Equilibrium Class 11 Chemistry Notes for NEET & JEE

We Need To Cover These Topics😊 –

S. No.	Topic	Subtopics
1	Introduction to Equilibrium	– Meaning of Equilibrium – Physical vs Chemical Equilibrium
2	Equilibrium in Physical Processes	– Solid ⇌ Liquid Equilibrium – Liquid ⇌ Vapour Equilibrium – Solid ⇌ Solution Equilibrium – Gas ⇌ Solution Equilibrium
3	Equilibrium in Chemical Processes	– Dynamic Nature of Chemical Equilibrium – Equilibrium in Reversible Reactions – Law of Mass Action
4	Law of Equilibrium	– Equilibrium Constant (Kc and Kp) – Relation between Kp and Kc – Units of Equilibrium Constant
5	Homogeneous and Heterogeneous Equilibrium	– Definition and Examples – Expressions for Kc and Kp for both cases
6	Applications of Equilibrium Constant	– Predicting Direction of Reaction – Calculation of Equilibrium Concentrations – Extent of Reaction
7	Ionic Equilibrium in Aqueous Solutions	– Weak and Strong Electrolytes – Arrhenius Concept – Brønsted-Lowry Concept – Lewis Concept (Brief Intro)
8	Ionization of Acids and Bases	– Ionization Constants of Acids (Ka) and Bases (Kb) – pH and pOH – Relationship between Ka and Kb
9	Buffer Solutions	– Definition and Working – Buffer Action – Henderson’s Equation
10	Common Ion Effect	– Definition and Mechanism – Applications in Salt Solubility and Buffer Systems
11	Salt Hydrolysis	– Hydrolysis of Salts of Strong Acid-Strong Base, Weak Acid-Strong Base, Strong Acid-Weak Base – Hydrolysis Constant
12	Solubility Product (Ksp)	– Definition and Expression – Common Ion Effect on Solubility
13	Le Chatelier’s Principle	– Principle Explanation – Effect of Change in Concentration, Temperature, and Pressure – Applications to Industrial Processes (Haber’s Process)

1. Introduction to Equilibrium

🔷 1. What is Equilibrium ?

In chemistry, equilibrium is a state in which the rate of the forward reaction becomes equal to the rate of the backward reaction. At this point, the concentration of reactants and products remains constant over time, though the reaction still continues at the molecular level. This is why equilibrium is called a dynamic state.

🔸 Real-Life Examples of Equilibrium:

A sealed water bottle with some water and water vapour inside. Here, water evaporates and condenses at equal rates, so the amount of water stays the same.
In a closed bottle of soda, carbon dioxide is in equilibrium between the gas above the liquid and the gas dissolved in the liquid.

🔷 2. Characteristics of Equilibrium

S. No.	Characteristic	Explanation
1	Dynamic State	Both forward and backward reactions are taking place at the same rate.
2	No Observable Change	Concentrations of reactants and products do not change over time.
3	Requires Closed System	No loss or gain of matter; must be a sealed or isolated system.
4	Reached from Either Direction	Whether you start with reactants or products, the system eventually reaches the same equilibrium.
5	Affects by External Conditions	Changes in pressure, temperature, or concentration can shift the equilibrium position.

🔷 3. Types of Equilibrium

Equilibrium can be divided into two main types:

✅ A. Physical Equilibrium

✅ B. Chemical Equilibrium

Let’s understand each of them in detail:

🔶 A. Physical Equilibrium

✳️ Definition:

Physical equilibrium is the equilibrium between different physical states (solid, liquid, gas) of the same substance.

In physical equilibrium, no chemical change occurs. Only physical changes like melting, boiling, or dissolving are involved.

✳️ Types of Physical Equilibrium:

1. Solid ⇌ Liquid Equilibrium

Example:
Ice ⇌ Water at 0°C
When ice melts to water and water freezes to ice at the same rate, an equilibrium is formed.
Explanation:
At 0°C and 1 atm pressure, both ice and water can exist in equilibrium.
The rate of melting = rate of freezing.

2. Liquid ⇌ Vapour Equilibrium

Example:
Water ⇌ Water vapour in a closed container at a given temperature.
Explanation:
Some water molecules evaporate, and some vapour molecules condense.
After a while, the rate of evaporation equals the rate of condensation.

3. Solid ⇌ Vapour Equilibrium

Example:
Iodine (s) ⇌ Iodine vapour in a closed container.
Explanation:
Some solid iodine changes to vapour while some vapour deposits back as solid.
At equilibrium, both processes occur at the same rate.

4. Gas ⇌ Solution Equilibrium

Example:
CO₂ (g) ⇌ CO₂ (aq) in a carbonated drink.
Explanation:
Gaseous CO₂ dissolves in water, and some of it escapes as gas.
When both processes occur at the same rate, equilibrium is reached.

🔸 Key Points of Physical Equilibrium:

Involves no chemical change.
Occurs in a closed system.
Equilibrium is reached when the rate of forward physical change = rate of reverse physical change.
It depends on temperature and pressure.

🔶 B. Chemical Equilibrium

✳️ Definition:

Chemical equilibrium is the state in a reversible chemical reaction when the rate of forward reaction equals the rate of backward reaction, and the concentrations of reactants and products remain constant.

It involves a chemical transformation with new products formed.

✳️ Reversible vs Irreversible Reactions

Type	Definition	Example
Irreversible	Proceeds in one direction only; products do not react back.	Na + Cl → NaCl
Reversible	Can proceed in both directions – forward and backward.	N₂ + 3H₂ ⇌ 2NH₃

✳️ Example of Chemical Equilibrium:

Let’s take the Haber’s Process (industrial synthesis of ammonia):

N_2(g)+ 3H_2(g)⇌ 2NH_3(g)

At equilibrium:
- The rate at which N₂ and H₂ form NH₃ = the rate at which NH₃ decomposes back into N₂ and H₂.
- Concentrations of all three gases remain constant (but not necessarily equal).

🔸 Important Terms in Chemical Equilibrium:

Term	Meaning
Dynamic Equilibrium	Forward and backward reactions continue, but concentrations remain unchanged.
Equilibrium Mixture	The mixture containing both reactants and products at equilibrium.
Equilibrium Constant	Denoted by K_c or K_p, it expresses the ratio of concentrations/pressures.

✳️ Graphical Representation of Chemical Equilibrium:

Imagine a graph with time on the x-axis and concentration on the y-axis.

Initially, reactants decrease and products increase.
At equilibrium, both lines flatten (become constant), but they don’t overlap.

This shows that even though molecules are reacting, their overall concentrations remain constant.

🔸 Comparison Table: Physical vs Chemical Equilibrium

Aspect	Physical Equilibrium	Chemical Equilibrium
Nature of Process	Involves physical changes (e.g., melting, boiling).	Involves chemical changes (breaking/making bonds).
Substances Involved	Same substance in different states.	Different substances – reactants and products.
Change in Composition	No change in chemical composition.	Chemical composition changes.
Reversibility	Easily reversible.	Reversible but may need external control (temperature, pressure, etc.).
Example	Ice ⇌ Water	N₂ + 3H₂ ⇌ 2NH₃
Affected By	Temperature and pressure.	Temperature, pressure, concentration, catalyst.

🔷 4. Dynamic Nature of Equilibrium

Even when a system is at equilibrium:

Reactions are still occurring.
Molecules of reactants continue to convert to products and vice versa.
But since both rates are equal, the macroscopic (overall) appearance remains unchanged.

That’s why equilibrium is not static, but dynamic.

🔸 How to Prove Chemical Equilibrium is Dynamic?

Consider this experiment:

Take labelled hydrogen gas (D₂) and nitrogen gas (N₂).
Carry out the Haber process.

Even at equilibrium, you’ll find labelled NH₃ molecules being formed, proving that the reaction still continues.

🔷 5. Conditions Required for Equilibrium

To achieve equilibrium, certain conditions must be met:

Condition	Explanation
Closed System	No addition or removal of substances (mass is conserved).
Reversible Reaction	The reaction should be reversible (⇌ symbol used).
Constant Temperature	Temperature must be maintained as it affects reaction rates.
Time	Sufficient time is needed to reach equilibrium.

🔷 6. Importance of Studying Equilibrium

Industrial Processes: Helps in optimizing reactions like ammonia synthesis, sulphuric acid production, etc.
Biology & Medicine: Blood pH and oxygen transport in the body rely on equilibrium.
Environmental Science: Equilibrium explains CO₂ balance in oceans and atmosphere.
Daily Life: Carbonation in drinks, salt solubility, and chemical preservatives rely on equilibrium principles.

🔷 7. Key Equilibrium Terminologies You Must Know

Term	Meaning
Forward Reaction	Reaction going from reactants to products.
Backward Reaction	Reaction going from products to reactants.
Equilibrium Mixture	Mixture of reactants and products at equilibrium.
Shifting of Equilibrium	Change in external conditions causes system to shift (Le Chatelier’s Principle).

🔷 8. Conclusion

The concept of equilibrium is a fundamental idea in chemistry, helping us understand how reactions behave in a reversible system. Whether it is a physical process like the melting of ice or a chemical reaction like ammonia formation, equilibrium shows us that balance is not the absence of change, but a perfect balance between two opposing processes.

Understanding the difference between physical and chemical equilibrium is the foundation to studying more advanced topics like ionic equilibrium, acid-base theories, solubility products, and Le Chatelier’s Principle, which are all essential for NEET/JEE success.

2. Equilibrium in Physical Processes

🔷 1. Introduction to Physical Equilibrium

Physical equilibrium refers to a state where a physical change is occurring in both directions at equal rates, resulting in no net observable change in the system.

🔸 Key Features:

No new substance is formed.
Involves phase changes like melting, boiling, dissolving etc.
Must occur in a closed system to maintain equilibrium.

🔷 2. Solid ⇌ Liquid Equilibrium

🔸 Example:

Ice_(solid) ⇌ Water_(liquid)at 0^°C

At 0°C, ice melts into water and water freezes back into ice at the same rate. This is an example of dynamic equilibrium.

🔸 Key Concepts:

Equilibrium is achieved only at melting/freezing point.
Rate of melting = rate of freezing.
Energy is exchanged but temperature remains constant.
Occurs in a closed system with both phases present.

🔸 Effect of Temperature:

Increasing temperature shifts equilibrium towards liquid.
Decreasing temperature shifts it towards solid.

🔸 Effect of Pressure:

For most substances, pressure has little effect on solid-liquid equilibrium.
Exception: Water expands on freezing, so increased pressure favors liquid phase.

🔷 3. Liquid ⇌ Vapour Equilibrium

🔸 Example:

Water_(liquid) ⇌ Water vapour_(gas)in a closed container

In a closed container, water evaporates and condenses simultaneously. After some time, rate of evaporation = rate of condensation → equilibrium is achieved.

🔸 Concepts to Know:

✅ Saturated Vapour Pressure (SVP):

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

✅ Boiling Point:

Temperature at which vapour pressure = atmospheric pressure.

🔸 Factors Affecting Liquid-Vapour Equilibrium:

Factor	Effect
Temperature	Higher temp → More evaporation → Higher SVP
Pressure	Higher external pressure → Condensation favored
Surface Area	Does not affect equilibrium position, only affects rate of evaporation.

🔸 Applications:

Pressure cookers
Refrigeration
Cloud formation and rain cycle

🔷 4. Solid ⇌ Solution Equilibrium

This refers to equilibrium between an undissolved solid and its saturated solution.

🔸 Example:

NaCl_(solid) ⇌ Na^⁺_(aq) + Cl^⁻_(aq)

When salt is added to water, it dissolves. But after a point, no more dissolves. This is the saturation point, and equilibrium is established.

🔸 Key Concepts:

Solubility = Amount of solute that dissolves in a given amount of solvent at a given temperature.
Dynamic Equilibrium: Dissolution = Crystallization
Only undissolved solid and ions in solution are involved.

🔸 Factors Affecting Solid-Liquid Equilibrium:

Factor	Effect
Temperature	Usually, ↑ temp → ↑ solubility (exceptions like CaSO₄ exist).
Nature of Solute & Solvent	Polar solutes dissolve better in polar solvents.
Common Ion Effect	Presence of a common ion ↓ solubility (based on Le Chatelier’s Principle).

🔸 Example Questions (NEET/JEE):

What happens to solubility when temperature is increased ?
How does NaCl behave in saturated solution at equilibrium ?

🔷 5. Gas ⇌ Solution Equilibrium

This equilibrium occurs when a gas is in dynamic balance with its dissolved form in a liquid.

🔸 Example:

CO₂_(g) ⇌ CO₂_(aq) in soda bottles

In carbonated drinks, CO₂ gas is dissolved under pressure. When opened, the pressure drops, gas escapes, disturbing equilibrium.

🔸 Henry’s Law:

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

C = k_P⋅P

Where:

C = concentration of dissolved gas
k_P = Henry’s constant (depends on gas and solvent)
P = partial pressure of gas

🔸 Factors Affecting Gas-Liquid Equilibrium:

Factor	Effect
Pressure	↑ Pressure → More gas dissolves → Equilibrium shifts toward dissolved phase
Temperature	↑ Temp → ↓ Solubility (exothermic process)
Nature of Gas	CO₂ is more soluble than O₂

🔸 Applications of Gas-Liquid Equilibrium:

Soda Bottling: CO₂ is pumped at high pressure → more dissolution.
Scuba Diving: Higher pressure underwater → more N₂ dissolves in blood. On rapid surfacing, it forms bubbles → “Decompression Sickness” (The Bends).
Aquatic Life: O₂ dissolves in water from air and supports fish and aquatic plants.

🧠 NEET/JEE Advanced Insights

🔸 High-Level Observations:

Concept	Advanced Understanding
Le Chatelier’s Principle	Predicts how equilibrium shifts with pressure/temp change.
Vapour Pressure Graphs	Used to determine boiling points and intermolecular force strength.
Henry’s Law Constants (k_H)	Lower kH → Higher solubility of gas.
Salt Solubility Equilibria	Linked with Solubility Product (Ksp) in ionic equilibrium.

✅ Summary Table: All Physical Equilibrium Types

Type of Equilibrium	Example	Dynamic Process	Main Controlling Factor
Solid ⇌ Liquid	Ice ⇌ Water	Melting ⇌ Freezing	Temperature
Liquid ⇌ Vapour	Water ⇌ Water Vapour	Evaporation ⇌ Condensation	Temperature, Pressure
Solid ⇌ Solution	NaCl ⇌ Na⁺ + Cl⁻ in water	Dissolution ⇌ Crystallization	Temperature, Common Ion
Gas ⇌ Solution	CO₂ (g) ⇌ CO₂ (aq)	Dissolving ⇌ Escaping	Pressure, Temperature

📚 Conclusion

Physical equilibrium is a dynamic balance between physical states or phases of a substance. Understanding these concepts lays a strong foundation for mastering chemical equilibrium, ionic equilibrium, and applications like Le Chatelier’s Principle, Ksp, and buffer solutions in the next parts of this chapter.

Mastering these topics also helps you:

Tackle NEET/JEE conceptual MCQs
Interpret graphs, data-based questions
Solve problems on solubility, pressure, and temperature dependence

3. Equilibrium in Chemical Processes

🔷 1. Introduction to Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products become constant over time in a reversible reaction.

“At equilibrium, the reaction has not stopped. It continues, but in both directions at the same rate.”

This equilibrium is not static but dynamic in nature.

🔷 2. Dynamic Nature of Chemical Equilibrium

✅ Definition:

At equilibrium, though the concentrations of reactants and products appear constant, the molecules continue to react – forward and backward – at equal rates. Hence, it is dynamic in nature.

✅ Molecular-Level Explanation:

Let’s take a reversible reaction: A + B ⇌ C + D

Initially, only A and B are present → forward reaction happens rapidly.
As C and D form, the backward reaction starts.
Eventually, the rate of forward reaction = rate of backward reaction.

At this point:

Concentrations of A, B, C, D become constant.
But molecular exchange continues (A+B → C+D and vice versa).
Hence, equilibrium is dynamic.

✅ Experimental Evidence:

Isotope Labelling:
- Use isotopes like D₂ (heavy hydrogen) in hydrogenation reactions.
- Even at equilibrium, newly formed products show isotope incorporation → proves ongoing reaction.
Constant Pressure & Concentration:
- Observing that measurable quantities remain unchanged despite ongoing microscopic activity.

🔷 3. Reversible and Irreversible Reactions

✅ Irreversible Reaction:

Proceeds only in one direction.
Reactants are completely converted into products.
Written as: A + B → C

✅ Reversible Reaction:

Proceeds in both directions.
Reactants ⇌ Products
A dynamic equilibrium is achieved in a closed system.

A + B ⇌ C + D

✅ Differences Table:

Feature	Irreversible Reaction	Reversible Reaction
Direction	One-way only	Both forward and backward
Completion	Goes to completion	Does not go to completion
Example	AgNO₃ + NaCl → AgCl + NaNO₃	N₂ + 3H₂ ⇌ 2NH₃
Equilibrium Possible?	❌ No	✅ Yes

🔷 4. Equilibrium in Reversible Reactions

When a reversible reaction proceeds in both directions, eventually the system reaches chemical equilibrium. N_2(g)+ 3H_2(g)⇌ 2NH_3(g)

✅ Rate Behavior:

Stage	Forward Rate	Backward Rate
Start	Maximum	Zero
Intermediate	Decreasing	Increasing
Equilibrium	Equal	Equal

✅ Graphical Representation:

A graph of rate vs time:

Forward rate decreases.
Backward rate increases.
At equilibrium, both curves meet → equal rates.

✅ Concentration Profile:

Another graph: concentration vs time.

[Reactants] decrease, [Products] increase.
At equilibrium, both become constant but not necessarily equal.

✅ Important Notes:

Equilibrium can be approached from both directions (reactants → products or products → reactants).
It is achieved only in a closed system.

🔷 5. Law of Mass Action

✅ Statement:

At constant temperature, the rate of a chemical reaction is proportional to the product of the active masses (concentrations) of the reactants, each raised to a power equal to its stoichiometric coefficient.

✅ Mathematical Expression:

For a general reaction: aA + bB ⇌ cC + dD

At equilibrium: K_c = [C]^c[D]^d/ [A]^a[B]^b

Where:

K_c = Equilibrium constant (concentration form)
= molar concentration
a, b, c, d = stoichiometric coefficients

🔷 6. Units of Kc

Depends on the reaction.

✅ Examples:

For a balanced reaction:
H₂ + I₂ ⇌ 2HI

K_c= [HI]²/[H₂][I₂] ⇒ Units = M

For a reaction where moles of reactants = moles of products, units cancel out → unitless.

🔷 7. Conditions for Law of Mass Action to Apply

Condition	Explanation
Closed system	No substance can enter/leave.
Constant temperature	Since K_c depends on temperature.
Equilibrium must be reached	Only then can you apply the Kc formula.

🔷 8. Applications of Law of Mass Action

✅ Predicting Direction of Reaction:

Use Reaction Quotient (Q): Q = [Products] / [Reactants]

If Q < K → Forward reaction dominates
If Q > K → Backward reaction dominates
If Q = K → Equilibrium achieved

✅ Calculating Unknown Concentrations:

Given K and some concentrations, you can:

Use ICE (Initial, Change, Equilibrium) tables
Solve for x (equilibrium concentrations)

🔷 9. NEET/JEE Level Concept Enhancements

✅ 1. Effect of Temperature:

K changes with temperature.
Exothermic Reactions: ↑ T → ↓ K
Endothermic Reactions: ↑ T → ↑ K

✅ 2. Pressure & Volume:

Affects gaseous equilibria.
More moles on one side? Pressure will shift equilibrium.

✅ 3. Catalyst:

Does not affect K.
Only helps to reach equilibrium faster.

🔷 10. Common Mistakes to Avoid

Mistake	Correction
Using partial pressure in Kc	Use Kp for gases in pressure terms; Kc is for molar concentration.
Assuming equilibrium means equal [ ]	It means equal rates, not equal concentrations.
Ignoring units of Kc	Always determine units based on stoichiometry.

📝 Summary Table

Concept	Key Point
Chemical Equilibrium	Rate of forward = backward; concentrations constant
Dynamic Nature	Reaction continues at molecular level
Reversible Reactions	Necessary for equilibrium
Law of Mass Action	K = [Products]^coeff / [Reactants]^coeff
Equilibrium Constant (Kc)	Temperature-dependent, units vary
Reaction Quotient (Q)	Used to predict direction before equilibrium

📚 Final Words

Understanding chemical equilibrium is central to mastering reaction behavior in chemistry. This section connects deeply with later topics like:

Le Chatelier’s Principle
Ionic Equilibrium
Acids and Bases
Solubility Products

A strong grasp on dynamic nature, mass action law, and equilibrium conditions will prepare you for both theory questions and numerical problems in NEET and JEE.

4. Law of Equilibrium

🔷 1. Equilibrium Constant – K_c (Concentration Form)

✅ Definition:

The equilibrium constant in terms of molar concentrations is denoted by K_c.

For a general reversible reaction: aA + bB ⇌ cC + dD

K_c = [C]^c[D]^d/ [A]^a[B]^b

Where:

= equilibrium concentrations in mol/L (M)
a, b, c, d = stoichiometric coefficients

✅ Key Features of K_c:

Depends only on temperature and reaction stoichiometry.
Independent of initial concentrations.
Units vary based on change in moles (Δn).
Only gases and aqueous species appear in Kc; solids and pure liquids are excluded.

🔷 2. Equilibrium Constant – K_p (Partial Pressure Form)

✅ Definition:

The equilibrium constant in terms of partial pressures of gases is called Kp.

For the gaseous reaction:

aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)

K_p= (P_A)^a⋅(P_B)^b / (P_C)^c⋅(P_D)^d

Where:

P = partial pressure of gases (usually in atm or bar)

✅ When to Use K_c:

For gaseous equilibria
When pressure data is given
Ideal gas behavior assumed

🔷 3. Relation Between K_p and K_c

This is one of the most important formulas in Equilibrium: K_p = K_c(RT)^Δn

✅ Where:

K_p = equilibrium constant in terms of partial pressure
K_c = equilibrium constant in terms of concentration
R = universal gas constant = 0.0821 L·atm/mol·K
T = temperature in Kelvin
Δn = (moles of gaseous products – moles of gaseous reactants)

✅ Understanding Δn:

Reaction Example	Δn Value
H₂ + I₂ ⇌ 2HI	2 – 2 = 0
N₂ + 3H₂ ⇌ 2NH₃	2 – 4 = –2
2SO₂ + O₂ ⇌ 2SO₃	2 – 3 = –1

If Δn = 0, then K_p = K_c

✅ Units of R:

Use R = 0.0821 when pressure is in atm
Use R = 8.314 when using SI units (Pa, m³)

🔷 4. Units of Equilibrium Constants

Equilibrium constants do not always have the same units. They depend on the net change in moles of gaseous substances (Δn).

✅ K_c> Units:

K_c = mol^Δn ⋅L^−Δn

Δn	K_c Units
0	Unitless
+1	mol·L⁻¹
+2	mol²·L⁻²
–1	mol⁻¹·L

✅ K_p Units:

K_p= (Pressure)^Δn

Δn	K_p Units
0	Unitless
+1	atm or bar
+2	atm² or bar²
–1	atm⁻¹ or bar⁻¹

⚠️ Note:

For heterogeneous equilibria, pure solids and liquids do not appear in Kc or Kp expressions.

🔷 6. NEET/JEE Level Tips & Tricks

Tip	Why It Matters
Always calculate Δn carefully	Small mistake changes K<sub>p</sub> calculation
Use R and T in correct units	Otherwise final K<sub>p</sub> value will be incorrect
Units of K_c and K_p depend on Δn	Not always unitless!
For solid/liquid reactions	Exclude solids/pure liquids in K expressions
Use log when needed	JEE often asks questions involving log(K) or ln(K)

🔷 7. Conclusion

The Law of Equilibrium (via equilibrium constants K_c and K_p) is the quantitative foundation of chemical equilibrium.

Mastering how to write K_c and K_p expressions
Understanding their relationship via K_p = K_c (RT)Δn
Knowing how to handle units and conversions

…are all crucial for NEET and JEE success.

5. Homogeneous and Heterogeneous Equilibrium

🔷 1. Homogeneous Equilibrium

✅ Definition:

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

Mostly observed in:

Gaseous systems (all reactants/products are gases)
Aqueous solutions (all species dissolved in water)

✅ Examples:

Gas-phase reaction:

N_2(g)+ 3H_2(g)⇌ 2NH_3(g)

All species are gases → homogeneous equilibrium

Solution-phase reaction:

CH₃COOH_(aq) + H₂O_(l) ⇌ CH₃COO⁻_(aq)+H₃O⁺_(aq)

All aqueous → homogeneous

🔹 Writing Expressions for K_c and K_p:

Let’s take the reaction: SO_2(g)+ NO_2(g)⇌ SO_3(g)+ NO_(g)

K_c expression:

K_c= [SO₃][NO] / [SO₂][NO₂]

K_p expression:

K_p= P_SO3 ⋅ P_NO/ P_SO2 ⋅ P_NO2

✔ All components appear in the equilibrium expression.

🔹 NEET/JEE Tip:

In homogeneous equilibrium, all species (gases or aqueous) are included in Kc and Kp expressions.

🔷 2. Heterogeneous Equilibrium

✅ Definition:

Equilibrium in which reactants and/or products are present in different phases (solid, liquid, gas) is called heterogeneous equilibrium.

✅ Key Features:

Concentration of pure solids/liquids = constant, so they do not appear in the equilibrium expression.
Only gases and aqueous species are considered in K_c and K_p.

🔹 Important Rule:

Pure solids and pure liquids are always omitted from the equilibrium expression because their molar concentration is constant.

🔹 General Form:

For reaction: aA_(s) + bB_(g) ⇌ cC_(g) + dD_(l)

Exclude solids and liquids:

K_c= [C]^c/[B]^b; K_p= (P_C)^c/(P_B)^b

🔷 3. Key Differences Table

Feature	Homogeneous Equilibrium	Heterogeneous Equilibrium
Phases involved	All species in same phase	Species in different phases
Examples	Gaseous or aqueous reactions	Solids, liquids, gases in the same system
Inclusion in K expressions	All species included	Only gases and aqueous species included
Common in	Acid-base, gas reactions	Decomposition, precipitation, phase change reactions

🔷 4. NEET/JEE Advanced Concepts

Concept	Explanation
Pure solids/liquids	Not included in expressions
Activity = 1	The “activity” of solids/liquids is taken as 1, hence omitted
Phase equilibrium	e.g., H₂O (l) ⇌ H₂O (g) – vapor pressure becomes equilibrium constant
K has no units if Δn = 0	Particularly in heterogeneous cases with no net gas mole change
Thermodynamic link	Equilibrium constant can relate to Gibbs free energy (ΔG° = –RT ln K)

🔷 6. Conclusion

Concept	Summary
Homogeneous Equilibrium	All substances in same phase; all appear in K expressions
Heterogeneous Equilibrium	Different phases; solids/pure liquids excluded
K<sub>c</sub> and K<sub>p</sub> Rules	Always write expressions based on gaseous/aqueous species only
Competitive Focus	Know when to exclude terms, and practice writing correct expressions

6. Applications of Equilibrium Constant

🔷 1. Predicting the Direction of Reaction

✅ Key Concept – Reaction Quotient (Q):

The reaction quotient (Q) is a value calculated using current concentrations or pressures, not necessarily at equilibrium.

For a general reaction: aA+bB⇌cC+dD

Q_c= [C]^c[D]^d/[A]^a[B]^b; Q_p= P_C^c⋅P_D^d/P_A^a⋅P_B^b

✅ Compare Q with K to predict direction:

Condition	Interpretation	Reaction Shifts
Q < K	More reactants, less products	Forward direction
Q > K	More products, less reactants	Backward direction
Q = K	System at equilibrium	No shift

🔷 2. Calculation of Equilibrium Concentrations

This is a very common and important application.

✅ Steps to Calculate:

Write the balanced equation
Use ICE table
(Initial, Change, Equilibrium)
Express all equilibrium terms in terms of x
Apply K_c or K_p formula
Solve the algebraic equation
Calculate concentrations

✅ ICE Table Format Example:

Reaction: A+B ⇌ C ; K_c = 4

Species	Initial (mol/L)	Change (mol/L)	Equilibrium (mol/L)
A	1.0	–x	1 – x
B	1.0	–x	1 – x
C	0	+x	x

Now apply: K_c = [C]/[A][B] = x/(1–x)²

Solve the equation to find x, then plug back to find equilibrium concentrations.

✅ Tips:

Use approximations if K is very small or very large.
Be careful with stoichiometric coefficients (e.g., if 2x, 3x instead of x).

🔷 3. Extent of Reaction

✅ What is Extent of Reaction ?

Extent of reaction refers to how far a reaction proceeds before reaching equilibrium.

✅ Use of K_c / K_p to Estimate Extent:

Value of K	Extent of Reaction
K ≫ 1	Reaction almost goes to completion (products favored)
K ≪ 1	Very little reaction occurs (reactants favored)
K ≈ 1	Significant amounts of both reactants and products

📌 Example 1:

For reaction: A ⇌ B, K_c= 10⁶

→ Almost complete reaction → products dominate

📌 Example 2:

For reaction:X ⇌ Y, K_c = 10⁻⁶

→ Very little product formed → reactants dominate

🔷 4. Advanced Concepts for NEET/JEE

✅ 1. Using ΔG° and K:

Free energy change and K are related by: ΔG^∘=−RT ln K

If K > 1 → ΔG° is negative → spontaneous
If K < 1 → ΔG° is positive → non-spontaneous

✅ 2. Equilibrium Position Shifts:

Using Q and K allows us to simulate conditions (adding/removing species) and predict system behavior without needing Le Chatelier’s principle explicitly.

✅ 3. Multiple Reactions:

If two or more reactions share intermediates, their equilibrium constants multiply when reactions are added.

✅ 4. Misconceptions to Avoid:

Mistake	Correct Concept
K tells you speed of reaction	❌ No! K tells extent, not rate
Q = K means equal concentrations	❌ No! It means equal ratio according to stoichiometry
Products always dominate	❌ Not true if K < 1

🔷 5. Summary Table

Concept	Key Takeaway
Reaction Quotient (Q)	Predicts direction by comparing with K
ICE Table	Helps calculate unknown equilibrium concentrations
K and Extent of Reaction	High K → almost complete; Low K → barely reacts
Q < K	Forward reaction favored
Q > K	Backward reaction favored
ΔG and K	Connected through thermodynamics: ΔG° = –RT ln K

📝 Final Words

Mastering the applications of equilibrium constant empowers you to:

Predict how systems respond to changes
Solve for unknown concentrations
Analyze the completeness of reactions
Handle both conceptual and numerical questions in NEET/JEE

7. Ionic Equilibrium in Aqueous Solutions

🔷 Key Topics Covered:

1. Weak and Strong Electrolytes
2. Arrhenius Concept of Acids and Bases
3. Brønsted–Lowry Concept
4. Lewis Concept (Brief Intro)

✅ PART 1: Strong and Weak Electrolytes

📌 What is an Electrolyte ?

An electrolyte is a substance that, when dissolved in water, conducts electricity by forming ions.

🔹 Classification of Electrolytes:

Type	Definition	Examples
Strong Electrolyte	Dissociates completely into ions in water	HCl, NaOH, NaCl, HNO₃, K₂SO₄
Weak Electrolyte	Dissociates partially in water, establishing ionic equilibrium	CH₃COOH, NH₄OH, H₂CO₃, HCN

🔹 Dissociation Equation:

Strong Electrolyte (HCl): HCl_(aq) → H^⁺_(aq) + Cl^⁻_(aq))
- (No equilibrium arrow; complete ionization)
Weak Electrolyte (CH₃COOH): CH₃COOH (aq)⇌CH₃COO⁻ (aq)+H⁺ (aq)
- (Reversible arrow; establishes equilibrium)

🔹 Degree of Ionization (α):

α = Number of molecules ionized / Total molecules present

Strong electrolyte: α ≈ 1
Weak electrolyte: α < 1 (depends on concentration and temperature)

🔹 Important for NEET/JEE:

Weak electrolytes follow Ostwald’s Dilution Law
Strong electrolytes obey conductivity laws, not equilibrium

✅ PART 2: Arrhenius Concept of Acids and Bases

🧪 Proposed by: Svante Arrhenius (1884)

📌 Definition:

Arrhenius Acid	Arrhenius Base
A substance that increases H⁺ ion concentration in water	A substance that increases OH⁻ ion concentration in water

🔹 Examples:

HCl → H⁺ + Cl⁻ (Acid)
NaOH → Na⁺ + OH⁻ (Base)

✅ Limitations of Arrhenius Concept:

Applicable only in aqueous medium
Does not explain acid–base behavior in non-aqueous solvents
Cannot explain basic nature of NH₃, which doesn’t have OH⁻

✅ PART 3: Brønsted–Lowry Concept

🧪 Proposed by: Johannes Brønsted and Thomas Lowry (1923)

📌 Definition:

Brønsted–Lowry Acid	Brønsted–Lowry Base
Proton (H⁺) donor	Proton (H⁺) acceptor

🔹 Examples:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
- NH₃ = base (accepts H⁺)
- H₂O = acid (donates H⁺)
HCl + H₂O → H₃O⁺ + Cl⁻
- HCl = acid (donates H⁺)
- H₂O = base (accepts H⁺)

✅ Conjugate Acid–Base Pair:

Acid and base differ by one proton (H⁺). HA⁺B ⇌ A⁻+HB⁺

Species	Role	Conjugate Pair
HCl	Acid	Cl⁻ (conjugate base)
NH₃	Base	NH₄⁺ (conjugate acid)

✅ Brønsted–Lowry: Key Points for NEET/JEE

Explains reactions in non-aqueous media
NH₃, H₂O can act as both acid and base → amphoteric behavior
Stronger acid = weaker conjugate base

✅ PART 4: Lewis Concept (Brief Introduction)

🧪 Proposed by: G.N. Lewis (1923)

📌 Definition:

Lewis Acid	Lewis Base
Electron pair acceptor	Electron pair donor

🔹 Examples:

BF₃ + NH₃ → F₃B←NH₃
BF₃ = Lewis acid (accepts lone pair)
NH₃ = Lewis base (donates lone pair)
Ag⁺ + :NH₃ → [Ag(NH₃)]⁺
Ag⁺ = Lewis acid
NH₃ = Lewis base

✅ Lewis Concept Highlights:

Most general concept of acid–base behavior
Applies even in absence of H⁺ or OH⁻
Useful in explaining complex formation, metal-ligand reactions

✅ BONUS: Comparison of Acid–Base Theories

Theory	Acid	Base	Medium
Arrhenius	H⁺ producer in water	OH⁻ producer in water	Aqueous only
Brønsted–Lowry	Proton donor	Proton acceptor	Any (including water)
Lewis	Electron pair acceptor	Electron pair donor	Any (most general)

✅ Summary Chart

Concept	Key Feature	NEET/JEE Tip
Strong Electrolyte	Complete ionization	No equilibrium formed
Weak Electrolyte	Partial ionization, reversible reaction	Establishes ionic equilibrium
Arrhenius Theory	H⁺/OH⁻ concept in water	Limited scope
Brønsted–Lowry Theory	Proton donor–acceptor	Explains acid–base in non-aqueous solutions
Lewis Theory	Electron pair donor–acceptor	Explains reactions beyond H⁺/OH⁻

🎯 NEET/JEE Level Add-Ons

Weak electrolytes obey Ostwald’s dilution law
Ionization equilibrium of weak acids is used to define Ka, Kb, and pKa, pKb
Polyprotic acids (e.g., H₂SO₄, H₃PO₄) ionize in steps – important in JEE
Brønsted–Lowry explains buffer solutions, conjugate pairs, and amphiprotic substances
Lewis concept is crucial for coordination chemistry and organic reaction mechanisms

✅ Final Notes

Understanding these acid–base theories is the foundation of ionic equilibrium. Whether it’s predicting the behavior of salts, calculating pH, or understanding buffer action — these concepts are directly applied in:

📘 Chapter: Ionic Equilibrium
🧪 Titration curves and pH changes
🔁 Solubility and common ion effect
⚗️ Redox and electrochemistry

8. Ionization of Acids and Bases

📘 Introduction:

Acids and bases undergo ionization in water, releasing H⁺ (protons) or OH⁻ ions. This ionization is equilibrium-based, and its extent determines the strength of an acid/base. The degree of ionization is quantified using equilibrium constants:

Ka (acid dissociation constant) for acids
Kb (base dissociation constant) for bases

These constants, along with pH, pOH, and their interrelationships, help us understand acid-base behaviour in chemical systems.

🔹 1. Ionization of Acids:

When an acid dissolves in water, it donates a proton (H⁺) to water molecules:

General Reaction: HA_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq)+ A⁻_(aq)

➤ Acid Dissociation Constant (Ka):

The equilibrium constant for ionization of a weak acid is:

Ka = [H₃O⁺][A⁻]/[HA]

➤ Ka and Acid Strength:

Ka Value	Acid Strength
Large Ka (>1)	Strong Acid (almost complete ionization)
Small Ka (<1)	Weak Acid (partial ionization)

🔹 2. Ionization of Bases:

Bases accept protons or release OH⁻ in aqueous solution.

General Reaction (for weak base B):

B_(aq) + H₂O_(l) ⇌ BH⁺_(aq)+ OH⁻_(aq)

➤ Base Dissociation Constant (Kb):

K_b= [BH⁺][OH⁻]/[B]

➤ Kb and Base Strength:

Kb Value	Base Strength
Large Kb (>1)	Strong Base
Small Kb (<1)	Weak Base

🔹 3. Degree of Ionization (α):

The extent to which an acid or base ionizes:

α = Number of molecules ionized / Total number of molecules present initially

For weak acids and bases: α≪1
Degree of ionization increases:
- In dilute solutions
- With weaker conjugates of the acid/base
- With higher temperature

🔹 4. pH and pOH – The Measure of Acidity and Basicity

🔷 pH (Power of Hydrogen):

pH = −log₁₀[OH⁺]

🔷 pOH (Power of Hydroxide):

pOH = −log₁₀[OH⁻]

📌 Important Relation:

At 25°C (298 K), pH + pOH = 14

🔷 Concentration to pH Conversion:

[H⁺] (mol/L)	pH	Nature of Solution
1	0	Strongly Acidic
10⁻⁷	7	Neutral
10⁻¹⁴	14	Strongly Basic

🔷 For Weak Acid/Base:

If Ka and initial concentration (C) of acid is known:

[H⁺] ≈ (K_a⋅C)^1/2 ⇒ pH = −log(K_a⋅C)

Similarly for base:[OH⁻] ≈ (K_b⋅C)^1/2 ⇒ pOH = −log[(K_b⋅C)]^1/2

🔹 5. Relationship between K_a and K_b

Let HA be a weak acid and A⁻ its conjugate base.

For acid: HA + H₂O ⇌ H₃O⁺ + A⁻(K_a)

For conjugate base: A⁻+ H₂O ⇌ HA + OH⁻(K_b)

Multiplying the two equilibrium expressions:

K_a⋅K_b = [H₃O⁺][A⁻]⋅[HA][OH⁻]/[HA][A⁻]=[H₃O⁺][OH⁻]=K_w

✅ Final Relation:

K_a⋅K_b = K_w = 1×10⁻¹⁴at 25°C

🔷 In Terms of pKa and pKb:

pK_a + pK_b= pK_w = 14

🔹 6. Effect of Dilution on Ka and Kb

Ka and Kb are constant for a given acid/base at a specific temperature.
Degree of ionization (α) increases with dilution (due to Le Chatelier’s Principle).
Strong acids/bases remain almost fully ionized even after dilution.

🔹 7. Strong vs Weak Acids and Bases

➤ Strong Acids (Completely ionize):

HCl, HNO₃, H₂SO₄, HBr, HI

➤ Weak Acids (Partially ionize):

CH₃COOH, HF, HCN, H₂CO₃

➤ Strong Bases:

NaOH, KOH, Ba(OH)₂, Ca(OH)₂

➤ Weak Bases:

NH₃, C₂H₅NH₂, (CH₃)₂NH

🔹 8. Application of K_a and K_b in Salt Hydrolysis

When salts of weak acids or weak bases are dissolved in water, they undergo hydrolysis.

Salt of Weak Acid + Strong Base (e.g., CH₃COONa):
- Hydrolysis by anion (CH₃COO⁻)
- Solution becomes basic.
Salt of Strong Acid + Weak Base (e.g., NH₄Cl):
- Hydrolysis by cation (NH₄⁺)
- Solution becomes acidic.
Salt of Weak Acid + Weak Base:
- Both ions hydrolyze.
- pH depends on Ka and Kb: pH=7+ ½ (pK_b−pK_a)

🔹 9. Polyprotic Acids and Stepwise Ionization

Some acids donate more than one proton:

Example: H₂SO₄, H₂CO₃, H₃PO₄

Each step has a separate Ka:

K_a1> K_a2>K_a3

🔹 10. Important Formulas Summary

Quantity	Formula
Ka	[A⁻][HA] / [H⁺]
Kb	[OH⁻][BH⁺]/[B]
pH	−log₁₀[H⁺]
pOH	−log₁₀[OH⁻]
pH + pOH	= 14 (at 25°C)
Ka × Kb	Kw = 1.0 × 10⁻¹⁴
pKa + pKb	= 14
[H⁺] (for weak acid)	(K_a⋅C)^1/2
[OH⁻] (for weak base)	(K_b⋅C)^1/2
pH (salt of weak acid/base)	7+ ½ (pK_b−pK_a)

🎯 Conclusion

Understanding ionization, Ka, Kb, pH, and their interrelations is crucial for solving NEET/JEE-level equilibrium, salt hydrolysis, buffer, and titration problems. Mastery of these concepts empowers you to:

Predict acidic/basic nature
Calculate pH/pOH of any solution
Understand salt behavior in water
Apply equilibrium concepts logically

9. Buffer Solutions

🔷 1. What is a Buffer Solution ?

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

In simple terms: Buffer ek aisa solution hota hai jo pH ko stable banaye rakhta hai, chahe aap usme thoda acid ya base daal do.

🔷 2. Types of Buffer Solutions

✅ a) Acidic Buffer:

Contains a weak acid + its salt with a strong base.
Example:
- CH₃COOH + CH₃COONa (Acetic acid + Sodium acetate)

✅ b) Basic Buffer:

Contains a weak base + its salt with a strong acid.
Example:
- NH₄OH + NH₄Cl (Ammonium hydroxide + Ammonium chloride)

🔷 3. Working Principle of Buffer Solution

🎯 Buffer Action – How Buffers Resist pH Change

Let’s take an acidic buffer example:

CH₃COOH + CH₃COONa

This mixture has:

Weak acid (CH₃COOH) ⇌ H⁺ + CH₃COO⁻
Salt (CH₃COONa) → Na⁺ + CH₃COO⁻ (provides extra acetate ions)

👉 When a Small Amount of Acid (H⁺) is Added:

Extra H⁺ ions react with CH₃COO⁻ (acetate ions):
- CH₃COO^– + H⁺ → CH₃COOH
So free H⁺ ions are consumed → pH remains almost same

👉 When a Small Amount of Base (OH⁻) is Added:

OH⁻ reacts with CH₃COOH:
- CH₃COOH + OH⁻ → CH₃COO⁻+H₂O
So OH⁻ gets neutralized → pH change is resisted.

📌 Conclusion: Buffer ka kaam hai H⁺ aur OH⁻ ions ko neutralize kar dena, taaki pH mein zyada badlaav na aaye.

🔷 4. Henderson’s Equation – pH of Buffer

Henderson–Hasselbalch Equation gives the pH of buffer solution:

✅ For Acidic Buffer:

pOH = pK_b + log₁₀([Salt]/[Acid])

Where:

pK_b = -log(Ka) of weak acid
[Salt] = concentration of salt (like CH₃COONa)
[Acid] = concentration of weak acid (like CH₃COOH)

✅ For Basic Buffer:

pOH = pK_b + log₁₀([Salt]/[Base])

Then: pH=14−pOH

🔷 5. Derivation of Henderson’s Equation (Acidic Buffer)

Let’s take:

Weak acid: HA
Its salt with strong base: A⁻

Equilibrium: HA ⇌ H⁺ + A⁻

Ka expression:

K_a = [H⁺][A⁻]/[HA]

Rearranging:

[H⁺] = K_a⋅ [HA]/[A⁻]

Taking log:

log[H⁺]= logK_a + log ([HA]/[A⁻])

Multiply by -1:

−log[H⁺]= −logK_a – log ([HA]/[A⁻])

pH = pK_a + log ([A⁻]/[HA])

Hence,

pH = pK_a + log([Salt]/[Acid])

🔷 6. Characteristics of Buffer Solutions

Resists change in pH when small acid/base is added.
Works best when [Salt] = [Acid] or [Salt] = [Base]
pH = pKa in that case (optimum buffering capacity)
Buffering capacity increases with concentration.

🔷 7. Applications of Buffer Solutions

Application Area	Use of Buffer
Biological Systems	Blood buffer maintains pH ~ 7.4
Pharmaceuticals	Stabilize drug pH in medicines
Chemical Reactions	Maintain constant pH for proper yield
Food Industry	Maintain taste and shelf life
Electroplating, Titrations	Used for maintaining medium pH

🔷 8. Blood as a Natural Buffer

Blood uses the carbonic acid–bicarbonate buffer system:

H₂CO₃ ⇌ H⁺+ HCO³⁻

Regulates blood pH around 7.35–7.45
Prevents sudden changes due to metabolic acids

🔷 9. Buffer Capacity (β)

Buffer capacity is the amount of acid/base a buffer can neutralize without significant pH change. Β = dB/d(pH)

Greater concentration → Greater buffer capacity
Maximum at pH = pKa

🔷 10. NEET/JEE Level Key Notes Recap

Concept	Formula/Key Point
Buffer Definition	Resists pH change
Acidic Buffer	Weak acid + Salt of that acid
Basic Buffer	Weak base + Salt of that base
Acidic Buffer pH	pH = pKa + log([Salt]/[Acid])
Basic Buffer pOH	pOH = pKb + log([Salt]/[Base])
Best Buffering	When [Salt] = [Acid/Base] ⇒ pH = pKa
Buffer Capacity	Maximum at pH = pKa
Blood Buffer	H₂CO₃ / HCO₃⁻

10. Common Ion Effect

🔷 1. Introduction – What is the Common Ion Effect?

📚 Definition:

Common Ion Effect is the suppression of the ionization of a weak electrolyte (weak acid/base) by the addition of a strong electrolyte having a common ion.

Simple Words Mein: Agar kisi weak acid ya base ke solution mein koi aisi salt daal di jaye jisme usi acid/base ka ion ho, toh uska ionization kam ho jaata hai. Is phenomenon ko Common Ion Effect kehte hain.

🔁 General Example:

Weak electrolyte: CH₃COOH ⇌ H⁺ + CH₃COO⁻
Add salt: CH₃COONa → Na⁺ + CH₃COO⁻

Now CH₃COO⁻ (common ion) increases, shifting the equilibrium left → suppressing ionization of CH₃COOH.

🔷 2. Mechanism – How Does It Work? (Le Chatelier’s Principle)

Common Ion Effect is governed by Le Chatelier’s Principle, which says:

If equilibrium is disturbed by adding more of a product, the system shifts to oppose the disturbance.

In the case of CH₃COOH + CH₃COONa:

Adding CH₃COONa increases CH₃COO⁻
Reaction shifts left to reduce [CH₃COO⁻]
Hence, less H⁺ is produced → ionization suppressed

📉 Result:

Decrease in [H⁺] → increase in pH
Acid becomes even weaker

🔷 3. Mathematical Understanding

For weak acid HA:

HA ⇌ H⁺+ A⁻

Ka expression:

K_a= [H⁺][A⁻]/[HA]

If [A⁻] increases due to added salt → [H⁺] decreases to keep Ka constant.

Similarly, for weak base:

B+ H₂O ⇌ BH⁺ + OH⁻

Adding BH⁺ (from salt) reduces [OH⁻] → base becomes weaker.

🔷 4. Common Ion Effect in Salt Solubility (Solubility Equilibria)

When a sparingly soluble salt is dissolved in water, it establishes an equilibrium between solid and its ions.

⚛️ Example:

AgCl ⇌ Ag⁺ + Cl⁻

Adding NaCl (source of common Cl⁻ ion):
- Increases [Cl⁻]
- Equilibrium shifts left
- Precipitation of AgCl increases, solubility decreases

🧠 Important Rule:

Presence of a common ion suppresses solubility of a salt (by shifting equilibrium to the left).

✅ Solubility vs Solubility Product (K_sp):

K_sp remains constant at a given temperature.
Solubility decreases due to added common ion.

🔷 5. Applications of Common Ion Effect

📌 a) In Buffer Solutions:

Buffer = Weak acid/base + Salt with common ion

CH₃COOH + CH₃COONa
→ CH₃COO⁻ is the common ion
→ Suppresses ionization of CH₃COOH
→ Maintains constant pH
NH₄OH + NH₄Cl
→ NH₄⁺ common ion suppresses NH₄OH ionization
→ pH remains stable

📌 b) In Salt Analysis (Qualitative Analysis):

Precipitation of Group I cations using HCl:
- Common Cl⁻ from HCl suppresses solubility of chlorides like AgCl, PbCl₂
In Group II analysis, H₂S is passed in acidic medium:
- Common H⁺ from HCl suppresses H₂S ionization
- Decreases S²⁻ concentration → only least soluble sulphides precipitate

📌 c) In Industrial and Laboratory Precipitation:

Purification of salts: Adding common ion helps in complete precipitation of impurities
Crystallization: Common ions can reduce solubility to help form pure crystals

🔷 6. NEET/JEE-Level Numerical Example

Q. The solubility of AgCl in water is 1.3 × 10⁻⁵ mol/L. Calculate its solubility in 0.1 M NaCl.

Given:

AgCl ⇌ Ag⁺ + Cl⁻
Ksp = [Ag⁺][Cl⁻] = (1.3 × 10⁻⁵)² = 1.69 × 10⁻¹⁰

Now in 0.1 M NaCl, [Cl⁻] = 0.1 M

So,

K_sp =[Ag⁺][Cl⁻] ⇒ 1.69 × 10⁻¹⁰ = [Ag^+]⋅0.1

[Ag⁺]= (1.69 × 10⁻¹⁰)/0.1 = 1.69 × 10⁻⁹ mol/L

📉 Conclusion:
Solubility of AgCl decreases drastically in presence of common ion Cl⁻.

🔷 7. Graphical Representation

System	Effect of Common Ion
Weak acid + common anion	Suppresses ionization, ↑ pH
Weak base + common cation	Suppresses ionization, ↓ pH
Sparingly soluble salt + common ion	↓ Solubility

🔷 8. Summary Table – NEET/JEE Snapshot

Topic	Key Points
Common Ion Effect	Suppression of ionization by a common ion
Mechanism	Le Chatelier’s principle: equilibrium shifts left
Effect on Weak Electrolytes	Decreases ionization
Effect on Solubility	Decreases solubility of sparingly soluble salts
Applications	Buffer solutions, Salt analysis, Precipitation
Equation	For HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA]

11. ✨ Introduction to Salt Hydrolysis:

🔷 What is Salt Hydrolysis ?

Salt hydrolysis is a process where a salt reacts with water to form either acid or base (or both). This changes the pH of the solution.

This happens because ions of the salt (either the cation or the anion or both) react with water molecules.

Some salts make the solution acidic
Some make it basic
Some make it neutral

🔷 Types of Salts and Their Hydrolysis Behavior

The behavior of a salt in water depends on which acid and base it comes from.

Acid (Parent)	Base (Parent)	Type of Salt	Nature in Water
Strong	Strong	Neutral Salt	No hydrolysis → Neutral
Weak	Strong	Basic Salt	Anion hydrolysis → Basic
Strong	Weak	Acidic Salt	Cation hydrolysis → Acidic
Weak	Weak	Depends	Both ions hydrolyze → Acidic/Basic/Neutral

✅ CASE 1: Salt of Strong Acid + Strong Base

Example: NaCl, KNO₃

These salts come from:

Strong Acid: HCl, HNO₃, etc.
Strong Base: NaOH, KOH, etc.

🔬 What happens in water ?

NaCl → Na⁺ + Cl⁻

Na⁺: Doesn’t react with water (comes from strong base)
Cl⁻: Doesn’t react with water (comes from strong acid)

🧾 Result:

No hydrolysis
pH = 7 (neutral solution)

✅ CASE 2: Salt of Weak Acid + Strong Base

Example: CH₃COONa, Na₂CO₃

These salts come from:

Weak Acid: CH₃COOH (acetic acid)
Strong Base: NaOH

🔬 What happens in water ?

CH₃COONa → CH₃COO⁻ + Na⁺

Na⁺: Doesn’t hydrolyze (comes from strong base)
CH₃COO⁻: Reacts with water
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

🧾 Result:

Hydrolysis of anion only
OH⁻ is produced, so the solution is basic
pH > 7

📘 Important Formulas:

Let:

K_a = Acid constant of weak acid
C = Concentration of salt
K_w = 1 × 10⁻¹⁴ (at 25°C)

Hydrolysis constant (Kh): K_h = K_w/K_a

Degree of hydrolysis (h): h = (K_w/CK_a)^1/2

[OH⁻] concentration: [OH⁻] = (CK_w/K_a)^1/2

pH calculation:

pOH = −log[OH⁻]

pH=14 − pOH

✅ CASE 3: Salt of Strong Acid + Weak Base

Example: NH₄Cl, FeCl₃

These salts come from:

Strong Acid: HCl
Weak Base: NH₄OH

🔬 What happens in water ?

NH₄Cl → NH₄⁺ + Cl⁻

Cl⁻: No hydrolysis (comes from strong acid)
NH₄⁺: Reacts with water
NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺

🧾 Result:

Hydrolysis of cation only
H₃O⁺ is produced, so the solution is acidic
pH < 7

📘 Important Formulas:

Let:

Kb = Base constant of weak base
C = Concentration of salt
K_w = 1 × 10⁻¹⁴ (at 25°C)

Hydrolysis constant (Kh): K_h= K_w/K_b

Degree of hydrolysis (h): h= (K_w/CK_b)^1/2

[H₃O⁺] concentration: [H₃O⁺]= (CK_w/K_b)^1/2

pH calculation: pH=−log[H₃O⁺]

✅ CASE 4: Salt of Weak Acid + Weak Base

Example: NH₄CH₃COO

These salts come from:

Weak Acid: CH₃COOH
Weak Base: NH₄OH

🔬 What happens in water ?

NH₄CH₃COO → NH₄⁺ + CH₃COO⁻

Both ions react with water:

NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

🧾 Result:

Both cation and anion hydrolyze
Final pH depends on:
- Ka (of acid)
- Kb (of base)

📘 Important Formulas:

Hydrolysis constant (Kh):

K_h = K_w/K_aK_b

Degree of hydrolysis (h):

h = (K_w/CK_aK_b)^1/2w

pH calculation:

pH=7+ ½ log(K_b/K_a)

If…	Then…
Ka = Kb	pH = 7 (neutral)
Ka < Kb	pH > 7 (basic)
Ka > Kb	pH < 7 (acidic)

🔷 Summary Table

Salt Type	Hydrolyzing Ion	Solution Nature	Kh	pH
Strong acid + Strong base	None	Neutral	–	7
Weak acid + Strong base	Anion only	Basic	Kw / Ka	>7
Strong acid + Weak base	Cation only	Acidic	Kw / Kb	<7
Weak acid + Weak base	Both	Depends	Kw / (Ka·Kb)	<7 / =7 / >7

🔷 Differences Between Neutralization and Hydrolysis

Feature	Neutralization	Hydrolysis
Reactants	Acid + Base	Salt + Water
Product	Salt + Water	Acid or Base
Energy	Exothermic	Endothermic
pH Change	Immediate	Gradual

🔷 Real Examples You Must Know

Salt	Type	Nature
NaCl	Strong acid + Strong base	Neutral
CH₃COONa	Weak acid + Strong base	Basic
NH₄Cl	Strong acid + Weak base	Acidic
NH₄CH₃COO	Weak acid + Weak base	Depends

🔷 Some Tricks for JEE/NEET

For weak acid + weak base salts:

If K_b>K_a ⇒ Basic Solution (pH > 7)

If K_b<K_a ⇒ Acidic Solution (pH < 7)

Strong acid or strong base part? → That ion will not hydrolyze.
The ion from weak part will hydrolyze and control the pH.

🔷 Practice MCQs (Important for NEET/JEE)

What is the pH of a solution of NaCl?
a) 5 b) 7 c) 9 d) Depends on temperature
✅ Answer: b) 7
Which of the following salts will form a basic solution?
a) NH₄Cl b) KNO₃ c) CH₃COONa d) FeCl₃
✅ Answer: c) CH₃COONa
The salt formed from a weak acid and weak base is:
a) Always neutral b) Always basic c) Can be acidic, basic or neutral d) Always acidic
✅ Answer: c) Can be acidic, basic or neutral
pH of NH₄Cl solution is < 7 because:
a) NH₄⁺ hydrolyzes to give OH⁻
b) Cl⁻ hydrolyzes
c) NH₄⁺ hydrolyzes to give H₃O⁺
d) NH₃ is strong base
✅ Answer: c) NH₄⁺ hydrolyzes to give H₃O⁺

12. Solubility Product (K_sp)

🔷 Introduction

In chemistry, solubility refers to how much of a solute (like a salt) can dissolve in a solvent (like water) to form a saturated solution.

When a sparingly soluble salt (like AgCl, BaSO₄, etc.) is added to water, it dissolves only a little. At equilibrium, a dynamic balance is established between the undissolved salt and its dissociated ions. This is known as solubility equilibrium.

To describe this equilibrium quantitatively, we use the term Solubility Product or Ksp.

🔷 What is Solubility Product (Ksp) ?

📘 Definition:

Solubility product (Ksp) is the equilibrium constant for the dissociation of a sparingly soluble salt in water.

It represents the product of the concentrations of the ions, each raised to the power of their respective stoichiometric coefficients, in a saturated solution of the salt.

Ksp = [Ag⁺][Cl⁻]

AgCl is slightly soluble, and at equilibrium, very small concentrations of Ag⁺ and Cl⁻ are present. Their product is constant at a given temperature.

🔷 Key Points About K_sp

K_sp is temperature dependent.
It applies only to sparingly soluble salts.
K_sp has no units when using molar concentrations and depending on the salt.
It is not the same as solubility (solubility = concentration of solute dissolved; Ksp = product of ion concentrations).

🔷 Expression of Ksp for Different Types of Salts

Salt Type	Dissociation	Ksp Expression
AB	A⁺ + B⁻	[A⁺][B⁻]
AB₂	A²⁺ + 2B⁻	[A²⁺][B⁻]²
A₂B	2A⁺ + B²⁻	[A⁺]²[B²⁻]
A₃B₂	3A²⁺ + 2B³⁻	[A²⁺]³[B³⁻]²
A₂B₃	2A³⁺ + 3B²⁻	[A³⁺]²[B²⁻]³

🔷 What is Solubility ?

📘 Definition:

Solubility is the amount of salt (in moles per litre) that can dissolve in water to form a saturated solution.

Let’s denote solubility by ‘S’.

🔢 Relationship Between Solubility and K_sp

We can relate Ksp with solubility ‘S’ depending on how the salt dissociates.

🧪 Case 1: AB ⇌ A⁺ + B⁻

Let solubility be S

At equilibrium:
[A⁺] = S, [B⁻] = S
So,

K_sp = S⋅S = S²

S= (K_sp)^1/2

🧪 Case 2: AB₂ ⇌ A²⁺ + 2B⁻

At equilibrium:
[A²⁺] = S, [B⁻] = 2S

K_sp = [A²⁺] [B⁻]²= S ⋅ (2S)² = 4S³

S = ^1/3(K_sp/4)^1/2

🧪 Case 3: A₂B₃ ⇌ 2A³⁺ + 3B²⁻

[A³⁺] = 2S, [B²⁻] = 3S

K_sp = (2S)² ⋅ (3S)³ = 4S² ⋅ 27S³ = 108S⁵

S= ^1/5(K_sp/108)^1/2

🔷 Common Ion Effect on Solubility

The common ion effect is an important concept in solubility equilibrium.

📘 Definition:

When a common ion (an ion already present in solution) is added to a saturated salt solution, the solubility of the salt decreases.

This is because of Le Chatelier’s Principle, which says that adding more of a product will shift equilibrium to the left (reduce dissociation).

🔬 Example:

AgCl ⇌ Ag⁺ + Cl⁻

If we add NaCl (which gives Cl⁻), the concentration of Cl⁻ increases → equilibrium shifts left → less AgCl dissolves → solubility decreases.

Note:

Solubility ↓
K_sp remains constant at a fixed temperature

🔷 Application of Common Ion Effect in Solubility Calculations

Suppose we have a saturated solution of AgCl, and we add 0.1 M NaCl. Let us find the new solubility.

Step-by-Step:

AgCl ⇌ Ag⁺ + Cl⁻
Let solubility be S.
Initial [Cl⁻] = 0.1 M from NaCl
Ag⁺ = S
Cl⁻ = S + 0.1 ≈ 0.1 (since S is very small)

K_sp= [Ag⁺][Cl⁻] = S × 0.1

S = K_sp/0.1

Clearly, solubility is now much less than when there was no common ion.

🔷 Solubility vs Ksp – What’s the Difference ?

Feature	Solubility	Ksp
Definition	How much salt dissolves	Product of ion concentrations at saturation
Unit	mol/L	No fixed unit
Changes with Common Ion?	Yes	No
Temperature Dependent?	Yes	Yes

🔷 Important NEET/JEE Concepts

🧠 Trick:

If Ksp is given and you are asked to compare solubility of two salts:

Find the expression of Ksp in terms of S
Solve for S and compare

📌 Precipitation Condition:

If we mix two ionic solutions:

If Ionic Product (IP) < Ksp → No precipitate
If IP = Ksp → Just saturated, no precipitate
If IP > Ksp → Precipitation occurs

Ionic Product (IP):
It’s the actual product of ion concentrations at any moment (not necessarily at equilibrium).

🔷 Solubility Product in Group Analysis (Qualitative Inorganic Analysis)

In qualitative salt analysis, the low solubility of some salts helps in precipitation-based identification of ions.

Examples:

Group I cations (Ag⁺, Pb²⁺, Hg₂²⁺) are precipitated as chlorides.
Group II cations as sulfides in presence of H₂S.

These group separations depend on the Ksp values of their salts.

🔷 Common Ion Effect in Buffer Solutions

Buffer solutions also use the common ion effect.

Acidic Buffer: CH₃COOH + CH₃COONa
Common ion: CH₃COO⁻
Prevents ionization of CH₃COOH → pH remains nearly constant

🔷 Temperature Effect on Ksp

Most salts have higher solubility at higher temperatures, so Ksp increases.
Some exceptions exist (like gases or salts with exothermic dissolution).

🔷 Previous Year NEET/JEE Questions (with Concept)

Q1: AgCl has a Ksp of 1.6 × 10⁻¹⁰. Find its solubility in pure water.

AgCl ⇌ Ag⁺ + Cl⁻
Ksp = S² ⇒ S = √(1.6 × 10⁻¹⁰)
S ≈ 1.26 × 10⁻⁵ mol/L

Q2: What happens to the solubility of BaSO₄ when Na₂SO₄ is added?

Answer: Na₂SO₄ adds SO₄²⁻ (common ion) → Equilibrium shifts left → Solubility decreases

Q3: A salt AB₃ dissociates as A³⁺ + 3B⁻. Express Ksp in terms of solubility S.

A³⁺ = S, B⁻ = 3S
Ksp = [A³⁺][B⁻]³ = S × (3S)³ = 27S⁴

🔷 Summary Points

Ksp is used for sparingly soluble salts to describe solubility equilibrium.
It is constant at a given temperature.
Solubility can be calculated from Ksp.
Common ion effect always decreases solubility, but doesn’t change Ksp.
Precipitation occurs when ionic product > Ksp.
Use Ksp expressions based on stoichiometry to solve problems.

🔷 Final Thoughts

Solubility product (Ksp) and common ion effect are crucial for understanding:

Salt solubility
Ionic equilibrium
Qualitative analysis
Buffer action
Precipitation

Mastering this topic will not only help in NEET/JEE problem-solving but also builds a strong foundation for inorganic chemistry.

14. Le Chatelier’s Principle

🔷 Introduction to Equilibrium

In a chemical equilibrium, the rate of forward reaction = rate of backward reaction, and the concentrations of reactants and products remain constant over time.

But what happens if we change something — like pressure, temperature, or concentration?

To predict how the system will behave, we use Le Chatelier’s Principle.

🔷 Le Chatelier’s Principle – Basic Explanation

📘 Definition:

Le Chatelier’s Principle states that:

“If a system at equilibrium is disturbed by changing temperature, pressure, or concentration, the system shifts its equilibrium position in such a way as to counteract the change and restore a new equilibrium.”

👉 In simple words: System opposes the disturbance.

🔷 Dynamic Nature of Equilibrium

Remember:

Equilibrium is dynamic, not static.
Shifting of equilibrium means the system changes the rate of forward or backward reaction to oppose the change.

🔷 Factors Affecting Equilibrium (According to Le Chatelier)

Change in Concentration
Change in Temperature
Change in Pressure or Volume
Addition of Catalyst (does not shift equilibrium)

🔶 1. Effect of Change in Concentration

🔬 Example:

N_2(g) + 3H_2(g)⇌ 2NH_3(g)

Forward Reaction: Exothermic

✅ Case 1: Increase in [N₂] or [H₂]

System has more reactants
It shifts forward to use them up
More NH₃ is formed

✅ Case 2: Increase in [NH₃]

System has more product
It shifts backward
More N₂ and H₂ are formed

✅ Case 3: Removing NH₃

NH₃ decreases
System shifts forward to make more

🔶 2. Effect of Change in Temperature

Temperature changes affect the rates and the equilibrium position
Depends on whether the reaction is endothermic or exothermic

🔬 Example:

N_2(g)+ 3H_2(g)⇌ 2NH_3(g)+ Heat

→ Exothermic Reaction

✅ Case 1: Increase in Temperature

System gets extra heat
Shifts backward to absorb heat
Less NH₃ formed

✅ Case 2: Decrease in Temperature

Less heat
System shifts forward to release heat
More NH₃ formed

📘 General Rule:

Type of Reaction	Increase Temperature	Decrease Temperature
Exothermic	Shift backward	Shift forward
Endothermic	Shift forward	Shift backward

🔶 3. Effect of Change in Pressure (or Volume)

Applies only to gaseous reactions where number of moles is different on both sides.

📘 Rule:

Increase Pressure → Shifts to side with fewer gas molecules
Decrease Pressure → Shifts to side with more gas molecules

🔬 Example:

N_2(g) + 3H_2(g)⇌ 2NH_3(g)

Left side: 1 + 3 = 4 moles
Right side: 2 moles

✅ Case 1: Increase in Pressure

System shifts to right (fewer moles)
More NH₃ formed

✅ Case 2: Decrease in Pressure

System shifts to left (more moles)
More N₂ and H₂

🔬 Special Case: Equal Moles

H_2(g)+ I_2(g)⇌ 2HI_(g)

1 + 1 = 2 moles ↔ 2 moles

👉 Change in pressure has no effect

🔶 4. Effect of Catalyst

Catalyst:

Increases rate of both forward and backward reactions
Does NOT shift equilibrium position
Equilibrium is reached faster

🔷 Summary Table of Effects on Equilibrium

Change Applied	System Response
Increase [Reactant]	Shifts Forward
Increase [Product]	Shifts Backward
Decrease [Reactant]	Shifts Backward
Decrease [Product]	Shifts Forward
Increase Temperature (Exothermic)	Shifts Backward
Increase Temperature (Endothermic)	Shifts Forward
Increase Pressure	Shifts to fewer gas moles
Decrease Pressure	Shifts to more gas moles
Catalyst Added	No shift; equilibrium reaches faster

🔷 Applications of Le Chatelier’s Principle

Let’s apply this principle to real-world chemical industries.

✅ A. Haber’s Process – Formation of Ammonia

N_2(g)+ 3H_2(g)⇌ 2NH_3(g) + Heat

Exothermic
4 moles → 2 moles

🎯 Industrial Goals:

Maximize NH₃ formation
Fast reaction

📘 Application of Le Chatelier:

Change	Effect	Result
High Pressure	Fewer gas moles	Shifts forward
Low Temperature	Favor exothermic	Shifts forward
Catalyst (Fe + Mo)	Faster equilibrium	Faster production

⚙️ Actual Conditions Used:

Pressure: 200–300 atm
Temperature: 450–500°C (compromise to keep rate fast)
Catalyst: Finely divided iron with molybdenum

✅ B. Contact Process – Manufacture of Sulfuric Acid

2SO_2(g) + O_2(g) ⇌ 2SO_3(g) + Heat

Exothermic
3 moles → 2 moles

📘 Application:

High pressure → Shifts forward
Low temperature → Shifts forward
Catalyst: V₂O₅ (vanadium pentoxide)

✅ C. Production of Nitric Acid (Ostwald Process)

4NH₃ + 5O₂ → 4NO + 6H₂O (Exothermic)

High pressure
Catalyst: Pt/Rh (Platinum-Rhodium)
Le Chatelier’s Principle used to increase yield

✅ D. Dissolution Equilibria

CaCO_3(s) ⇌ Ca²⁺ + CO₃²⁻

Adding Ca²⁺ or CO₃²⁻ shifts left (precipitation)
Removing Ca²⁺ shifts right (more dissolution)

🔷 Advanced NEET/JEE-Level Notes

📌 Van’t Hoff’s Prediction:

Equilibrium constant K is related to temperature by:

d(ln K)/dT = ΔH^∘/ RT²

If ΔH > 0 → K increases with T (endothermic)
If ΔH < 0 → K decreases with T (exothermic)

📌 Limitations of Le Chatelier’s Principle:

It gives only qualitative prediction
Does not tell how much shift occurs
Cannot predict rate of reaction

🔷 NEET/JEE Previous Year Questions

🧪 NEET 2022:

Q: In Haber’s process, which condition favors maximum yield of ammonia?
Answer: High pressure and low temperature

🧪 JEE Main 2021:

Q: What happens if temperature increases for the following reaction?

A+B⇌C+D+Heat

Answer: Reaction shifts backward

🔷 Practice MCQs

Which change increases yield of NH₃ in Haber’s process?
a) Low pressure
b) High temperature
c) High pressure
d) Addition of N₂O
✅ Answer: c)

What happens if pressure is increased for this reaction?

✅ Answer: Shifts to left (fewer moles)

Which of the following does not affect equilibrium position ?
a) Pressure
b) Catalyst
c) Concentration
d) Temperature
✅ Answer: b)

🔷 Summary

Factor	Direction of Shift
↑ Reactant	Forward
↑ Product	Backward
↑ Temp (Exo)	Backward
↑ Temp (Endo)	Forward
↑ Pressure	Side with fewer gas moles
↓ Pressure	Side with more gas moles
Catalyst	No shift (only speed)