A comprehensive, examination-ready reference covering core principles, quantitative analysis, Le Chatelier's Principle, Ionic Equilibrium, and beyond.
Part I — Chemical Equilibrium
Section 01
Core Principles of Chemical Equilibrium
Chemical equilibrium is the dynamic state in a reversible reaction where the rate of forward reaction equals the rate of reverse reaction, so the concentrations of reactants and products remain constant over time — though both reactions continue to occur simultaneously.
Fig 1. Dynamic equilibrium — both reactions proceed at equal rates.
Key Characteristics
⇌Only possible in a closed system
📊Concentrations remain constant, not equal
⚡Both reactions continue — it is dynamic
🌡️Attained at constant temperature
⚖️Forward and reverse rates are equal
🔄Reversible reaction: aA + bB ⇌ cC + dD
Section 02
Quantitative Analysis — The Equilibrium Constant
The Law of Mass Action (Guldberg & Waage, 1864) states that the rate of a reaction is proportional to the product of the molar concentrations of reactants, each raised to the power equal to their stoichiometric coefficient.
Equilibrium Constant Expression — Kc
aA + bB ⇌ cC + dD
Kc = [C]ᶜ [D]ᵈ / [A]ᵃ [B]ᵇ
[ ] denotes molar concentration (mol L⁻¹); pure solids & liquids are excluded.
Equilibrium Constant in Partial Pressures — Kp
Kp = (Pc)ᶜ (Pd)ᵈ / (Pa)ᵃ (Pb)ᵇ
Kp = Kc (RT)^Δn
Δn = (c+d) − (a+b) = change in moles of gas; R = 0.0821 L·atm·mol⁻¹·K⁻¹
Reaction Quotient Q vs. Equilibrium Constant K
Q is calculated using instantaneous concentrations at any point in the reaction. Comparing Q with K predicts the direction of reaction:
Condition
Meaning
Direction of Shift
Q < K
Reaction has not reached equilibrium
→ Forward (more products form)
Q = K
System is at equilibrium
No net change
Q > K
Products exceed equilibrium
← Reverse (reactants reform)
Significance of K Value
Fig 2. Interpretation of equilibrium constant magnitude.
Section 03
Le Chatelier's Principle
"If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium will shift in the direction that tends to counteract the imposed change." — Henri Le Chatelier (1884)
Fig 3. Stresses on equilibrium and the predicted response.
Stress Applied
Shift Direction
Effect on K
↑ Concentration of reactant
Forward →
No change
↑ Concentration of product
Reverse ←
No change
↑ Pressure (gas phase)
Side with fewer gas moles
No change
↑ Temperature (exothermic rxn)
Reverse ←
K decreases
↑ Temperature (endothermic rxn)
Forward →
K increases
Adding inert gas (constant V)
No shift
No change
Catalyst added
No shift (equilibrium faster)
No change
Section 04
Essential Formulas & Concepts
Gibbs Free Energy & Equilibrium
The relationship between Gibbs free energy and the equilibrium constant reveals the spontaneity and equilibrium position of a reaction.
Standard Gibbs Free Energy Change
ΔG° = −RT ln K
ΔG = ΔG° + RT ln Q
R = 8.314 J·mol⁻¹·K⁻¹; T = temperature in Kelvin; Q = reaction quotient
ΔG° Value
K Value
Reaction Tendency
ΔG° < 0
K > 1
Spontaneous (product-favoured)
ΔG° = 0
K = 1
Equilibrium (neither direction favoured)
ΔG° > 0
K < 1
Non-spontaneous (reactant-favoured)
Van't Hoff Equation
Describes how the equilibrium constant K varies with temperature T, allowing prediction of K at different temperatures when the enthalpy change ΔH° is known.
Van't Hoff Equation (Integrated Form)
ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)
log(K₂/K₁) = −(ΔH°/2.303R)(1/T₂ − 1/T₁)
ΔH° = standard enthalpy change; K₁ at T₁, K₂ at T₂
Key insight: For an exothermic reaction (ΔH° < 0), increasing T decreases K. For an endothermic reaction (ΔH° > 0), increasing T increases K. This is consistent with Le Chatelier's Principle.
Henry's Law
Henry's Law governs the solubility of gases in liquids — the amount of dissolved gas is directly proportional to its partial pressure above the liquid at constant temperature.
Henry's Law
p = Kₕ × x
or C = Kₕ' × p
p = partial pressure of gas; x = mole fraction; Kₕ = Henry's constant; C = concentration
🫧Carbonated beverages use high CO₂ pressure
🤿Decompression sickness — N₂ dissolved in blood
🌡️Solubility of gas decreases with temperature rise
⚠️Does NOT apply to reactive gases (HCl, NH₃ in water)
Part II — Ionic Equilibrium
Section 05
Ionic Equilibrium — Introduction
Ionic equilibrium deals with the partial or complete ionisation of electrolytes in aqueous solution and the equilibria established between ions and undissociated molecules. Strong electrolytes ionise completely; weak electrolytes establish a dynamic equilibrium.
Degree of Ionisation (α)
α = (moles ionised) / (total moles dissolved)
α → 1 for strong electrolytes; α << 1 for weak electrolytes
Section 06
Acids & Bases: Three Theories
Fig 4. Comparison of three acid-base theories.
Scope hierarchy: Lewis > Brønsted-Lowry > Arrhenius. Every Arrhenius acid/base is a Brønsted-Lowry acid/base, and every Brønsted-Lowry acid/base is a Lewis acid/base — but not vice versa.
Section 07
Acid–Base Equilibria & Ionisation Constants
Weak acids and bases partially ionise in water, establishing an equilibrium characterised by the ionisation constant Kₐ (acid) or K_b (base).
Weak Acid Equilibrium
HA ⇌ H⁺ + A⁻
Kₐ = [H⁺][A⁻] / [HA]
Weak Base Equilibrium
B + H₂O ⇌ BH⁺ + OH⁻
K_b = [BH⁺][OH⁻] / [B]
Relationship: Kₐ × K_b = Kw
Kₐ × K_b = Kw = 1.0 × 10⁻¹⁴ (at 25°C)
pKₐ + pK_b = pKw = 14
Applies to a conjugate acid-base pair.
Ostwald's Dilution Law
Relates the degree of ionisation α to the ionisation constant and concentration C:
Ostwald's Dilution Law (weak electrolyte)
Kₐ = Cα² / (1−α) ≈ Cα² (when α << 1)
α = √(Kₐ/C)
Ionisation increases on dilution (↓C → ↑α)
Section 08
Ionisation of Water & the pH Scale
Pure water undergoes self-ionisation (autoprotolysis), establishing an equilibrium expressed by the ionic product of water, Kw.
Ionic Product of Water
H₂O ⇌ H⁺ + OH⁻
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ (at 25°C)
Kw increases with temperature (endothermic ionisation).
pH & pOH Definitions
pH = −log₁₀[H⁺] pOH = −log₁₀[OH⁻]
pH + pOH = 14 (at 25°C)
Fig 5. The pH scale — 0 (strongly acidic) to 14 (strongly basic).
Solution
[H⁺] (mol L⁻¹)
pH
Gastric acid
~ 0.1
~ 1
Pure water (25°C)
10⁻⁷
7
Blood plasma
~ 4 × 10⁻⁸
~ 7.4
0.1 M NaOH
10⁻¹³
13
Section 09
Common Ion Effect
The common ion effect is the suppression of ionisation of a weak electrolyte when a strong electrolyte sharing a common ion is added to the solution. This is a direct application of Le Chatelier's Principle.
Practical importance: The common ion effect is the basis of buffer solutions and is used industrially to precipitate ions selectively from solution.
↓Suppresses ionisation of weak acid/base
⚗️Reduces solubility of sparingly soluble salts
🧪Used in qualitative analysis (group separation)
💊Controls pH in pharmaceutical formulations
Section 10
Hydrolysis of Salts & pH of Solutions
Salt hydrolysis is the reaction of a salt's ions with water, producing an acidic or basic solution depending on the strength of the parent acid and base.
The solubility product Kₛₚ is the equilibrium constant for the dissolution of a sparingly soluble ionic compound in water. It is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient.
Ksp Expression
AmBn(s) ⇌ mAⁿ⁺(aq) + nBᵐ⁻(aq)
Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Pure solid is excluded from the expression.
Solubility (s) from Ksp
For AgCl: Ksp = s² → s = √Ksp
For Ca₃(PO₄)₂: Ksp = (3s)³(2s)² = 108s⁵
Ionic Product (Q) vs. Ksp — Precipitation Criterion
Condition
Outcome
Q < Ksp
Unsaturated — more solid can dissolve
Q = Ksp
Saturated — equilibrium; no net change
Q > Ksp
Supersaturated — precipitation occurs
Common ion effect on solubility: Adding a common ion reduces the solubility of a sparingly soluble salt (Q increases → precipitation). Example: AgCl is less soluble in NaCl solution than in pure water.
Section 12
Buffer Solutions
A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer).
Buffer capacity (β) is the amount of strong acid or base that can be added per litre before a significant pH change occurs. Maximum buffer capacity is achieved when pH = pKₐ (i.e., [HA] = [A⁻]).
Buffer Capacity (Van Slyke)
β = 2.303 × C × (Kₐ[H⁺]) / (Kₐ + [H⁺])²
C = total buffer concentration; max when [H⁺] = Kₐ
🩸Blood: CO₂/HCO₃⁻ buffer (pH 7.35–7.45)
🧬Intracellular: phosphate buffer system
🏭Industry: acetate buffer in electroplating
💉IV fluids: buffered to physiological pH
Quick Reference Formula Sheet
Summary
All Key Formulas at a Glance
Kc / Kp
Kp = Kc(RT)^Δn
Gibbs Free Energy
ΔG° = −RT ln K
Van't Hoff
ln(K₂/K₁)=−(ΔH°/R)(1/T₂−1/T₁)
Henry's Law
p = Kₕ × x
Kw & pH
Kw = [H⁺][OH⁻] = 10⁻¹⁴
pH + pOH = 14
Ostwald Dilution
α = √(Kₐ/C)
Henderson–Hasselbalch
pH = pKₐ + log([A⁻]/[HA])
Ksp & Solubility
Ksp = [Aⁿ⁺]ᵐ [Bᵐ⁻]ⁿ
Kₐ × K_b Relation
Kₐ × K_b = Kw = 10⁻¹⁴
Part III — Worked Examples & ICE Tables
Worked Examples
Step-by-Step Solved Problems
Click any example to expand the full solution with an ICE table and commentary.
EX 01Find Kc for: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) given equilibrium concentrations▼
Given: [N₂] = 0.50 M, [H₂] = 0.30 M, [NH₃] = 0.20 M at equilibrium.
1
Write the equilibrium expression from the balanced equation.
Kc = [NH₃]² / ([N₂][H₂]³)
2
Substitute equilibrium concentrations.
Kc = (0.20)² / (0.50 × (0.30)³)
= 0.04 / (0.50 × 0.027)
= 0.04 / 0.0135 = 2.96
3
Result: Kc ≈ 2.96. Since K > 1, products are somewhat favoured under these conditions.
✅
This is the Haber process for ammonia synthesis. Industrially, conditions are optimised using high pressure and moderate temperature (400–500 °C) to balance yield and rate.
EX 02ICE Table: Find equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g)▼
Given: Initial: [H₂] = 1.00 M, [I₂] = 1.00 M, [HI] = 0. Kc = 54.3 at 430 °C.
1
Set up the ICE (Initial–Change–Equilibrium) table. Let x = moles reacted.
pH change = only 0.04 units — buffer action confirmed!
💡
Without the buffer, adding 0.01 mol HCl to 1 L pure water would drop pH from 7 to 2 — a change of 5 pH units! The buffer limits this to just 0.04 units.
EX 05Ksp of PbCl₂: find molar solubility and effect of common ion▼
Ksp(PbCl₂) = 1.6 × 10⁻⁵ at 25°C
1
Write the dissolution and Ksp expression.
PbCl₂(s) ⇌ Pb²⁺(aq) + 2Cl⁻(aq)
Ksp = [Pb²⁺][Cl⁻]² = (s)(2s)² = 4s³
2
Solve for molar solubility s in pure water.
4s³ = 1.6×10⁻⁵ → s³ = 4.0×10⁻⁶
s = ∛(4.0×10⁻⁶) = 1.587×10⁻² M ≈ 0.016 M
3
Solubility in 0.10 M NaCl (common ion Cl⁻).
Ksp = s(2s + 0.10)² ≈ s(0.10)² = 0.01s
s = Ksp / 0.01 = 1.6×10⁻⁵ / 0.01 = 1.6×10⁻³ M
Solubility drops from 0.016 M → 0.0016 M in NaCl — 10× reduction!
⚠️
The common ion approximation (2s + 0.10 ≈ 0.10) is valid only when s << 0.10. Always verify after solving.
EX 06Van't Hoff: Calculate K₂ at 500 K given K₁ at 300 K and ΔH°▼
Endothermic reaction: K increases dramatically with temperature.
✅
This confirms Le Chatelier's Principle quantitatively — raising temperature favours the endothermic direction (forward reaction here), greatly increasing K.
Part IV — Deep Dive Concepts
Section 13
Conjugate Acid–Base Pairs
In every Brønsted-Lowry acid-base reaction, a proton transfer creates a conjugate pair: the species formed when an acid loses a proton is its conjugate base, and vice versa.
Fig 7. Conjugate acid-base pairs in the HCl–water reaction.
Strong Acids (Ka large)
HCl, HBr, HI
HNO₃, H₂SO₄, HClO₄
Near complete ionisation in water
Conjugate base is very weak
Weak Acids (Ka small)
CH₃COOH (Ka = 1.8×10⁻⁵)
HF (Ka = 6.8×10⁻⁴)
H₂CO₃ (Ka = 4.3×10⁻⁷)
Partially ionise; equilibrium established
Section 14
Amphoteric (Amphiprotic) Substances
An amphoteric substance can act as either an acid or a base depending on the reaction partner. Water is the classic example — it donates a proton to strong bases and accepts one from strong acids.
Water as Acid and Base
H₂O + NH₃ → NH₄⁺ + OH⁻ (H₂O acts as acid)
H₂O + HCl → H₃O⁺ + Cl⁻ (H₂O acts as base)
💧Water (H₂O) — amphiprotic
🧪HCO₃⁻ — amphoteric anion
⚗️HSO₄⁻ — amphoteric anion
🔬Al(OH)₃, ZnO — amphoteric metal oxides
🧬Amino acids — amphoteric (zwitterions)
📐H₂PO₄⁻, HPO₄²⁻ — amphiprotic
Section 15
Polyprotic Acids
Polyprotic acids have more than one ionisable proton. Each ionisation step has its own Kₐ, and successive values always decrease: Ka₁ > Ka₂ > Ka₃.
Acid
Step
Equilibrium
Kₐ
H₂SO₃
Ka₁
H₂SO₃ ⇌ H⁺ + HSO₃⁻
1.5 × 10⁻²
Ka₂
HSO₃⁻ ⇌ H⁺ + SO₃²⁻
6.2 × 10⁻⁸
H₃PO₄
Ka₁
H₃PO₄ ⇌ H⁺ + H₂PO₄⁻
7.5 × 10⁻³
Ka₂
H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻
6.2 × 10⁻⁸
Ka₃
HPO₄²⁻ ⇌ H⁺ + PO₄³⁻
4.8 × 10⁻¹³
H₂CO₃
Ka₁
H₂CO₃ ⇌ H⁺ + HCO₃⁻
4.3 × 10⁻⁷
Ka₂
HCO₃⁻ ⇌ H⁺ + CO₃²⁻
4.7 × 10⁻¹¹
📌
For a diprotic acid H₂A, [H⁺] at equilibrium is approximated by √(Ka₁ × C) since Ka₁ >> Ka₂. The second ionisation is negligible for most purposes.
Section 16
Industrial Applications of Equilibrium
🏭 Haber Process (NH₃ Synthesis)
The Haber-Bosch process synthesises ammonia from nitrogen and hydrogen. The reaction is exothermic, so low temperature favours yield — but the rate would be impractically slow. A compromise of 400–500 °C with an iron catalyst and 200 atm pressure is used.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH° = −92 kJ/mol
High pressure favours NH₃ (fewer moles of gas on product side). Low temp favours NH₃ (exothermic). Catalyst (Fe/K₂O/Al₂O₃) increases rate without shifting K.
🏭 Contact Process (H₂SO₄ Synthesis)
The key equilibrium step converts SO₂ to SO₃ over a vanadium pentoxide catalyst. Operating at ~450 °C gives ~98% conversion with good reaction rate.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH° = −197 kJ/mol
Catalyst: V₂O₅; T = 450°C; P = 1–2 atm. SO₃ absorbed in H₂SO₄ → oleum, then diluted.
🩸 Blood Buffer System
Blood is maintained at pH 7.35–7.45 by the carbonic acid–bicarbonate buffer. The lungs regulate CO₂ (acid side) and the kidneys regulate HCO₃⁻ (base side).
CO₂(g) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
pH = 6.1 + log([HCO₃⁻]/[H₂CO₃])
Normal ratio [HCO₃⁻]/[H₂CO₃] ≈ 20:1 gives pH ≈ 7.4. Acidosis: pH < 7.35; Alkalosis: pH > 7.45.
Part V — Interactive Calculators
🧮 pH Calculator
Calculate pH of strong/weak acids, bases, and buffer solutions instantly.
🔬 Ksp & Molar Solubility Calculator
Find molar solubility from Ksp or predict precipitation from ion concentrations.
⚗️ Kp ↔ Kc Converter
Convert between Kp and Kc using Kp = Kc(RT)^Δn.
Part VI — Self-Test Quiz
🎯 Chemical & Ionic Equilibrium — 15-Question Self Test
Test your understanding. Click an option to check your answer and read the explanation.
Question 1 of 15
At equilibrium, which statement is CORRECT?
A Concentrations of reactants and products are equal
B Rate of forward reaction equals rate of reverse reaction
C The reaction has completely stopped
D Kc = 1 always
Question 2 of 15
For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the value of Kp relates to Kc by:
A Kp = Kc × RT
B Kp = Kc × (RT)²
C Kp = Kc × (RT)⁻¹
D Kp = Kc
Question 3 of 15
When pressure is increased on a gaseous equilibrium, the reaction shifts toward:
A The side with more moles of gas
B The side with fewer moles of gas
C No shift occurs
D The exothermic side
Question 4 of 15
If Q > K for a reaction, then:
A The reaction proceeds in the forward direction
B The reaction proceeds in the reverse direction
C The system is at equilibrium
D Concentrations remain unchanged
Question 5 of 15
ΔG° = −RT ln K. When K > 1, ΔG° is:
A Positive (non-spontaneous)
B Negative (spontaneous)
C Zero
D Cannot be determined
Question 6 of 15
Henry's Law (p = Kₕ × x) applies to:
A All gases at any temperature
B Reactive gases like HCl in water
C Gases that do not chemically react with the solvent at low pressures
D Only to CO₂ in carbonated water
Question 7 of 15
The Brønsted-Lowry theory defines an acid as:
A A substance that produces OH⁻ in water
B A proton donor
C An electron pair acceptor
D A substance with pH < 7 only
Question 8 of 15
For a weak acid HA with Ka = 4 × 10⁻⁵ and C = 0.10 M, the degree of ionisation α ≈
A 0.4%
B 2.0%
C 20%
D 40%
Question 9 of 15
pH of pure water at 25°C is 7 because:
A Water is a neutral solvent with no ions
B Kw = 10⁻⁷
C [H⁺] = [OH⁻] = 10⁻⁷ M, so pH = −log(10⁻⁷) = 7
D All aqueous solutions have pH 7
Question 10 of 15
A solution of NH₄Cl in water is expected to be:
A Basic (pH > 7)
B Acidic (pH < 7)
C Neutral (pH = 7)
D Depends on concentration
Question 11 of 15
Adding NaCl to a saturated solution of AgCl will:
A Increase the solubility of AgCl
B Decrease the solubility of AgCl (common ion effect)
C Have no effect
D Change the value of Ksp
Question 12 of 15
Henderson–Hasselbalch: pH = pKₐ when:
A [A⁻] = 0
B [HA] = 0
C [A⁻] = [HA] (equal concentrations)
D [A⁻] = 10 × [HA]
Question 13 of 15
Van't Hoff equation predicts that for an exothermic reaction, increasing temperature will:
A Increase K
B Decrease K
C Keep K unchanged
D Make K = 1
Question 14 of 15
BF₃ reacts with NH₃ to form F₃B←NH₃. According to Lewis theory, BF₃ is:
A A Brønsted acid
B A Lewis base
C A Lewis acid (electron pair acceptor)
D An Arrhenius acid
Question 15 of 15
For Ca₃(PO₄)₂ ⇌ 3Ca²⁺ + 2PO₄³⁻, if molar solubility = s, then Ksp =
A s⁵
B 36s⁵
C 108s⁵
D 6s⁵
Score: 0 / 15
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