[H^+][OH^-]$$Kw​=[H+][OH−]

Taking logarithm:$$\boxed{pH+pOH=14}$$pH+pOH=14​

(at 25°C)

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pH Scale

The pH scale generally ranges from 0 to 14.

pH Value	Nature
Less than 7	Acidic
Equal to 7	Neutral
More than 7	Basic

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Acidic Solution

$$[H^+]>[OH^-]$$[H+]>[OH−]

Example:

HCl solution

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Basic Solution

$$[OH^-]>[H^+]$$[OH−]>[H+]

Example:

NaOH solution

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3. Strong Acids and Strong Bases

Strong Acid

Completely ionises in water.

Example:$$HCl\rightarrow H^++Cl^-$$HCl→H++Cl−

For strong acids:$$[H^+]=\text{acid concentration}$$[H+]=acid concentration

Example:

If HCl concentration = $10^{-3}M$10−3M$$pH=3$$pH=3

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Strong Base

Completely dissociates.

Example:$$NaOH\rightarrow Na^++OH^-$$NaOH→Na++OH−

For strong bases:$$[OH^-]=\text{base concentration}$$[OH−]=base concentration

Then:$$pOH=-\log[OH^-]$$pOH=−log[OH−]

and$$pH=14-pOH$$pH=14−pOH

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4. Weak Acids and Weak Bases

Weak electrolytes ionise partially.

Example:$$CH_3COOH\rightleftharpoons H^++CH_3COO^-$$CH3​COOH⇌H++CH3​COO−

For weak acids:$$K_a=\frac{[H^+][A^-]}{[HA]}$$Ka​=[HA][H+][A−]​

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For weak bases:$$K_b=\frac{[BH^+][OH^-]}{[B]}$$Kb​=[B][BH+][OH−]​

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Acid Strength and $pK_a$pKa​

To make calculations easier:$$\boxed{pK_a=-\log K_a}$$pKa​=−logKa​​

Relationship:

• Larger $K_a$Ka​ → Smaller $pK_a$pKa​
• Stronger acid → Smaller $pK_a$pKa​

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Base Strength and $pK_b$pKb​

$$\boxed{pK_b=-\log K_b}$$pKb​=−logKb​​

Relationship:

• Larger $K_b$Kb​ → Smaller $pK_b$pKb​
• Stronger base → Smaller $pK_b$pKb​

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5. Common Ion Effect

Definition

The decrease in ionisation of a weak electrolyte due to the addition of a strong electrolyte containing a common ion is called the common ion effect.

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Example:

Acetic acid:$$CH_3COOH\rightleftharpoons H^++CH_3COO^-$$CH3​COOH⇌H++CH3​COO−

Adding sodium acetate:$$CH_3COONa\rightarrow Na^++CH_3COO^-$$CH3​COONa→Na++CH3​COO−

The concentration of $CH_3COO^-$CH3​COO− increases.

According to Le Chatelier’s principle:

Equilibrium shifts backward.

Therefore:

• Ionisation of acetic acid decreases.
• More CH₃COOH remains unionised.

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Applications of Common Ion Effect

1. Preparation of buffer solutions.
2. Control of solubility of salts.
3. Qualitative analysis of ions.

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6. Buffer Solutions

Definition

A buffer solution is a solution that resists change in pH when a small amount of acid or base is added.

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Types of Buffers

(A) Acidic Buffer

Contains:

• Weak acid
• Salt of its conjugate base

Example:$$CH_3COOH+CH_3COONa$$CH3​COOH+CH3​COONa

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(B) Basic Buffer

Contains:

• Weak base
• Salt of its conjugate acid

Example:$$NH_4OH+NH_4Cl$$NH4​OH+NH4​Cl

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How Buffer Works

Acidic Buffer

Example:$$CH_3COOH/CH_3COO^-$$CH3​COOH/CH3​COO−

When acid is added:

Extra $H^+$H+ reacts with acetate ions.$$CH_3COO^-+H^+\rightarrow CH_3COOH$$CH3​COO−+H+→CH3​COOH

pH changes very little.

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When base is added:

OH⁻ reacts with acid:$$OH^-+CH_3COOH\rightarrow CH_3COO^-+H_2O$$OH−+CH3​COOH→CH3​COO−+H2​O

Again pH remains almost constant.

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Henderson-Hasselbalch Equation

For acidic buffer:$$\boxed{pH=pK_a+\log\frac{[Salt]}{[Acid]}}$$pH=pKa​+log[Acid][Salt]​​

or$$pH=pK_a+\log\frac{[A^-]}{[HA]}$$pH=pKa​+log[HA][A−]​

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For basic buffer:$$\boxed{pOH=pK_b+\log\frac{[Salt]}{[Base]}}$$pOH=pKb​+log[Base][Salt]​​

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7. Solubility Equilibrium

Some ionic solids dissolve only slightly in water.

Example:$$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$$AgCl(s)⇌Ag+(aq)+Cl−(aq)

At equilibrium:

Rate of dissolution = Rate of precipitation

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Solubility Product ($K_{sp}$Ksp​)

The equilibrium constant for a sparingly soluble salt is called the solubility product constant.

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For:$$AB(s)\rightleftharpoons A^+ + B^-$$AB(s)⇌A++B− $$\boxed{K_{sp}=[A^+][B^-]}$$Ksp​=[A+][B−]​

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Example:

For calcium fluoride:$$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$$CaF2​(s)⇌Ca2++2F− $$\boxed{K_{sp}=[Ca^{2+}][F^-]^2}$$Ksp​=[Ca2+][F−]2​

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Factors Affecting Solubility

1. Common Ion Effect

Adding a common ion decreases solubility.

Example:

AgCl solubility decreases when NaCl is added.

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2. Temperature

Solubility may increase or decrease depending on whether dissolution is endothermic or exothermic.

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Important Formula Sheet

Ionic Product of Water

$$K_w=[H^+][OH^-]$$Kw​=[H+][OH−]

pH

$$pH=-\log[H^+]$$pH=−log[H+]

pOH

$$pOH=-\log[OH^-]$$pOH=−log[OH−]

Relationship

$$pH+pOH=14$$pH+pOH=14

Acid Constant

$$K_a=\frac{[H^+][A^-]}{[HA]}$$Ka​=[HA][H+][A−]​

Base Constant

$$K_b=\frac{[BH^+][OH^-]}{[B]}$$Kb​=[B][BH+][OH−]​

Buffer Equation

$$pH=pK_a+\log\frac{Salt}{Acid}$$pH=pKa​+logAcidSalt​

Solubility Product

$$K_{sp}=[ions]^{coefficients}$$Ksp​=[ions]coefficients

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Complete Chapter Quick Revision

✅ Equilibrium is dynamic.
✅ $K_c$Kc​ shows extent of reaction.
✅ $K_p=K_c(RT)^{\Delta n}$Kp​=Kc​(RT)Δn.
✅ Catalyst does not change equilibrium position.
✅ Temperature changes equilibrium constant.
✅ Acids donate H⁺.
✅ Bases accept H⁺.
✅ $K_a$Ka​ measures acid strength.
✅ pH measures hydrogen ion concentration.
✅ Buffers resist pH change.
✅ $K_{sp}$Ksp​ represents solubility of sparingly soluble salts.