Class 11 · Chemistry · Chapter 06

Equilibrium

Every reaction that looks "finished" is secretly still running both ways. This chapter is about reading that quiet tug-of-war — in beakers, in acids and bases, and in the pH of your own blood.

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Chapter Snapshot

The chapter, at a glance

18
Core sub-topics, from dynamic equilibrium to buffer solutions
3
Equilibrium types — physical, ionic & chemical
7–8
Marks typically weighted in board examinations
4
Landmark laws: Le Chatelier, Ostwald, Henderson, Arrhenius
Why This Chapter Matters

It quietly decides your Physical Chemistry score

  • Numerical-heavy: Kc, Kp, degree of dissociation and pH calculations appear almost every year — and they're method-based, so marks are easy once the pattern clicks.
  • Feeds forward: Class 12 Electrochemistry, Chemical Kinetics and Ionic Equilibrium all lean on the concepts built here.
  • High recall value: Le Chatelier's Principle alone shows up across JEE, NEET and board papers as both theory and application questions.
  • Common mistake zone: sign errors in ΔG° = −RT ln K and confusing Kc with Qc cost students avoidable marks.
Typical mark distribution
Numericals (Kc/Kp/pH)
Conceptual (Le Chatelier, factors)
Ionic equilibrium & buffers
Key Concept Highlights

Six ideas the whole chapter balances on

Physical

Dynamic Equilibrium

Forward and reverse rates become equal — nothing stops, but nothing visibly changes.

Law

Law of Mass Action

Rate depends on active masses of reactants, giving rise to the equilibrium constant K.

Principle

Le Chatelier's Principle

A system at equilibrium shifts to counter any change in concentration, pressure or temperature.

Ionic

Acids, Bases & pH

Arrhenius, Bronsted-Lowry and Lewis concepts, plus the pH scale for measuring acidity.

Salts

Hydrolysis & Buffers

Salts of weak acids/bases hydrolyse; buffers resist pH change on small additions of acid/base.

Solubility

Ksp & Common Ion Effect

Predicting precipitation and the effect of a shared ion on solubility equilibria.

Important Formulae & Reactions

The equation shelf — keep these within reach

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Equilibrium constant for aA + bB ⇌ cC + dD

Kp = Kc(RT)^Δn

Relation between pressure- and concentration-based constants

pH = −log[H⁺]

Definition of pH; pH + pOH = 14 at 25°C

Ka·Kb = Kw

Links conjugate acid–base pair constants at a given temperature

ΔG° = −RT ln K

Connects thermodynamics with the equilibrium constant

pH = pKa + log([salt]/[acid])

Henderson–Hasselbalch equation for buffer solutions

Ksp = [Mⁿ⁺]ˣ[Aˣ⁻]ⁿ

Solubility product for a sparingly soluble salt MₓAₙ

What You Will Learn

Your path through the chapter

1 · Reversible reactions & the equilibrium state

How systems reach a point where concentrations stop changing without reactions stopping.

2 · Equilibrium constants — Kc, Kp & Kx

Writing expressions correctly and relating them under different conditions.

3 · Factors affecting equilibrium

Applying Le Chatelier's Principle to concentration, pressure, volume and temperature changes.

4 · Acid–base theories

Comparing Arrhenius, Bronsted-Lowry and Lewis definitions with real examples.

5 · Ionisation of acids, bases & salts

Degree of dissociation, ionisation constants, and hydrolysis of salts.

6 · Buffers & solubility equilibria

How buffers work and how to use Ksp to predict precipitation.

Chapter Resources

Jump straight to what you need

📖Full NotesComplete theory 🎯MCQs PracticeTopic-wise quiz 🗂️PYQ BankPast-year questions ⚖️True / FalseQuick-fire check ✍️ExercisesNCERT back-questions
Exam Strategy & Preparation Tips

How to actually score well here

01

Master ICE tables (Initial–Change–Equilibrium) — nearly every numerical is built on this one structure.

02

Memorise the Kc–Kp relation and practise Δn calculations until they're automatic, not looked-up.

03

Draw the shift direction with an arrow whenever you apply Le Chatelier — examiners reward shown reasoning.

04

Keep a one-page formula sheet of Ka, Kb, Kw, Ksp and Henderson's equation for last-minute revision.

05

Solve at least 3 years of PYQs on pH and buffer numericals — the question style repeats closely.

06

Don't skip True/False sets — this chapter is a favourite for statement-based conceptual traps.

Chapter 6 · CBSE · Class XI
⚖️

Equilibrium

Equilibrium NCERT Class 11 Chemistry Chemical Equilibrium Ionic Equilibrium Reversible Reactions Irreversible Reactions Dynamic Equilibrium Law of Chemical Equilibrium Equilibrium Constant Law of Mass Action Homogeneous Equilibrium Heterogeneous Equilibrium Le Chatelier Principle Factors Affecting Equilibrium Effect of Temperature Effect of Pressure Effect of Concentration Catalyst Effect Acids Bases and Salts Arrhenius Concept Bronsted Lowry Theory Lewis Acids and Bases Ionization of Water pH Scale Ionization Constant of Weak Acids Ionization Constant of Weak Bases Common Ion Effect Buffer Solutions Solubility Product Salt Hydrolysis
📋
Table of Contents
16 sections
🗺️ Overview
Equilibrium is one of the most important concepts in Chemistry because almost every natural, industrial, and biological process involves a state of equilibrium. Chemical industries such as ammonia manufacturing (Haber Process), sulphuric acid production (Contact Process), blood buffering, respiration, photosynthesis, dissolution of salts, and even atmospheric reactions operate under equilibrium conditions.

NCERT Class 11 introduces equilibrium as a dynamic state rather than a state in which reactions stop. Understanding this chapter is essential for Board examinations, JEE Main, JEE Advanced, NEET, CUET and other competitive examinations.
📘 Definition
Definition of Equilibrium
⚖️ Representation Of Equilibrium
A reversible process is represented by a double half-arrow.
Example
\[\mathrm{H_2O(l)\; \rightleftharpoons \;H_2O(g)}\] The forward arrow represents evaporation whereas the backward arrow represents condensation. Both processes continue simultaneously.
The double arrow (⇌) indicates that neither process has stopped. Instead, both continue with equal rates once equilibrium is established.
📌 Equilibrium Mixture
📊 Macroscopic and Microscopic View of Equilibrium
Macroscopic View Microscopic View
Macroscopic View Microscopic View
No visible change occurs. Molecules continue to react continuously.
Colour remains constant. Forward and reverse reactions continue.
Pressure remains constant. Molecules collide continuously.
Concentration remains constant. Equal reaction rates maintain constant concentration.
🗂️ Types of Equilibrium
Static Equilibrium
Static equilibrium is the state in which a system remains at rest and no opposing process takes place.
Characteristics
  • Occurs mainly in Physics.
  • No molecular movement responsible for change.
  • Forward and reverse processes do not occur.
  • The system appears completely stationary.
Examples
  • A book resting on a table.
  • A hanging pendulum at rest.
  • A balanced beam.
Dynamic Equilibrium
Dynamic equilibrium is the state of a reversible reaction in which the forward and reverse reactions continue simultaneously at equal rates, resulting in no net change in the concentration of reactants and products.
Concept
Although the system appears unchanged from outside, molecules are continuously converting into products and products are continuously converting back into reactants.
Therefore, equilibrium is called dynamic rather than static.
Mathematical Representation
At equilibrium, \[\color{cyan}\text{Rate of Forward Reaction}=\text{Rate of Reverse Reaction}\] However, \[\color{cyan}\text{Reactant Concentration}\neq\text{Product Concentration}\] They merely remain constant with time.
⚖️ Difference Between Static and Dynamic Equilibrium
Static Equilibrium Dynamic Equilibrium
No molecular activity. Molecules continuously react.
Occurs mainly in Physics. Occurs in Chemistry and Physical processes.
Forward and reverse processes do not occur. Forward and reverse reactions occur simultaneously.
System remains completely at rest. System appears unchanged but reactions continue.
No reaction rate exists. Forward rate equals reverse rate.
📊 Physical and Chemical Equilibrium
Physical Equilibrium Chemical Equilibrium
Only physical change occurs. Chemical reaction occurs.
No new substance is formed. New substances are formed.
Example: Ice ⇌ Water Example: H₂ + I₂ ⇌ 2HI
✏️ Examples of Physical Equilibrium
  • \( \mathrm{Ice(s)\rightleftharpoons Water(l)} \)
  • \( \mathrm{Water(l)\rightleftharpoons Water(g)} \)
  • Dissolution of sugar in saturated solution.
  • Dissolution of iodine in alcohol.
Examples of Chemical Equilibrium
  • \( \mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)} \)
  • \( \mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)} \)
  • \( \mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)} \)
⚖️ Classification of Reactions Based on Extent of Equilibrium
Depending upon the amount of products formed at equilibrium, reversible reactions are divided into three categories.
Type Description Equilibrium Position
Reaction proceeds almost to completion Nearly all reactants are converted into products. Far towards products.
Reaction proceeds only slightly Very little product is formed. Far towards reactants.
Moderately reversible reaction Both reactants and products exist in appreciable amounts. Middle position.
🔷 Characteristics of Chemical Equilibrium
🔷 Characteristics
  • Occurs only in reversible reactions.
  • Can be established only in a closed system.
  • Forward and reverse reaction rates become equal.
  • Concentrations remain constant.
  • Macroscopic properties remain unchanged.
  • Equilibrium can be attained from either direction.
  • The system possesses minimum Gibbs free energy under given conditions.
  • Catalyst does not change equilibrium position; it only helps attain equilibrium faster.
🎨 SVG Diagram
Concept Diagram
Reactants Products Forward Reaction Reverse Reaction At Equilibrium: Forward Rate = Reverse Rate
✏️ Example
Solved Example
A student observes that after some time the concentration of reactants and products no longer changes. Does this mean the reaction has stopped?
  • Dynamic equilibrium
  • Equal reaction rates
Understand whether molecules continue reacting after equilibrium is achieved.
No. The reaction has not stopped. At equilibrium, the forward and reverse reactions continue simultaneously with equal rates. Therefore, there is no net change in concentration although molecular transformations continue continuously.
Why is equilibrium called dynamic rather than static?
Molecules continuously change into products while products simultaneously convert back into reactants. Since both reactions continue with equal rates, equilibrium is dynamic even though no observable change occurs.
⚡ Exam Tip
❌ Common Mistakes
  • Writing concentration of reactants equals concentration of products.
  • Thinking equilibrium means reaction has stopped.
  • Using a single arrow instead of reversible arrows.
  • Confusing physical equilibrium with chemical equilibrium.
  • Assuming catalyst changes equilibrium composition.
📋 CBSE Competency-Based (HOTS) Question

A sealed flask contains liquid water and water vapour at constant temperature. After several hours, the amount of liquid water remains unchanged. Explain whether evaporation has stopped.

Concept Tested

  • Physical equilibrium
  • Dynamic equilibrium
  • Equal rates of evaporation and condensation

Answer

Evaporation has not stopped. Water molecules continuously escape from the liquid surface, while an equal number of vapour molecules condense back into the liquid. Since both processes occur at equal rates, the amount of water remains constant and the system is in dynamic equilibrium.

🌟 Why This Topic is Important
⚖️

Equilibrium in Physical Processes

📋
Table of Contents
14 sections
📘 Definition
📌 Phase Transformations
📘 Solid-Liquid Equilibrium
🔷 Characteristics of Dynamic Solid-Liquid Equilibrium
🔷 Characteristics
  • Melting and freezing occur simultaneously.
  • Rate of melting equals rate of freezing.
  • Mass of ice remains constant.
  • Mass of water remains constant.
  • Temperature remains constant at the melting point.
  • The system must be perfectly insulated so that no heat enters or leaves.
📘 Normal Melting Point
📘 Normal Freezing Point
📘 Liquid-Vapour Equilibrium
📘 Equilibrium Vapour Pressure
📘 Normal Boiling Point
📘 Solid-Vapour Equilibrium
⚖️ Comparison of Physical Equilibria
Equilibrium Forward Process Reverse Process
Solid ⇌ Liquid Melting Freezing
Liquid ⇌ Vapour Evaporation Condensation
Solid ⇌ Vapour Sublimation Deposition
🎨 SVG Diagram
Concept Diagram: Physical Equilibria
Solid Liquid Vapour Melting Freezing Evaporation Condensation Dynamic Equilibrium: Forward Rate = Reverse Rate
✏️ Example
Solved Example
Why does the violet colour inside a closed vessel containing iodine crystals become constant after some time?
  • Solid-vapour equilibrium
  • Dynamic equilibrium
Compare the rates of sublimation and deposition.
Initially, iodine sublimes continuously, increasing the concentration of violet vapours. As the vapour concentration increases, deposition also increases. Eventually, the rate of sublimation becomes equal to the rate of deposition. The amount of vapour then remains constant, so the intensity of the violet colour no longer changes. This is the state of dynamic equilibrium.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming evaporation stops at equilibrium.
  • Confusing evaporation with boiling.
  • Thinking vapour pressure depends on the amount of liquid.
  • Ignoring the requirement of a closed system.
  • Writing that boiling point is always 100°C (it depends on pressure).
📋 CBSE Competency-Based (HOTS) Question

Two identical beakers containing pure water are placed at sea level and on a mountain. Which one will boil first when heated, and why?

Answer

Water on the mountain boils first because atmospheric pressure is lower at higher altitudes. Since boiling occurs when the vapour pressure equals atmospheric pressure, this condition is reached at a lower temperature on mountains, resulting in a lower boiling point.

⚖️

Equilibrium in Chemical Processes (Dynamic Equilibrium)

📋
Table of Contents
14 sections
🗺️ Overview
Just like physical processes, many chemical reactions are reversible. Such reactions proceed simultaneously in both the forward and reverse directions until a state of dynamic equilibrium is established.

In a chemical equilibrium, reactants are continuously converted into products, while products simultaneously convert back into reactants. Although molecular changes continue throughout the reaction, there is no observable change in the composition of the reaction mixture because both reactions occur at equal rates.

Chemical equilibrium forms the foundation of numerous industrial and biological processes, including the Haber process for ammonia synthesis, Contact process for sulphuric acid manufacture, blood buffering, photosynthesis, respiration, and enzyme-catalysed reactions.
📘 Definition of Chemical Equilibrium
📌 Variation of Reaction Rates During Equilibrium
🗒️ Molecular Interpretation Of Chemical Equilibrium

At the molecular level, reactant molecules continuously collide to form products, while product molecules simultaneously collide to regenerate reactants.

Since both processes occur with identical rates, the overall composition of the reaction mixture does not change even though millions of molecular collisions occur every second.

Thus,

Chemical equilibrium is a dynamic balance between two opposing reactions.

🤔 Did You Know?
Why is Chemical Equilibrium Dynamic?
Chemical equilibrium is called dynamic because:
  • Both forward and reverse reactions continue continuously.
  • Molecules are constantly transformed into one another.
  • No visible change occurs because both reactions occur at identical rates.
  • The reaction never comes to a complete stop.
🗒️ Necessary Conditions For Chemical Equilibrium
  • The reaction must be reversible.
  • The system must be closed.
  • Temperature should remain constant.
  • No reactant or product should escape from the system.
  • The forward and reverse reaction rates must become equal.
🔷 Characteristics of Chemical Equilibrium
🔷 Characteristics
  • Chemical equilibrium is dynamic in nature.
  • It can be established only in reversible reactions.
  • It is attained in a closed system.
  • The rates of forward and reverse reactions are equal.
  • Concentrations of all species remain constant.
  • Macroscopic properties such as pressure, colour and density remain unchanged.
  • Equilibrium can be attained from either the reactant side or the product side.
  • A catalyst does not change the equilibrium position; it only helps attain equilibrium faster.
🗒️ Difference Between Physical and Chemical Equilibrium
Physical Equilibrium Chemical Equilibrium
Only physical state changes. Chemical composition changes.
No new substance is formed. New substances are formed.
Example: Ice ⇌ Water Example: N₂ + 3H₂ ⇌ 2NH₃
Governed by physical processes. Governed by chemical reactions.
📌 Stepwise Development of Chemical Equilibrium
✏️ Example
Solved Example
During a reversible reaction, the concentrations of reactants and products become constant after some time. Does this mean the reaction has stopped?
  • Dynamic chemical equilibrium
  • Equality of reaction rates
Compare molecular activity with macroscopic observations.
No. The reaction continues in both directions even after equilibrium is established. The concentrations remain constant because the forward and reverse reactions occur at equal rates. Hence, equilibrium is dynamic rather than static.
At equilibrium, are the concentrations of reactants and products always equal?
No. At equilibrium, the rates of the forward and reverse reactions are equal, but the concentrations of reactants and products are merely constant. Their values depend on the equilibrium constant and need not be equal.
⚡ Exam Tip
❌ Common Mistakes
  • Writing that forward and reverse reactions stop at equilibrium.
  • Assuming equal concentrations at equilibrium.
  • Ignoring the requirement of a closed system.
  • Using a single arrow instead of a reversible arrow.
  • Confusing reaction rate with concentration.
📋 CBSE Competency-Based (HOTS) Question

A student observes that the brown colour of nitrogen dioxide in a sealed flask remains unchanged after some time. Can it be concluded that the reaction has stopped? Justify your answer.

Answer

No. The constant brown colour indicates that chemical equilibrium has been established. Nitrogen dioxide molecules continue to combine to form dinitrogen tetroxide, while dinitrogen tetroxide simultaneously decomposes back into nitrogen dioxide. Since both reactions occur at equal rates, the concentrations remain constant and no visible change is observed.

🌟 Why This Topic is Important for Board and Competitive Exams
⚖️

Laws of Chemical Equilibrium and Equilibrium Constant

📋
Table of Contents
16 sections
🗺️ Overview

After understanding how chemical equilibrium is established, the next important question is:

Can the composition of an equilibrium mixture be predicted mathematically?

The answer is Yes. The relationship between the concentrations of reactants and products at equilibrium is given by the Law of Mass Action, from which the concept of the Equilibrium Constant (K) is derived.

The equilibrium constant is one of the most important quantitative tools in Chemistry. It predicts the extent of a reaction and helps determine whether products or reactants are favoured at equilibrium.

📌 Law of Mass Action
🗒️ Guldberg And Waage's Contribution
The Norwegian scientists Cato Maximilian Guldberg and Peter Waage proposed the Law of Mass Action in 1864.
They showed experimentally that the concentrations of reactants and products at equilibrium are related by a constant value known as the Equilibrium Constant.
🔢 General Reversible Reaction
📌 Equilibrium Constant for a General Reaction
🔄 Steps for Writing an Equilibrium Constant Expression
  • 1
    Write the balanced chemical equation.
  • 2
    Write concentrations of all products in the numerator.
  • 3
    Write concentrations of all reactants in the denominator.
  • 4
    Raise each concentration to its stoichiometric coefficient.
  • 5
    Ignore pure solids and pure liquids (their concentration is constant).
🌟 Importance of Stoichiometric Coefficients
📐 Derivation of the Equilibrium Constant Expression
Consider the reaction \[\mathrm{aA+bB\rightleftharpoons cC+dD}\] Apply the Law of Mass Action
Forward reaction rate: \[R_f=k_f[A]^a[B]^b\] Reverse reaction rate: \[R_r=k_r[C]^c[D]^d\] R_r=k_r[C]^c[D]^d \[R_f=R_r\] Therefore, \[k_f[A]^a[B]^b=k_r[C]^c[D]^d\] Rearranging, \[\frac{k_f}{k_r}=\frac{[C]^c[D]^d}{[A]^a[B]^b}\] Since \[\frac{k_f}{k_r}=K_c\] Therefore, \[\boxed{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}\]
🌟 Physical Significance of the Equilibrium Constant
Value of Kc Interpretation
\(K_c\gg1\) Products are highly favoured. Reaction proceeds almost to completion.
\(K_c>1\) Products are favoured.
\(K_c\approx1\) Reactants and products are present in comparable amounts.
\(K_c<1\) Reactants are favoured.
\(K_c\ll1\) Very little product is formed.
✏️ Examples of Equilibrium Constant Expressions
1
Example
\[\begin{aligned}\mathrm{H_2+I_2\rightleftharpoons 2HI}\\\\ K_c=\dfrac{[HI]^2}{[H_2][I_2]} \end{aligned} \]
2
Example
\[ \begin{aligned}\mathrm{N_2+3H_2 \rightleftharpoons 2NH_3 }\\\\ K_c=\dfrac{[NH_3]^2}{[N_2][H_2]^3} \end{aligned} \]
3
Example
\[ \begin{aligned} \mathrm{SO_2+O_2\rightleftharpoons 2SO_3}\\\\ K_c=\dfrac{[SO_3]^2}{[SO_2]^2[O_2]} \end{aligned} \]
🌟 Important Rules for Writing \(K_c\)
✏️ Example
Solved Examples
Write the equilibrium constant expression for: \[\mathrm{2NO(g)+O_2(g) \rightleftharpoons 2NO_2(g) }\]
  • Law of Mass Action
  • Stoichiometric coefficients
  1. 1
    Write products in the numerator.
  2. 2
    Write reactants in the denominator.
  3. 3
    Use coefficients as powers.
\[K_c=\frac{[NO_2]^2}{[NO]^2[O_2]}\]
A reaction has \(K_c=5\times10^8\). Which side of the reaction is favoured?
Since \(K_c\) is much greater than 1, the equilibrium lies far towards the products. Almost all reactants are converted into products before equilibrium is established.
⚡ Exam Tip
❌ Common Mistakes
  • Using initial instead of equilibrium concentrations.
  • Ignoring stoichiometric coefficients.
  • Including pure solids and liquids in Kc.
  • Writing reactants in the numerator.
  • Assuming Kc changes with concentration or pressure.
📋 CBSE Competency-Based (HOTS) Question

Two reversible reactions have equilibrium constants of \(2\times10^{-6}\) and \(4\times10^{7}\), respectively. Which reaction forms more products at equilibrium? Explain your answer.

Answer

The reaction with \(K_c=4\times10^{7}\) forms considerably more products because its equilibrium lies far towards the product side. The reaction with \(K_c=2\times10^{-6}\) is reactant-favoured and produces only a very small amount of products.

🌟 Importance of the Equilibrium Constant
⚖️

Homogeneous Equilibrium

📋
Table of Contents
13 sections
🗺️ Overview
Equilibrium reactions can be classified based on the physical states (phases) of the reactants and products into:
  • Homogeneous Equilibrium
  • Heterogeneous Equilibrium

The classification is important because the expression of the equilibrium constant, equilibrium composition and the treatment of different phases depend upon whether the system is homogeneous or heterogeneous.
📘 Definition
💡 Concept of Homogeneous Equilibrium
🗂️ Phases of Homogeneous Equilibrium
Homogeneous Equilibrium in Gaseous Phase
The Haber process for the manufacture of ammonia is a classic example of homogeneous equilibrium. \[\mathrm{N_2(g)+3H_2(g) \rightleftharpoons 2NH_3(g)}\] Here,
  • Nitrogen is a gas.
  • Hydrogen is a gas.
  • Ammonia is also a gas.
Since every substance is present in the gaseous state, the reaction represents a homogeneous equilibrium. \[K_c=\frac{[NH_3]^2}{[N_2][H_2]^3}\]
Homogeneous Equilibrium in Aqueous Solution
Hydrolysis of ethyl acetate is another example of homogeneous equilibrium. \[\mathrm{CH_3COOC_2H_5(aq)+H_2O(l) \rightleftharpoons CH_3COOH(aq)+C_2H_5OH(aq)}\] Here,
  • Ethyl acetate is present in solution.
  • Water acts as the solvent.
  • Acetic acid is dissolved in water.
  • Ethanol is also present in solution.
Since all reacting species are present in a single liquid phase, the reaction is considered a homogeneous equilibrium.
✏️ More Examples of Homogeneous Equilibrium
Reaction Phase
\( \mathrm{H_2(g)+I_2(g)\rightleftharpoons2HI(g)} \) Gaseous
\( \mathrm{2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)} \) Gaseous
\( \mathrm{N_2O_4(g)\rightleftharpoons2NO_2(g)} \) Gaseous
\( \mathrm{CH_3COOH(aq)\rightleftharpoons H^+(aq)+CH_3COO^-(aq)} \) Aqueous
🔷 Characteristics of Homogeneous Equilibrium
🔷 Characteristics
  • All reactants and products exist in the same physical phase.
  • Only one phase is present throughout the system.
  • The composition of the reaction mixture is uniform.
  • The reaction remains dynamic in nature.
  • The rates of forward and reverse reactions become equal at equilibrium.
  • Concentrations of all species remain constant at equilibrium.
  • The equilibrium constant can be written directly using concentrations or partial pressures.
🌟 Importance of Homogeneous Equilibrium
⚖️ Difference Between Homogeneous and Heterogeneous Equilibrium
Homogeneous Equilibrium Heterogeneous Equilibrium
Only one phase is present. Two or more phases are present.
All reactants and products have the same physical state. Reactants and products have different physical states.
All species generally appear in the equilibrium expression. Pure solids and pure liquids are omitted from the equilibrium expression.
Examples: Gaseous and aqueous equilibria. Examples: Solid-liquid and solid-gas equilibria.
✏️ Example
Solved Example
Classify the following reaction as homogeneous or heterogeneous equilibrium: \[\mathrm{H_2(g)+I_2(g)\rightleftharpoons 2HI(g)}\]
Classification based on physical states.
Hydrogen, iodine vapour and hydrogen iodide are all present in the gaseous state. Therefore, the reaction establishes a homogeneous equilibrium.
Why is the Haber process considered a homogeneous equilibrium?
In the Haber process, \[\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\] nitrogen, hydrogen and ammonia are all gases. Since every reactant and product is present in the same gaseous phase, the equilibrium is homogeneous.
⚡ Exam Tip
❌ Common Mistakes
  • Thinking that aqueous species and water are different phases.
  • Ignoring the physical state symbols while classifying reactions.
  • Assuming every reversible reaction is homogeneous.
  • Confusing homogeneous equilibrium with homogeneous mixtures.
📋 CBSE Competency-Based (HOTS) Question

Consider the following reactions:

(i) \[\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}\]

(ii) \[\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}\]

Identify which reaction represents homogeneous equilibrium and justify your answer.

Answer

Reaction (i) is a homogeneous equilibrium because all reactants and products are gases. Reaction (ii) is a heterogeneous equilibrium because solids and a gas are present simultaneously, resulting in more than one phase.

🌟 Significance for Board and Competitive Examinations
  • Frequently tested in conceptual and assertion-reason questions.
  • Forms the basis for writing equilibrium constant expressions.
  • Essential for understanding industrial equilibrium processes.
  • Helpful in studying Le Chatelier's Principle and gaseous equilibria.
  • Important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Equilibrium Constant in Gaseous Systems

📋
Table of Contents
13 sections
🗺️ Overview
For gaseous reactions, the composition of an equilibrium mixture can be expressed either in terms of molar concentration or partial pressure. Hence, two equilibrium constants are commonly used:
  • Kc — Equilibrium constant expressed in terms of molar concentration.
  • Kp — Equilibrium constant expressed in terms of partial pressures.

Both constants describe the same equilibrium but differ in the way equilibrium composition is represented.
📌 Ideal Gas Equation
🔗 Relationship Between Pressure and Concentration
Rearranging the Ideal Gas Equation, \[P=\frac{n}{V}RT\] Since, \[\frac{n}{V}=C\] where \(C\) is the molar concentration, \[\boxed{P=CRT}\] Therefore, \[C=\frac{P}{RT}\] Hence, at constant temperature, \[P\propto C\] This important relation allows conversion between concentration and pressure while deriving the relationship between \(K_c\) and \(K_p\).
📌 Equilibrium Constant in Terms of Partial Pressure \(K_p\)
📐 Derivation
\(K_p = K_c\) for H₂ + I₂ ⇌ 2HI
Using\[P=CRT\]Therefore, \[P_{HI}=[HI]RT\] \[P_{H_2}=[H_2]RT\] \[P_{I_2}=[I_2]RT\] Substitute these values into the expression of \(K_p\): \[K_p=\dfrac{([HI]RT)^2}{([H_2]RT)([I_2]RT)}\] Simplifying, \[K_p=\dfrac{[HI]^2(RT)^2}{[H_2][I_2](RT)^2}\] \[\boxed{K_p=K_c}\] This happens because the total number of gaseous moles remains unchanged.
✏️ Example
Ammonia Synthesis
Consider the Haber process: \[\mathrm{N_2(g)+3H_2(g) \rightleftharpoons 2NH_3(g)}\] The equilibrium constant in terms of concentration is: \[K_c=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\] The equilibrium constant in terms of pressure is: \[K_p=\dfrac{(P_{NH_3})^2}{P_{N_2}(P_{H_2})^3}\] Replace each partial pressure using \[P=CRT\] Then, \[K_p=\dfrac{([NH_3]RT)^2}{([N_2]RT)([H_2]RT)^3}\] Rearranging, \[K_p=\dfrac{[NH_3]^2}{[N_2][H_2]^3}(RT)^{-2}\] Since, \[\dfrac{[NH_3]^2}{[N_2][H_2]^3}=K_c\] Therefore, \[\boxed{K_p=K_c(RT)^{-2}}\]
🔗 Relations
General Relationship Between \(K_p\text{ and } K_c\)
Consider the general gaseous reaction: \[\mathrm{aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)}\] The equilibrium constant in terms of concentration is: \[K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}\] The equilibrium constant in terms of pressure is: \[K_p=\dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\] The relationship between them is: \[\boxed{K_p=K_c(RT)^{\Delta n}}\]
Meaning of Δn
\[\boxed{\Delta n=(\text{Total moles of gaseous products}) - (\text{Total moles of gaseous reactants})}\]
✏️ Example
1
Example
\[\mathrm{H_2+I_2\rightleftharpoons 2HI}\] Products = 2 moles
Reactants = 2 moles \[\Delta n=2-2=0\] Therefore, \[K_p=K_c(RT)^0=K_c\]
2
Example
\[\mathrm{N_2+3H_2\rightleftharpoons 2NH_3}\] Products = 2 moles
Products = 2 moles \[\Delta n=2-4=-2\] Hence, \[K_p=K_c(RT)^{-2}\]
⭐ Special Case
🔢 Formula Summary
✏️ Example
Solved Examples
Find the relationship between \(K_p\) and \(K_c\) for: \[\mathrm{H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g)}\]
  1. 1
    Calculate Δn.
  2. 2
    Use the formula \(K_p=K_c(RT)^{\Delta n}\).
Products = 2 moles
Reactants = 2 moles \[\Delta n=2-2=0\] Therefore, \[\boxed{K_p=K_c}\]
For the reaction \[\mathrm{N_2+3H_2\rightleftharpoons 2NH_3}\] if \(K_c=0.50\) and \(\Delta n=-2\), write the expression for \(K_p\).
\[K_p=K_c(RT)^{-2}\] Substituting \(K_c=0.50\), \[ K_p=0.50(RT)^{-2}\]
⚡ Exam Tip
❌ Common Mistakes
  • Including solids or liquids while calculating Δn.
  • Using total moles instead of gaseous moles.
  • Writing \(K_p=K_cRT\) instead of \(K_p=K_c(RT)^{\Delta n}\).
  • Ignoring stoichiometric coefficients.
  • Using an unbalanced equation to calculate Δn.
📋 CBSE Competency-Based (HOTS) Question

For the equilibrium \[\mathrm{PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)}\] determine Δn and write the relationship between \(K_p\) and \(K_c\).

Answer

Products = 2 gaseous moles
Reactants = 1 gaseous mole

\[ \Delta n=2-1=1 \]

Therefore,

\[ \boxed{ K_p=K_c(RT) } \]

⚖️

Heterogeneous Equilibrium

📋
Table of Contents
15 sections
🗺️ Overview
Equilibrium reactions are classified into homogeneous and heterogeneous equilibria based on the number of physical phases present in the reaction system.

When reactants and products exist in two or more different physical phases, the equilibrium established is called heterogeneous equilibrium.
📘 Definition of Heterogeneous Equilibrium
💡 Concept of Heterogeneous Equilibrium
🗂️ Examples
Water–Vapour Equilibrium
Consider water kept in a closed container. \[\mathrm{H_2O(l)\rightleftharpoons H_2O(g)}\] In this system:
  • Liquid water continuously evaporates.
  • Water vapour continuously condenses.
  • Both processes occur at equal rates at equilibrium.
Since liquid and gaseous phases coexist, this is a heterogeneous equilibrium.
Thermal Decomposition of Calcium Carbonate
One of the most important examples of heterogeneous equilibrium is the thermal dissociation of calcium carbonate. \[\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}\] Here,
  • Calcium carbonate is a solid.
  • Calcium oxide is also a solid.
  • Carbon dioxide is a gas.
Since two different phases (solid and gas) are present, the reaction establishes a heterogeneous equilibrium.
🔗 Equilibrium Constant Expression
Applying the Law of Mass Action, \[K_c=\frac{[CaO][CO_2]}{[CaCO_3]}\] However,
  • Calcium carbonate is a pure solid.
  • Calcium oxide is also a pure solid.
The concentration of every pure solid remains constant because its density does not change during the reaction. Therefore, \[[CaCO_3]=\text{Constant}\] \[[CaO]=\text{Constant}\] These constant terms are incorporated into the equilibrium constant. Hence, [\[boxed{K_c=[CO_2]}\]
Expression for \(K_p\)
Since only carbon dioxide exists in the gaseous phase, [\[boxed{K_c=[CO_2]}\] Thus, the equilibrium pressure of carbon dioxide alone determines the equilibrium constant.
🤔 Did You Know?
Why Are Pure Solids and Pure Liquids Omitted?
The concentration of a pure solid or a pure liquid remains constant irrespective of the amount present, provided some of the substance is available. Mathematically, \[\text{Density}=\frac{\text{Mass}}{\text{Volume}}\] Since the density of a pure solid or liquid is constant, \[\text{Concentration}=\text{Constant}\] Therefore, their concentrations are merged into the equilibrium constant and do not appear explicitly in the equilibrium expression.
✏️ More Examples of Heterogeneous Equilibrium
Reaction Phases Present
\[ \mathrm{H_2O(l)\rightleftharpoons H_2O(g)} \] Liquid + Gas
\[ \mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)} \] Solid + Gas
\[ \mathrm{NH_4HS(s)\rightleftharpoons NH_3(g)+H_2S(g)} \] Solid + Gas
\[ \mathrm{I_2(s)\rightleftharpoons I_2(g)} \] Solid + Gas
⚖️ Comparison Between Homogeneous and Heterogeneous Equilibrium
Homogeneous Equilibrium Heterogeneous Equilibrium
Only one physical phase exists. Two or more physical phases coexist.
Composition is uniform. Composition is not uniform.
All species generally appear in the equilibrium expression. Pure solids and pure liquids are omitted.
Examples: Gaseous and aqueous equilibria. Examples: Solid-gas and liquid-gas equilibria.
🗒️ Rules For Writing Equilibrium Constant In Heterogeneous Equilibrium
  1. Write the balanced chemical equation.
  2. Write products in the numerator.
  3. Write reactants in the denominator.
  4. Use stoichiometric coefficients as exponents.
  5. Omit all pure solids.
  6. Omit all pure liquids.
  7. Include only gaseous and aqueous species.
📌 Species Included in Equilibrium Constant Expression
✏️ Example
Solved Examples
Write the equilibrium constant expression for: \[\mathrm{CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)}\]
  • Law of Mass Action
  • Pure solids are omitted.
  1. 1
    Write the complete equilibrium expression.
  2. 2
    Remove pure solids.
  3. 3
    Simplify the expression.
Complete expression:\[K_c=\frac{[CaO][CO_2]}{[CaO][CO_2]}\] Since both calcium carbonate and calcium oxide are pure solids, \[\boxed{K_c=[CO_2]}\] Similarly \[\boxed{K_p=P_{CO_2}}\]
Why are pure solids omitted from the equilibrium constant expression?
The concentration of a pure solid remains constant because its density does not change during the reaction. Since this value is constant, it is combined with the equilibrium constant. Therefore, pure solids do not appear explicitly in the equilibrium expression.
⚡ Exam Tip
❌ Common Mistakes
  • Including pure solids in Kc.
  • Including pure liquids in Kp.
  • Forgetting to use stoichiometric coefficients as exponents.
  • Confusing heterogeneous equilibrium with heterogeneous mixtures.
  • Ignoring the physical state symbols while writing the equilibrium expression.
📋 CBSE Competency-Based (HOTS) Question

Write the equilibrium constant expression for the following reaction:

\[ \mathrm{ NH_4HS(s) \rightleftharpoons NH_3(g)+H_2S(g) } \]

Answer

Since ammonium hydrogen sulphide is a pure solid, it is omitted from the equilibrium expression.

Therefore,

\[ \boxed{ K_c=[NH_3][H_2S] } \]

and

\[ \boxed{ K_p=P_{NH_3}\times P_{H_2S} } \]

🌟 Significance for Board and Competitive Examinations
  • Essential for writing correct equilibrium constant expressions.
  • Frequently tested in assertion-reason and competency-based questions.
  • Important for Le Chatelier's Principle and thermal decomposition reactions.
  • Forms the basis of solubility equilibrium and ionic equilibrium.
  • Highly relevant for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Applications of the Equilibrium Constant

📋
Table of Contents
13 sections
🗺️ Overview
The equilibrium constant is one of the most useful quantitative tools in Chemistry. It not only describes the composition of an equilibrium mixture but also helps chemists predict whether a reaction is feasible, the direction in which it will proceed, and the amount of product formed at equilibrium.

Knowledge of equilibrium constants is essential in chemical industries, environmental chemistry, analytical chemistry, biochemistry, and pharmaceutical sciences. It is also one of the most frequently tested concepts in CBSE Board examinations, JEE Main, NEET and CUET.
🌟 Important Features of the Equilibrium Constant
🗂️ Types / Category
Applications
Predicting the Extent of a Reaction
The magnitude of the equilibrium constant indicates the extent to which a reaction proceeds before equilibrium is established.

It tells whether the equilibrium mixture contains mostly reactants or mostly products.

The equilibrium constant gives information about the extent of reaction, but it does not indicate how fast equilibrium is attained.
Interpretation of Different Values of K
Value of K Interpretation
\(K \gg 1\) Reaction proceeds almost to completion. Products predominate.
\(K > 1\) Products are favoured.
\(K \approx 1\) Reactants and products are present in comparable amounts.
\(K < 1\) Reactants are favoured.
\(K \ll 1\) Very little product is formed.
Predicting the Direction of a Reaction
Reaction Quotient (Q)
Before equilibrium is established, the composition of the reaction mixture is described by the Reaction Quotient (Q).
For the reaction \[\mathrm{aA+bB\rightleftharpoons cC+dD}\] the reaction quotient is \[\boxed{Q_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}}\] Notice that the mathematical expression of \(Q_c\) is identical to that of \(K_c\). The only difference is:
  • \(Q_c\) uses concentrations at any instant.
  • \(K_c\) uses concentrations only at equilibrium.
🔗 Relationship Between \(Q_c\text{ and }K_c\)
Condition Direction of Reaction
\(Q_c Forward reaction occurs to produce more products.
\(Q_c>K_c\) Reverse reaction occurs to produce more reactants.
\(Q_c=K_c\) The system is already at equilibrium.
🤔 Did You Know?
Why Does the Reaction Move in a Particular Direction?
  • If \(Q_c < K_c\), the reaction has fewer products than required at equilibrium, so the forward reaction is favoured.
  • If \(Q_c>K_c\), excess products are present, so the reverse reaction is favoured.
  • If \(Q_c=K_c\), the composition is already at equilibrium and no net reaction occurs.
🗒️ ICE Table Method (Initial–Change–Equilibrium)
The easiest method for solving equilibrium concentration problems is the ICE Table.
Step Description
1 Write the balanced chemical equation.
2 Prepare an ICE table.
3 Write the initial concentration of every species.
4 Represent concentration changes using \(+x\), \(-x\), etc.
5 Calculate equilibrium concentrations.
6 Substitute these values into the equilibrium constant expression.
7 Solve for \(x\).
8 Verify the obtained concentrations by substituting them into the equilibrium expression.
📌 General ICE Table Format
📝 Summary of Important Properties of the Equilibrium Constant
✏️ Example
Solved Example
A reaction has \(K_c=100\). At a certain instant, \(Q_c=10\). Predict the direction in which the reaction will proceed.
Comparison of \(Q_c\) and \(K_c\).
Since\[Q_c < K_c\] the reaction contains fewer products than required at equilibrium. Therefore, it will proceed in the forward direction to produce more products.
If the equilibrium constant of a reaction is 25, calculate the equilibrium constant of the reverse reaction.
Using\[K_{\text{reverse}}=\frac{1}{K_{\text{forward}}}\] Therefore, \[ \begin{aligned} K_{\text{reverse}} &= \frac{1}{25}\\ &= 0.04 \end{aligned} \]
Why is an ICE table used while solving equilibrium problems?
The ICE table systematically records the initial concentrations, changes occurring during the reaction, and equilibrium concentrations. It simplifies the substitution of equilibrium values into the equilibrium constant expression and helps avoid calculation errors.
⚡ Exam Tip
❌ Common Mistakes
  • Using initial concentrations instead of equilibrium concentrations while calculating \(K\).
  • Assuming that a large value of \(K\) means a fast reaction.
  • Confusing reaction rate with the extent of reaction.
  • Comparing \(Q\) and \(K\) incorrectly.
  • Ignoring stoichiometric changes while preparing the ICE table.
📋 CBSE Competency-Based (HOTS) Question

A reversible reaction has \(K_c=2.5\times10^5\). At a certain instant, \(Q_c=5\times10^6\). Predict the direction in which the reaction will proceed and explain your answer.

Answer

Since \[ Q_c>K_c, \] the reaction mixture contains more products than required at equilibrium. Therefore, the reaction will proceed in the reverse direction until equilibrium is restored by converting some products back into reactants.

🌟 Importance of This Topic
⚖️

Relationship Between Equilibrium Constant (K), Reaction Quotient (Q) and Gibbs Free Energy (ΔG)

📋
Table of Contents
15 sections
🗺️ Overview
Thermodynamics and chemical equilibrium are closely related. While the equilibrium constant (K) indicates the extent of a reaction, the Gibbs Free Energy (ΔG) predicts whether a reaction is spontaneous under given conditions.

The relationship between ΔG, Q and K enables us to determine:
  • Whether a reaction is spontaneous.
  • Whether equilibrium has been established.
  • The direction in which a reaction will proceed.
  • The effect of reaction composition on spontaneity.
📘 Gibbs Free Energy (ΔG)
🌟 Physical Significance of ΔG
Value of ΔG Interpretation
\(\Delta G<0\) The reaction is spontaneous in the forward direction.
\(\Delta G>0\) The forward reaction is non-spontaneous; the reverse reaction is spontaneous.
\(\Delta G=0\) The reaction has reached equilibrium.
🗒️ Reaction Quotient (Q)
The reaction quotient (Q) has the same mathematical expression as the equilibrium constant, but it is calculated using concentrations or partial pressures at any stage of the reaction, not necessarily at equilibrium.

For the reaction \[\mathrm{aA+bB\rightleftharpoons cC+dD}\] \[Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]
🔗 Relationship Between ΔG and Q
At any stage of the reaction, \[\boxed{\Delta G=\Delta G^\Theta+RT\ln Q}\] where,
Symbol Meaning
\(\Delta G\) Gibbs free energy under actual conditions.
\(\Delta G^\Theta\) Standard Gibbs free energy change.
\(R\) Universal gas constant (8.314 J mol-1 K-1).
\(T\) Absolute temperature (K).
\(Q\) Reaction quotient.
📐 Derivation of the Relationship Between ΔG° and K
The general Gibbs equation is: \[\Delta G=\Delta G^\Theta+RT\ln Q\] At equilibrium,
  • \(\Delta G=0\)
  • \(Q=K\)
Therefore, \[0=\Delta G^\Theta+RT\ln K\] Rearranging, \[\boxed{\Delta G^\Theta=-RT\ln K}\] This is the most important thermodynamic relation in chemical equilibrium.
Alternative Form of the Equation
Rearranging further, \[\ln K=-\frac{\Delta G^\Theta}{RT}\] Therefore, \[\boxed{K=e^{-\Delta G^\Theta/RT}}\] This equation shows that the equilibrium constant depends directly on the standard Gibbs free energy change.
🔗 Relations
Relationship Between \(K\) and \(\Delta G^\Theta\)
Value of \(\Delta G^\Theta\) Value of K Interpretation
Negative \(K>1\) Products are favoured.
Positive \(K<1\) Reactants are favoured.
Zero \(K=1\) Reactants and products are equally favoured.
🔗 Combined Relationship Between ΔG, Q and K
Condition Comparison Direction of Reaction
\(\Delta G<0\) \(Q < K\) Forward reaction proceeds spontaneously.
\(\Delta G>0\) \(Q > K\) Reverse reaction proceeds spontaneously.
\(\Delta G=0\) \(Q = K\) System is at equilibrium.
🌟 Importance of the Relationship
🔢 Formula Summary
✏️ Example
Solved Example
The standard Gibbs free energy change for a reaction is negative. Predict the value of the equilibrium constant.
\(\Delta G^\Theta=-RT\ln K\)
Since \(\Delta G^\Theta\) is negative, \(\ln K\) becomes positive. Therefore, \(K>1\), indicating that the reaction favours product formation at equilibrium.
A reaction mixture has \(Q > K\). Predict the sign of ΔG and the direction in which the reaction proceeds.
Since \(Q>K\), the reaction mixture contains excess products. Therefore, \(\Delta G>0\) for the forward reaction, making it non-spontaneous. The reaction proceeds in the reverse direction until equilibrium is re-established.
What are the values of ΔG and Q when a reaction reaches equilibrium?
At equilibrium,\[\boxed{\Delta G=0}\] and \[\boxed{Q=K}\]
⚡ Exam Tip
❌ Common Mistakes
  • Confusing ΔG with \(\Delta G^\Theta\).
  • Using \(Q=K\) before equilibrium is established.
  • Writing \(\Delta G=-RT\ln K\) instead of \(\Delta G^\Theta=-RT\ln K\).
  • Assuming a spontaneous reaction is always fast.
  • Ignoring the sign of ΔG while predicting reaction direction.
📋 CBSE Competency-Based (HOTS) Question

A reversible reaction has a positive value of \(\Delta G^\Theta\). Predict the value of K and explain whether the reaction is product-favoured or reactant-favoured.

Answer

Since \[ \Delta G^\Theta=-RT\ln K, \] a positive value of \(\Delta G^\Theta\) gives a negative value of \(\ln K\), which means \(K<1\). Therefore, the equilibrium lies towards the reactants, and only a small amount of products is formed under standard conditions.

🌟 Importance for Board and Competitive Examinations
⚖️

Factors Affecting Chemical Equilibrium

📋
Table of Contents
12 sections
🗺️ Overview
Once a reversible reaction reaches equilibrium, its position can be altered by changing external conditions such as concentration, pressure, temperature or volume. The prediction of these changes is based on the Le Chatelier's Principle.

"Whenever a system at equilibrium is subjected to a change in concentration, pressure, temperature or volume, the equilibrium shifts in a direction that tends to oppose the imposed change and establishes a new equilibrium."

This principle was proposed by the French scientist Henri Louis Le Chatelier and is extensively applied in chemical industries to maximise product yield.
📌 Note

Effect of Change in Concentration

📌 Note

Effect of Pressure (or Volume)

📌 Note

Effect of Adding an Inert Gas

📌 Note

Effect of Temperature

📘 Definition

Effect of a Catalyst

📝 Summary of Factors Affecting Equilibrium
✏️ Example
Solved Example
Predict the effect of increasing pressure on the equilibrium: \[\mathrm{N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)}\]
Le Chatelier's Principle
The reactant side has four moles of gas, while the product side has two moles. Increasing pressure shifts the equilibrium towards the side with fewer gaseous molecules. Therefore, the equilibrium shifts towards ammonia formation.
An exothermic reaction is heated. Predict the effect on the equilibrium constant.
Increasing temperature favours the endothermic (reverse) direction. Consequently, the value of the equilibrium constant decreases for an exothermic reaction.
Does a catalyst increase the yield of products at equilibrium?
No. A catalyst accelerates both the forward and reverse reactions equally. It only helps the system reach equilibrium more rapidly and does not alter the equilibrium composition or the value of the equilibrium constant.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming catalysts increase the equilibrium yield.
  • Thinking pressure affects all reactions.
  • Believing concentration changes alter the equilibrium constant.
  • Confusing reaction rate with equilibrium position.
  • Ignoring the sign of ΔH while predicting temperature effects.
📋 CBSE Competency-Based (HOTS) Question

The reaction \[ \mathrm{ N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)+Heat } \] is at equilibrium. Explain the effect of increasing (i) pressure, (ii) temperature, and (iii) adding a catalyst on the yield of ammonia.

Answer

  • Increasing pressure: Shifts equilibrium towards the products because fewer gaseous moles are present on the product side, increasing the yield of ammonia.
  • Increasing temperature: Since the reaction is exothermic, equilibrium shifts towards the reactants and the yield of ammonia decreases.
  • Adding a catalyst: Does not change the equilibrium position or the yield of ammonia. It only decreases the time required to attain equilibrium.
🌟 Importance of Le Chatelier's Principle
⚖️

Ionic Equilibrium in Solution

📋
Table of Contents
14 sections
🗺️ Overview
Most chemical reactions occurring in living organisms, natural water bodies and industrial processes involve ions dissolved in water. The equilibrium established between ions and unionised molecules in aqueous solution is known as ionic equilibrium.

Ionic equilibrium forms the foundation of several important concepts such as acid-base equilibrium, buffer solutions, pH, hydrolysis of salts, common ion effect, solubility product (Ksp) and electrochemistry.

It is one of the most important topics in Class XI Chemistry and serves as the basis for many numerical problems in Class XII as well as competitive examinations like JEE Main, NEET and CUET.
📘 Definition of Ionic Equilibrium
💡 Concept of Ionic Equilibrium
📘 Definition

Electrolytes

📘 Definition
Electrolytes are substances that produce ions when dissolved in water or in the molten state, thereby conducting electricity.

Classification of Electrolytes

Type Degree of Ionisation Examples
Strong Electrolytes Almost complete ionisation HCl, HNO₃, H₂SO₄, NaOH, KOH, NaCl
Weak Electrolytes Partial ionisation CH₃COOH, NH₄OH, HF, H₂CO₃
⚖️ Difference Between Strong and Weak Electrolytes
Strong Electrolytes Weak Electrolytes
Ionise almost completely. Ionise only partially.
Very high electrical conductivity. Low electrical conductivity.
No significant ionic equilibrium. Ionic equilibrium is established.
Almost no unionised molecules remain. Both ions and molecules coexist.
🗂️ Examples of Ionic Equilibrium
  • Acetic Acid \(\mathrm{CH_3COOH(aq)\rightleftharpoons H^+(aq)+CH_3COO^-(aq)}\)
  • Ammonium Hydroxide \(\mathrm{NH_4OH(aq)\rightleftharpoons NH_4^+(aq)+OH^-(aq)}\)
  • Hydrofluoric Acid \(\mathrm{HF(aq)\rightleftharpoons H^+(aq)+F^-(aq)}\)
📌 >Role of Ions in Aqueous Solutions
🔷 Characteristics of Ionic Equilibrium
🔷 Characteristics

  • Occurs only in weak electrolytes.

  • Both ionisation and recombination occur simultaneously.

  • The system is dynamic in nature.

  • Concentrations of ions remain constant at equilibrium.

  • The equilibrium obeys the Law of Mass Action.

  • Temperature influences the degree of ionisation.

🗒️ Importance of Ionic Equilibrium
Area Application
Acid-Base Chemistry Determination of pH and strength of acids and bases.
Buffer Solutions Maintains constant pH.
Medicine Maintains blood pH near 7.4.
Biological Systems Enzyme activity and metabolic reactions.
Analytical Chemistry Qualitative and quantitative analysis.
Industrial Chemistry Electroplating, batteries and electrolysis.
✏️ Solved Example
Why does acetic acid establish ionic equilibrium whereas hydrochloric acid does not?
Strong and weak electrolytes.
Acetic acid is a weak electrolyte and ionises only partially. Therefore, both ions and unionised molecules coexist in solution, establishing ionic equilibrium.

Hydrochloric acid is a strong electrolyte and ionises almost completely. Since almost no unionised molecules remain, no significant ionic equilibrium exists.
Identify the weak electrolyte among the following:
  • HCl
  • NaOH
  • CH₃COOH
  • NaCl
CH₃COOH is a weak electrolyte because it undergoes only partial ionisation in water and establishes ionic equilibrium.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming all electrolytes establish ionic equilibrium.
  • Considering HCl as a weak electrolyte.
  • Confusing strong electrolytes with concentrated solutions.
  • Ignoring the reversible nature of ionisation in weak electrolytes.
  • Using the terms ionisation and dissociation interchangeably without context.
📋 CBSE Competency-Based (HOTS) Question

Two aqueous solutions contain HCl and CH₃COOH of the same concentration. Explain why the HCl solution conducts electricity better than the CH₃COOH solution.

Answer

Hydrochloric acid is a strong electrolyte and ionises almost completely, producing a large number of ions. Acetic acid is a weak electrolyte and ionises only partially, producing fewer ions. Since electrical conductivity depends on the number of mobile ions, the HCl solution conducts electricity much better than the CH₃COOH solution.

🌟 Significance for Board and Competitive Examinations
  • Serves as the foundation for the entire Ionic Equilibrium chapter.
  • Essential for understanding acid-base equilibria, pH and buffer solutions.
  • Forms the basis of Ka, Kb, Kw and pH derivations.
  • Frequently appears in CBSE competency-based and assertion-reason questions.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Arrhenius Concept of Acids and Bases

📋
Table of Contents
14 sections
🗺️ Overview
The first scientific explanation of the acidic and basic nature of substances was proposed by the Swedish scientist Svante Arrhenius in 1884. His theory explains the behaviour of acids and bases in aqueous solutions on the basis of the ions they produce.

Although modern acid-base theories are more comprehensive, the Arrhenius concept remains the foundation for understanding acid-base chemistry and ionic equilibrium.
📘 Definition
✏️ Examples of Arrhenius Acids
Acid Ionisation Equation
Hydrochloric Acid (HCl) \[ \mathrm{ HCl+H_2O \rightarrow H_3O^++Cl^- } \]
Nitric Acid (HNO₃) \[ \mathrm{ HNO_3+H_2O \rightarrow H_3O^++NO_3^- } \]
Acetic Acid (CH₃COOH) \[ \mathrm{ CH_3COOH+H_2O \rightleftharpoons H_3O^++CH_3COO^- } \]
📘 Arrhenius Base
✏️ Examples of Arrhenius Bases
Base Ionisation Equation
Sodium Hydroxide \[ \mathrm{ NaOH \rightarrow Na^++OH^- } \]
Potassium Hydroxide \[ \mathrm{ KOH \rightarrow K^++OH^- } \]
Calcium Hydroxide \[ \mathrm{ Ca(OH)_2 \rightarrow Ca^{2+}+2OH^- } \]
📌 Neutralisation According to Arrhenius Theory
🔷 Characteristics of Arrhenius Acids and Bases
🔷 Characteristics

  • Applicable only to aqueous solutions.

  • Acids produce H+ (or H3O+) ions.

  • Bases produce OH- ions.

  • Acids and bases conduct electricity because they produce ions.

  • Neutralisation forms water and a salt.

⚠️ Limitations of Arrhenius Theory
Limitation Explanation
Applicable only in aqueous solution. Cannot explain acid-base reactions occurring in non-aqueous solvents or gases.
Cannot explain ammonia as a base. NH₃ has no OH⁻ group but behaves as a base.
Cannot explain acidic or basic oxides. Many oxides show acidic or basic behaviour without producing H⁺ or OH⁻ directly.
Cannot explain proton-transfer reactions. This limitation led to the Brönsted–Lowry theory.
⚖️ Comparison of Arrhenius Acids and Bases
Arrhenius Acid Arrhenius Base
Produces H+ ions. Produces OH- ions.
pH less than 7. pH greater than 7.
Turns blue litmus red. Turns red litmus blue.
Examples: HCl, HNO₃, CH₃COOH. Examples: NaOH, KOH, Ca(OH)₂.
✏️ Example
Solved Examples
Explain why hydrochloric acid is called an Arrhenius acid.
Arrhenius definition of acids.
Hydrochloric acid ionises completely in water to produce hydronium ions. \[ \mathrm{ HCl+H_2O \rightarrow H_3O^++Cl^- } \] Since it increases the concentration of H₃O⁺ ions in water, it is an Arrhenius acid.
Why is sodium hydroxide considered an Arrhenius base?
Sodium hydroxide dissociates completely in water: \[ \mathrm{ NaOH \rightarrow Na^++OH^- } \] Since it increases the concentration of hydroxide ions, it is an Arrhenius base.
Why is ammonia not explained satisfactorily by the Arrhenius theory?
Ammonia (NH₃) behaves as a base in water but does not contain an OH⁻ group. Therefore, the Arrhenius theory cannot explain its basic nature satisfactorily. This limitation was overcome by the Brönsted–Lowry concept of acids and bases.
⚡ Exam Tip
❌ Common Mistakes
  • Writing free H⁺ ions instead of H₃O⁺ in aqueous solution.
  • Assuming all bases must contain OH⁻ groups.
  • Ignoring that the theory is valid only in water.
  • Confusing strong acids with concentrated acids.
  • Ignoring the limitations of the Arrhenius concept.
📋 CBSE Competency-Based (HOTS) Question

Ammonia increases the concentration of OH⁻ ions in water but contains no hydroxyl group. Explain why this observation cannot be completely explained by the Arrhenius theory.

Answer

According to Arrhenius, a base should directly produce OH⁻ ions upon dissociation. Ammonia does not contain an OH⁻ group; instead, it reacts with water to produce OH⁻ ions. Therefore, its basic behaviour cannot be explained completely by the Arrhenius theory, leading to the development of the Brönsted–Lowry concept.

🌟 Significance
Significance for Board and Competitive Examinations
  • Forms the foundation of acid-base chemistry.
  • Frequently asked in CBSE theory questions and competency-based assessments.
  • Provides the basis for understanding pH and ionic equilibrium.
  • Essential before studying the Brönsted–Lowry and Lewis theories.
  • Important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Brönsted–Lowry Concept of Acids and Bases

📋
Table of Contents
16 sections
🗺️ Overview
The Arrhenius theory successfully explained the behaviour of acids and bases in aqueous solutions, but it could not explain acid-base reactions occurring in non-aqueous media or the basic nature of substances such as ammonia (NH₃).

To overcome these limitations, the Danish chemist Johannes Nicolaus Brönsted and the English chemist Thomas Martin Lowry independently proposed a broader acid-base theory in 1923, now known as the Brönsted–Lowry Theory.
🗂️ Types / Category
Definition of a Brönsted Acid and Base
Definition of a Brönsted Acid
According to the Brönsted–Lowry theory, an acid is a substance that donates a proton (H+) to another substance.
Brönsted Acid = Proton (H+) Donor
Definition of a Brönsted Base
According to the Brönsted–Lowry theory, a base is a substance that accepts a proton (H+) from another substance.

Brönsted Base = Proton (H+) Acceptor
💡 Basic Concept of the Brönsted–Lowry Theory
🗂️ Examples
Hydrochloric Acid and Water
Consider the reaction: \[\mathrm{HCl+H_2O\rightarrow H_3O^++Cl^-}\]
Species Role
HCl Brönsted Acid (proton donor)
H₂O Brönsted Base (proton acceptor)
Cl⁻ Conjugate Base
H₃O⁺ Conjugate Acid
Ammonia and Water
One of the greatest advantages of the Brönsted theory is that it explains the basic nature of ammonia.\[\mathrm{NH_3+H_2O \rightleftharpoons NH_4^++OH^-}\] Here,
  • NH₃ accepts a proton and behaves as a Brönsted base.
  • H₂O donates a proton and behaves as a Brönsted acid.
📘 Conjugate Acid–Base Pair
✏️ Examples of Conjugate Acid–Base Pairs
Acid Conjugate Base Base Conjugate Acid
HCl Cl⁻ H₂O H₃O⁺
H₂O OH⁻ NH₃ NH₄⁺
CH₃COOH CH₃COO⁻ H₂O H₃O⁺
HF F⁻ NH₃ NH₄⁺
🏷️ Properties of Conjugate Acid–Base Pairs
Properties
One-proton-difference
Conjugate acid–base pairs differ by exactly one proton (H⁺); removing or adding that proton converts the acid into its conjugate base and vice versa.
Acid-has-more-proton
A conjugate acid is the species formed when a base accepts a proton, so it carries one additional H⁺ compared with the original base and may have a higher positive charge.
Base-has-less-proton
A conjugate base results when an acid donates a proton, leaving the species with one fewer H⁺ and often with a greater negative character or lower formal charge.
Acid-always-has-conjugate-base
After an acid donates a proton it becomes its conjugate base; this is true for every acid, whether the proton transfer is complete (strong acids) or partial (weak acids).
Base-always-has-conjugate-acid
When a base accepts a proton it forms its conjugate acid; the strength of the original base influences how readily this protonation occurs.
🔗 Relationship Between Acid Strength and Conjugate Base Strength
The strengths of an acid and its conjugate base are inversely related.
Acid Strength Conjugate Base Strength
Strong Acid Very Weak Base
Weak Acid Relatively Strong Base
Similarly,
Base Strength Conjugate Acid Strength
Strong Base Very Weak Acid
Weak Base Relatively Strong Acid
✅ Advantages of the Brönsted–Lowry Theory
  • Applicable to aqueous as well as non-aqueous solutions.
  • Explains the basic nature of ammonia.
  • Explains proton-transfer reactions.
  • Introduces the concept of conjugate acid-base pairs.
  • Provides a broader definition than the Arrhenius theory.
⚠️ Limitations of the Brönsted–Lowry Theory
  • Cannot explain acid-base reactions in which no proton transfer occurs.
  • Cannot explain the acidic behaviour of electron-pair acceptors such as BF₃ and AlCl₃.
  • Led to the development of the Lewis theory of acids and bases.
⚖️ Comparison Between Arrhenius and Brönsted–Lowry Concepts
Arrhenius Theory Brönsted–Lowry Theory
Applicable only in aqueous solutions. Applicable in aqueous and non-aqueous media.
Acid produces H⁺ ions. Acid donates a proton.
Base produces OH⁻ ions. Base accepts a proton.
Cannot explain NH₃ completely. Successfully explains NH₃ as a base.
✏️ Example
Solved Example
Identify the acid, base, conjugate acid and conjugate base in the reaction: \[\mathrm{NH_3+H_2O \rightleftharpoons NH_4^++OH^-}\]
>Brönsted–Lowry proton-transfer concept
Acid H₂O
Base NH₃
Conjugate Acid NH₄⁺
Conjugate Base OH⁻
Identify the conjugate base of H₂O and the conjugate acid of NH₃.
Water loses one proton to form OH⁻. Therefore, OH⁻ is the conjugate base of H₂O.

Ammonia gains one proton to form NH₄⁺. Therefore, NH₄⁺ is the conjugate acid of NH₃.
Why is HCl considered a strong acid but Cl⁻ a weak base?
HCl donates its proton almost completely in water and is therefore a strong acid. Its conjugate base, Cl⁻, has very little tendency to accept a proton. Hence, Cl⁻ is an extremely weak base. This illustrates the inverse relationship between the strengths of an acid and its conjugate base.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing conjugate acid with conjugate base.
  • Thinking proton donation means loss of an electron.
  • Assuming only substances containing OH⁻ can behave as bases.
  • Ignoring that water can act as both an acid and a base.
  • Forgetting that conjugate acid-base pairs differ by exactly one proton.
📋 CBSE Competency-Based (HOTS) Question

In the reaction \[\mathrm{HF+H_2O \rightleftharpoons H_3O^++F^-}\] identify the two conjugate acid-base pairs and explain why fluoride ion (F⁻) is called the conjugate base of HF.

Answer

The two conjugate acid-base pairs are:

  • HF / F⁻
  • H₂O / H₃O⁺

Fluoride ion is called the conjugate base because it is formed when HF donates one proton (H⁺). Since the two species differ by exactly one proton, they form a conjugate acid-base pair.

🌟 Significance for Board and Competitive Examinations
  • One of the most frequently asked conceptual topics in Ionic Equilibrium.
  • Forms the basis for understanding conjugate acid-base pairs and amphoteric substances.
  • Essential before studying the Lewis concept of acids and bases.
  • Frequently appears in assertion-reason, competency-based and matching-type questions.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Lewis Concept of Acids and Bases

📋
Table of Contents
16 sections
🗺️ Overview
The most comprehensive theory of acids and bases was proposed by the American chemist Gilbert Newton Lewis (G. N. Lewis) in 1923. Unlike the Arrhenius and Brönsted–Lowry theories, the Lewis concept does not require the presence of hydrogen ions (H+) or proton transfer.

Instead, Lewis explained acid-base reactions in terms of the transfer of an electron pair. This theory successfully explains the behaviour of many substances that cannot be classified as acids or bases by earlier concepts.
🗂️ Definition of a Lewis Acid and Base
Lewis Acid
According to the Lewis theory, a Lewis acid is a species that accepts an electron pair from another species to form a coordinate (dative) covalent bond.

Lewis Acid = Electron Pair Acceptor
Lewis Base
According to the Lewis theory, a Lewis base is a species that donates an electron pair to another species.

Lewis Base = Electron Pair Donor
💡 Basic Concept of Lewis Theory
✏️ Reaction Between BF₃ and NH₃
Boron trifluoride (BF₃) is an electron-deficient molecule because the boron atom has only six electrons in its outermost shell. Therefore, BF₃ readily accepts an electron pair.
Ammonia (NH₃) possesses a lone pair of electrons on the nitrogen atom, which it can donate.

The reaction is represented as: \[\boxed{\mathrm{BF_3+{:NH_3}\rightarrow BF_3{:NH_3}}}\] In this reaction:
  • BF₃ acts as the Lewis acid.
  • NH₃ acts as the Lewis base.
  • A coordinate covalent bond is formed.
✏️ Formation of the Coordinate Covalent Bond
The nitrogen atom donates its lone pair to the vacant orbital of boron.
This electron-pair donation forms a coordinate (dative) bond.
Representation:
\[\mathrm{:NH_3 \rightarrow BF_3}\]
✏️ Examples of Lewis Acids
Lewis Acid Reason
BF₃ Electron-deficient boron atom.
AlCl₃ Incomplete octet around aluminium.
Fe³⁺ Accepts electron pairs from ligands.
Co³⁺ Forms coordination compounds.
Mg²⁺ Accepts electron pairs.
H⁺ Accepts an electron pair to form a bond.
✏️ Examples of Lewis Bases
Lewis Base Reason
NH₃ Contains a lone pair on nitrogen.
H₂O Contains lone pairs on oxygen.
OH⁻ Donates an electron pair.
Cl⁻ Possesses lone pairs.
CN⁻ Acts as a ligand by donating electrons.
⚖️ Comparison of Arrhenius, Brönsted–Lowry and Lewis Concepts
Theory Acid Base
Arrhenius Produces H⁺ ions. Produces OH⁻ ions.
Brönsted–Lowry Proton donor. Proton acceptor.
Lewis Electron-pair acceptor. Electron-pair donor.
✅ Advantages of Lewis Theory
  • Applicable to reactions occurring without protons.
  • Explains reactions in non-aqueous media.
  • Explains the acidic behaviour of electron-deficient molecules.
  • Forms the basis of coordination chemistry.
  • Applicable to complex formation reactions.
🗒️ Limitations of Lewis Theory
  • Does not provide a quantitative measure of acid or base strength.
  • Cannot easily predict the direction of acid-base reactions.
  • Very broad definition; many reactions may be classified as acid-base reactions.
⚖️ Comparison
Brönsted–Lowry Theory Lewis Theory
Based on proton transfer. Based on electron-pair transfer.
Requires H⁺ transfer. No proton is necessary.
Applicable to proton-transfer reactions. Applicable to a much wider range of reactions.
Example: HCl + NH₃. Example: BF₃ + NH₃.
✏️ Example
Solved Example
Explain why BF₃ behaves as a Lewis acid.
Lewis acid-base concept.
The boron atom in BF₃ possesses only six valence electrons and has an incomplete octet. Therefore, it readily accepts an electron pair from another species. Hence, BF₃ acts as a Lewis acid.
Why is NH₃ considered a Lewis base?
Nitrogen in NH₃ possesses one lone pair of electrons. This lone pair can be donated to an electron-deficient species such as BF₃ or H⁺. Therefore, NH₃ acts as a Lewis base.
Classify the following as Lewis acids or Lewis bases:
  • AlCl₃
  • OH⁻
  • Mg²⁺
  • H₂O
Species Classification
AlCl₃ Lewis Acid
OH⁻ Lewis Base
Mg²⁺ Lewis Acid
H₂O Lewis Base
⚡ Exam Tip
❌ Common Mistakes
  • Confusing proton acceptors with electron-pair acceptors.
  • Assuming every Lewis acid contains hydrogen.
  • Thinking all Lewis bases contain OH⁻ ions.
  • Ignoring the role of lone pairs while identifying Lewis bases.
  • Failing to recognise metal cations as Lewis acids.
📋 CBSE Competency-Based (HOTS) Question

Explain why BF₃ behaves as a Lewis acid even though it does not contain a proton. Also, identify the Lewis acid and Lewis base in the reaction:

\[\mathrm{BF_3+NH_3\rightarrow BF_3{:}NH_3}\]

Answer

Boron in BF₃ has an incomplete octet and can accept a lone pair of electrons. Therefore, BF₃ acts as a Lewis acid despite containing no proton. Ammonia donates its lone pair and acts as the Lewis base. The product is a Lewis acid-base adduct containing a coordinate covalent bond.

🌟 Significance for Board and Competitive Examinations
  • Represents the broadest acid-base concept in Class XI Chemistry.
  • Forms the foundation of coordination chemistry and complex ion formation.
  • Frequently asked in conceptual, assertion-reason and competency-based questions.
  • Essential for understanding transition metal complexes and chemical bonding.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Ionisation of Water and Ionic Product of Water (Kw)

📋
Table of Contents
14 sections
📖 Introduction
📘 Self-Ionisation (Auto-Ionisation) of Water
🤔 Did You Know?
Why is Water Amphoteric?
A substance that can act both as a Brönsted acid and as a Brönsted base is called an amphoteric or amphiprotic substance.
Behaviour Reaction
Water as an Acid Donates H⁺ to NH₃ to form NH₄⁺.
Water as a Base Accepts H⁺ from HCl to form H₃O⁺.
🗒️ Ionisation Constant of Water
Applying the Law of Mass Action to the equilibrium, \[\mathrm{H_2O(l)+H_2O(l) \rightleftharpoons H_3O^+(aq)+OH^-(aq)}\] the equilibrium constant is \[K=\dfrac{[H_3O^+][OH^-]}{[H_2O]^2}\]
📐 Derivation
Derivation of the Ionic Product of Water \(K_w\)
In pure liquid water, the concentration of water remains practically constant because only a very small fraction of molecules ionises. Therefore, \[[H_2O]=\text{constant}\] Multiplying the equilibrium constant by the constant concentration of water, \[K[H_2O]^2=[H_3O^+][OH^-]\] The product \[K[H_2O]^2\] is defined as the Ionic Product of Water, denoted by Kw. Hence, \[\boxed{K_w=[H_3O^+][OH^-]}\]
Value of the Ionic Product of Water
At 25°C (298 K), \[\boxed{K_w=1.0\times10^{-14}}\]
Unit
\[mol^2L^{-2}\]
Concentration of H₃O⁺ and OH⁻ in Pure Water
Since pure water is electrically neutral, \[[H_3O^+]=[OH^-]\] Let each concentration be x. Then, \[x^2=1.0\times10^{-14}\] Therefore, \[\boxed{[H_3O^+]=[OH^-]=1.0\times10^{-7}\;M}\]
🌟 Significance of the Ionic Product of Water
  • Forms the basis for the pH scale.
  • Used to calculate the concentration of H₃O⁺ and OH⁻ ions.
  • Helps distinguish acidic, basic and neutral solutions.
  • Provides the foundation for buffer solutions and salt hydrolysis.
  • Essential for understanding acid-base equilibrium.
📌 Note
Effect of Temperature on \(K_w\)
🔎 Nature of Aqueous Solutions Using \(K_w\)
🔢 Important Formulae
✏️ Example
Solved Example
Calculate the hydroxide ion concentration in pure water at 25°C.
\(K_w=[H_3O^+][OH^-]\)
In pure water, \[[H_3O^+]=[OH^-]\] Therefore, \[[OH^-]=10^{-7}M\]
Why is the concentration of water omitted from the expression for Kw?
Water is a pure liquid whose concentration remains practically constant throughout the reaction (approximately 55.5 mol L-1). Therefore, its concentration is incorporated into the equilibrium constant, giving the ionic product of water, \(K_w=[H_3O^+][OH^-]\).
Explain why Kw increases with temperature.
The ionisation of water is an endothermic process. According to Le Chatelier's Principle, increasing temperature favours the forward reaction, producing more H₃O⁺ and OH⁻ ions. Consequently, the value of Kw increases with temperature.
⚡ Exam Tip
❌ Common Mistakes
  • Writing H⁺ instead of H₃O⁺ in aqueous solutions.
  • Including the concentration of liquid water in the final expression of Kw.
  • Assuming Kw is constant at all temperatures.
  • Confusing ionisation constant (K) with ionic product (Kw).
  • Forgetting that water behaves as both an acid and a base.
📋 CBSE Competency-Based (HOTS) Question

Water behaves simultaneously as an acid and a base during its self-ionisation. Explain this statement using the Brønsted–Lowry theory and derive the expression for the ionic product of water.

Answer

During self-ionisation, one water molecule donates a proton to another water molecule. Therefore, the first acts as a Brønsted acid, while the second acts as a Brønsted base. Applying the law of mass action to the equilibrium \[ \mathrm{H_2O+H_2O\rightleftharpoons H_3O^++OH^-} \] gives \[ K=\frac{[H_3O^+][OH^-]}{[H_2O]^2}. \] Since the concentration of liquid water remains constant, it is incorporated into the equilibrium constant to obtain \[ \boxed{K_w=[H_3O^+][OH^-]}. \]

🌟 Significance for Board and Competitive Examinations
  • Forms the foundation for the pH and pOH scales.
  • Essential for understanding acid-base equilibria and buffer solutions.
  • Frequently tested in derivations and numerical problems.
  • Important for salt hydrolysis and solubility product calculations.
  • Highly relevant for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

The pH Scale

📋
Table of Contents
14 sections
🗺️ Overview
The concentration of hydronium ions in aqueous solutions is usually very small and is often expressed in exponential form. To simplify calculations and comparison of acidic and basic solutions, the Danish chemist Søren P. L. Sørensen introduced the pH scale in 1909.

The pH scale is a logarithmic scale that expresses the acidity or basicity of a solution in terms of the concentration (or more accurately, the activity) of hydronium ions \((\mathrm{H_3O^+})\).
📘 Definition of pH
🤔 Did You Know?
Why is the pH Scale Logarithmic?
The concentration of hydronium ions in aqueous solutions varies over a very wide range, approximately from \(1\) mol L-1 to \(10^{-14}\) mol L-1.

Using ordinary numbers becomes inconvenient. A logarithmic scale compresses this enormous range into manageable values from approximately 0 to 14.

A change of one pH unit corresponds to a tenfold change in the hydronium ion concentration.
🔗 Relationship Between pH and Hydronium Ion Concentration
Hydronium Ion Concentration pH
\(1\times10^{-1}\) 1
\(1\times10^{-2}\) 2
\(1\times10^{-4}\) 4
\(1\times10^{-7}\) 7
\(1\times10^{-10}\) 10
\(1\times10^{-14}\) 14
⚖️ Classification of Solutions Based on pH
pH Value Nature of Solution
Less than 7 Acidic
Equal to 7 Neutral
Greater than 7 Basic (Alkaline)
🔗 Relationship Between pH, pOH and \(K_w\)
Since \[K_w=[H_3O^+][OH^-]\] Taking the negative logarithm of both sides, \[pK_w=pH+pOH\] At 25°C, \[\boxed{pH+pOH=14}\]
🌟 Importance of pH
📌 Approximate pH of Common Substances
🔢 Important Formulae
🎨 SVG Diagram
THE pH SCALE
THE pH SCALE Bioluminescent Aquatic Reference Guide 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ACIDIC High $H^+$ Concentration NEUTRAL Pure Water | $[H^+] = [OH^-]$ ALKALINE (BASE) High $OH^-$ Concentration pH 1 Battery Acid pH 2.5 Soda / Vinegar pH 5.6 Clean Rain Slightly Acidic pH 8.1 Ocean Water Crucial for Marine Life pH 13 Bleach pH 14 Liquid Drain Cleaner Note: The pH scale is logarithmic; each whole unit change represents a tenfold change in acidity/alkalinity.
✏️ Example
Solved Example
Calculate the pH of a solution having \[ [H_3O^+]=1.0\times10^{-3}\;M. \]
\(pH=-\log[H_3O^+]\)
\[ pH=-\log(10^{-3})=3 \] Hence, the solution is acidic.
The pH of a solution is 9. Calculate the concentration of hydronium ions. \[pH=-\log[H_3O^+]\] \[[H_3O^+]=10^{-9}\;M\]
Calculate the pOH of a solution whose pH is 4.
At 25°C, \[pH+pOH=14\] Therefore, \[pOH=14-4=10\]
⚡ Exam Tip
❌ Common Mistakes
  • Using H⁺ instead of H₃O⁺ without mentioning aqueous solution.
  • Assuming pH is always between 0 and 14; concentrated solutions may lie outside this range.
  • Confusing pH with acid strength.
  • Forgetting that the pH scale is logarithmic.
  • Using \(pH+pOH=14\) at temperatures other than 25°C without considering the change in \(K_w\).
📋 CBSE Competency-Based (HOTS) Question

Two solutions have pH values of 3 and 5, respectively. Which solution is more acidic? How many times is it more acidic than the other?

Answer

The solution with pH 3 has a higher concentration of hydronium ions.

Difference in pH = 2

Therefore, \[ 10^2=100 \]

Hence, the solution with pH 3 is 100 times more acidic than the solution with pH 5.

🌟 Significance for Board and Competitive Examinations
  • Forms the basis of all pH and pOH numerical problems.
  • Essential for understanding buffer solutions and salt hydrolysis.
  • Frequently tested in competency-based and assertion-reason questions.
  • Important for biological, environmental and industrial applications.
  • Highly relevant for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Ionisation Constant of Weak Acids/Bases \(K_a\)

📋
Table of Contents
14 sections
🗺️ Overview
Strong acids ionise almost completely in water, whereas weak acids ionise only partially. As a result, a dynamic equilibrium is established between the unionised acid molecules and the ions produced in solution.

The extent of ionisation of a weak acid is measured by its acid ionisation constant (Ka). The larger the value of Ka, the stronger is the acid.
📌 Ionisation of a Weak Acid

Ionisation Constant of Weak Acids

📐 Derivation of the Acid Ionisation Constant \(K_a\)
Let the initial concentration of the weak acid be \[c\;mol\,L^{-1}\] Let \(\alpha\) be the degree of ionisation.
Species HX H₃O⁺ X⁻
Initial \(c\) 0 0
Change \(-c\alpha\) \(+c\alpha\) \(+c\alpha\)
Equilibrium \(c(1-\alpha)\) \(c\alpha\) \(c\alpha\)
Expression for \(K_a\)
Applying the Law of Mass Action, \[K_a=\frac{[H_3O^+][X^-]}{[HX]}\] Substituting equilibrium concentrations, \[K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}\] Therefore, \[\boxed{K_a=\frac{c\alpha^2}{1-\alpha}}\]
Physical Significance of Ka
Value of Ka Interpretation
Large Greater ionisation; stronger acid.
Small Less ionisation; weaker acid.
✏️ Examples of Weak Acid
Weak Acid Formula
Acetic Acid CH₃COOH
Carbonic Acid H₂CO₃
Hydrofluoric Acid HF
Formic Acid HCOOH
📘 Definition

Ionisation Constant of Weak Bases \(K_b\)

📌 Note
Ionisation of a Weak Base
📐 Derivation of the Base Ionisation Constant \(K_b\)
Applying the Law of Mass Action, \[K_b=\frac{[M^+][OH^-]}{[MOH]}\] Substituting equilibrium concentrations, \[K_b=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}\] Hence,\[\boxed{K_b=\frac{c\alpha^2}{1-\alpha}}\]
Physical Significance of Kb
Value of Kb Interpretation
Large Greater ionisation; stronger base.
Small Less ionisation; weaker base.
⚖️ Comparison
Comparison Between \(K_a\) and \(K_b\)
Ka Kb
Measures acid strength. Measures base strength.
Applicable to weak acids. Applicable to weak bases.
Larger Ka → Stronger acid. Larger Kb → Stronger base.
Depends on temperature. Depends on temperature.
🔗 Relationship Between \(K_a\), \(K_b\) and \(K_w\)
For a conjugate acid-base pair, \[\boxed{K_a\times K_b=K_w}\] At 25°C, \[\boxed{K_aK_b=1.0\times10^{-14}}\] This relation shows that a strong acid has a weak conjugate base, and a strong base has a weak conjugate acid.
✏️ Example
Solved Example
Two weak acids have ionisation constants \(1.8\times10^{-5}\) and \(4.5\times10^{-4}\). Which acid is stronger?
The acid having the larger value of Ka undergoes greater ionisation. Therefore, \[4.5\times10^{-4}>1.8\times10^{-5}\] Hence, the second acid is stronger.
What does a very small value of Kb indicate?
A very small value of Kb indicates that only a small fraction of the base ionises in water. Therefore, the base is weak.
Explain why hydrochloric acid has no measurable Ka, whereas acetic acid has a definite value of Ka.
Hydrochloric acid ionises almost completely in water; hence no significant equilibrium exists, and Ka is extremely large. Acetic acid ionises only partially, forming a dynamic equilibrium between ions and molecules. Therefore, it possesses a measurable ionisation constant, Ka.
⚡ Exam Tip
❌ Common Mistakes
  • Using Ka for strong acids.
  • Confusing degree of ionisation (α) with ionisation constant.
  • Ignoring the assumption that water concentration remains constant.
  • Writing incorrect ICE table values.
  • Forgetting that Ka and Kb depend on temperature.
📋 CBSE Competency-Based (HOTS) Question

Two weak acids have the same concentration, but one has a larger Ka value than the other. Predict which solution has the lower pH and justify your answer.

Answer

The acid with the larger Ka ionises to a greater extent, producing a higher concentration of hydronium ions. Since pH is inversely related to hydronium ion concentration, this solution has the lower pH and is therefore the stronger acid.

🌟 Significance for Board and Competitive Examinations
  • Forms the basis of acid-base equilibrium calculations.
  • Essential for pH calculations involving weak acids and weak bases.
  • Frequently tested in derivations and numerical problems.
  • Provides the foundation for buffer solutions and salt hydrolysis.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Di- and Polybasic (Polyprotic) Acids

📋
Table of Contents
13 sections
🗺️ Overview
Some acids contain more than one ionisable hydrogen atom (proton) in each molecule. These acids lose their protons one after another in separate steps instead of releasing all protons simultaneously.

Such acids are known as polybasic acids or polyprotic acids.

Polyprotic Acid = An acid capable of donating two or more ionisable protons (H⁺) per molecule.
🗒️ Classification Of Polyprotic Acids
Type of Acid Number of Ionisable H⁺ Examples
Monoprotic Acid 1 HCl, HNO₃, HF
Diprotic (Dibasic) Acid 2 H₂SO₄, H₂C₂O₄, H₂CO₃
Triprotic (Tribasic) Acid 3 H₃PO₄
📌 Note
🔗 Relations
Relationship Between \(K_{a_1}\) and \(K_{a_2}\)
The first proton is always removed more easily than the second proton because, after the first ionisation, the negatively charged ion (HX⁻) attracts the remaining proton more strongly.
For every polyprotic acid, \[K_{a1}>K_{a2}>K_{a3}\] Thus, successive ionisation constants decrease rapidly.
Why Does Ka Decrease in Successive Ionisations?
  • The first proton is removed from a neutral molecule.
  • The second proton is removed from a negatively charged ion.
  • The negative charge holds the remaining proton more strongly.
  • Hence, each successive ionisation becomes progressively more difficult.
✏️ Examples of Polyprotic Acids
Acid Successive Ionisations
Sulphuric Acid (H₂SO₄) H₂SO₄ → HSO₄⁻ → SO₄²⁻
Carbonic Acid (H₂CO₃) H₂CO₃ → HCO₃⁻ → CO₃²⁻
Oxalic Acid (H₂C₂O₄) H₂C₂O₄ → HC₂O₄⁻ → C₂O₄²⁻
Phosphoric Acid (H₃PO₄) H₃PO₄ → H₂PO₄⁻ → HPO₄²⁻ → PO₄³⁻
📘 Definition

Di- and Polyacidic Bases

✏️ Examples of Polyacidic Bases
Base Number of Protons Accepted
CO₃²⁻ 2
PO₄³⁻ 3
S²⁻ 2
🔢 Important Formulae
✏️ Example
Solved Example
Why is the first ionisation constant of a dibasic acid always greater than the second ionisation constant?
During the first ionisation, a proton is removed from a neutral molecule. During the second ionisation, the proton is removed from a negatively charged ion, which holds the remaining proton more strongly due to electrostatic attraction. Therefore, \[ K_{a1}>K_{a2}. \]
Identify the conjugate base formed after the first ionisation of phosphoric acid.
The first ionisation is \[\mathrm{H_3PO_4\rightleftharpoons H^++H_2PO_4^-}\] Hence, the conjugate base formed is H₂PO₄⁻.
Classify the following as monoprotic, diprotic or triprotic acids: HCl, H₂SO₄, H₃PO₄.
Acid Classification
HCl Monoprotic
H₂SO₄ Diprotic
H₃PO₄ Triprotic
⚡ Exam Tip
❌ Common Mistakes
  • Assuming all hydrogen atoms in a molecule are ionisable.
  • Writing only one equilibrium constant for polyprotic acids.
  • Forgetting that each ionisation step has a different Ka.
  • Assuming \(K_{a1}=K_{a2}\).
  • Ignoring the intermediate ions (HX⁻, H₂PO₄⁻, etc.).
📋 CBSE Competency-Based (HOTS) Question

Oxalic acid (H₂C₂O₄) is a diprotic acid. Explain why its first ionisation constant is much greater than its second ionisation constant.

Answer

The first proton is removed from a neutral oxalic acid molecule, whereas the second proton is removed from the negatively charged hydrogen oxalate ion. The negative charge strongly attracts the remaining proton, making its removal more difficult. Consequently, \[ K_{a1}>K_{a2}. \]

🌟 Significance for Board and Competitive Examinations
  • Essential for understanding stepwise ionisation equilibria.
  • Frequently tested in derivations and conceptual questions.
  • Important for pH calculations involving polyprotic acids.
  • Provides the basis for buffer solutions containing hydrogen salts.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Factors Affecting Acid Strength

📋
Table of Contents
9 sections
📌 Note
🗂️ Factors governing Acid strength
Effect of H–A Bond Strength
The energy required to break the H–A bond is called the bond dissociation energy.

If the H–A bond is weak, it breaks more easily and releases H+ ions readily. Therefore, the acid becomes stronger. Weaker H–A bond → Easier proton release → Stronger acid
Conversely,
Stronger H–A bond → Difficult proton release → Weaker acid
Effect of Bond Polarity
Bond polarity depends on the difference in electronegativity between hydrogen and the atom A. When atom A is highly electronegative, the H–A bond becomes more polar, resulting in a greater separation of charges. This weakens the attraction between hydrogen and A, facilitating the release of H+.
Greater bond polarity → Easier ionisation → Stronger acid
Acid Strength Down a Group of the Periodic Table
When comparing hydrides of elements belonging to the same group, the bond strength plays a more important role than bond polarity.

As we move down a group:
  • Atomic size increases.
  • The H–A bond becomes longer.
  • Bond strength decreases.
  • Proton release becomes easier.
  • Acid strength increases.
Example: Hydrogen Halides
\[\boxed{HF < HCl < HBr < HI}\]
Acid Strength Across a Period
When comparing hydrides of elements belonging to the same period, bond polarity becomes the deciding factor.
Across a period:
  • Electronegativity increases.
  • Bond polarity increases.
  • Release of H+ becomes easier.
  • Acid strength increases.
Example: Second Period Hydrides
Acid strength increases from left to right because electronegativity increases. \[\boxed{CH_4 < NH_3 < H_2O < HF}\]
📝 Summary of Acid Strength Trends
🧠 Important Points to Remember
✏️ Example
Solved Example
Arrange the following hydrogen halides in increasing order of acid strength: HF, HCl, HBr and HI.
Bond strength decreases down the group.
As atomic size increases down the group, the H–X bond weakens.
Therefore, \[HF < HCl < HBr < HI\]
Arrange CH₄, NH₃, H₂O and HF in increasing order of acid strength.
Across a period, electronegativity increases, increasing bond polarity.
Hence, \[CH_4 < NH_3 < H_2O < HF\]
Why is H₂S a stronger acid than H₂O?
Sulphur is larger than oxygen. Consequently, the H–S bond is weaker than the H–O bond and breaks more readily to release H⁺ ions. Therefore, H₂S is a stronger acid than H₂O.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming higher electronegativity always means a stronger acid, even down a group.
  • Ignoring bond strength when comparing elements within the same group.
  • Confusing acid strength with bond polarity alone.
  • Believing that all compounds containing hydrogen are acidic.
  • Using periodic trends without considering whether the comparison is within a group or across a period.
📋 CBSE Competency-Based (HOTS) Question

Although fluorine is more electronegative than iodine, hydroiodic acid (HI) is a much stronger acid than hydrofluoric acid (HF). Explain this observation.

Answer

While comparing acids within the same group, bond strength is more important than bond polarity. The H–I bond is much weaker than the H–F bond because iodine has a larger atomic size. As a result, HI releases H⁺ ions more easily and is therefore a much stronger acid than HF.

🌟 Significance for Board and Competitive Examinations
  • Frequently asked in assertion-reason and competency-based questions.
  • Essential for comparing the strengths of inorganic and organic acids.
  • Forms the conceptual basis for acid dissociation constants (Ka).
  • Useful in predicting reaction feasibility and equilibrium direction.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Common Ion Effect

📋
Table of Contents
14 sections
🗺️ Overview
The Common Ion Effect is one of the most important applications of Le Chatelier's Principle in Ionic Equilibrium.

When an electrolyte containing an ion already present in an equilibrium system is added, the equilibrium shifts in the direction that reduces the concentration of that common ion. Consequently, the ionisation of a weak electrolyte is suppressed.
Common Ion Effect is the suppression of the ionisation of a weak electrolyte due to the addition of another electrolyte containing a common ion.
📘 Definition
📌 Principle Behind the Common Ion Effect
✏️ Example 1: Common Ion Effect in a Weak Acid
Consider the equilibrium: \[\boxed{\mathrm{CH_3COOH\rightleftharpoons H^+ + CH_3COO^- }}\] On adding sodium acetate: \[\mathrm{CH_3COONa\rightarrow Na^+ + CH_3COO^- }\] Since acetate ion is the common ion, the equilibrium shifts towards the left.
Result:
  • Ionisation of CH₃COOH decreases.
  • Hydrogen ion concentration decreases.
  • Acidic strength decreases.
✏️ Example 2: Common Ion Effect in a Weak Base
Consider ammonium hydroxide: \[\boxed{\mathrm{NH_4OH\rightleftharpoons NH_4^+ + OH^-}}\] Addition of ammonium chloride causes complete dissociation: \[\mathrm{NH_4Cl\rightarrow NH_4^+ + Cl^-}\] The increase in NH₄⁺ concentration shifts the equilibrium towards the left.
Therefore,
  • Ionisation of NH₄OH decreases.
  • OH⁻ ion concentration decreases.
  • Basic strength decreases.
🔗 Relation with Le Chatelier's Principle
The Common Ion Effect is an excellent application of Le Chatelier's Principle.

According to the principle: When the concentration of one component of an equilibrium system is increased, the equilibrium shifts in the direction that reduces the effect of the increase.

Therefore, adding a common ion suppresses the ionisation of the weak electrolyte.
🛠️ Applications of the Common Ion Effect
Application Importance
Buffer Solutions Maintains nearly constant pH.
Qualitative Analysis Selective precipitation of ions.
Salt Hydrolysis Controls hydrolysis of salts.
Solubility Control Reduces solubility of sparingly soluble salts.
Industrial Chemistry Used in purification and crystallisation processes.
🌟 Importance in Ionic Equilibrium
🗒️ Key Equilibrium Equations
Weak Acid \[\mathrm{CH_3COOH\rightleftharpoons H^+ + CH_3COO^-}\]
Weak Base \[\mathrm{NH_4OH\rightleftharpoons NH_4^+ + OH^-}\]
Le Chatelier's Shift Addition of common ion \[ \Longrightarrow \] Equilibrium shifts towards reactants.
✏️ Example
Solved Examples
Explain why adding sodium acetate decreases the ionisation of acetic acid.
Sodium acetate completely dissociates and provides acetate ions (CH₃COO⁻), which are already present in the ionisation equilibrium of acetic acid. The increased concentration of acetate ions shifts the equilibrium towards the left according to Le Chatelier's Principle. Hence, the ionisation of acetic acid decreases.
Why does the addition of NH₄Cl reduce the basic strength of NH₄OH?
NH₄Cl dissociates completely and increases the concentration of NH₄⁺ ions. Since NH₄⁺ is a common ion, the equilibrium shifts towards the reactants, suppressing the ionisation of NH₄OH. Consequently, fewer OH⁻ ions are produced, reducing the basic strength.
Name one important application of the Common Ion Effect in analytical chemistry.
The Common Ion Effect is used for selective precipitation of ions during qualitative inorganic analysis, allowing different metal ions to be separated efficiently.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming the Common Ion Effect applies only to acids.
  • Confusing common ion effect with the common-ion itself.
  • Forgetting that the added electrolyte should be a strong electrolyte.
  • Ignoring the shift in equilibrium predicted by Le Chatelier's Principle.
  • Assuming it increases ionisation instead of suppressing it.
📋 CBSE Competency-Based (HOTS) Question

Equal amounts of acetic acid are taken in two beakers. Sodium acetate is added to one beaker, while distilled water is added to the other. Which solution will have the lower concentration of H⁺ ions? Explain using Le Chatelier's Principle.

Answer

The beaker containing sodium acetate will have the lower concentration of H⁺ ions. Sodium acetate supplies acetate ions (CH₃COO⁻), which shift the equilibrium of acetic acid towards the reactants. This suppresses ionisation and decreases the concentration of H⁺ ions. The addition of distilled water does not introduce a common ion and therefore does not produce the common ion effect.

🌟 Significance for Board and Competitive Examinations
  • One of the most frequently tested concepts in Ionic Equilibrium.
  • Forms the basis for understanding buffer solutions.
  • Essential for studying the Solubility Product (Ksp).
  • Frequently appears in assertion-reason, competency-based and numerical questions.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Hydrolysis of Salts and the pH of Their Solutions

📋
Table of Contents
12 sections
🗺️ Overview

Salts are ionic compounds formed by the neutralization reaction between an acid and a base. When dissolved in water, they dissociate completely into their constituent cations and anions.

These ions may either remain as hydrated ions or react with water molecules to form the corresponding acid or base. This reaction between the ions of a salt and water is called salt hydrolysis.

Salt Hydrolysis is the reaction of the cation and/or anion of a salt with water, producing H₃O⁺ or OH⁻ ions and thereby changing the pH of the solution.

📘 Definition of Salt Hydrolysis
🤔 Did You Know?
Why Does Salt Hydrolysis Occur?
Whether hydrolysis occurs or not depends upon the strength of the parent acid and base from which the salt is formed.
  • Ions originating from strong acids and strong bases do not hydrolyse.
  • Ions derived from weak acids or weak bases undergo hydrolysis because they react with water.
  • The greater the weakness of the parent acid or base, the greater is the tendency for hydrolysis.
⚖️ Hydration and Hydrolysis
Hydration Hydrolysis
Ions become surrounded by water molecules. Ions chemically react with water.
No change in pH. pH changes.
Physical interaction. Chemical reaction.
Occurs for all ions. Occurs mainly for ions of weak acids or weak bases.
🗒️ Classification Of Salts Based On Hydrolysis
Parent Acid Parent Base Hydrolysis Nature of Solution
Strong Strong No Neutral (pH = 7)
Strong Weak Cation only Acidic
Weak Strong Anion only Basic
Weak Weak Both ions Depends on Ka and Kb
📌 Salts of Strong Acids and Strong Bases
🔢 Important Formulae and Facts
✏️ Example
Solved Example
Why is an aqueous solution of sodium chloride neutral?
Sodium chloride is formed from the strong acid HCl and the strong base NaOH. Its ions (Na⁺ and Cl⁻) do not react with water; they only become hydrated. Hence, no hydrolysis occurs, and the solution has a pH of 7.
Distinguish between hydration and hydrolysis.
Hydration Hydrolysis
Physical interaction with water. Chemical reaction with water.
No change in pH. pH changes.
Which type of salts undergo hydrolysis?
Salts containing ions derived from weak acids, weak bases, or both undergo hydrolysis. Salts formed from a strong acid and a strong base generally do not undergo hydrolysis.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing hydration with hydrolysis.
  • Assuming every salt undergoes hydrolysis.
  • Forgetting that Na⁺ and Cl⁻ are spectator ions in water.
  • Ignoring the strength of the parent acid and base.
  • Assuming all salt solutions have pH 7.
📋 CBSE Competency-Based (HOTS) Question

Three students prepare aqueous solutions of NaCl, NH₄Cl and CH₃COONa. One student claims that all three solutions should be neutral because they are salts. Do you agree? Justify your answer.

Answer

No. NaCl is formed from a strong acid and a strong base, so it does not undergo hydrolysis and its solution is neutral (pH = 7). NH₄Cl is formed from a strong acid and a weak base, so its solution is acidic due to cation hydrolysis. CH₃COONa is formed from a weak acid and a strong base, so its solution is basic due to anion hydrolysis. Therefore, the nature of the parent acid and base determines the pH of the salt solution.

🌟 Significance for Board and Competitive Examinations
  • Foundation for understanding acidic, basic and neutral salt solutions.
  • Essential for pH calculations involving salts.
  • Frequently tested in assertion-reason and competency-based questions.
  • Forms the basis for buffer solutions and solubility equilibria.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Buffer Solutions

📋
Table of Contents
14 sections
🗺️ Overview

In many chemical, biological and industrial processes, it is essential to maintain the pH of a solution within a narrow range. Even a slight change in pH may alter the rate of a reaction, affect enzyme activity or change the properties of a chemical system.

Solutions that resist changes in pH when small quantities of an acid or a base are added, or even upon dilution, are known as buffer solutions.

A buffer solution is a solution that resists changes in its pH when small amounts of acid, base or water are added.

📘 Definition of Buffer Solution
🔷 Characteristics of Buffer Solutions
🗂️ Types of Buffer Solutions
Acidic Buffer Solution
An acidic buffer consists of a weak acid and its salt with a strong base.

Common examples are:
Weak Acid Salt
CH₃COOH CH₃COONa
H₂CO₃ NaHCO₃
Example equilibrium:
\[\mathrm{CH_3COOH\rightleftharpoons H^+ + CH_3COO^-}\]
Basic Buffer Solution
A basic buffer consists of a weak base and its salt with a strong acid.

Common examples are:
Weak Base Salt
NH₄OH NH₄Cl
NH₃(aq) NH₄Cl
Example equilibrium:
\[\mathrm{NH_4OH\rightleftharpoons NH_4^+ + OH^-}\]
🗂️ Working of Buffer
Working of an Acidic Buffer
Consider an acidic buffer containing acetic acid and sodium acetate.

Acetic acid establishes the equilibrium: \[\mathrm{CH_3COOH\rightleftharpoons H^+ + CH_3COO^-}\] Sodium acetate ionises completely: \[\mathrm{CH_3COONa\rightarrow Na^+ + CH_3COO^-}\]
When a Small Amount of Acid is Added
The added H⁺ ions combine with acetate ions. \[\mathrm{H^+ + CH_3COO^-\rightarrow CH_3COOH}\] Hence, the increase in H⁺ concentration is prevented and the pH changes only slightly.
When a Small Amount of Base is Added
The added OH⁻ ions react with acetic acid. \[\mathrm{OH^-+CH_3COOH\rightarrow CH_3COO^- + H_2O}\] Thus, OH⁻ ions are removed and the pH remains nearly constant.
Working of a Basic Buffer
Consider a mixture of ammonium hydroxide and ammonium chloride.

Equilibrium: \[\mathrm{NH_4OH\rightleftharpoons NH_4^+ + OH^-}\]
On Addition of Acid
\[\mathrm{H^+ + OH^-\rightarrow H_2O}\] More NH₄OH ionises to restore equilibrium.
On Addition of Base
Excess OH⁻ ions combine with NH₄⁺ ions. \[\mathrm{NH_4^+ + OH^- \rightarrow NH_4OH}\] Hence, the pH changes only slightly.
🗒️ How Does a Buffer Resist pH Change?
How Does a Buffer Resist pH Change?
Added Substance Buffer Response
Acid (H⁺) Neutralised by the conjugate base.
Base (OH⁻) Neutralised by the weak acid.
Water (Dilution) Ratio of acid and salt changes very little; pH remains almost constant.
🔢 Henderson–Hasselbalch Equation
🛠️ Application
Applications of Buffer Solutions
Field Application
Human Body Blood maintains a pH of about 7.4 using buffer systems.
Biochemistry Maintains optimum pH for enzyme activity.
Medicine Preparation of injections and pharmaceutical formulations.
Industry Dyeing, electroplating and fermentation processes.
Analytical Chemistry Used in titrations and pH calibration.
📌 Buffer Capacity
✏️ Example
Solved Example
Identify whether the following is an acidic or a basic buffer: CH₃COOH + CH₃COONa.
It contains a weak acid (CH₃COOH) and its salt with a strong base (CH₃COONa). Therefore, it is an acidic buffer.
Why does the pH of an acidic buffer change only slightly when a small amount of HCl is added?
The added H⁺ ions combine with acetate ions to form undissociated acetic acid. Since most of the added H⁺ ions are consumed, the pH changes only slightly.
State one biological importance of buffer solutions.
Buffer systems maintain the pH of blood (about 7.4), which is essential for the proper functioning of enzymes and metabolic reactions.
⚡ Exam Tip
❌ Common Mistakes
  • Assuming every mixture of an acid and a salt forms a buffer.
  • Confusing acidic buffers with acidic solutions.
  • Ignoring the common ion effect in explaining buffer action.
  • Believing that buffers prevent any change in pH.
  • Forgetting that buffer capacity has a finite limit.
📋 CBSE Competency-Based (HOTS) Question

Two beakers contain equal volumes of acetic acid. Sodium acetate is added to one beaker, while sodium chloride is added to the other. On adding a few drops of dilute HCl, predict which solution will show a smaller change in pH. Justify your answer.

Answer

The solution containing sodium acetate will show a much smaller change in pH because it forms an acidic buffer with acetic acid. The acetate ions react with the added H⁺ ions, preventing a significant increase in hydronium ion concentration. Sodium chloride does not provide a conjugate base and therefore does not form a buffer.

🌟 Significance for Board and Competitive Examinations
  • One of the most important topics in Ionic Equilibrium.
  • Frequently asked in derivations, numerical problems and competency-based questions.
  • Essential for understanding blood chemistry and biological systems.
  • Provides the basis for pH control in analytical and industrial chemistry.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Solubility Product Constant \(K_{sp}\)

📋
Table of Contents
15 sections
🗺️ Overview

Many ionic compounds dissolve readily in water, whereas others dissolve only to a very small extent. Such compounds are called sparingly soluble salts. Although their solubility is very low, a dynamic equilibrium is established between the undissolved solid and the dissolved ions in a saturated solution.

The equilibrium constant associated with the dissolution of a sparingly soluble salt is called the Solubility Product Constant (Ksp).

📘 Definition of Solubility Product
📌 Dissolution of Barium Sulphate
📐 Derivation of the Solubility Product Expression
Applying the Law of Mass Action, \[K=\frac{[Ba^{2+}][SO_4^{2-}]}{[BaSO_4]}\] Since BaSO₄ is a pure solid, its concentration remains constant throughout the equilibrium.
Therefore, this constant concentration is incorporated into the equilibrium constant.
Hence, \[\boxed{K_{sp}=[Ba^{2+}][SO_4^{2-}]}\]
🗒️ Physical Meaning Of \(K {sp}\)
The value of Ksp indicates the extent to which a sparingly soluble salt dissolves in water.
Value of Ksp Interpretation
Large Higher solubility
Small Lower solubility

For barium sulphate, \[\boxed{K_{sp}=1.1\times10^{-10}\quad\text{at }298\,K}\]
🔢 General Expression for Solubility Product
✏️ Examples of Solubility Product Expressions
Salt Dissociation Ksp Expression
AgCl \[\mathrm{AgCl\rightleftharpoons Ag^+ + Cl^-}\] \[[Ag^+][Cl^-]\]
BaSO₄ \[\mathrm{BaSO_4\rightleftharpoons Ba^{2+}+SO_4^{2-}}\] \[[Ba^{2+}][SO_4^{2-}]\]
CaF₂ \[\mathrm{CaF_2\rightleftharpoons Ca^{2+}+2F^-}\] \[Ca^{2+}][F^-]^2\]
Al(OH)₃ \[\mathrm{Al(OH)_3\rightleftharpoons Al^{3+}+3OH^-}\] \[[Al^{3+}][OH^-]^3\]
🔗 Relationship Between Solubility and \(K_{sp}\)
Let the molar solubility of BaSO₄ be S mol L-1. Then, \[[Ba^{2+}]=S\] \[[SO_4^{2-}]=S\] Therefore, \[\boxed{K_{sp}=S^2}\] Hence, \[\boxed{S=\sqrt{K_{sp}}}\]
🗒️ Factors Affecting Solubility Product
  • Ksp depends only on temperature.
  • It is independent of the amount of solid present.
  • It is independent of the volume of the solution.
  • The presence of a common ion decreases the solubility but does not change Ksp.
  • Different salts have different Ksp values.
🛠️ Applications of the Solubility Product Constant
Application Importance
Calculating Solubility Determines the molar solubility of sparingly soluble salts.
Predicting Precipitation Used to determine whether a precipitate will form.
Qualitative Analysis Selective precipitation of metal ions.
Common Ion Effect Explains reduction in solubility.
Industrial Chemistry Water purification and separation processes.
✏️ Example
Solved Example
Write the solubility product expression for calcium fluoride (CaF₂).
Dissociation: \[\mathrm{CaF_2\rightleftharpoons Ca^{2+}+2F^-}\] Therefore, \[\boxed{K_{sp}=[Ca^{2+}][F^-]^2}\]
The Ksp of AgCl is very small. What does this indicate?
A very small Ksp means that AgCl is only sparingly soluble in water and only a very small amount dissolves to produce Ag⁺ and Cl⁻ ions.
Why is the concentration of solid BaSO₄ not included in the Ksp expression?
BaSO₄ is a pure solid, and its concentration remains constant throughout the equilibrium. Hence, it is incorporated into the equilibrium constant, giving \[ K_{sp}=[Ba^{2+}][SO_4^{2-}]. \]
⚡ Exam Tip
❌ Common Mistakes
  • Including the concentration of the solid in the equilibrium expression.
  • Confusing solubility with the solubility product.
  • Ignoring stoichiometric coefficients while writing Ksp.
  • Assuming Ksp changes when more solid is added.
  • Confusing Ksp with the common equilibrium constant (K).
📋 CBSE Competency-Based (HOTS) Question

A student adds excess solid BaSO₄ to water and stirs the mixture until equilibrium is established. The student then adds more solid BaSO₄. Predict whether the value of Ksp changes and justify your answer.

Answer

The value of Ksp does not change because it depends only on temperature. Adding more solid BaSO₄ increases the amount of undissolved solid but does not change the equilibrium concentrations of Ba²⁺ and SO₄²⁻ ions in the saturated solution.

🌟 Significance for Board and Competitive Examinations
  • Forms the foundation of precipitation equilibria.
  • Essential for numerical problems on solubility.
  • Closely related to the common ion effect and qualitative analysis.
  • Frequently tested in CBSE competency-based questions and derivations.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
⚖️

Common Ion Effect on the Solubility of Ionic Salts

📋
Table of Contents
12 sections
🗺️ Overview

The solubility of a sparingly soluble ionic salt is significantly affected by the presence of one of its constituent ions in the solution. This phenomenon is known as the Common Ion Effect on Solubility.

According to Le Chatelier's Principle, if the concentration of one of the ions produced during dissolution is increased, the equilibrium shifts towards the undissolved solid, thereby reducing the solubility of the salt.

The addition of a common ion suppresses the solubility of a sparingly soluble salt by shifting the dissolution equilibrium towards the solid state.

📘 Definition
🗒️ Principle Behind The Common Ion Effect
Consider the equilibrium between solid barium sulphate and its saturated solution. \[\boxed{\mathrm{BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)}}\] The solubility product is \[\boxed{K_{sp}=[Ba^{2+}][SO_4^{2-}]}\] If additional sulphate ions are introduced by adding sodium sulphate (Na₂SO₄), the concentration of SO₄²⁻ increases.
According to Le Chatelier's Principle, the equilibrium shifts towards the left, producing more solid BaSO₄.
Thus, the solubility of BaSO₄ decreases.
📌 Relation Between \(\mathrm{Q_{sp}\text{ and }K_{sp}
✏️ Example

Example 1: Silver Chloride

Dissolution equilibrium: \[\mathrm{AgCl(s)\rightleftharpoons Ag^+ + Cl^-}\] Addition of NaCl supplies Cl⁻ ions and decreases the solubility of AgCl.

Example 2: Calcium Fluoride

\[\mathrm{CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-}\] Addition of NaF increases fluoride ion concentration, reducing the solubility of CaF₂.
🛠️ Application
Application Importance
Qualitative Analysis Selective precipitation of metal ions.
Water Treatment Removal of unwanted ions.
Analytical Chemistry Prediction of precipitation reactions.
Industrial Separation Purification of chemicals.
Buffer Systems Closely related to the common ion effect.
🌟 Important Point
✏️ Example
Solved Example
What happens when NaCl is added to a saturated AgCl solution?
NaCl provides Cl⁻ ions, which are common ions for AgCl. Consequently, \(Q_{sp}>K_{sp}\), causing AgCl to precipitate until \(Q_{sp}=K_{sp}\).
Why does the solubility of BaSO₄ decrease in the presence of Na₂SO₄?
Na₂SO₄ supplies sulphate ions, increasing the ionic product. According to Le Chatelier's Principle, the dissolution equilibrium shifts towards the solid state, reducing the solubility of BaSO₄.
State the condition under which precipitation begins.
Precipitation starts whenever the ionic product exceeds the solubility product. \[\boxed{Q_{sp}>K_{sp}}\]
⚡ Exam Tip
❌ Common Mistakes
  • Confusing Ksp with Qsp.
  • Assuming Ksp changes after adding a common ion.
  • Ignoring stoichiometric coefficients while calculating Qsp.
  • Believing precipitation occurs when Qsp equals Ksp.
  • Using concentrations instead of activities for highly concentrated electrolyte solutions.
📋 CBSE Competency-Based (HOTS) Question

Two test tubes contain saturated solutions of AgCl. Sodium chloride is added to the first test tube, while distilled water is added to the second. Predict the changes in solubility and justify your answer using Qsp and Ksp.

Answer

Addition of NaCl increases the concentration of Cl⁻ ions, making \(Q_{sp} > K_{sp}\). Consequently, AgCl precipitates and its solubility decreases until \(Q_{sp} = K_{sp}\). In the second test tube, dilution lowers the ionic product so that \(Q_{sp} < K_{sp}\), causing more AgCl to dissolve until equilibrium is restored.

🌟 Significance for Board and Competitive Examinations
  • Direct application of Le Chatelier's Principle and Ksp.
  • Essential for predicting precipitation reactions.
  • Frequently used in qualitative inorganic analysis.
  • Important for numerical problems involving solubility and ionic product.
  • Highly important for CBSE Board, JEE Main, NEET and CUET examinations.
· Updated
NCERT CLASS XI · CHEMISTRY · CHAPTER 6

Equilibrium

A deep-sea dive into dynamic balance — from physical and chemical equilibria to ionic equilibrium, acids & bases, pH, buffers and solubility. Explore concepts, solve problems step-by-step, and test your mastery.

Core Concepts

Fourteen building blocks of Chemical Equilibrium — click any card to expand.

1

What Is Equilibrium?

▾

Equilibrium is a state in which no net change occurs in the measurable properties of a system over time, even though the process has not actually stopped. At the molecular level, forward and reverse processes continue to occur — but at equal rates, so their effects cancel out. This is why equilibrium is called dynamic, not static.

Equilibrium can be established in both physical processes (like ice melting into water) and chemical processes (like the synthesis of ammonia).

Defining ConditionRateforward = Ratereverse  →  concentrations of all species remain constant
2

Equilibrium in Physical Processes

▾

Physical equilibria involve no chemical change — only a change of state or phase distribution.

  • Solid ⇌ Liquid: e.g. ice ⇌ water, established at the melting point where rate of melting = rate of freezing.
  • Liquid ⇌ Vapour: e.g. water ⇌ water vapour in a closed vessel; the equilibrium vapour pressure is constant at a given temperature.
  • Solid ⇌ Vapour: e.g. sublimation of iodine or camphor in a closed container.
  • Gas dissolved in liquid: governed by Henry's Law — the mass of a gas dissolved per unit volume of solvent is proportional to the pressure of the gas in equilibrium with the solution.

General characteristics: physical equilibrium is dynamic, attainable only in a closed system, and every equilibrium is characterised by a constant value of some measurable property at a given temperature (vapour pressure, solubility, etc.).

3

Law of Chemical Equilibrium & Equilibrium Constant (Kc)

▾

For a general reversible reaction at a given temperature, aA + bB ⇌ cC + dD, at equilibrium the ratio of the product of equilibrium concentrations of products (raised to their stoichiometric powers) to that of reactants is a constant, called the equilibrium constant, Kc.

Law of Mass ActionKc = [C]c[D]d ÷ [A]a[B]b

This is the mathematical statement of the Law of Chemical Equilibrium, derived from the Law of Mass Action (rate of a reaction is proportional to the product of active masses of reactants). Kc is defined in terms of molar concentrations.

4

Homogeneous & Heterogeneous Equilibria

▾
  • Homogeneous equilibrium: all reactants and products are in the same phase. e.g. N2(g) + 3H2(g) ⇌ 2NH3(g)
  • Heterogeneous equilibrium: reactants and products exist in more than one phase. e.g. CaCO3(s) ⇌ CaO(s) + CO2(g)

For heterogeneous equilibria, the molar concentrations of pure solids and pure liquids are taken as constant (since their concentration doesn't change) and are omitted from the Kc expression — only gaseous and dissolved species appear.

ExampleFor CaCO₃(s) ⇌ CaO(s) + CO₂(g): Kc = [CO2]
5

Relationship Between Kp and Kc

▾

For reactions involving gases, the equilibrium constant may also be expressed in terms of partial pressures, Kp. Using the ideal gas equation, Kp and Kc are related by:

Kp – Kc RelationKp = Kc(RT)Δn

where Δn = (moles of gaseous products) − (moles of gaseous reactants), R = 0.0831 L bar K−1 mol−1, and T is the absolute temperature in Kelvin.

  • If Δn = 0, Kp = Kc
  • If Δn > 0, Kp > Kc
  • If Δn < 0, Kp < Kc
6

Characteristics of Equilibrium Constant

▾
  • The value of K is constant at a given temperature regardless of the initial concentrations used to reach equilibrium.
  • K changes only when temperature changes — never with concentration, pressure, volume, or catalyst.
  • The equilibrium constant for the reverse reaction is the reciprocal: K'c = 1/Kc.
  • If a reaction equation is multiplied by n, the new equilibrium constant becomes Kc raised to the power n: K'c = (Kc)n.
  • K is dimensionless in practice (activities are used), though concentration/pressure units are often carried through calculations.
7

Applications of Equilibrium Constant

▾

Reaction Quotient (Qc): has the same expression as Kc but uses concentrations at any point in time, not just at equilibrium. Comparing Qc to Kc predicts the direction a reaction will shift:

  • If Qc < Kc: reaction proceeds forward (more products form)
  • If Qc > Kc: reaction proceeds in reverse (more reactants form)
  • If Qc = Kc: the system is already at equilibrium

Extent of reaction: a very large Kc (>>1) indicates the reaction goes nearly to completion (mostly products); a very small Kc (<<1) indicates the reaction barely proceeds (mostly reactants remain).

8

Le Chatelier's Principle

▾

When a system at equilibrium is subjected to a change (stress) in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that tends to counteract (reduce) the effect of that change.

  • Increase concentration of a reactant → equilibrium shifts forward (towards products) to consume it.
  • Increase pressure / decrease volume → equilibrium shifts towards the side with fewer moles of gas.
  • Increase temperature → equilibrium shifts in the endothermic direction (favours the reaction that absorbs heat); decreasing temperature favours the exothermic direction.
  • Adding an inert gas at constant volume → no effect on equilibrium (partial pressures of reactants/products unchanged).
  • Catalyst → speeds up both forward and reverse reactions equally; equilibrium is reached faster but its position (K value) is unaffected.
9

Ionic Equilibrium in Solution

▾

Substances that conduct electricity in molten or dissolved state are electrolytes. Strong electrolytes (e.g. NaCl, HCl, NaOH) ionise almost completely; weak electrolytes (e.g. CH3COOH, NH4OH) ionise only partially, establishing an ionic equilibrium between unionised molecules and their ions.

Degree of Ionisation (α)α = (moles ionised) ÷ (total moles dissolved)

The extent of ionisation depends on the nature of solute/solvent, dilution (weak electrolytes ionise more on dilution), and temperature.

10

Acids, Bases & Theories

▾
  • Arrhenius Theory: acids furnish H+ ions and bases furnish OH− ions in aqueous solution. Limited — applies only to aqueous systems.
  • Brønsted–Lowry Theory: an acid is a proton (H+) donor; a base is a proton acceptor. Every acid has a conjugate base and every base a conjugate acid (differing by one proton). Water is amphoteric — it can act as both.
  • Lewis Theory: an acid is an electron-pair acceptor; a base is an electron-pair donor. This is the broadest definition, covering species with no transferable protons (e.g. BF3, AlCl3 as Lewis acids).
11

Ionization Constants: Ka and Kb

▾

For a weak acid HA ⇌ H+ + A−, the acid ionisation constant Ka measures the strength of the acid: a larger Ka means a stronger acid. Similarly, for a weak base, Kb measures base strength.

Ka, pKa and the Ka–Kb linkKa = [H+][A−] ÷ [HA]  ·  pKa = −log Ka  ·  Ka × Kb = Kw (for a conjugate pair)

A smaller pKa means a stronger acid. For a weak acid of initial concentration C and degree of ionisation α (small), [H+] ≈ Cα, and Ka ≈ Cα², giving the useful approximation α = √(Ka/C).

12

The pH Scale & Ionic Product of Water

▾

Water undergoes self-ionisation: 2H2O ⇌ H3O+ + OH−, with ionic product Kw = [H+][OH−] = 1.0 × 10−14 at 25°C.

pH / pOHpH = −log[H+]  ·  pOH = −log[OH−]  ·  pH + pOH = 14 (at 25°C)
  • pH < 7 → acidic solution
  • pH = 7 → neutral solution
  • pH > 7 → basic (alkaline) solution
13

Buffer Solutions

▾

A buffer solution resists changes in pH upon addition of small amounts of acid or base. An acidic buffer is made from a weak acid and its salt with a strong base (e.g. CH3COOH + CH3COONa); a basic buffer is made from a weak base and its salt with a strong acid (e.g. NH4OH + NH4Cl).

Henderson–Hasselbalch EquationpH = pKa + log([Salt] ÷ [Acid])

Buffering works best when [Salt] ≈ [Acid], i.e. pH ≈ pKa, which gives the buffer its maximum capacity to resist pH change.

14

Solubility Equilibria & Common Ion Effect

▾

For a sparingly soluble salt, e.g. AgCl(s) ⇌ Ag+(aq) + Cl−(aq), the solubility product Ksp is the product of molar ionic concentrations at saturation, each raised to its stoichiometric power.

Ksp for AB-type and AB2-type saltsAB: Ksp = s2   |   AB2: Ksp = 4s3

Common ion effect: adding an ion already present in the equilibrium (e.g. adding NaCl to a saturated AgCl solution) shifts the equilibrium backward, decreasing the solubility of the sparingly soluble salt — a direct consequence of Le Chatelier's Principle.

  • If ionic product > Ksp: precipitation occurs
  • If ionic product < Ksp: solution is unsaturated, no precipitate
  • If ionic product = Ksp: solution is exactly saturated

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Step-by-Step Solution

Formula Reference

Every key relationship from the chapter, organised by category.

⚖ Chemical Equilibrium

Equilibrium constant (concentration)
Kc = [C]c[D]d / [A]a[B]b
Equilibrium constant (pressure)
Kp = (pC)c(pD)d / (pA)a(pB)b
Kp–Kc relation
Kp = Kc(RT)Δn
Reverse reaction
K'c = 1 / Kc
Reaction multiplied by n
K'c = (Kc)n

📊 Reaction Quotient

Reaction quotient
Qc = [C]c[D]d / [A]a[B]b (at any instant)
Direction check
Q<K → forward  |  Q>K → reverse  |  Q=K → at equilibrium
Degree of ionisation
α = moles ionised / total moles taken

💧 Water & pH

Ionic product of water
Kw = [H+][OH−] = 1.0 × 10−14 (25°C)
pH and pOH
pH = −log[H+]  ·  pOH = −log[OH−]
pH–pOH relation
pH + pOH = 14 (at 25°C)
pKw
pKw = pH + pOH = 14

⚗ Acids & Bases

Acid ionisation constant
Ka = [H+][A−] / [HA]
Base ionisation constant
Kb = [BH+][OH−] / [BOH]
Ka–Kb conjugate relation
Ka × Kb = Kw
pKa / pKb
pKa = −log Ka  ·  pKa + pKb = 14
Weak acid [H+] approximation
[H+] ≈ √(Ka × C)  ·  α = √(Ka/C)

⚖ Buffers

Henderson–Hasselbalch (acidic buffer)
pH = pKa + log([Salt]/[Acid])
Basic buffer
pOH = pKb + log([Salt]/[Base])

❄ Solubility Equilibria

AB-type salt (e.g. AgCl)
Ksp = s2  →  s = √Ksp
AB2-type salt (e.g. CaF₂)
Ksp = 4s3  →  s = (Ksp/4)1/3
A2B-type salt (e.g. Ag₂CrO₄)
Ksp = 4s3
Precipitation condition
Ionic Product > Ksp → precipitate forms

Tips & Tricks

Shortcuts and habits that make equilibrium problems faster and more reliable.

💡
Neglect x when Ka/Kb is tiny compared to concentration
If C/Ka > 100 (roughly), you can safely approximate (C − x) ≈ C in the denominator, letting you solve linearly instead of with the quadratic formula. Always sanity-check that the computed x is indeed <5% of C afterward.
💡
Memorise the Kp–Kc shortcut using Δn's sign
Don't recompute from scratch each time — just remember: more gas moles on the product side (Δn > 0) means Kp > Kc, and fewer gas moles (Δn < 0) means Kp < Kc. This lets you sanity-check your final numeric answer instantly.
💡
Build the ICE table every single time
Even for "simple" problems, write out Initial, Change, Equilibrium rows explicitly. It prevents sign errors (products increase by +x per their coefficient, reactants decrease by −x per their coefficient) and makes the final substitution into K almost automatic.
💡
Le Chatelier shortcut: "the system opposes the push"
Whatever change is imposed, the system shifts to undo it. Add reactant → system consumes it (shifts forward). Raise pressure → system shifts to the side with fewer gas moles (to reduce total moles & pressure). Raise temperature → system shifts to absorb the extra heat (endothermic direction).
💡
pH quick estimates without a calculator
Learn log values of small integers: log 2 ≈ 0.3, log 5 ≈ 0.7. So if [H+] = 2 × 10−3, pH = 3 − log 2 ≈ 3 − 0.3 = 2.7. This mental-math trick is invaluable in timed exams.
💡
Buffer pH = pKa when [Salt] = [Acid]
In the Henderson–Hasselbalch equation, when salt and acid concentrations are equal, the log term is log(1) = 0, so pH simply equals pKa. This is the point of maximum buffering capacity — useful for quickly choosing the right buffer system for a target pH.
💡
For Ksp problems, always identify the salt "type" first
Write the dissociation equation and note the ion ratio (1:1, 1:2, 2:1, etc.) before substituting into Ksp — this determines whether you use s2, 4s3, or a more complex expression, and skipping this step is the single biggest source of Ksp errors.

Common Mistakes

Errors students frequently make — and how to avoid them.

❌
Including pure solids/liquids in the Kc expression
Students often write [CaCO3] or [CaO] in the equilibrium constant expression for heterogeneous equilibria.
Fix: only gaseous and aqueous (dissolved) species appear in Kc/Kp. Pure solids and pure liquids have constant "concentration" (their density) and are absorbed into the value of K itself.
❌
Thinking a catalyst changes the value of K
A common assumption is that catalysts increase yield or shift equilibrium.
Fix: a catalyst speeds up both forward and reverse reactions equally. It helps equilibrium get reached faster, but the equilibrium constant and equilibrium concentrations remain unchanged.
❌
Forgetting that K depends only on temperature
Students sometimes recalculate K after a concentration or pressure change, assuming it should be different.
Fix: Kc and Kp are constant at a fixed temperature no matter what concentrations, pressures, or volumes you start with. Only a temperature change alters K.
❌
Using mole fractions or masses directly in Kc
Plugging in grams, moles, or mole fractions without converting to molar concentration (mol/L) or partial pressure.
Fix: always convert to molarity (moles ÷ volume in litres) for Kc, or partial pressure for Kp, before substituting into the expression.
❌
Ignoring stoichiometric coefficients as exponents
Writing Kc = [NH3]/[N2][H2] instead of [NH3]2/[N2][H2]3 for N2 + 3H2 ⇌ 2NH3.
Fix: every species' equilibrium concentration must be raised to the power of its balanced stoichiometric coefficient — never omit or approximate these exponents.
❌
Confusing pH and pOH direction
Assuming a higher pOH means a more basic solution, or mixing up which one decreases as basicity increases.
Fix: higher pH = more basic; higher pOH = more acidic (since pOH = 14 − pH). Always double check which quantity the question asks for before reporting your final answer.
❌
Applying the Henderson–Hasselbalch equation to non-buffer systems
Using pH = pKa + log([Salt]/[Acid]) for a plain weak acid solution with no added salt/conjugate base.
Fix: this equation is valid only when there is a genuine buffer — comparable amounts of a weak acid (or base) and its conjugate present together. For a lone weak acid, use [H+] = √(KaC) instead.
❌
Wrong exponent in Ksp for AB2 or A2B salts
Writing Ksp = s2 for a salt like CaF2 that actually dissociates into one Ca2+ and two F− ions.
Fix: for CaF2(s) ⇌ Ca2+ + 2F−, if solubility is s, then [F−] = 2s, so Ksp = (s)(2s)2 = 4s3. Always derive the exponents from the actual dissociation equation.

Practice Questions

Original, concept-building problems — organised by topic, each with a full step-by-step solution. Click a question to reveal its solution.

Interactive Zone

Six hands-on modules to build and test your intuition for equilibrium.

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    Equilibrium | Chemistry Class 11 | Academia Aeternum
    Equilibrium | Chemistry Class 11 | Academia Aeternum — Complete Notes & Solutions · academia-aeternum.com
    Equilibrium is one of the most fundamental concepts in chemistry, explaining how reversible physical and chemical processes attain a state of balance. This chapter covers physical and chemical equilibrium, equilibrium constants (Kc and Kp), Le Chatelier's Principle, ionic equilibrium, acids and bases, pH, buffer solutions, salt hydrolysis, and solubility product. Understanding these concepts helps in predicting the direction and extent of chemical reactions, calculating equilibrium…
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    EQUILIBRIUM — Learning Resources

    🧠 Practice MCQs
    ✔️ True / False

    Frequently Asked Questions

    Equilibrium is the state of a reversible process where the forward and reverse reactions occur at equal rates and the macroscopic properties remain constant.

    Dynamic equilibrium is the state in which forward and reverse reactions continue simultaneously at equal rates without any net change in concentrations.

    In static equilibrium, no process occurs, whereas in dynamic equilibrium, opposite processes continue at equal rates.

    Equilibrium requires a reversible process, a closed system and constant temperature.

    Physical equilibrium exists between different physical states such as solid-liquid, liquid-vapour and solid-vapour.

    Chemical equilibrium is the state where the rates of the forward and reverse chemical reactions become equal.

    The equilibrium constant (K) is the ratio of the equilibrium concentrations of products to reactants, each raised to their stoichiometric coefficients.

    Kc is the equilibrium constant expressed in terms of molar concentrations.

    Kp is the equilibrium constant expressed in terms of the partial pressures of gaseous species.

    \(K_p = K_c(RT)^{\Delta n}\)

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