State Le Chatelier's Principle. Apply it to the industrial synthesis of ammonia by Haber's process: N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g); ΔH = −92.38 kJ mol -1 with respect to (a) Pressure, (b) Temperature, (c) Catalyst, and (d) Removal of NH 3 .

Statement: When a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts in a direction that reduces or counteracts the effect of that change. (a) Pressure: 4 moles of gaseous reactants give 2 moles of gaseous product (Δn g = −2). High pressure (~200 atm) shifts equilibrium forward, increasing NH 3 yield. (b) Temperature: Forward reaction is exothermic (ΔH < 0), so low temperature favours yield. Because very low T slows rate excessively, an optimum temperature of ~500 °C (773 K) is chosen. (c) Catalyst: Finely divided Iron (Fe) speeds up both forward and reverse rates equally without altering K c or equilibrium yield. (d) Removal of NH 3 : Continuously liquefying and removing NH 3 keeps Q c < K c , driving the forward reaction.

Write the relationship between K p and K c . For which of the following equilibria is K p = K c ? (i) N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) (ii) H 2 (g) + I 2 (g) ⇌ 2HI(g). Also write K c for CaCO 3 (s) ⇌ CaO(s) + CO 2 (g).

Relation: K p = K c (RT) Δn g , where Δn g = (moles of gaseous products) − (moles of gaseous reactants). K p = K c for (ii) H 2 (g) + I 2 (g) ⇌ 2HI(g) because Δn g = 2 − 2 = 0. For CaCO 3 (s) ⇌ CaO(s) + CO 2 (g), concentrations of pure solids are constant (unity), so K c = [CO 2 (g)] .

NCERT Plus One (Class 11) Chemistry • Chapter 6

Equilibrium

Q: Write the relationship between K p and K c . For which of the following equilibria is K p = K c ? (i) N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) (ii) H 2 (g) + I 2 (g) ⇌ 2HI(g). Also write K c for CaCO 3 (s) ⇌ CaO(s) + CO 2 (g).

Relation: K p = K c (RT) Δn g , where Δn g = (moles of gaseous products) − (moles of gaseous reactants). K p = K c for (ii) H 2 (g) + I 2 (g) ⇌ 2HI(g) because Δn g = 2 − 2 = 0. For CaCO 3 (s) ⇌ CaO(s) + CO 2 (g), concentrations of pure solids are constant (unity), so K c = [CO 2 (g)] .

Q: For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl 2 (g), the value of K p is 1.8 × 10 -2 bar at 500 K. Calculate K c for this reaction at 500 K. [R = 0.0831 bar·L mol -1 K -1 ]

Given: K p = 1.8 × 10 -2 bar, T = 500 K, R = 0.0831 bar·L mol -1 K -1 . Δn g = n p − n r = (2 + 1) − 2 = +1 . Using K p = K c (RT) Δn g ⇒ K c = K p / (RT) 1 K c = (1.8 × 10 -2 ) / (0.0831 × 500) = 0.018 / 41.55 = 4.33 × 10 -4 mol L -1 .

Q: What is meant by a conjugate acid–base pair? Write the conjugate acid and conjugate base of (i) H 2 O and (ii) HCO 3 - .

Conjugate Acid–Base Pair: An acid–base pair that differs from each other by only one proton (H + ) . H 2 O: Conjugate acid = H 3 O + ; Conjugate base = OH - . HCO 3 - : Conjugate acid = H 2 CO 3 ; Conjugate base = CO 3 2- .

Q: Distinguish between Reaction Quotient (Q c ) and Equilibrium Constant (K c ). How does Q c predict the direction of a chemical reaction?

K c is calculated using equilibrium concentrations (constant at fixed T), whereas Q c uses concentrations at any arbitrary time t . If Q c K c : Net reaction proceeds in the reverse direction (products → reactants). If Q c = K c : Reaction is at equilibrium .

Q: Define Lewis acids and Lewis bases. Classify the following into Lewis acids and Lewis bases: H 2 O, BF 3 , H + , OH - .

Lewis Acid: An electron-pair acceptor. | Lewis Base: An electron-pair donor. Lewis Acids: BF 3 (incomplete octet), H + (vacant 1s orbital). Lewis Bases: H 2 O , OH - (possess lone pairs on oxygen).

Q: What are buffer solutions? Give one example each for an acidic buffer and a basic buffer. Write the Henderson–Hasselbalch equation for an acidic buffer and explain why dilution does not alter its pH.

Buffer Solution: A solution that resists change in pH on dilution or with the addition of small amounts of strong acid or alkali. Acidic Buffer: Equimolar mixture of CH 3 COOH + CH 3 COONa (pH ≈ 4.76). Basic Buffer: Mixture of NH 4 OH + NH 4 Cl (pH ≈ 9.25). Henderson–Hasselbalch Equation: pH = pK a + log([Salt] / [Acid]) . Effect of Dilution: Both [Salt] and [Acid] decrease by the same factor, so their ratio [Salt]/[Acid] inside the logarithm stays unchanged, leaving pH constant.

Physical & Dynamic Chemical Equilibrium, Kp & Kc, Le Chatelier Principle, Acids-Bases, pH, Buffers & Solubility Product

📄 Official NCERT Textbook PDF

Topic 6.1: Physical Equilibrium: Phase Changes, Vapour Pressure & Henry's Law

Picture a chilly morning in Munnar with a glass thermos holding ice cubes floating in cold water at 0 °C (273 K), and a sealed bottle of fizzy soda sitting beside it. Nothing appears to be changing in either container—yet at the molecular borderline, water molecules melt off the ice at the exact speed liquid molecules freeze onto it, while pressurised CO₂ gas dives into the soda at the exact speed dissolved CO₂ bubbles out. Pop the soda cap once, and that hidden tug-of-war collapses into a loud fizz!

Equilibrium can exist in physical phase transformations and dissolution processes when opposing processes proceed simultaneously at equal rates inside a closed system at constant temperature.

1. Types of Physical Equilibria

Solid ⇌ Liquid (H₂O(s) ⇌ H₂O(l)): In an insulated thermos at 273 K and 1 atm (1.013 bar), mass of ice and water remain constant because Rate of melting = Rate of freezing. The temperature at which both phases coexist at 1 atm is the normal melting/freezing point.
Liquid ⇌ Vapour (H₂O(l) ⇌ H₂O(vap)): In a closed vessel connected to a mercury manometer, Rate of evaporation = Rate of condensation. The constant pressure exerted by vapour at a given temperature is equilibrium vapour pressure. Higher vapour pressure ⇒ more volatile liquid ⇒ lower boiling point (Ether > Acetone > Ethanol > Water). Water boils at 100 °C at 1.013 bar (normal boiling point; decreases at high altitudes).
Solid ⇌ Vapour (Sublimation): Solid iodine in a closed flask fills with violet vapour until colour intensity becomes steady: I₂(s) ⇌ I₂(vap). Other examples: Camphor, NH₄Cl(s).
Dissolution of Solids in Liquids: In a saturated solution, Rate of dissolution = Rate of crystallisation (Sugar(s) ⇌ Sugar(aq))—proved dynamic using radioactive sugar!
Dissolution of Gases in Liquids (Henry's Law): For CO₂(g) ⇌ CO₂(aq), Henry's Law states that the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent (m ∝ p). Gas solubility decreases with increasing temperature.

Physical Equilibrium Process	Example	Constant Quantity at Fixed T (Table 6.1)
Liquid ⇌ Vapour	H₂O(l) ⇌ H₂O(g)	p_H2O (Vapour pressure) is constant
Solid ⇌ Liquid	H₂O(s) ⇌ H₂O(l)	Melting point is fixed at constant pressure
Solute(s) ⇌ Solute(solution)	Sugar(s) ⇌ Sugar(aq)	Concentration (solubility) of solute is constant
Gas(g) ⇌ Gas(aq)	CO₂(g) ⇌ CO₂(aq)	[gas(aq)] / [gas(g)] is constant (Henry's Law)

2. General Characteristics of Physical Equilibria

(i) Possible ONLY in a closed system at a given temperature; (ii) Opposing rates are equal (dynamic, not static); (iii) Measurable properties remain constant; (iv) Characterised by a constant value of one intensive parameter at fixed T.

Equilibrium is NEVER reached in an open container (e.g., an open watch glass of acetone evaporates completely because vapour escapes into the room). Also, unlike most solid solutes, gas solubility in water decreases when temperature increases—warm soda turns flat faster!

Topic 6.2 – 6.5: Dynamic Chemical Equilibrium, Law of Mass Action (Kc), Kp = Kc(RT)^Δng & Heterogeneous Equilibria

Stand at the Kochi Metro platform during rush hour: 15 commuters step inside the coach every 30 seconds while 15 step out onto the platform. To a satellite camera, the number of passengers inside the train never changes—even though people keep flowing both ways! Chemical reactions behave identically, and Fritz Haber proved it by mixing ordinary hydrogen (H₂) and heavy deuterium (D₂) in ammonia synthesis at equilibrium—finding scrambled molecules like NH₂D and NHD₂ formed continuously!

In a reversible reaction A + B ⇌ C + D, initially forward rate is high while reverse rate is zero. As reactants deplete and products build up, Rate_forward = Rate_reverse ≠ 0, and concentrations become constant. Equilibrium can be reached from either side (pure reactants or pure products, e.g., H₂ + I₂ ⇌ 2HI).

1. Law of Chemical Equilibrium (Guldberg & Waage, 1864)

For a A + b B ⇌ c C + d D: K_c = [C]^c [D]^d / ([A]^a [B]^b)
• Reverse reaction (cC + dD ⇌ aA + bB): K'_c = 1 / K_c
• Multiply equation by n (naA + nbB ⇌ ncC + ndD): K''_c = (K_c)ⁿ (Halving gives √K_c)

NCERT Problem 6.1: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K: [N₂] = 1.5×10^-2 M, [H₂] = 3.0×10^-2 M, [NH₃] = 1.2×10^-2 M.
K_c = [NH₃]² / ([N₂][H₂]³) = (1.2×10^-2)² / [(1.5×10^-2)(3.0×10^-2)³] = 3.55 × 10².

2. Homogeneous Gaseous Equilibria: K_p and K_c

Using ideal gas equation p = (n/V)RT = [gas]RT (with R = 0.0831 bar·L mol^-1 K^-1):

K_p = K_c (RT)^Δn_g where Δn_g = (moles of gaseous products) − (moles of gaseous reactants)

If Δn_g = 0 (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)) ⇒ K_p = K_c (no units).
If Δn_g < 0 (e.g., N₂ + 3H₂ ⇌ 2NH₃, Δn_g = −2) ⇒ K_p < K_c (at T > 12 K).
If Δn_g > 0 (e.g., PCl₅ ⇌ PCl₃ + Cl₂, Δn_g = +1) ⇒ K_p > K_c.

NCERT Problem 6.8 & Exercise 6.5:
• For N₂O₄(g) ⇌ 2NO₂(g) at 400 K, K_p = 85.87. Here Δn_g = 2 − 1 = +1 ⇒ K_c = K_p / (RT)¹ = 85.87 / (0.0831 × 400) = 2.58 ≈ 2.6.
• For 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) at 500 K, K_p = 1.8 × 10^-2. Here Δn_g = (2 + 1) − 2 = +1 ⇒ K_c = (1.8 × 10^-2) / (0.0831 × 500) = 4.33 × 10^-4.

3. Heterogeneous Equilibria (Multiple Phases)

The molar concentration of any pure solid (s) or pure liquid (l) stays constant regardless of amount taken, so it is treated as unity and omitted from K_c and K_p (though some solid/liquid MUST be present for equilibrium to exist!):

CaCO₃(s) ⇌ CaO(s) + CO₂(g): K'_c = [CO₂(g)], K_p = p_CO2, Δn_g = 1 − 0 = +1.
Ni(s) + 4CO(g) ⇌ Ni(CO)₄(g): K_c = [Ni(CO)₄] / [CO]⁴.

Two giant exam traps! (1) In K_p = K_c(RT)^Δn_g, count ONLY gaseous species in Δn_g—never count solids (s) or pure liquids (l)! (2) Use R = 0.0831 bar·L mol^-1 K^-1 when pressure is in bar (0.0821 for atm), NEVER 8.314!

Topic 6.6 – 6.7: Reaction Quotient (Qc vs Kc), Extent of Reaction, ICE Tables & ΔG° = −RT ln K

Imagine checking your score at the 15th over of a T20 chase: your current run rate (Q_c) tells you immediately whether you are behind the target equilibrium rate (K_c) and must attack forward (Q_c < K_c), cruising ahead and need to ease back (Q_c > K_c), or sitting dead-even (Q_c = K_c). Chemists use the exact same live speedometer—called the Reaction Quotient (Q_c)—to forecast which way any chemical mixture will shift!

Key Facts about K: K depends ONLY on temperature and reaction stoichiometry. It is completely independent of initial concentrations, pressure changes, or catalysts.

1. Predicting Extent of Reaction from K_c (Fig. 6.6)

K_c > 10³: Products heavily predominate; reaction proceeds nearly to completion (e.g., H₂ + Cl₂ ⇌ 2HCl has K_c = 4.0×10³¹ at 300 K).
K_c < 10^-3: Reactants heavily predominate; barely proceeds (e.g., N₂ + O₂ ⇌ 2NO has K_c = 4.8×10^-31 at 298 K).
10^-3 ≤ K_c ≤ 10³: Appreciable concentrations of BOTH reactants and products coexist (H₂ + I₂ ⇌ 2HI, K_c = 57.0 at 700 K; N₂O₄ ⇌ 2NO₂, K_c = 4.64×10^-3 at 25 °C).

2. Predicting Direction using Reaction Quotient (Q_c or Q_p)

Defined identically to K_c using concentrations at any arbitrary time t: Q_c = [C]_t^c [D]_t^d / ([A]_t^a [B]_t^b).

• Q_c < K_c ⇒ Net reaction moves FORWARD (Left → Right, Reactants → Products)
• Q_c = K_c ⇒ System is at EQUILIBRIUM (No net shift)
• Q_c > K_c ⇒ Net reaction moves REVERSE (Right → Left, Products → Reactants)

NCERT Problem 6.7: For 2A ⇌ B + C, K_c = 2 × 10^-3. At a given moment, [A] = [B] = [C] = 3 × 10^-4 M.
Q_c = ([B][C]) / [A]² = (3×10^-4)(3×10^-4) / (3×10^-4)² = 1.0. Since Q_c (1) > K_c (2×10^-3), the reaction proceeds in the reverse direction!

NCERT Problem 6.9 (ICE Table): 3.00 mol PCl₅ in 1 L vessel at 380 K (K_c = 1.80).
PCl₅ (3 − x) ⇌ PCl₃ (x) + Cl₂ (x) ⇒ x²/(3 − x) = 1.80 ⇒ x² + 1.8x − 5.4 = 0 ⇒ x = 1.59 M.
At equilibrium: [PCl₃] = [Cl₂] = 1.59 M, [PCl₅] = 3.00 − 1.59 = 1.41 M.

3. Relationship Between K, Q & Gibbs Energy (ΔG)

ΔG = ΔG° + RT ln Q | At equilibrium (ΔG = 0, Q = K): ΔG° = −RT ln K = −2.303 RT log K ⇔ K = e^−ΔG°/RT

• ΔG° < 0 ⇒ K > 1 (spontaneous forward) | • ΔG° > 0 ⇒ K < 1 (non-spontaneous forward).

NCERT Problem 6.10: Glycolysis step ΔG° = +13.8 kJ mol^-1 at 298 K.
ln K_c = −13800 / (8.314 × 298) = −5.569 ⇒ K_c = e^-5.569 = 3.81 × 10^-3.

A huge K_c does NOT mean a fast reaction! Hydrogen and oxygen have K_c = 2.4 × 10⁴⁷ at 500 K, yet sit unreactive at room temperature without a spark because equilibrium constants say nothing about reaction rate (activation energy). Also, always divide moles by vessel volume V to get molarity BEFORE using ICE tables!

Topic 6.8: Le Chatelier's Principle & Industrial Applications (Haber & Contact Processes)

Imagine standing inside a city bus when the driver suddenly slams the brakes—your body automatically leans forward to oppose the sudden stop! In 1884, French chemist Henri Le Chatelier discovered that every physical and chemical equilibrium behaves just as stubbornly: disturb it by squeezing pressure, adding chemicals, or changing heat, and it immediately shifts in the exact direction that counteracts your disturbance!

Le Chatelier's Principle: If a system at equilibrium is subjected to a change in any of the factors determining equilibrium conditions (concentration, pressure, temperature), the system shifts in such a way as to reduce or counteract the effect of the change.

Factor Changed	How Equilibrium Shifts	Does K_c / K_p Change?
1. Concentration	Adding reactant (or removing product) ⇒ Forward. Adding product (or removing reactant) ⇒ Reverse. (Demo: Fe³⁺ + SCN^- ⇌ [Fe(SCN)]²⁺ deep red)	NO
2. Pressure (↑ P / ↓ V)	Shifts toward fewer moles of gas! (e.g., CO + 3H₂ ⇌ CH₄ + H₂O: 4→2 mol shifts forward). If Δn_g = 0 ⇒ No shift.	NO
3. Temperature	• Exothermic (ΔH < 0): ↑T shifts Reverse (K ↓) • Endothermic (ΔH > 0): ↑T shifts Forward (K ↑)	YES! (ONLY T changes K!)
4. Inert Gas (e.g., Ar)	• At Constant Volume: NO shift (partial pressures unchanged) • At Constant Pressure: Shifts toward more gas moles	NO
5. Catalyst	Lowers activation energy E_a for BOTH forward & reverse equally ⇒ equilibrium reached faster, composition unchanged.	NO

Classic NCERT Experiments & Industrial Synthesis

NO₂–N₂O₄ Temperature Demo: 2NO₂(g, brown) ⇌ N₂O₄(g, colourless); ΔH = −57.2 kJ mol^-1. Cooling in ice favours N₂O₄ (brown fades); warming in hot water (363 K) favours NO₂ (deep brown!).
Cobalt Complex Demo: [Co(H₂O)₆]³⁺(pink) + 4Cl^- ⇌ [CoCl₄]^2-(blue) + 6H₂O (endothermic; cooling shifts to pink).
Haber Process for Ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92.38 kJ mol^-1 (Δn_g = −2). Exothermic & fewer gas moles on right ⇒ high yield needs low T, high P, continuous liquefaction/removal of NH₃. Too-low T slows rate, so optimum industrial conditions are ~500 °C, 200 atm, Iron catalyst!
Contact Process for H₂SO₄: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g); K_c = 1.7 × 10²⁶. Huge K_c yet sluggish speed ⇒ Pt or V₂O₅ catalyst accelerates reaction!

Top Board Exam Trap: ONLY Temperature changes K_c or K_p! Adding reactants, compressing volume, adding inert gas, or adding a catalyst NEVER changes K! Also, adding an inert gas at constant volume causes ZERO shift because partial pressures of reacting gases don't change!

Topic 6.9 – 6.10: Ionic Equilibrium & Acid–Base Theories: Arrhenius, Brönsted–Lowry Conjugates & Lewis

Drop table salt (NaCl) into pure water and a light bulb glows brightly; drop sugar or weak vinegar (CH₃COOH) in water and it stays dim or flickers faintly. Why? Water's astonishing dielectric constant of 80 slashes electrostatic pull between Na⁺ and Cl^- ions 80-fold so they swim freely (strong electrolyte, ~100% ionized), while vinegar barely ionizes (<5%), creating a busy ionic equilibrium. Across a century, chemists widened the definition of "acid" and "base" through three classic lenses—water ions, proton hand-offs, and electron-pair sharing!

Michael Faraday classified aqueous solutes into non-electrolytes (sugar, urea), strong electrolytes (~100% ionized: HCl, NaOH, NaCl), and weak electrolytes (<5% ionized: CH₃COOH, NH₄OH, establishing ionic equilibrium between ions and unionized molecules).

Three Definitions of Acids and Bases

Theory	Acid Definition	Base Definition	Key Examples & Scope
1. Arrhenius (1884)	Produces H⁺(aq) / H₃O⁺ in water	Produces OH^-(aq) in water	HCl, HNO₃; NaOH, KOH. Limitation: Aqueous only; fails for NH₃ (no OH group!).
2. Brönsted–Lowry (1923)	Proton (H⁺) Donor	Proton (H⁺) Acceptor	Explains NH₃ + H₂O ⇌ NH₄⁺ + OH^- and dual/amphoteric role of H₂O & HCO₃^-.
3. G. N. Lewis (1923)	Electron-Pair Acceptor	Electron-Pair Donor	Explains proton-free acids like BF₃, AlCl₃, BCl₃, Mg²⁺, Co³⁺; bases: :NH₃, H₂O, F^-, OH^-.

Conjugate Acid–Base Pairs (Brönsted–Lowry)

A pair differing by only ONE proton (H⁺) is a conjugate acid–base pair:

Conjugate Acid = Species + H⁺ | Conjugate Base = Species − H⁺
Rule: Strong Acid ⇔ Weak Conjugate Base (e.g., HCl → Cl^-) | Weak Acid ⇔ Strong Conjugate Base (e.g., CH₃COOH → CH₃COO^-)

NCERT Problem 6.14 (Amphoteric Species):
• H₂O: Conjugate Acid = H₃O⁺ | Conjugate Base = OH^-
• HCO₃^-: Conjugate Acid = H₂CO₃ | Conjugate Base = CO₃^2-
• HSO₄^-: Conjugate Acid = H₂SO₄ | Conjugate Base = SO₄^2-
• NH₃: Conjugate Acid = NH₄⁺ | Conjugate Base = NH₂^-
NCERT Problem 6.15 (Lewis): HO^- & F^- donate lone pairs (Lewis bases); H⁺ & BCl₃ accept lone pairs (Lewis acids).

Two classic traps! (1) Always add or remove STRICTLY ONE H⁺ for conjugate pairs—the conjugate base of H₂SO₄ is HSO₄^- (NOT SO₄^2-!). (2) Be careful with NH₄⁺: it IS a Brönsted acid (donates H⁺), but it is NOT a Lewis acid because nitrogen already has a full octet with no empty orbital! Meanwhile BF₃ and AlCl₃ ARE Lewis acids despite having zero protons.

Topic 6.11.1 – 6.11.7: Ionization of Water (Kw), pH Scale, Ostwald's Dilution Law & Ka × Kb = Kw

Even ultra-pure distilled water isn't 100% neutral molecules—about 2 in every billion water molecules collide and hand off a proton: H₂O + H₂O ⇌ H₃O⁺ + OH^-! Because hydrogen ion concentrations span from 1 M battery acid down to 10^-14 M drain cleaner, Danish chemist Søren Sørensen invented the neat pH scale so a billion-fold change fits tidily between 0 and 14!

1. Ionic Product of Water (K_w) & the pH Scale

K_w = [H₃O⁺][OH^-] = 1.0 × 10^-14 M² (at 298 K; increases with T)
pH = −log [H⁺] | pOH = −log [OH^-] | pH + pOH = pK_w = 14 (at 298 K)

At 298 K: Acidic (pH < 7), Neutral (pH = 7, [H⁺] = [OH^-] = 10^-7 M), Basic (pH > 7). Note: 1 unit change in pH = 10× change in [H⁺]. Table 6.5 examples: Gastric juice (~1.2), Lemon juice (~2.2), Vinegar (~3.0), Black coffee (5.0), Milk (6.8), Human blood (7.4), Milk of magnesia (10.0), 0.1 M NaOH (13).

NCERT Problem 6.17 (Classic Dilute Acid Trap): Calculate pH of 1.0 × 10^-8 M HCl.
Must include [H⁺] from H₂O! Let [OH^-] = x from water ⇒ [H₃O⁺] = 10^-8 + x.
(10^-8 + x)(x) = 10^-14 ⇒ x² + 10^-8x − 10^-14 = 0 ⇒ x = 9.5 × 10^-8 M ⇒ [H⁺] = 1.05 × 10^-7 M ⇒ pH = 6.98!

2. Weak Acids, Weak Bases & Ostwald's Dilution Law

For a weak acid HX (conc. c, degree of ionization α): K_a = [H⁺][X^-] / [HX] = cα² / (1 − α). When α ≪ 1 (α < 0.05):

Ostwald's Dilution Law: K_a ≈ cα² ⇒ α = √(K_a / c) | [H⁺] = cα = √(K_a · c)
(For weak base: α = √(K_b / c), [OH^-] = √(K_b · c)) — Diluting increases α!

NCERT Problem 6.20: 0.08 M HOCl (K_a = 2.5 × 10^-5).
[H⁺] = √(K_a · c) = √(2.5×10^-5 × 0.08) = √(2.0×10^-6) = 1.41 × 10^-3 M ⇒ pH = 2.85; % ionization = (1.41×10^-3/0.08)×100 = 1.76%.

3. Relation Between K_a and K_b, Polyprotic Acids & Acid Strength

For any conjugate acid–base pair: K_a × K_b = K_w = 1.0 × 10^-14 | pK_a + pK_b = 14 (at 298 K)

Polyprotic Acids (H₂SO₄, H₂CO₃, H₃PO₄): Always K_a1 >> K_a2 >> K_a3 (harder to remove +proton from a −ve anion!).
Acid Strength Trends: Down a group (bond strength dominates): HF ≪ HCl ≪ HBr ≪ HI (H₂O < H₂S). Across a period (electronegativity/polarity dominates): CH₄ < NH₃ < H₂O < HF.

NEVER answer pH = 8 for 10^-8 M HCl! An acid can NEVER make neutral water alkaline—always add water's H⁺ giving pH = 6.98! Also remember neutral water at 310 K (K_w = 2.7×10^-14) has pH ≈ 6.78, yet it is STILL neutral because [H⁺] = [OH^-]!

Topic 6.11.8 – 6.12: Common Ion Effect, Salt Hydrolysis & Buffer Solutions (Henderson–Hasselbalch)

Every second, your human blood maintains a razor-sharp pH of 7.4 whether you drink sour lime juice or take alkaline antacids—even though a single drop of acid would crash pure water's pH below 3! How does blood pull off this miracle? It uses a chemical shock absorber called a Buffer Solution powered by the Common Ion Effect!

1. Common Ion Effect in Weak Electrolytes

Common Ion Effect is the suppression in the degree of dissociation of a weak electrolyte upon adding a strong electrolyte that furnishes an ion already present in the equilibrium (Le Chatelier's Principle). E.g., adding CH₃COONa (Ac^-) to CH₃COOH shifts HAc ⇌ H⁺ + Ac^- left, suppressing ionization.

2. Hydrolysis of Salts and pH of Their Solutions

Cations of strong bases (Na⁺, K⁺) and anions of strong acids (Cl^-, Br^-, NO₃^-) only hydrate and do NOT hydrolyse. Salts containing weak-acid anions or weak-base cations react with water:

Salt Type	Examples	Which Ion Hydrolyses?	pH & Formula
Strong Acid + Strong Base	NaCl, KBr, NaNO₃	Neither (No hydrolysis)	Neutral (pH = 7)
Weak Acid + Strong Base	CH₃COONa, NaCN, KF, NaNO₂	Anion: A^- + H₂O ⇌ HA + OH^-	Basic (pH > 7)
Strong Acid + Weak Base	NH₄Cl, NH₄NO₃	Cation: BH⁺ + H₂O ⇌ B + H₃O⁺	Acidic (pH < 7)
Weak Acid + Weak Base	CH₃COONH₄	Both cation & anion	pH = 7 + ½(pK_a − pK_b) (Indep. of c!)

NCERT Problem 6.25: For CH₃COONH₄, pK_a(CH₃COOH) = 4.76 and pK_b(NH₄OH) = 4.75.
pH = 7 + ½(4.76 − 4.75) = 7 + 0.005 = 7.005 (independent of concentration!).

3. Buffer Solutions & Henderson–Hasselbalch Equation

A Buffer Solution resists change in pH upon dilution or addition of small amounts of strong acid or alkali.

Acidic Buffer (Weak Acid + its Salt with Strong Base, e.g., CH₃COOH + CH₃COONa, pH ≈ 4.76):
pH = pK_a + log([Salt] / [Acid])

Basic Buffer (Weak Base + its Salt with Strong Acid, e.g., NH₄OH + NH₄Cl, pH ≈ 9.25):
pOH = pK_b + log([Salt] / [Base]) ⇒ pH = 14 − pOH = pK_a(BH⁺) + log([Base] / [Salt])

NCERT Problem 6.22: Calculate pH of a buffer containing 0.20 M NH₄Cl and 0.10 M NH₃ (pK_b = 4.75).
pOH = 4.75 + log(0.20 / 0.10) = 4.75 + 0.301 = 5.05 ⇒ pH = 14 − 5.05 = 8.95.
Note: Diluting a buffer leaves its pH unchanged because the ratio [Salt]/[Acid] inside the log stays identical!

Watch TWO classic Henderson–Hasselbalch errors! (1) It is PLUS log([Salt] / [Acid]) with Salt on TOP! (2) For a basic buffer, pK_b + log([Salt]/[Base]) gives pOH, NOT pH—remember to subtract from 14! Also, maximum buffer capacity occurs when [Salt] = [Acid] (where pH = pK_a).

Topic 6.13: Solubility Equilibria of Sparingly Soluble Salts: Ksp & Common Ion Precipitation

Before taking a stomach X-ray, doctors often ask patients to drink a thick white milkshake of Barium Sulphate (BaSO₄)—even though free Ba²⁺ ions are deadly poison! How is that safe? Because BaSO₄ is a sparingly soluble salt with an ultra-tiny solubility product K_sp = 1.1 × 10^-10, releasing virtually harmless traces of Ba²⁺. Meanwhile, bubbling HCl gas through saturated NaCl brine forces pure white NaCl crystals to snow down instantly via the Common Ion Effect!

A salt dissolves only when ion solvation enthalpy exceeds lattice enthalpy. NCERT classifies salts as: Soluble (>0.1 M), Slightly soluble (0.01–0.1 M), and Sparingly soluble (<0.01 M, e.g., BaSO₄, AgCl, PbCl₂).

1. Solubility Product Constant (K_sp) & Molar Solubility (S)

For a general sparingly soluble salt M_xX_y(s) ⇌ x M^p+(aq) + y X^q-(aq) with molar solubility S mol L^-1:

K_sp = [M^p+]^x [X^q-]^y = (xS)^x (yS)^y = x^x · y^y · S^(x+y) | S = (K_sp / (x^x y^y))^1/(x+y)

Salt Type	NCERT Examples	Dissociation	K_sp in terms of S
AB (1:1)	BaSO₄, AgCl, AgBr, AgCN	M⁺ + X^-	K_sp = S² ⇒ S = √K_sp
AB₂ / A₂B (1:2)	Ni(OH)₂, PbCl₂, Ag₂CrO₄	M²⁺ + 2X^-	K_sp = (S)(2S)² = 4S³
AB₃ (1:3)	Fe(OH)₃, Al(OH)₃	M³⁺ + 3OH^-	K_sp = (S)(3S)³ = 27S⁴
A₂B₃ (2:3)	A₂X₃ (Prob 6.26)	2A³⁺ + 3X^2-	K_sp = (2S)²(3S)³ = 108S⁵
A₃B₄ (3:4)	Zr₃(PO₄)₄	3Zr⁴⁺ + 4PO₄^3-	K_sp = (3S)³(4S)⁴ = 6912 S⁷

2. Predicting Precipitation (Q_sp vs K_sp) & Common Ion Effect

Q_sp < K_sp: Unsaturated solution — NO precipitate; more solid can dissolve.
Q_sp = K_sp: Saturated dynamic equilibrium.
Q_sp > K_sp: Supersaturated — Precipitation occurs until Q_sp drops back to K_sp!
Common Ion Effect on Solubility: Adding a common ion raises Q_sp > K_sp, causing precipitation and slashing solubility S! Used to purify NaCl by passing HCl(g), and in qualitative group precipitation of sulphides/hydroxides. Also, salts of weak acids (e.g., phosphates, AgC₆H₅COO) dissolve more in acidic pH because H⁺ protonates the anion.

NCERT Problem 6.26: K_sp of A₂X₃ is 1.1 × 10^-23. K_sp = 108 S⁵ = 1.1×10^-23 ⇒ S⁵ ≈ 10^-25 ⇒ S = 1.0 × 10^-5 mol L^-1.
NCERT Problem 6.28 (Common Ion): Molar solubility of Ni(OH)₂ (K_sp = 2.0 × 10^-15) in 0.10 M NaOH:
[OH^-] = 0.10 + 2S ≈ 0.10 M ⇒ K_sp = S (0.10)² = 2.0×10^-15 ⇒ S = 2.0 × 10^-13 M (slashed by over 10 million times vs pure water's 7.94×10^-6 M!).

NEVER compare K_sp numbers directly unless both salts have the SAME formula type! For instance, AgBr (K_sp = 5.0×10^-13) has S = √(5×10^-13) = 7.1×10^-7 M, whereas Ag₂CrO₄ (K_sp = 1.1×10^-12) has S = (1.1×10^-12 / 4)^1/3 = 6.5×10^-5 M! Also, when mixing equal volumes of two solutions, ALWAYS halve every concentration before calculating Q_sp!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 4 Marks): State Le Chatelier's Principle. Apply it to the industrial synthesis of ammonia by Haber's process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = −92.38 kJ mol^-1 with respect to (a) Pressure, (b) Temperature, (c) Catalyst, and (d) Removal of NH₃.

Statement: When a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts in a direction that reduces or counteracts the effect of that change.

(a) Pressure: 4 moles of gaseous reactants give 2 moles of gaseous product (Δn_g = −2). High pressure (~200 atm) shifts equilibrium forward, increasing NH₃ yield.
(b) Temperature: Forward reaction is exothermic (ΔH < 0), so low temperature favours yield. Because very low T slows rate excessively, an optimum temperature of ~500 °C (773 K) is chosen.
(c) Catalyst: Finely divided Iron (Fe) speeds up both forward and reverse rates equally without altering K_c or equilibrium yield.
(d) Removal of NH₃: Continuously liquefying and removing NH₃ keeps Q_c < K_c, driving the forward reaction.

Q2 (Kerala HSE SAY / Model (Repeated) • 2 Marks): Write the relationship between K_p and K_c. For which of the following equilibria is K_p = K_c? (i) N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ii) H₂(g) + I₂(g) ⇌ 2HI(g). Also write K_c for CaCO₃(s) ⇌ CaO(s) + CO₂(g).

Relation: K_p = K_c (RT)^Δn_g, where Δn_g = (moles of gaseous products) − (moles of gaseous reactants).

K_p = K_c for (ii) H₂(g) + I₂(g) ⇌ 2HI(g) because Δn_g = 2 − 2 = 0.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), concentrations of pure solids are constant (unity), so K_c = [CO₂(g)].

Q3 (Kerala HSE March (Standard Numerical) • 3 Marks): For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl₂(g), the value of K_p is 1.8 × 10^-2 bar at 500 K. Calculate K_c for this reaction at 500 K. [R = 0.0831 bar·L mol^-1 K^-1]

Given: K_p = 1.8 × 10^-2 bar, T = 500 K, R = 0.0831 bar·L mol^-1 K^-1.
Δn_g = n_p − n_r = (2 + 1) − 2 = +1.
Using K_p = K_c (RT)^Δn_g ⇒ K_c = K_p / (RT)¹
K_c = (1.8 × 10^-2) / (0.0831 × 500) = 0.018 / 41.55 = 4.33 × 10^-4 mol L^-1.

Q4 (Kerala HSE March (Repeated) • 2 Marks): What is meant by a conjugate acid–base pair? Write the conjugate acid and conjugate base of (i) H₂O and (ii) HCO₃^-.

Conjugate Acid–Base Pair: An acid–base pair that differs from each other by only one proton (H⁺).

H₂O: Conjugate acid = H₃O⁺; Conjugate base = OH^-.
HCO₃^-: Conjugate acid = H₂CO₃; Conjugate base = CO₃^2-.

Q5 (Kerala HSE SAY • 2 Marks): Distinguish between Reaction Quotient (Q_c) and Equilibrium Constant (K_c). How does Q_c predict the direction of a chemical reaction?

K_c is calculated using equilibrium concentrations (constant at fixed T), whereas Q_c uses concentrations at any arbitrary time t.

If Q_c < K_c: Net reaction proceeds in the forward direction (reactants → products).
If Q_c > K_c: Net reaction proceeds in the reverse direction (products → reactants).
If Q_c = K_c: Reaction is at equilibrium.

Q6 (Kerala HSE March (Classic Trap) • 2 Marks): Calculate the pH of a 1.0 × 10^-8 M solution of HCl at 298 K. Why is its pH not equal to 8?

Aqueous HCl cannot have pH 8 (basic!) because at extreme dilution, H₃O⁺ from water's auto-ionization (2H₂O ⇌ H₃O⁺ + OH^-) cannot be neglected.

Let [OH^-] = x from water ⇒ [H₃O⁺] = 10^-8 + x.
K_w = (10^-8 + x)(x) = 10^-14 ⇒ x² + 10^-8x − 10^-14 = 0 ⇒ x = 9.5 × 10^-8 M.
[H₃O⁺] = 1.05 × 10^-7 M ⇒ pH = −log(1.05 × 10^-7) = 6.98.

Q7 (Kerala HSE Model • 2 Marks): Define Lewis acids and Lewis bases. Classify the following into Lewis acids and Lewis bases: H₂O, BF₃, H⁺, OH^-.

Lewis Acid: An electron-pair acceptor. | Lewis Base: An electron-pair donor.

Lewis Acids: BF₃ (incomplete octet), H⁺ (vacant 1s orbital).
Lewis Bases: H₂O, OH^- (possess lone pairs on oxygen).

Q8 (Kerala HSE March • 3 Marks): What are buffer solutions? Give one example each for an acidic buffer and a basic buffer. Write the Henderson–Hasselbalch equation for an acidic buffer and explain why dilution does not alter its pH.

Buffer Solution: A solution that resists change in pH on dilution or with the addition of small amounts of strong acid or alkali.

Acidic Buffer: Equimolar mixture of CH₃COOH + CH₃COONa (pH ≈ 4.76).
Basic Buffer: Mixture of NH₄OH + NH₄Cl (pH ≈ 9.25).
Henderson–Hasselbalch Equation: pH = pK_a + log([Salt] / [Acid]).
Effect of Dilution: Both [Salt] and [Acid] decrease by the same factor, so their ratio [Salt]/[Acid] inside the logarithm stays unchanged, leaving pH constant.

Q9 (Kerala HSE SAY • 2 Marks): Explain the effect of (a) adding an inert gas at constant volume, and (b) adding a catalyst on a chemical equilibrium.

(a) Inert gas at constant volume: No effect on equilibrium, because partial pressures and molar concentrations of reactants and products remain unchanged.

(b) Catalyst: Lowers activation energy for both forward and reverse reactions equally; equilibrium is achieved faster, without changing K_c or equilibrium composition.

Q10 (Kerala HSE Improvement • 1 Marks): Explain why pure water is neutral even when K_w changes with temperature, and write the mathematical relation between K_a and K_b for a conjugate acid–base pair at 298 K.

In pure water, dissociation always produces equal concentrations of H₃O⁺ and OH^- ions ([H₃O⁺] = [OH^-]), so it remains strictly neutral at any temperature.

Relation: K_a × K_b = K_w = 1.0 × 10^-14 (or pK_a + pK_b = 14) at 298 K.

Q11 (Kerala HSE March • 2 Marks): Predict whether aqueous solutions of the following salts are neutral, acidic, or basic, giving reasons: (i) NaCl, (ii) CH₃COONa, (iii) NH₄Cl.

(i) NaCl: Neutral (pH = 7) — derived from strong acid (HCl) and strong base (NaOH); neither Na⁺ nor Cl^- hydrolyses.
(ii) CH₃COONa: Basic (pH > 7) — acetate ion hydrolyses (CH₃COO^- + H₂O ⇌ CH₃COOH + OH^-), raising [OH^-].
(iii) NH₄Cl: Acidic (pH < 7) — ammonium ion hydrolyses (NH₄⁺ + H₂O ⇌ NH₄OH + H⁺), raising [H⁺].

Q12 (Kerala HSE March / NCERT Problem 6.28 • 3 Marks): Define Common Ion Effect and Solubility Product (K_sp). Calculate the molar solubility of Ni(OH)₂ in 0.10 M NaOH given K_sp of Ni(OH)₂ = 2.0 × 10^-15.

Common Ion Effect: Shift in equilibrium (suppression of ionization/solubility) caused by adding a substance providing an ionic species already present in the equilibrium.

Solubility Product (K_sp): Product of molar concentrations of ions in a saturated solution of a sparingly soluble salt, each raised to its stoichiometric coefficient.

Calculation: Ni(OH)₂(s) ⇌ Ni²⁺(aq) + 2OH^-(aq).
Let solubility = S ⇒ [Ni²⁺] = S, [OH^-] = 0.10 + 2S ≈ 0.10 M (since K_sp is very small).
K_sp = S × (0.10)² = 2.0 × 10^-15 ⇒ S = 2.0 × 10^-15 / 0.01 = 2.0 × 10^-13 mol L^-1.