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Class 11 Chemistry Chapter 6 Equilibrium — Formulas & Key Points

Equilibrium is the foundation of physical and inorganic chemistry in Class 11, bridging thermodynamics and kinetics. NCERT Chapter 6 introduces the law of mass action, equilibrium constants in concentration (Kc) and pressure (Kp) terms, Le Chatelier's principle, ionic equilibrium including pH and buffer calculations, and solubility equilibria. Mastering these formulas is non-negotiable for CBSE board exams and competitive tests like NEET and JEE.

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Key takeaways

✓Equilibrium constant Kc depends on temperature alone; pressure and catalyst do not change its value.
✓Relation Kp = Kc(RT)^Δn applies only to gaseous equilibria; Δn = (moles of gaseous products) − (moles of gaseous reactants).
✓pH = −log[H⁺] and pOH = −log[OH⁻]; at 25°C, pH + pOH = 14 for all aqueous solutions.
✓For weak acids, Ka = [H⁺][A⁻]/[HA]; degree of dissociation α = √(Ka/C) when α ≪ 1.
✓Henderson-Hasselbalch equation: pH = pKa + log([Salt]/[Acid]) for acidic buffers; pOH = pKb + log([Salt]/[Base]) for basic buffers.
✓Common-ion effect suppresses ionization; adding CH₃COONa to CH₃COOH shifts equilibrium left, reducing [H⁺].
✓Solubility product Ksp = [Cation]^m[Anion]^n; precipitation occurs when ionic product > Ksp.

Core Equilibrium Formulas and Constants

Chemical equilibrium is the state where the rate of the forward reaction equals the rate of the reverse reaction, and concentrations remain constant. The equilibrium constant quantifies this balance. For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration Kc and pressure Kp are central. The law of mass action states that at equilibrium, the ratio of product concentrations to reactant concentrations (each raised to their stoichiometric coefficients) is constant at a given temperature. Understanding when to use Kc versus Kp and the interconversion formula is critical for solving numerical problems in CBSE exams.

Law of Mass Action: For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b at constant temperature
Kp expression: Kp = (Pc)^c(Pd)^d / (Pa)^a(Pb)^b where P denotes partial pressures in atm or bar
Relation between Kc and Kp: Kp = Kc(RT)^Δn; Δn = (c + d) − (a + b); R = 0.0831 bar·L·mol⁻¹·K⁻¹
When Δn = 0, Kp = Kc; when Δn > 0, Kp > Kc at usual temperatures; when Δn < 0, Kp < Kc
Equilibrium constant is dimensionless when expressed in activities; in exams, use mol/L for Kc and bar for Kp

Le Chatelier's Principle and Reaction Quotient

Le Chatelier's principle predicts the direction of shift when an equilibrium system is disturbed by changes in concentration, pressure, or temperature. The reaction quotient Q has the same mathematical form as K but uses instantaneous concentrations or pressures. Comparing Q with K tells us whether the system is at equilibrium or which direction it will proceed. This principle is extensively tested in CBSE board subjective and MCQ questions, especially for predicting the effect of adding or removing reactants, changing volume, or altering temperature on yield of products.

Reaction Quotient: Q = [C]^c[D]^d / [A]^a[B]^b at any instant (not necessarily at equilibrium)
If Q < K, forward reaction proceeds to reach equilibrium; if Q > K, reverse reaction proceeds; if Q = K, system is at equilibrium
Increase in concentration of reactant shifts equilibrium to the right (more product)
Increase in pressure (decrease in volume) shifts equilibrium toward the side with fewer moles of gas
Increase in temperature favours endothermic direction; decrease in temperature favours exothermic direction
Catalysts do not change K or the equilibrium position; they only speed up attainment of equilibrium

Ionic Equilibrium: pH, pOH, and Water Ionization

Water undergoes auto-ionization: H₂O ⇌ H⁺ + OH⁻ with ionic product Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. The pH scale is logarithmic: pH = −log₁₀[H⁺] and pOH = −log₁₀[OH⁻]. At 25°C, pH + pOH = 14. For strong acids and bases, complete dissociation is assumed; for weak acids and bases, equilibrium constants Ka and Kb are used. NCERT Class 11 Chemistry emphasizes numerical problems on pH calculation for various solutions, making this section high-yield for exams.

Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C
pH = −log[H⁺]; pOH = −log[OH⁻]; pKw = pH + pOH = 14 at 25°C
For strong acid of concentration C: [H⁺] = C, so pH = −log C
For strong base of concentration C: [OH⁻] = C, so pOH = −log C, pH = 14 + log C
Neutral solution: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ M, pH = 7 at 25°C
Acidic solution: pH < 7; basic solution: pH > 7

Weak Acid and Weak Base Equilibria

Weak acids (HA) and weak bases (BOH or B) only partially ionize in water. Their ionization is governed by the acid dissociation constant Ka and the base dissociation constant Kb. The degree of dissociation α is the fraction of molecules that ionize. For dilute solutions where α ≪ 1, Ostwald's dilution law gives simplified formulas for [H⁺], pH, and α. These formulas are repeatedly tested in CBSE Class 11 Chemistry practicals and board theory papers, especially in the context of calculating pH of buffer solutions or comparing strengths of acids.

Weak acid ionization: HA ⇌ H⁺ + A⁻; Ka = [H⁺][A⁻]/[HA]
For initial concentration C and degree of dissociation α: [H⁺] = Cα, [A⁻] = Cα, [HA] = C(1−α)
Ka = Cα²/(1−α); if α ≪ 1, Ka ≈ Cα², so α ≈ √(Ka/C)
[H⁺] = Cα = √(Ka·C); pH = ½[pKa − log C]
Weak base ionization: B + H₂O ⇌ BH⁺ + OH⁻; Kb = [BH⁺][OH⁻]/[B]
Similarly, [OH⁻] = √(Kb·C); pOH = ½[pKb − log C]; pH = 14 − pOH
Relation: Ka × Kb = Kw for a conjugate acid-base pair

Buffer Solutions and Henderson-Hasselbalch Equation

Buffers resist changes in pH upon addition of small amounts of acid or base. An acidic buffer is a mixture of a weak acid and its salt with a strong base (e.g., CH₃COOH + CH₃COONa); a basic buffer is a weak base plus its salt with a strong acid (e.g., NH₄OH + NH₄Cl). The Henderson-Hasselbalch equation provides a logarithmic relation between pH (or pOH), pKa (or pKb), and the ratio of salt to acid (or base) concentrations. This formula is indispensable for buffer pH calculations in CBSE board exams and NEET, where students are often asked to design a buffer of a target pH.

Acidic buffer: pH = pKa + log([Salt]/[Acid]) = pKa + log([A⁻]/[HA])
Basic buffer: pOH = pKb + log([Salt]/[Base]) = pKb + log([BH⁺]/[B]); then pH = 14 − pOH
Buffer capacity is maximum when [Salt] = [Acid] or [Salt] = [Base], i.e., pH = pKa or pOH = pKb
pKa = −log Ka; pKb = −log Kb
Common acidic buffers: CH₃COOH/CH₃COONa, HCOOH/HCOONa; common basic buffers: NH₄OH/NH₄Cl

Common-Ion Effect and Salt Hydrolysis

The common-ion effect is the suppression of ionization of a weak electrolyte by adding a strong electrolyte that shares a common ion. For instance, adding sodium acetate to acetic acid decreases the degree of ionization of acetic acid because acetate ion from the salt shifts the equilibrium backward. Salt hydrolysis occurs when salts of weak acids or weak bases react with water to produce acidic or basic solutions. Salts of strong acid and strong base (e.g., NaCl) do not hydrolyze and yield neutral solutions. Understanding these concepts helps predict solution pH and is tested in NCERT Class 11 Chemistry exercises and CBSE sample papers.

Common-ion effect: Addition of CH₃COONa to CH₃COOH suppresses ionization; [H⁺] decreases, pH increases
Salt of weak acid + strong base (e.g., CH₃COONa): A⁻ + H₂O ⇌ HA + OH⁻; solution is basic; pH > 7
Hydrolysis constant: Kh = Kw/Ka for anion hydrolysis; [OH⁻] = √(Kh·C); pOH = ½[pKw − pKa − log C]
Salt of strong acid + weak base (e.g., NH₄Cl): BH⁺ + H₂O ⇌ B + H₃O⁺; solution is acidic; pH < 7
Kh = Kw/Kb for cation hydrolysis; [H⁺] = √(Kh·C); pH = ½[pKw − pKb − log C]
Salt of weak acid + weak base (e.g., CH₃COONH₄): pH = 7 + ½(pKa − pKb)
Salt of strong acid + strong base (e.g., NaCl): no hydrolysis; pH ≈ 7

Solubility Equilibrium and Solubility Product (Ksp)

Sparingly soluble salts establish equilibrium between undissolved solid and dissolved ions. The solubility product constant Ksp is the product of ionic concentrations at saturation, each raised to its stoichiometric power. For a salt AmBn, Ksp = [A⁺]^m[B⁻]^n. The relationship between solubility S (in mol/L) and Ksp depends on stoichiometry. Precipitation occurs when the ionic product (calculated from actual concentrations) exceeds Ksp. Common-ion effect reduces solubility when a salt sharing an ion is added. These principles are vital for qualitative inorganic analysis in CBSE practicals and for solving numerical problems on precipitation and solubility in Class 11 Chemistry solutions.

For AB-type salt (e.g., AgCl): Ksp = [A⁺][B⁻]; if solubility = S, Ksp = S²
For AB₂-type salt (e.g., PbCl₂): Ksp = [A²⁺][B⁻]²; if solubility = S, Ksp = S(2S)² = 4S³
For A₂B-type salt (e.g., Ag₂CrO₄): Ksp = [A⁺]²[B²⁻]; if solubility = S, Ksp = (2S)²S = 4S³
Ionic product Q = [A⁺]^m[B⁻]^n from actual concentrations; precipitation if Q > Ksp
Common-ion effect: solubility of AgCl decreases in presence of NaCl because [Cl⁻] increases

Important Definitions, Terms, and Constants

Mastering terminology is as important as formulas. Equilibrium is dynamic, not static—forward and reverse reactions continue at equal rates. The equilibrium constant K is temperature-dependent; increasing temperature shifts equilibrium in the endothermic direction, changing K. A catalyst accelerates both forward and reverse reactions equally, so it does not affect the equilibrium position or K value. The degree of dissociation α ranges from 0 (no dissociation) to 1 (complete dissociation); strong electrolytes have α ≈ 1, weak electrolytes α ≪ 1. pKa and pKb are negative logarithms of Ka and Kb; lower pKa means stronger acid. These definitions frequently appear in CBSE Class 11 Chemistry Chapter 6 theory questions and MCQs, so clarity here boosts marks.

Dynamic equilibrium: forward and reverse rates equal; macroscopic properties constant but microscopic processes ongoing
Homogeneous equilibrium: all species in same phase (e.g., all gases or all in solution)
Heterogeneous equilibrium: species in different phases (e.g., solid ⇌ gas)
Active mass: molar concentration [mol/L] for species in solution or gas; activity for pure solids and liquids = 1
Degree of dissociation α = (amount dissociated)/(initial amount); ranges 0 to 1
pKa = −log Ka; pKb = −log Kb; lower pKa → stronger acid; lower pKb → stronger base
Standard constants at 25°C: Kw = 1.0×10⁻¹⁴; R = 0.0831 bar·L·mol⁻¹·K⁻¹ or 8.314 J·mol⁻¹·K⁻¹

Memory Tricks, Mnemonics, and Common Mistakes

Students often confuse Ka with Kb or forget to square terms in Ksp expressions. A mnemonic for pH and pOH: 'pH Plus pOH = Fourteen' at 25°C. To remember the Henderson-Hasselbalch equation, note that the log term is always [base form]/[acid form]—for acidic buffers, [A⁻]/[HA]; for basic buffers, [B]/[BH⁺]. Common errors include using concentrations instead of partial pressures in Kp, neglecting the (RT)^Δn factor in Kp = Kc(RT)^Δn, or assuming α ≈ 1 for weak acids (it is ≪ 1). In solubility problems, students forget that for PbCl₂, the Cl⁻ concentration is 2S, not S. Careful unit handling is crucial: Kc in M, Kp in bar or atm, and R in matching units. These tips save precious marks in CBSE board exams and competitive tests, where small sign or calculation errors cascade into wrong answers.

Mnemonic for pH: 'Power of Hydrogen'; pOH: 'Power of Hydroxide'; sum = 14 at room temperature
To avoid sign errors: pH and pOH are always positive; log of a number < 1 is negative, so −log gives positive pH
Common mistake: writing Kp = Kc without the (RT)^Δn term; always calculate Δn first
Remember: in Henderson-Hasselbalch, log([Salt]/[Acid]) for acidic buffer, log([Salt]/[Base]) for basic buffer
For AB₂ salts, Ksp = 4S³, not S²; write out the stoichiometry explicitly to avoid errors
Units trap: if Kp is given in atm and you use R = 0.0831 bar·L·mol⁻¹·K⁻¹, convert pressure to bar or use R = 0.0821 atm·L·mol⁻¹·K⁻¹
Sign convention: exothermic reactions have ΔH < 0; raising temperature favours endothermic (ΔH > 0) side

Three Solved Mini-Examples Applying Key Formulas

Worked examples cement formula application. Example 1 demonstrates Kp calculation from Kc. Example 2 shows pH calculation for a weak acid using Ostwald's dilution law. Example 3 applies the Henderson-Hasselbalch equation to design an acidic buffer. These mirror the pattern of CBSE board numerical problems in NCERT Class 11 Chemistry Chapter 6 and help students gain confidence for exams. Step-by-step solutions reveal how to substitute values, handle units, and round answers appropriately. Practising such examples regularly on platforms like CBSETUTOR.ai—where students upload photo snapshots of homework problems and receive instant, detailed solutions for just ₹999/month across all subjects and classes (6-12) with a 3-day free trial—ensures mastery and speed in board exams.

One-Glance Last-Minute Revision Box

In the final hour before the CBSE board exam, students need a quick snapshot of all critical formulas and facts. This revision box consolidates the essential equations, constants, and tips from the entire chapter into a compact checklist. Memorize these and you cover 80 percent of the numerical and theory marks in Chapter 6. Keep this list on your phone or printed as a flashcard. Review it during metro commutes, before entering the exam hall, or during the 15-minute reading time. Platforms like CBSETUTOR.ai allow students to generate personalized last-minute revision sheets by uploading their class notes and receiving AI-curated key points, all within the ₹999/month subscription covering every CBSE class from 6 to 12 with a 3-day free trial to experience the benefit risk-free.

Kc = [Products]^coeff / [Reactants]^coeff; Kp = Kc(RT)^Δn where Δn = Σ(coeff products) − Σ(coeff reactants) for gases
Q vs K: Q < K → forward; Q > K → reverse; Q = K → equilibrium
Kw = 1.0×10⁻¹⁴ at 25°C; pH + pOH = 14; pH = −log[H⁺]; pOH = −log[OH⁻]
Weak acid: Ka = [H⁺][A⁻]/[HA]; [H⁺] = √(Ka·C); pH = ½(pKa − log C)
Weak base: Kb = [BH⁺][OH⁻]/[B]; [OH⁻] = √(Kb·C); pOH = ½(pKb − log C)
Buffer (acidic): pH = pKa + log([Salt]/[Acid]); buffer (basic): pOH = pKb + log([Salt]/[Base])
Hydrolysis: salt of weak acid & strong base → basic (pH>7); salt of strong acid & weak base → acidic (pH<7)
Solubility: AB → Ksp=S²; AB₂ → Ksp=4S³; A₂B → Ksp=4S³; ionic product Q > Ksp → precipitation
Le Chatelier: add reactant → shift right; increase pressure → shift to fewer moles; raise temp → shift endothermic side
Common mistakes: forget Δn in Kp formula; confuse [Salt]/[Acid] order in Henderson-Hasselbalch; use wrong R units

Frequently asked questions

What is the difference between Kc and Kp in equilibrium?+

Kc is the equilibrium constant in terms of molar concentrations (mol/L), while Kp uses partial pressures (atm or bar). They are related by Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. Use Kc for reactions in solution and Kp for gaseous reactions.

How do I calculate pH of a weak acid like acetic acid?+

For a weak acid HA with concentration C and dissociation constant Ka, use [H⁺] = √(Ka·C) assuming α ≪ 1. Then pH = −log[H⁺]. For example, 0.1 M CH₃COOH with Ka = 1.8×10⁻⁵ gives [H⁺] ≈ 1.34×10⁻³ M, so pH ≈ 2.87.

What is the Henderson-Hasselbalch equation and when should I use it?+

Henderson-Hasselbalch: pH = pKa + log([Salt]/[Acid]) for acidic buffers and pOH = pKb + log([Salt]/[Base]) for basic buffers. Use it to calculate buffer pH or to design a buffer of desired pH by choosing the correct salt-to-acid ratio.

Why does adding common ion reduce solubility of a salt?+

By Le Chatelier's principle, adding a common ion (e.g., Cl⁻ to AgCl solution) shifts the dissolution equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left, decreasing Ag⁺ concentration and thus reducing solubility. This is the common-ion effect, crucial in qualitative analysis.

How is the degree of dissociation α related to Ka?+

For a weak acid at concentration C, Ka = Cα²/(1−α). If α ≪ 1, Ka ≈ Cα², so α ≈ √(Ka/C). As C decreases, α increases (Ostwald's dilution law). Knowing α helps calculate [H⁺] = Cα and hence pH.

What does it mean if Q is greater than K?+

If the reaction quotient Q > K, the system has too many products relative to equilibrium. The reverse reaction will proceed to consume products and form reactants until Q = K. This tells you the direction of spontaneous change.

How do I remember which salts produce acidic or basic solutions?+

Mnemonic: 'Weak Acid + Strong Base = Basic; Strong Acid + Weak Base = Acidic; Strong + Strong = Neutral; Weak + Weak = depends on Ka vs Kb'. For example, CH₃COONa (weak acid salt) is basic; NH₄Cl (weak base salt) is acidic; NaCl is neutral.

Why does a catalyst not change the equilibrium constant K?+

A catalyst lowers activation energy for both forward and reverse reactions equally, speeding up attainment of equilibrium but not altering the position. K depends only on temperature, so adding a catalyst leaves K unchanged—this is a common CBSE exam conceptual question.

What is the ionic product Kw and why is it important?+

Kw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 25°C is the auto-ionization constant of water. It links pH and pOH (pH + pOH = 14) and is the basis for calculating pH of all aqueous solutions. Kw changes with temperature; at higher temps, Kw increases.

How can CBSETUTOR.ai help me master Equilibrium formulas quickly?+

CBSETUTOR.ai offers 24×7 AI-powered doubt solving: upload a photo of any Equilibrium problem and get step-by-step solutions instantly. At ₹999/month for all subjects and classes 6–12, with a 3-day free trial, it is the most affordable way to practice NCERT exercises, sample papers, and custom numericals, ensuring exam-ready confidence.

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