Chemical and Ionic Equilibrium

Physical Chemistry Weightage: 3-4 Questions (12-16 Marks) JEE Unit 6

“Welcome back! These notes assume you have already read the NCERT chapters on equilibrium, so you know what K, Q, pH, pK and Ksp mean and can state Le Chatelier's principle. Here we keep what JEE Main actually asks. Equilibrium is one of the highest-weightage physical chapters, with three or four questions a paper: an ICE table with a shifted equilibrium, Kp with a degree of dissociation, a buffer or pH calculation, salt hydrolysis, or a solubility-product precipitation. Learn the constant rules, the ICE patterns, the pH toolkit and the buffer tricks below and most of these become mechanical.”

— SCORECHEM ACADEMIC TEAM

1. K, Q and the Rules of Constants

Fig. 2: Equilibrium constant rules. K depends only on temperature (not on P, V, catalyst or inert gas).

K_p = K_c(RT)^Δn_g, with Δn_g = gaseous products − gaseous reactants (solids and liquids are left out). If Δn_g = 0, K_p = K_c. With R = 0.0821 and T = 400 K, RT = 32.8, so a K_p/K_c ratio near 33 means Δn_g = +1 and near 1/33 (0.03) means Δn_g = −1.
Pure solids and liquids have activity 1. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_p = p(CO₂); for NH₄HS(s) ⇌ NH₃ + H₂S, K_p = p(NH₃) × p(H₂S) = (P/2)².
Direction test: compare Q (same expression as K, evaluated now) with K. Q < K forward, Q > K backward.
Independent of: initial amounts, catalyst, total pressure, volume and inert gas. Depends only on temperature (and on how you write the equation).

⚠️ JEE Trap: The equation decides K. Halving the coefficients gives √K, not K/2, and reversing gives 1/K. Adding equations multiplies the constants. A catalyst speeds both directions equally and never changes K or the composition at equilibrium.

2. ICE Tables: Dissociation and Shifted Equilibria

General dissociation: A_n ⇌ nA starting with 1 mol: moles = (1 − α) and nα, total = 1 + (n − 1)α. Vapour density: D/d = 1 + (n − 1)α. For n = 1 (2HI ⇌ H₂ + I₂) the density does not change, so the formula cannot be used.
K_p in terms of α and P: partial pressure = (moles/total moles) × P. For NH₃ ⇌ ½N₂ + 3/2 H₂: K_p = (3√3/4) α²P/(1 − α²). Worked pattern 1: P = √3 atm and K_p = 9 gives (9/4)α²/(1 − α²) = 9, so α² = 0.8 and α = 0.8^1/2 = (125 × 10⁻²)^−1/2.
For A₂ ⇌ 2A at total pressure P: K_p = 4α²P/(1 − α²). Worked pattern 2 (K from ΔG°): ΔG° = 2(−50.832) − (−100) = −1.664 kJ, so ln K_p = 1664/(8 × 300) = 0.693 and K_p = 2. Then 4α² = 2(1 − α²) at 1 bar gives α² = 1/3 and α = 0.577. The problem says "degree not negligible", so do not drop 1 − α².
Adding a reactant, isomerisation (A ⇌ B): K = 0.375/0.5 = 0.75. After adding 0.1 mol A the total is 0.975 M and each species is a share: [A] = 0.975/(1 + 0.75) = 0.557 M and [B] = 0.418 M.
Adding a reactant, Δn_g = 0 (perfect square): P₂ + Q₂ ⇌ 2PQ with 2 mol each at equilibrium has K = 4/4 = 1. Add 1 mol each of P₂ and Q₂ (start 3, 3, 2) and let x mol react: (2 + 2x)²/(3 − x)² = 1 gives 2 + 2x = 3 − x, so x = 1/3 and all three species become 2.67 mol.
Same trick with a product added: X₂ + Y₂ ⇌ 2Z at 3, 3, 9 mol in 1 L has K = 81/9 = 9. Add 10 mol Z: (19 − x)/(3 + x/2) = 3 (square root of K), so 19 − x = 9 + 1.5x and x = 4. Z at the new equilibrium = 19 − 4 = 15 mol.
Simultaneous equilibria: a common species is counted from both. For A(s) ⇌ X + Y and B(s) ⇌ Z + Y with [X] = t, [Z] = u: [Y] = t + u, K₁ = t(t + u), K₂ = u(t + u), so K₁/K₂ = t/u. Two gas equilibria (N₂ + 3H₂ ⇌ 2NH₃ and N₂ + 2H₂ ⇌ N₂H₄): use a separate unknown for each reaction and track the shared N₂ and H₂ with both.

⚠️ JEE Trap: Moles versus concentration, and the "x" bookkeeping. K_c needs concentrations, so divide by V unless Δn_g = 0 (then volume cancels and moles work directly). After adding a reactant the new starting composition includes the added amount; do not restart from the old equilibrium. Also decide the direction with Q before you set up the algebra.

3. Le Chatelier's Principle

Fig. 1: Le Chatelier's principle. Temperature is the only stress that changes K.

Industrial cases: Haber (N₂ + 3H₂ ⇌ 2NH₃, exothermic, Δn_g = −2): high pressure, moderate T (about 450 °C), Fe with Mo promoter. Contact (2SO₂ + O₂ ⇌ 2SO₃): moderate T, V₂O₅. Birkeland–Eyde (N₂ + O₂ ⇌ 2NO, endothermic, Δn_g = 0): high T; pressure has no effect.
Physical equilibria: the vapour pressure of a liquid depends on the liquid and T only. Ice ⇌ water is favoured by pressure (water is denser), so ice melts below 0 °C at high pressure. Graphite → diamond needs high P and high T.

⚠️ JEE Trap: Inert gas. At constant volume an inert gas changes nothing (partial pressures are unchanged). At constant pressure the volume grows, and the shift is towards more gas moles, just like reducing pressure. Also, a catalyst never increases the yield, only the speed.

4. Thermodynamics of Equilibrium

Fig. 3: Left: van't Hoff plot. Right: free energy along the reaction. ΔG° is the gap between the pure ends, not the depth of the minimum.

ΔG = ΔG° + RT ln Q, and at equilibrium ΔG = 0, so ΔG° = −RT ln K = −2.303 RT log K. K > 1 when ΔG° < 0.
van't Hoff: log K = −ΔH°/(2.303RT) + ΔS°/(2.303R). Slope = −ΔH°/2.303R, intercept = ΔS°/2.303R. Two temperatures: log(K₂/K₁) = (ΔH°/2.303R)(1/T₁ − 1/T₂).
Reading the G–extent graph: the minimum is the equilibrium mixture. ΔG° = G(pure products) − G(pure reactants), so for N₂O₄ ⇌ 2NO₂ with ends at 0 and +5.40 kJ, ΔG° = +5.40 kJ even though the reaction still proceeds partly. Going from pure N₂O₄ to the mixture, ΔG = −0.84 − 0 = −0.84 kJ; from pure 2NO₂, ΔG = −0.84 − 5.40 = −6.24 kJ.

⚠️ JEE Trap: Positive ΔG° does not mean "no reaction". It means K < 1. Some product still forms until G is minimum, and the reverse reaction never goes to completion either. Use ΔG° for K and ΔG for direction.

5. Acids, Bases and Ostwald's Law

Ostwald's dilution law: K = Cα²/(1 − α), so α = √(K/C) = √(KV) when α < 0.1. Dilution raises α; temperature raises it (ionisation is endothermic); a common ion lowers it. Weak electrolytes have K ≤ 10⁻³ or so.
Concepts: Brønsted acid = proton donor; conjugate pair differs by one H⁺; a strong acid has a very weak conjugate base. Lewis acid = electron-pair acceptor (BF₃, AlCl₃, H⁺, Fe²⁺); Lewis base = donor (NH₃, OH⁻, F⁻). Amphiprotic: H₂O, HCO₃⁻, HS⁻, H₂PO₄⁻. Boric acid is a Lewis acid (it accepts OH⁻), not an Arrhenius acid.
Water: K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 25 °C and rises with T. If a question gives K_w = 10⁻¹³, neutral pH = 6.5 and pH + pOH = 13. Pure water is 55.5 M, so its degree of ionisation is 10⁻⁷/55.5 = 1.8 × 10⁻⁹, and the K_a of water is 1.8 × 10⁻¹⁶ (pK_a = 15.74).

6. pH Calculations

Fig. 4: pH toolkit. Check the condition before you choose the formula.

Very dilute strong acid: for 10⁻⁸ M HCl, [H⁺] = 10⁻⁸ + x with x(10⁻⁸ + x) = 10⁻¹⁴, so [H⁺] = 1.05 × 10⁻⁷ and pH = 6.98.
Worked pattern 3 (weak acid): pK_a = 4 and C = 10 mM = 0.01 M, with α negligible: pH = ½(4 + 2) = 3.
Worked pattern 4 (diprotic): 0.1 M H₂X with K_a1 = 2.5 × 10⁻⁸, K_a2 = 1.0 × 10⁻¹³. [H⁺] = √(2.5 × 10⁻⁹) = 5 × 10⁻⁵ = [HX⁻], so [X²⁻] = K_a2 = 1.0 × 10⁻¹³ M = 100 × 10⁻¹⁵.
H₃PO₄, 0.1 M (K₁ = 7.5 × 10⁻³, K₂ = 6.2 × 10⁻⁸, K₃ = 3.6 × 10⁻¹³): K₁ is not small enough here, so solve x²/(0.1 − x) = 7.5 × 10⁻³, x = 0.024 M. Then [HPO₄²⁻] = K₂ = 6.2 × 10⁻⁸ and [PO₄³⁻] = K₃[HPO₄²⁻]/[H⁺] = 9.3 × 10⁻¹⁹ M.
Two weak acids: 0.1 M HCOOH (K_a = 1.8 × 10⁻⁴) and 0.1 M HOCN (3.3 × 10⁻⁴) give [H⁺] = √(1.8 × 10⁻⁵ + 3.3 × 10⁻⁵) = 7.1 × 10⁻³ M.
Mixing strong acid and strong base: find (N₁V₁ − N₂V₂)/(V₁ + V₂). Stoichiometry example: 25 mL of 0.1 M Ba(OH)₂ neutralises 2 × 2.5 mmol = 5 mmol HCl, which is 182.5 mg.

⚠️ JEE Trap: Check α before using √(K_aC). If the answer gives α ≥ 0.1 the approximation fails and you must solve the quadratic. And an acid at 10⁻⁸ M is never basic: water supplies the rest, so pH is just under 7.

7. Salt Hydrolysis

Fig. 5: Salt hydrolysis. The pH of a WA + WB salt and of an amphiprotic anion does not depend on concentration.

Cationic (NH₄Cl) gives pH < 7, anionic (CH₃COONa) gives pH > 7, and a WA + WB salt is near 7 if K_a ≈ K_b (acidic if K_a > K_b).
pH of amphiprotic anions is independent of C: HCO₃⁻ and H₂PO₄⁻ have pH = ½(pK_a1 + pK_a2); HPO₄²⁻ has ½(pK_a2 + pK_a3).
Polyvalent anions: only the first hydrolysis step matters (K_h1 >> K_h2).

⚠️ JEE Trap: Use the acid's K_a for its anion. For sodium acetate, K_h = K_w/K_a(acetic acid). Using K_b of the wrong species is the commonest slip, and the salt of a strong acid with a strong base never hydrolyses.

8. Buffers, Titrations and Indicators

Henderson–Hasselbalch: pH = pK_a + log([salt]/[acid]) for an acidic buffer; pOH = pK_b + log([salt]/[base]) for a basic buffer. Capacity is greatest at [salt] = [acid], where pH = pK_a; the working range is pK_a ± 1, and β = 2.303 ab/(a + b). Dilution alone does not change pH.
Worked pattern 5 (partial neutralisation makes a buffer): which mixture gives pH 9.25 with pK_b(NH₄OH) = 4.75? That needs pOH = pK_b, i.e. equal moles of salt and base. 0.2 M × 0.5 L = 100 mmol NH₄OH plus 0.1 M × 0.5 L = 50 mmol HCl leaves 50 mmol base and forms 50 mmol NH₄Cl, so it works.
Worked pattern 6 (design a buffer): weak base with pK_b = 5.699, pH 9 (pOH = 5), from x mL of 0.02 M HCl and y mL of 0.02 M base to make 100 mL. 5 = 5.699 + log(salt/base) gives salt/base = 1/5. Salt = 0.02x, base left = 0.02(y − x), so y = 6x and x + y = 100: x = 14.3 mL, y = 85.7 mL.

Fig. 6: Titration of 50 mL of 0.1 M acid with 0.1 M NaOH (weak acid K_a = 10⁻⁵). Pick an indicator whose range sits on the steep part.

Titration and indicator choice: strong acid vs strong base: equivalence pH 7 and any indicator works. Weak acid vs strong base: equivalence in the basic region, use phenolphthalein (8.3–10). Strong acid vs weak base: acidic equivalence, use methyl orange (3.1–4.4). Weak acid vs weak base: no sharp change and no simple indicator. At half-equivalence a weak-acid titration has pH = pK_a.
Indicators: pH = pK_In + log([In⁻]/[HIn]), colour range pK_In ± 1.
Mixed-indicator titration: NaOH reacts 100% at both end points. Na₂CO₃ is 50% neutralised at phenolphthalein (to NaHCO₃) and 100% at methyl orange. NaHCO₃ is 0% at phenolphthalein and 100% at methyl orange.

⚠️ JEE Trap: Subtract moles before using Henderson. A base plus HCl becomes a buffer only if some base is left over. If the acid is equal to the base you get the pure salt (a hydrolysis problem), and if the acid is in excess you have a strong-acid problem.

9. Solubility Product

Salt type	Ksp in terms of s	Example
AB	s²	AgCl, BaSO₄
AB₂ or A₂B	4s³	CaF₂, Ag₂CrO₄
AB₃	27s⁴	Fe(OH)₃
A₂B₃	108s⁵	Bi₂S₃, As₂S₃

Precipitation: ionic product Q > K_sp precipitates; Q < K_sp stays dissolved. A common ion lowers solubility, and complex formation (AgCl in NH₃, giving [Ag(NH₃)₂]⁺) raises it.
Simultaneous solubility: two salts sharing an ion, K_sp1 = x(x + y) and K_sp2 = y(x + y), so x/y = K_sp1/K_sp2. If CaSO₄ (K_sp = p) dissolves b M, then [Ba²⁺] = bq/p in the presence of BaSO₄ (K_sp = q).
Worked pattern 7 (selective precipitation with H₂S): X²⁺ and Y²⁺ are both 0.01 M, K_sp(XS) = 10⁻²², K_sp(YS) = 4 × 10⁻¹⁶, saturated H₂S is 0.1 M and K_a1K_a2 = 10⁻²¹. YS starts to precipitate when [S²⁻] = 4 × 10⁻¹⁶/0.01 = 4 × 10⁻¹⁴. Since [S²⁻] = K_a1K_a2[H₂S]/[H⁺]², [H⁺]² = 10⁻²²/(4 × 10⁻¹⁴) = 2.5 × 10⁻⁹, so [H⁺] = 5 × 10⁻⁵ and pH = 4.3, i.e. 4. Lower pH keeps [S²⁻] low, so the less soluble XS precipitates first.

⚠️ JEE Trap: Comparing solubilities from K_sp. K_sp values can be compared directly only for salts of the same type (AB with AB). Otherwise convert to s first. And in a common-ion solution, use the total ion concentration, not just the ion from the salt.

10. How JEE Frames Questions

Numerical (integer answer): α from K_p and P, new equilibrium moles after adding a species, pH of a weak acid or diprotic species, concentration of the second-step anion, precipitation pH.
Buffer questions: find which mixture gives a target pH, or design volumes of acid and base for a given pH. Subtract moles first.
Statements and graphs: ln K or log K against 1/T, G against extent (ΔG° sign and equilibrium), titration curves and indicators.
Conceptual: Le Chatelier with inert gases, catalysts and T; which salt hydrolyses; Lewis versus Brønsted acids.

⚠️ JEE Trap: Units and the 10^−x scale factors. Read whether the blank is "× 10⁻²" or "× 10⁻¹⁵" before you type the number, and remember 10 mM = 0.01 M.

11. Quick Sheet and Checklist

Idea	Rule
K_p and K_c	K_p = K_c(RT)^Δn_g
Direction	Q < K forward, Q > K backward
Free energy	ΔG° = −RT ln K; ΔG = ΔG° + RT ln Q
van't Hoff plot	slope −ΔH°/2.303R, intercept ΔS°/2.303R
Weak acid	pH = ½(pK_a − log C)
Buffer	pH = pK_a + log(salt/acid)
Anionic hydrolysis	pH = ½(pK_w + pK_a + log C)
Amphiprotic anion	pH = ½(pK_a1 + pK_a2)
Precipitation	Q > K_sp

Before the exam, check you can:

Convert between K_p and K_c and combine or reverse equations without slips.
Set up an ICE table for a shifted equilibrium and decide direction with Q.
Solve for α from K_p and total pressure, including the "degree not negligible" case.
Pick the right pH formula for a weak acid, a diprotic acid, a salt or a buffer, and subtract moles before Henderson.
Find the pH at which a sulphide, hydroxide or other salt starts to precipitate.