Chemical & Ionic Equilibrium

Equilibrium is a heavyweight Advanced topic: Kp/Kc and Le Chatelier for gaseous systems, then the ionic side — pH, buffers, hydrolysis, solubility product and common-ion effects. The quantitative pH and Ksp problems, and the reasoning behind indicators and buffer action, decide marks.

Key results

• Gaseous: Kp = Kc(RT)^Δn_g; reaction quotient Q vs K tells direction (Q<K forward). Le Chatelier: system shifts to oppose change.
• Water/pH: Kw = [H⁺][OH⁻] = 1×10⁻¹⁴ (25°C); pH = −log[H⁺]; pH + pOH = 14.
• Weak acid: Ka = Cα²/(1−α); for small α, [H⁺] = √(KaC). pH = ½(pKa − logC).
• Buffer (Henderson): pH = pKa + log([salt]/[acid]); best buffering at pH = pKa.
• Solubility: Ksp = product of ion concentrations (with powers). Common ion suppresses solubility; precipitation when ionic product > Ksp.

Quantity	Relation
Kp–Kc	Kc(RT)^Δn
Weak-acid [H⁺]	√(KaC)
Buffer pH	pKa + log(salt/acid)
Precipitation	Q_ionic > Ksp

Exam Tricks & Tips

• 🎯 Compare Q with K to predict direction: Q < K shifts forward, Q > K backward, Q = K is equilibrium — no need to solve fully.
• 🎯 A buffer resists pH change and is most effective at pH = pKa (equal salt and acid); choose the acid whose pKa is near the target pH.
• 🎯 Common-ion effect suppresses ionisation/solubility — adding NaCl to a saturated AgCl solution lowers Ag⁺ and precipitates AgCl.
• 🎯 Kp = Kc(RT)^Δn_g: if Δn_g = 0 they are equal; watch units and use R in the pressure units of Kp.
• 🎯 Salt hydrolysis sets pH: salt of strong acid + weak base is acidic; weak acid + strong base is basic — predict pH sign instantly.
• ❌ Common mistake: including pure solids and liquids in the equilibrium expression — their activities are 1 and they are omitted from Kc/Kp/Ksp.

Expected exam pattern

2 questions likely: a Kp/Kc or Le-Chatelier gaseous-equilibrium problem and an ionic-equilibrium pH/buffer/Ksp numerical. Precipitation and common-ion conceptuals appear as multiple-correct.

Quick recap

Q vs K gives direction; Kp = Kc(RT)^Δn. Weak acid [H⁺] = √(KaC); buffer pH = pKa + log(salt/acid). Ksp with ion powers; precipitate when ionic product exceeds Ksp. Omit pure solids/liquids from expressions.

Cover the answer, recall, then check. 12 cards on equilibrium for JEE Advanced.

Q1. Relation between Kp and Kc.
A1. Kp = Kc(RT)^Δn_g, Δn_g = change in moles of gas.

Q2. How does the reaction quotient Q predict direction?
A2. Q < K shifts forward, Q > K shifts backward, Q = K is equilibrium.

Q3. Ionic product of water and pH relation at 25°C.
A3. Kw = [H⁺][OH⁻] = 10⁻¹⁴; pH + pOH = 14.

Q4. [H⁺] of a weak acid of concentration C and constant Ka.
A4. [H⁺] ≈ √(KaC) for small dissociation.

Q5. Henderson–Hasselbalch equation for a buffer.
A5. pH = pKa + log([salt]/[acid]).

Q6. At what pH is an acidic buffer most effective?
A6. pH = pKa (equal salt and acid concentrations).

Q7. Condition for precipitation from a solution.
A7. Ionic product exceeds the solubility product (Q > Ksp).

Q8. Effect of a common ion on solubility.
A8. It suppresses solubility (shifts equilibrium toward the solid).

Q9. Nature (acidic/basic) of a salt of a strong acid and weak base.
A9. Acidic (the cation hydrolyses to give H⁺).

Q10. Which species are omitted from equilibrium expressions?
A10. Pure solids and pure liquids (their activity is 1).

Q11. Ostwald dilution law for a weak electrolyte.
A11. Ka = Cα²/(1 − α); α increases on dilution.

Q12. Le Chatelier response to increased pressure on a gaseous equilibrium.
A12. Shifts toward the side with fewer moles of gas.

Equilibrium governs how far a reaction goes; ionic equilibrium specialises it to acids, bases, salts and sparingly soluble solids. JEE Advanced tests Le Chatelier reasoning, the K_c/K_p relationship, and pH/buffer/solubility calculations that demand careful approximation.

Core concept: at equilibrium forward and reverse rates are equal, giving a constant K; disturbing the system shifts it to partly undo the change (Le Chatelier).

Chemical equilibrium

Beginner — the equilibrium constant

For aA + bB ⇌ cC + dD, K_c = [C]^c[D]^d/([A]^a[B]^b). For gases K_p = K_c(RT)^Δn_g. The reaction quotient Q compares to K: Q < K shifts forward, Q > K shifts backward, Q = K is equilibrium. K depends only on temperature (via ΔG° = −RT ln K); it is unaffected by pressure, concentration or catalyst.

Intermediate — Le Chatelier

Disturb equilibrium and it shifts to oppose: adding reactant shifts forward; increasing pressure shifts toward fewer gas moles; raising temperature shifts in the endothermic direction (favouring the side that absorbs heat). A catalyst speeds both directions equally — it does not shift the position, only shortens the time to reach it. Adding an inert gas at constant volume does nothing; at constant pressure it shifts toward more moles.

Ionic equilibrium

Acids, bases and pH

pH = −log[H⁺]; pH + pOH = 14 at 25 °C. Strong acids/bases ionise fully. Weak acids: K_a = [H⁺][A⁻]/[HA], and for a weak acid of concentration C, [H⁺] ≈ √(K_aC) (valid when ionisation is small). pK_a = −log K_a; lower pK_a = stronger acid. Buffers resist pH change: Henderson–Hasselbalch pH = pK_a + log([salt]/[acid]); a buffer works best when pH ≈ pK_a (equal salt and acid).

Advanced — salt hydrolysis and solubility

Salts of weak acids/bases hydrolyse: e.g. CH₃COONa gives a basic solution, NH₄Cl acidic. Solubility product K_sp = product of ion concentrations (each to its stoichiometric power) for a saturated solution. Precipitation occurs when the ionic product > K_sp. The common-ion effect suppresses solubility (adding a shared ion shifts equilibrium back), used in qualitative analysis to control sulfide/hydroxide precipitation.

Worked example

Calculate the pH of 0.1 M acetic acid (K_a = 1.8×10⁻⁵). Using [H⁺] ≈ √(K_aC) = √(1.8×10⁻⁵ × 0.1) = √(1.8×10⁻⁶) = 1.34×10⁻³ M. So pH = −log(1.34×10⁻³) ≈ 2.87. Check the approximation: ionisation ≈ 1.34×10⁻³/0.1 = 1.3%, small enough to justify neglecting the change in [HA]. Knowing when the √(K_aC) shortcut is valid — and when you must solve the quadratic — is the graded skill.

How JEE Advanced tests this

K_p/K_c interconversion and degree-of-dissociation problems; Le Chatelier shifts (including inert-gas subtleties); pH of weak acids/bases and buffers; salt hydrolysis and the resulting pH; K_sp precipitation and common-ion effect; and combining buffer action with titration curves.

Exam tricks & shortcuts

• K_p = K_c(RT)^Δn_g; Q vs K tells the shift direction.
• Weak acid [H⁺] ≈ √(K_aC); buffer pH = pK_a + log([salt]/[acid]).
• Common ion suppresses ionisation/solubility.
• Mnemonic: "Q less than K, go forward's the way."

Key formulas

• Equilibrium constant: Kp = Kc(RT)^Δn; compare Q to K for direction.
• Water: Kw = [H⁺][OH⁻] = 10⁻¹⁴; pH = −log[H⁺]; pH + pOH = 14.
• Weak acid: Ka ≈ Cα²; [H⁺] = √(Ka C); pH = ½(pKa − log C).
• Buffer (Henderson): pH = pKa + log([salt]/[acid]).
• Solubility: AB → s = √Ksp; AB₂ → s = (Ksp/4)^(1/3).

Le Chatelier predicts shifts qualitatively; K, Ka, Kw and Ksp quantify equilibria and pH.

Worked Example

Problem: For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), the equilibrium constant is K_p = 0.66 atm at a certain temperature. If the total pressure at equilibrium is 1.0 atm, find the degree of dissociation α of N₂O₄.

Solution:
Start with 1 mole of N₂O₄; let α dissociate. At equilibrium: N₂O₄ = (1 − α), NO₂ = 2α, total moles = 1 + α.

Mole fractions × total pressure P give partial pressures. The standard result for this reaction is:
K_p = (4α²/(1 − α²)) P.

Substitute K_p = 0.66, P = 1.0:
0.66 = 4α²/(1 − α²)
0.66(1 − α²) = 4α²
0.66 − 0.66α² = 4α²
0.66 = 4.66α²
α² = 0.1416 → α = 0.376.

So about 38% of the N₂O₄ dissociates.

Answer: Degree of dissociation α ≈ 0.38 (38%).