Chemical Equilibrium and Ionic Equilibrium
Why This Chapter Matters
Equilibrium is a major JEE Chemistry chapter — 8-12 marks. Kc, Kp, Le Chatelier's principle, pH calculations, buffer solutions, and Ksp are all tested every year.
Core Concepts
1. Equilibrium Constant
For: aA + bB ⇌ cC + dD
Kc = [C]^c[D]^d / [A]^a[B]^b
Kp = Kc(RT)^Δn where Δn = (c+d) - (a+b) = change in moles of gas
If Kc > 10³: reaction essentially complete (products favoured)
If Kc < 10⁻³: reaction barely occurs (reactants favoured)
If 10⁻³ < Kc < 10³: significant amounts of both
Reaction quotient Q: Same expression as Kc but at any concentration.
Q < Kc: reaction proceeds forward | Q > Kc: reaction goes backward
2. Le Chatelier's Principle
"System at equilibrium shifts to oppose any change imposed on it."
| Change | Shift |
|---|
|---|---|
| Add reactant | Forward (→) |
|---|---|
| Add product | Backward (←) |
| Increase pressure (gas reaction) | Toward fewer moles of gas |
| Increase temperature | Toward endothermic direction |
| Add catalyst | No shift (only speeds equilibration) |
| Add inert gas at constant V | No shift |
| Add inert gas at constant P | Toward more moles of gas |
3. pH and Acid-Base Equilibria
pH = -log[H⁺]. Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C.
pH + pOH = 14.
Weak acid HA: Ka = [H⁺][A⁻]/[HA]. pH = ½(pKa - log C)
Degree of dissociation: α = √(Ka/C) (for weak acid, α<<1)
Weak base: Kb. pOH = ½(pKb - log C). pH = 14 - pOH.
Ka × Kb = Kw (for conjugate acid-base pair)
4. Buffer Solutions
Resist change in pH on adding small amounts of acid/base.
Henderson-Hasselbalch equation:
pH = pKa + log([Salt]/[Acid]) (acidic buffer: weak acid + its salt)
pOH = pKb + log([Salt]/[Base]) (basic buffer)
Buffer capacity = ability to resist pH change. Maximum at pH = pKa.
5. Solubility Product (Ksp)
For sparingly soluble salt MxNy:
MxNy ⇌ xM^y+ + yN^x-
Ksp = [M^y+]^x [N^x-]^y
If IP (ionic product) < Ksp: unsaturated, more can dissolve
If IP > Ksp: precipitate forms
Common ion effect: adding common ion decreases solubility.
PYQs
2024: Kc = 4×10⁻² for H₂+I₂⇌2HI at 400°C. Initial [H₂]=0.1M, [I₂]=0.1M. Find equilibrium [HI].
Kc = (2x)²/((0.1-x)(0.1-x)) = (2x/(0.1-x))² = 4×10⁻²
2x/(0.1-x) = 0.2. 2x = 0.02 - 0.2x. 2.2x = 0.02. x = 0.009. [HI] = 0.018M.
2023: pH of 0.1M CH₃COOH (Ka = 1.8×10⁻⁵)?
[H⁺] = √(Ka×C) = √(1.8×10⁻⁶) = 1.34×10⁻³ M
pH = -log(1.34×10⁻³) = 2.87
2022: Ksp of AgCl = 1.8×10⁻¹⁰. Solubility in pure water?
AgCl ⇌ Ag⁺ + Cl⁻. Ksp = s² = 1.8×10⁻¹⁰. s = 1.34×10⁻⁵ mol/L.