Equilibrium Class 11 Chemistry Notes for NEET & JEE

We Need To Cover These Topics😊 –

S. No.TopicSubtopics
1Introduction to Equilibrium– Meaning of Equilibrium – Physical vs Chemical Equilibrium
2Equilibrium in Physical Processes– Solid ⇌ Liquid Equilibrium – Liquid ⇌ Vapour Equilibrium – Solid ⇌ Solution Equilibrium – Gas ⇌ Solution Equilibrium
3Equilibrium in Chemical Processes– Dynamic Nature of Chemical Equilibrium – Equilibrium in Reversible Reactions – Law of Mass Action
4Law of Equilibrium– Equilibrium Constant (Kc and Kp) – Relation between Kp and Kc – Units of Equilibrium Constant
5Homogeneous and Heterogeneous Equilibrium– Definition and Examples – Expressions for Kc and Kp for both cases
6Applications of Equilibrium Constant– Predicting Direction of Reaction – Calculation of Equilibrium Concentrations – Extent of Reaction
7Ionic Equilibrium in Aqueous Solutions– Weak and Strong Electrolytes – Arrhenius Concept – Brønsted-Lowry Concept – Lewis Concept (Brief Intro)
8Ionization of Acids and Bases– Ionization Constants of Acids (Ka) and Bases (Kb) – pH and pOH – Relationship between Ka and Kb
9Buffer Solutions– Definition and Working – Buffer Action – Henderson’s Equation
10Common Ion Effect– Definition and Mechanism – Applications in Salt Solubility and Buffer Systems
11Salt Hydrolysis– Hydrolysis of Salts of Strong Acid-Strong Base, Weak Acid-Strong Base, Strong Acid-Weak Base – Hydrolysis Constant
12Solubility Product (Ksp)– Definition and Expression – Common Ion Effect on Solubility
13Le 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.CharacteristicExplanation
1Dynamic StateBoth forward and backward reactions are taking place at the same rate.
2No Observable ChangeConcentrations of reactants and products do not change over time.
3Requires Closed SystemNo loss or gain of matter; must be a sealed or isolated system.
4Reached from Either DirectionWhether you start with reactants or products, the system eventually reaches the same equilibrium.
5Affects by External ConditionsChanges 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

TypeDefinitionExample
IrreversibleProceeds in one direction only; products do not react back.Na + Cl → NaCl
ReversibleCan 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):

N2​(g) + 3H2​(g) ⇌ 2NH3​(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:

TermMeaning
Dynamic EquilibriumForward and backward reactions continue, but concentrations remain unchanged.
Equilibrium MixtureThe mixture containing both reactants and products at equilibrium.
Equilibrium ConstantDenoted by Kc or Kp, 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

AspectPhysical EquilibriumChemical Equilibrium
Nature of ProcessInvolves physical changes (e.g., melting, boiling).Involves chemical changes (breaking/making bonds).
Substances InvolvedSame substance in different states.Different substances – reactants and products.
Change in CompositionNo change in chemical composition.Chemical composition changes.
ReversibilityEasily reversible.Reversible but may need external control (temperature, pressure, etc.).
ExampleIce ⇌ WaterN₂ + 3H₂ ⇌ 2NH₃
Affected ByTemperature 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:

ConditionExplanation
Closed SystemNo addition or removal of substances (mass is conserved).
Reversible ReactionThe reaction should be reversible (⇌ symbol used).
Constant TemperatureTemperature must be maintained as it affects reaction rates.
TimeSufficient 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

TermMeaning
Forward ReactionReaction going from reactants to products.
Backward ReactionReaction going from products to reactants.
Equilibrium MixtureMixture of reactants and products at equilibrium.
Shifting of EquilibriumChange 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:

FactorEffect
TemperatureHigher temp → More evaporation → Higher SVP
PressureHigher external pressure → Condensation favored
Surface AreaDoes 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:

FactorEffect
TemperatureUsually, ↑ temp → ↑ solubility (exceptions like CaSO₄ exist).
Nature of Solute & SolventPolar solutes dissolve better in polar solvents.
Common Ion EffectPresence 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 = kP​⋅P

Where:

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

🔸 Factors Affecting Gas-Liquid Equilibrium:

FactorEffect
Pressure↑ Pressure → More gas dissolves → Equilibrium shifts toward dissolved phase
Temperature↑ Temp → ↓ Solubility (exothermic process)
Nature of GasCO₂ 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:

ConceptAdvanced Understanding
Le Chatelier’s PrinciplePredicts how equilibrium shifts with pressure/temp change.
Vapour Pressure GraphsUsed to determine boiling points and intermolecular force strength.
Henry’s Law Constants (kH)Lower kH → Higher solubility of gas.
Salt Solubility EquilibriaLinked with Solubility Product (Ksp) in ionic equilibrium.

Summary Table: All Physical Equilibrium Types

Type of EquilibriumExampleDynamic ProcessMain Controlling Factor
Solid ⇌ LiquidIce ⇌ WaterMelting ⇌ FreezingTemperature
Liquid ⇌ VapourWater ⇌ Water VapourEvaporation ⇌ CondensationTemperature, Pressure
Solid ⇌ SolutionNaCl ⇌ Na⁺ + Cl⁻ in waterDissolution ⇌ CrystallizationTemperature, Common Ion
Gas ⇌ SolutionCO₂ (g) ⇌ CO₂ (aq)Dissolving ⇌ EscapingPressure, 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, K<sub>sp</sub>, 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:

  1. Isotope Labelling:
    • Use isotopes like D₂ (heavy hydrogen) in hydrogenation reactions.
    • Even at equilibrium, newly formed products show isotope incorporation → proves ongoing reaction.
  2. 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:

FeatureIrreversible ReactionReversible Reaction
DirectionOne-way onlyBoth forward and backward
CompletionGoes to completionDoes not go to completion
ExampleAgNO₃ + NaCl → AgCl + NaNO₃N₂ + 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. N2​(g) + 3H2​(g) ⇌ 2NH3​(g)

Rate Behavior:

StageForward RateBackward Rate
StartMaximumZero
IntermediateDecreasingIncreasing
EquilibriumEqualEqual

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: Kc​ = [C]c[D]d​ / [A]a[B]b

Where:

  • Kc = 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

Kc​= [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

ConditionExplanation
Closed systemNo substance can enter/leave.
Constant temperatureSince Kc​ depends on temperature.
Equilibrium must be reachedOnly 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

MistakeCorrection
Using partial pressure in KcUse 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 KcAlways determine units based on stoichiometry.

📝 Summary Table

ConceptKey Point
Chemical EquilibriumRate of forward = backward; concentrations constant
Dynamic NatureReaction continues at molecular level
Reversible ReactionsNecessary for equilibrium
Law of Mass ActionK = [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 – Kc (Concentration Form)

Definition:

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

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

Kc​ =  [C]c[D]d​ / [A]a[B]b

Where:

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

Key Features of Kc:

  • 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 K<sub>c</sub>; solids and pure liquids are excluded.

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

Definition:

The equilibrium constant in terms of partial pressures of gases is called K<sub>p</sub>.

For the gaseous reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

Kp​=  (PA​)a⋅(PB​)b  / (PC​)c⋅(PD​)d​

Where:

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

When to Use Kc:

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

🔷 3. Relation Between Kp and Kc

This is one of the most important formulas in Equilibrium: Kp ​= Kc​(RT)Δn

✅ Where:

  • Kp = equilibrium constant in terms of partial pressure
  • Kc = 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 ExampleΔn Value
H₂ + I₂ ⇌ 2HI2 – 2 = 0
N₂ + 3H₂ ⇌ 2NH₃2 – 4 = –2
2SO₂ + O₂ ⇌ 2SO₃2 – 3 = –1

If Δn = 0, then Kp​ = Kc


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).


Kc> Units:

Kc​ = molΔn ⋅L−Δn

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

Kp Units:

Kp​= (Pressure)Δn

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

⚠️ Note:

For heterogeneous equilibria, pure solids and liquids do not appear in K<sub>c</sub> or K<sub>p</sub> expressions.


🔷 6. NEET/JEE Level Tips & Tricks

TipWhy It Matters
Always calculate Δn carefullySmall mistake changes K<sub>p</sub> calculation
Use R and T in correct unitsOtherwise final K<sub>p</sub> value will be incorrect
Units of Kc and Kp depend on ΔnNot always unitless!
For solid/liquid reactionsExclude solids/pure liquids in K expressions
Use log when neededJEE often asks questions involving log(K) or ln(K)

🔷 7. Conclusion

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

  • Mastering how to write Kc and Kp expressions
  • Understanding their relationship via Kp = Kc (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:

  1. Gas-phase reaction:

N2​(g) + 3H2​(g) ⇌ 2NH3​(g)

  • All species are gases → homogeneous equilibrium
  1. Solution-phase reaction:

CH3​COOH(aq) + H2​O(l) ⇌ CH3​COO(aq) +H3​O+(aq)

  • All aqueous → homogeneous

🔹 Writing Expressions for Kc and Kp :

Let’s take the reaction: SO2​(g) + NO2​(g) ⇌ SO3​(g) + NO(g)

  • Kc expression:

Kc​=  [SO3​][NO] / [SO2​][NO2​] ​

  • Kp expression:

Kp​= ​​PSO3​​ ⋅ PNO/ PSO2 ​​⋅ PNO2

✔ All components appear in the equilibrium expression.


🔹 NEET/JEE Tip:

In homogeneous equilibrium, all species (gases or aqueous) are included in K<sub>c</sub> and K<sub>p</sub> 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 Kc and Kp.

🔹 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:

Kc​= [C]c​/[B]b; Kp​= (PC​)c​/(PB​)b


🔷 3. Key Differences Table

FeatureHomogeneous EquilibriumHeterogeneous Equilibrium
Phases involvedAll species in same phaseSpecies in different phases
ExamplesGaseous or aqueous reactionsSolids, liquids, gases in the same system
Inclusion in K expressionsAll species includedOnly gases and aqueous species included
Common inAcid-base, gas reactionsDecomposition, precipitation, phase change reactions

🔷 4. NEET/JEE Advanced Concepts

ConceptExplanation
Pure solids/liquidsNot included in expressions
Activity = 1The “activity” of solids/liquids is taken as 1, hence omitted
Phase equilibriume.g., H₂O (l) ⇌ H₂O (g) – vapor pressure becomes equilibrium constant
K has no units if Δn = 0Particularly in heterogeneous cases with no net gas mole change
Thermodynamic linkEquilibrium constant can relate to Gibbs free energy (ΔG° = –RT ln K)

🔷 6. Conclusion

ConceptSummary
Homogeneous EquilibriumAll substances in same phase; all appear in K expressions
Heterogeneous EquilibriumDifferent phases; solids/pure liquids excluded
K<sub>c</sub> and K<sub>p</sub> RulesAlways write expressions based on gaseous/aqueous species only
Competitive FocusKnow 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

​​Qc​= [C]c[D]d​/[A]a[B]b  ; Qp​=  PCc​⋅PDd​​/PAa​⋅PBb


Compare Q with K to predict direction:

ConditionInterpretationReaction Shifts
Q < KMore reactants, less productsForward direction
Q > KMore products, less reactantsBackward direction
Q = KSystem at equilibriumNo shift

🔷 2. Calculation of Equilibrium Concentrations

This is a very common and important application.


Steps to Calculate:

  1. Write the balanced equation
  2. Use ICE table
    (Initial, Change, Equilibrium)
  3. Express all equilibrium terms in terms of x
  4. Apply Kc or Kp formula
  5. Solve the algebraic equation
  6. Calculate concentrations

ICE Table Format Example:

Reaction: A+B ⇌ C ; Kc​ = 4

SpeciesInitial (mol/L)Change (mol/L)Equilibrium (mol/L)
A1.0–x1 – x
B1.0–x1 – x
C0+xx

Now apply: Kc ​= [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 Kc / Kp to Estimate Extent:

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

📌 Example 1:

For reaction: A ⇌ B, Kc ​= 106

→ Almost complete reaction → products dominate


📌 Example 2:

For reaction:X ⇌ Y, Kc​ = 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:

MistakeCorrect 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

ConceptKey Takeaway
Reaction Quotient (Q)Predicts direction by comparing with K
ICE TableHelps calculate unknown equilibrium concentrations
K and Extent of ReactionHigh K → almost complete; Low K → barely reacts
Q < KForward reaction favored
Q > KBackward reaction favored
ΔG and KConnected 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:

TypeDefinitionExamples
Strong ElectrolyteDissociates completely into ions in waterHCl, NaOH, NaCl, HNO₃, K₂SO₄
Weak ElectrolyteDissociates partially in water, establishing ionic equilibriumCH₃COOH, 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 AcidArrhenius Base
A substance that increases H⁺ ion concentration in waterA 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 AcidBrønsted–Lowry Base
Proton (H⁺) donorProton (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+

SpeciesRoleConjugate Pair
HClAcidCl⁻ (conjugate base)
NH₃BaseNH₄⁺ (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 AcidLewis Base
Electron pair acceptorElectron 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

TheoryAcidBaseMedium
ArrheniusH⁺ producer in waterOH⁻ producer in waterAqueous only
Brønsted–LowryProton donorProton acceptorAny (including water)
LewisElectron pair acceptorElectron pair donorAny (most general)

✅ Summary Chart

ConceptKey FeatureNEET/JEE Tip
Strong ElectrolyteComplete ionizationNo equilibrium formed
Weak ElectrolytePartial ionization, reversible reactionEstablishes ionic equilibrium
Arrhenius TheoryH⁺/OH⁻ concept in waterLimited scope
Brønsted–Lowry TheoryProton donor–acceptorExplains acid–base in non-aqueous solutions
Lewis TheoryElectron pair donor–acceptorExplains 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) + H2​O(l) ⇌ H3​O+(aq) + A(aq)

Acid Dissociation Constant (Ka):

The equilibrium constant for ionization of a weak acid is:

Ka ​= [H3​O+][A]​/[HA]

Ka and Acid Strength:

Ka ValueAcid 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) + H2​O(l) ⇌ BH+(aq)+ OH(aq)

Base Dissociation Constant (Kb):

Kb​= [BH+][OH]​/[B] 

Kb and Base Strength:

Kb ValueBase 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)pHNature of Solution
10Strongly Acidic
10⁻⁷7Neutral
10⁻¹⁴14Strongly Basic

🔷 For Weak Acid/Base:

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

[H+] ≈ (Ka​⋅C)1/2​ ⇒ pH = −log(Ka​⋅C​)

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


🔹 5. Relationship between Ka and Kb

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

For acid: HA + H2​O ⇌ H3​O+ + A(Ka​)

For conjugate base: A+ H2​O ⇌ HA + OH−  (Kb​)

Multiplying the two equilibrium expressions:

Ka​⋅Kb​ = [H3​O+][A]⋅[HA][OH]/[HA][A]=[H3​O+][OH]=Kw​​

✅ Final Relation:

Ka​⋅Kb​ = Kw ​= 1×10−14 at 25°C

🔷 In Terms of pKa and pKb:

pKa ​+ pKb ​= pKw​ = 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 Ka and Kb 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 (pKb​−pKa​)

🔹 9. Polyprotic Acids and Stepwise Ionization

Some acids donate more than one proton:

  • Example: H₂SO₄, H₂CO₃, H₃PO₄

Each step has a separate Ka:

  • Ka1​ > Ka2​>Ka3​

🔹 10. Important Formulas Summary

QuantityFormula
Ka[A][HA] / [H+
Kb[OH][BH+]​/[B]
pH−log10​[H+]
pOH−log10​[OH]
pH + pOH= 14 (at 25°C)
Ka × KbKw = 1.0 × 10−14
pKa + pKb= 14
[H⁺] (for weak acid)(Ka⋅C​)1/2
[OH⁻] (for weak base)(Kb⋅C​)1/2
pH (salt of weak acid/base)7+ ½ (pKb−pKa)

🎯 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):
    • CH3​COO + H+ → CH3​COOH
  • So free H⁺ ions are consumed → pH remains almost same

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

  • OH⁻ reacts with CH₃COOH:
    • CH3​COOH + OH → CH3​COO+H2​O
  • 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 = pKb ​+ log10​([Salt]/[Acid]​)

Where:

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

✅ For Basic Buffer:

pOH = pKb ​+ 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:

Ka ​= [H+][A]/[HA] ​

Rearranging:

[H+] = Ka​⋅ [HA]/[A]​

Taking log:

log[H+]= logKa​ + log ([HA]/[A])

Multiply by -1:

−log[H+]= −logKa ​– log ([HA]/[A]​)

pH = pKa​ + log ([A]/[HA]​)

Hence,

pH = pKa​ + 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 AreaUse of Buffer
Biological SystemsBlood buffer maintains pH ~ 7.4
PharmaceuticalsStabilize drug pH in medicines
Chemical ReactionsMaintain constant pH for proper yield
Food IndustryMaintain taste and shelf life
Electroplating, TitrationsUsed for maintaining medium pH

🔷 8. Blood as a Natural Buffer

Blood uses the carbonic acid–bicarbonate buffer system:

H2​CO3​ ⇌ 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

ConceptFormula/Key Point
Buffer DefinitionResists pH change
Acidic BufferWeak acid + Salt of that acid
Basic BufferWeak base + Salt of that base
Acidic Buffer pHpH = pKa + log([Salt]/[Acid])
Basic Buffer pOHpOH = pKb + log([Salt]/[Base])
Best BufferingWhen [Salt] = [Acid/Base] ⇒ pH = pKa
Buffer CapacityMaximum at pH = pKa
Blood BufferH₂CO₃ / 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:

Ka​= [H+][A]​/[HA] 

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

Similarly, for weak base:

B+ H2​O ⇌ 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 (Ksp):

  • Ksp 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,

Ksp ​=[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

SystemEffect of Common Ion
Weak acid + common anionSuppresses ionization, ↑ pH
Weak base + common cationSuppresses ionization, ↓ pH
Sparingly soluble salt + common ion↓ Solubility

🔷 8. Summary Table – NEET/JEE Snapshot

TopicKey Points
Common Ion EffectSuppression of ionization by a common ion
MechanismLe Chatelier’s principle: equilibrium shifts left
Effect on Weak ElectrolytesDecreases ionization
Effect on SolubilityDecreases solubility of sparingly soluble salts
ApplicationsBuffer solutions, Salt analysis, Precipitation
EquationFor 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 SaltNature in Water
StrongStrongNeutral SaltNo hydrolysis → Neutral
WeakStrongBasic SaltAnion hydrolysis → Basic
StrongWeakAcidic SaltCation hydrolysis → Acidic
WeakWeakDependsBoth 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:

  • Ka​ = Acid constant of weak acid
  • C = Concentration of salt
  • Kw​ = 1 × 10⁻¹⁴ (at 25°C)

Hydrolysis constant (Kh): Kh​ = Kw​​/Ka

Degree of hydrolysis (h): h = (Kw​​​/CKa​)1/2

[OH⁻] concentration: [OH] = (CKw​​​/Ka​)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
  • Kw = 1 × 10⁻¹⁴ (at 25°C)

Hydrolysis constant (Kh): Kh​= Kw​​/Kb

Degree of hydrolysis (h): h= (Kw​​​/CKb​)1/2

[H₃O⁺] concentration: [H3​O+]= (CKw​/Kb​)1/2 ​​​​

pH calculation: pH=−log[H3​O+]


✅ 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):

Kh ​= Kw​​/Ka​Kb

Degree of hydrolysis (h):

h = (Kw​​​/CKa​Kb​)1/2w​​​

pH calculation:

pH=7+ ½  log(Kb/Ka​​)

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

🔷 Summary Table

Salt TypeHydrolyzing IonSolution NatureKhpH
Strong acid + Strong baseNoneNeutral7
Weak acid + Strong baseAnion onlyBasicKw / Ka>7
Strong acid + Weak baseCation onlyAcidicKw / Kb<7
Weak acid + Weak baseBothDependsKw / (Ka·Kb)<7 / =7 / >7

🔷 Differences Between Neutralization and Hydrolysis

FeatureNeutralizationHydrolysis
ReactantsAcid + BaseSalt + Water
ProductSalt + WaterAcid or Base
EnergyExothermicEndothermic
pH ChangeImmediateGradual

🔷 Real Examples You Must Know

SaltTypeNature
NaClStrong acid + Strong baseNeutral
CH₃COONaWeak acid + Strong baseBasic
NH₄ClStrong acid + Weak baseAcidic
NH₄CH₃COOWeak acid + Weak baseDepends

🔷 Some Tricks for JEE/NEET

  • For weak acid + weak base salts:

If Kb​>Ka​ ⇒ Basic Solution (pH > 7)

If Kb​<Ka​ ⇒ 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)

  1. What is the pH of a solution of NaCl?
     a) 5 b) 7 c) 9 d) Depends on temperature
     ✅ Answer: b) 7
  2. Which of the following salts will form a basic solution?
     a) NH₄Cl b) KNO₃ c) CH₃COONa d) FeCl₃
     ✅ Answer: c) CH₃COONa
  3. 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
  4. 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 (Ksp)

🔷 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 Ksp

  • Ksp is temperature dependent.
  • It applies only to sparingly soluble salts.
  • Ksp 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 TypeDissociationKsp Expression
ABA⁺ + B⁻[A⁺][B⁻]
AB₂A²⁺ + 2B⁻[A²⁺][B⁻]²
A₂B2A⁺ + B²⁻[A⁺]²[B²⁻]
A₃B₂3A²⁺ + 2B³⁻[A²⁺]³[B³⁻]²
A₂B₃2A³⁺ + 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 Ksp

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,

​​Ksp​ = S⋅S = S2

S= (Ksp​)1/2


🧪 Case 2: AB₂ ⇌ A²⁺ + 2B⁻

At equilibrium:
[A²⁺] = S, [B⁻] = 2S

Ksp ​= [A2+] [B]2 = S ⋅ (2S)2 = 4S3

S = 1/3(Ksp​/4)1/2​​


🧪 Case 3: A₂B₃ ⇌ 2A³⁺ + 3B²⁻

[A³⁺] = 2S, [B²⁻] = 3S

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

S=  1/5(Ksp​​​/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 ↓
  • Ksp 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)

Ksp​= [Ag+][Cl] = S × 0.1

S = Ksp​​/0.1

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


🔷 Solubility vs Ksp – What’s the Difference ?

FeatureSolubilityKsp
DefinitionHow much salt dissolvesProduct of ion concentrations at saturation
Unitmol/LNo fixed unit
Changes with Common Ion?YesNo
Temperature Dependent?YesYes

🔷 Important NEET/JEE Concepts

🧠 Trick:

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

  1. Find the expression of Ksp in terms of S
  2. 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 > KspPrecipitation 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)

  1. Change in Concentration
  2. Change in Temperature
  3. Change in Pressure or Volume
  4. Addition of Catalyst (does not shift equilibrium)

🔶 1. Effect of Change in Concentration

🔬 Example:

N2​(g) + 3H2​(g) ⇌ 2NH3​(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:

N2​(g) + 3H2​(g) ⇌ 2NH3​(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 ReactionIncrease TemperatureDecrease Temperature
ExothermicShift backwardShift forward
EndothermicShift forwardShift 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:

N2​(g) + 3H2​(g) ⇌ 2NH3​(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

H2​(g) + I2​(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 AppliedSystem 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 PressureShifts to fewer gas moles
Decrease PressureShifts to more gas moles
Catalyst AddedNo 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

N2​(g) + 3H2​(g) ⇌ 2NH3​(g) + Heat

  • Exothermic
  • 4 moles → 2 moles

🎯 Industrial Goals:

  • Maximize NH₃ formation
  • Fast reaction

📘 Application of Le Chatelier:

ChangeEffectResult
High PressureFewer gas molesShifts forward
Low TemperatureFavor exothermicShifts forward
Catalyst (Fe + Mo)Faster equilibriumFaster 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

2SO2​(g)  + O2​(g) ⇌ 2SO3​(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 + 6H2​O  (Exothermic)

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

✅ D. Dissolution Equilibria

CaCO3​(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

  1. 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)

  1. What happens if pressure is increased for this reaction?

Answer: Shifts to left (fewer moles)


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

🔷 Summary

FactorDirection of Shift
↑ ReactantForward
↑ ProductBackward
↑ Temp (Exo)Backward
↑ Temp (Endo)Forward
↑ PressureSide with fewer gas moles
↓ PressureSide with more gas moles
CatalystNo shift (only speed)

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