1. Chapter at a Glance
Many reactions do not go to completion; instead they reach a state of dynamic equilibrium in which the forward and backward reactions occur at the same rate, so concentrations stop changing. This chapter covers chemical equilibrium (Kc, Kp, Le Chatelier's principle) and ionic equilibrium (acids and bases, pH, buffers, hydrolysis and solubility product). It is a numerical-heavy chapter; JEE Main usually asks one or two questions.
- Relation Kp = Kc(RT)Δn and the effect of changing conditions (Le Chatelier).
- pH of strong and weak acids/bases, and of their mixtures.
- Buffer pH using the Henderson equation.
- Solubility product Ksp, solubility and the common ion effect.
2. Chemical Equilibrium
For aA + bB ⇌ cC + dD:
Δn = (moles of gaseous products) − (moles of gaseous reactants). Pure solids and liquids are not written in the expression.
- Reversing a reaction: K → 1/K. Multiplying the equation by n: K → Kⁿ. Adding reactions: multiply their K values.
- Reaction quotient Q has the same form as K but uses current concentrations. Q < K: reaction moves forward; Q > K: backward; Q = K: at equilibrium.
- K depends only on temperature. For an exothermic reaction, K decreases as temperature rises; for an endothermic reaction, K increases.
Le Chatelier's principle
If an equilibrium is disturbed, it shifts in the direction that reduces the disturbance.
| Change | Shift of N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0) |
|---|---|
| Add N₂ or H₂ | Forward |
| Remove NH₃ | Forward |
| Increase pressure (decrease volume) | Forward (fewer gas moles) |
| Increase temperature | Backward (reaction is exothermic); K decreases |
| Add a catalyst | No shift; equilibrium is reached faster |
| Add inert gas at constant volume | No shift |
| Add inert gas at constant pressure | Towards more gas moles (backward here) |
3. Acids, Bases and pH
- Arrhenius: acids give H⁺, bases give OH⁻ in water. Brønsted–Lowry: acids donate protons, bases accept them (conjugate pairs differ by one H⁺). Lewis: acids accept an electron pair, bases donate one.
- Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10−14 at 298 K, so pH + pOH = 14. Kw increases with temperature.
- pH = −log[H⁺]. Neutral water at 298 K has pH 7.
- Weak acid (Ostwald's dilution law): α = √(Ka/C), [H⁺] = √(KaC), pH = ½(pKa − log C).
- For a conjugate pair, Ka × Kb = Kw.
The pH of 10−8 M HCl is not 8. Water itself contributes 10−7 M H⁺, so the total [H⁺] is slightly above 10−7 M and the pH is about 6.98 — still slightly acidic.
4. Buffers and Salt Hydrolysis
A buffer resists changes in pH when a small amount of acid or base is added. Acidic buffer: weak acid + its salt (CH₃COOH + CH₃COONa). Basic buffer: weak base + its salt (NH₄OH + NH₄Cl).
When [salt] = [acid], pH = pKa. Buffer capacity is greatest near this point.
| Salt of | Example | Solution | pH (298 K) |
|---|---|---|---|
| Strong acid + strong base | NaCl | Neutral | 7 |
| Weak acid + strong base | CH₃COONa | Basic | 7 + ½(pKa + log C) |
| Strong acid + weak base | NH₄Cl | Acidic | 7 − ½(pKb + log C) |
| Weak acid + weak base | CH₃COONH₄ | Depends on Ka, Kb | 7 + ½(pKa − pKb) |
5. Solubility Product and Common Ion Effect
For a sparingly soluble salt AxBy ⇌ xAy+ + yBx− with molar solubility s:
- AB (AgCl): Ksp = s². AB₂ or A₂B (CaF₂, Ag₂CrO₄): Ksp = 4s³. AB₃ (Fe(OH)₃): Ksp = 27s⁴.
- Precipitation occurs when the ionic product Qsp exceeds Ksp.
- Common ion effect: adding an ion already present decreases the solubility of a salt (or the ionisation of a weak electrolyte). E.g. AgCl is less soluble in NaCl solution: s = Ksp/[Cl⁻].
6. Formula Master Table
| Quantity | Formula |
|---|---|
| Kp and Kc | Kp = Kc(RT)Δn |
| Ionic product of water | Kw = 10−14; pH + pOH = 14 |
| Weak acid | α = √(Ka/C); [H⁺] = √(KaC) |
| Conjugate pair | KaKb = Kw |
| Acidic buffer | pH = pKa + log([salt]/[acid]) |
| Basic buffer | pOH = pKb + log([salt]/[base]) |
| Salt of WA + SB | pH = 7 + ½(pKa + log C) |
| Solubility product | Ksp = xxyysx+y |
| Free energy link | ΔG° = −2.303RT log K |
7. Mind Map
- Kc, Kp, Q
- Le Chatelier
- K changes only with T
- Arrhenius, Brønsted, Lewis
- pH, Kw
- Ostwald dilution
- Henderson equation
- Hydrolysis pH
- Ksp = xxyysx+y
- Common ion effect
- Precipitation Q > Ksp
8. Tips and Tricks
- When mixing an acid and a base, first find which is in excess (in moles), then divide by the total volume.
- Diluting a strong acid 10 times raises its pH by 1; diluting a weak acid 10 times raises it by only about 0.5.
- If the ratio [salt]/[acid] is 10, pH = pKa + 1; if it is 1/10, pH = pKa − 1.
- For Ksp problems, write s for the salt and multiply by the stoichiometric numbers for each ion.
- Including solids or pure liquids in the equilibrium expression.
- Thinking a catalyst changes K or the equilibrium position.
- Forgetting to square the stoichiometric coefficient inside Ksp: for CaF₂, [F⁻] = 2s, so Ksp = s(2s)² = 4s³.
9. Important One-liners
- At equilibrium, the rates of forward and backward reactions are equal, but both reactions continue.
- Kp = Kc when Δn = 0 (e.g. H₂ + I₂ ⇌ 2HI).
- A large K (> 10³) means products are strongly favoured.
- BF₃ and AlCl₃ are Lewis acids; NH₃ is a Lewis base.
- The conjugate base of a strong acid is very weak.
- Blood is buffered mainly by the H₂CO₃/HCO₃⁻ system at pH about 7.4.
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main. (log 2 = 0.30, log 3 = 0.48.)
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kp and Kc are related by:
- Kp = Kc(RT)²
- Kp = Kc(RT)−2
- Kp = Kc
- Kp = Kc(RT)−1
Show solution
Δn = 2 − 4 = −2, so Kp = Kc(RT)−2.
Answer: (B)
The pH of a 10−8 M aqueous HCl solution at 298 K is approximately:
- 8
- 6
- 6.98
- 7.02
Show solution
Water contributes about 10−7 M H⁺, which cannot be ignored. Solving [H⁺]² − 10−8[H⁺] − 10−14 = 0 gives [H⁺] ≈ 1.05 × 10−7 M, so pH ≈ 6.98. An acid solution cannot have pH above 7.
Answer: (C)
The pH of a 0.1 M acetic acid solution (Ka = 1.8 × 10−5) is about:
- 1.0
- 2.87
- 4.74
- 3.87
Show solution
[H⁺] = √(KaC) = √(1.8 × 10−6) ≈ 1.34 × 10−3 M ⇒ pH ≈ 2.87.
Answer: (B)
A buffer contains 0.1 M CH₃COOH and 0.1 M CH₃COONa (pKa = 4.74). Its pH is:
- 4.74
- 5.74
- 3.74
- 7.00
Show solution
pH = pKa + log(0.1/0.1) = 4.74 + 0 = 4.74.
Answer: (A)
Ksp of CaF₂ is 3.2 × 10−11. If its molar solubility is s = x × 10−4 mol L−1, find x.
Show solution
CaF₂ ⇌ Ca²⁺ + 2F⁻: Ksp = s(2s)² = 4s³ = 3.2 × 10−11 ⇒ s³ = 8 × 10−12 ⇒ s = 2 × 10−4.
Answer: 2
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0, the yield of ammonia increases when:
- Temperature is increased
- Pressure is increased
- A catalyst is added
- NH₃ is added
Show solution
Increasing pressure favours the side with fewer gas moles (2 versus 4), i.e. the forward direction. Raising temperature favours the backward (endothermic) direction; a catalyst does not shift equilibrium; adding NH₃ pushes the reaction backward.
Answer: (B)
50 mL of 0.2 M HCl is mixed with 50 mL of 0.1 M NaOH. The pH of the resulting solution is:
- 1.0
- 1.3
- 2.0
- 0.7
Show solution
Moles of H⁺ = 0.05 × 0.2 = 0.010; moles of OH⁻ = 0.05 × 0.1 = 0.005. Excess H⁺ = 0.005 mol in 0.1 L ⇒ [H⁺] = 0.05 M.
pH = −log(5 × 10−2) = 2 − log 5 = 2 − 0.70 = 1.30.
Answer: (B)
A weak monobasic acid has Ka = 1.0 × 10−5. Find its percentage degree of dissociation in a 0.1 M solution.
Show solution
α = √(Ka/C) = √(10−5/0.1) = √(10−4) = 0.01 = 1%.
Answer: 1
11. Quick Revision Summary
- Kp = Kc(RT)Δn; Q vs K decides the direction; only temperature changes K.
- Le Chatelier: the system opposes the change; catalysts do not shift equilibrium.
- pH + pOH = 14; weak acid [H⁺] = √(KaC); α = √(Ka/C).
- Buffer: pH = pKa + log([salt]/[acid]).
- Ksp: AB = s², AB₂ = 4s³, AB₃ = 27s⁴; common ions reduce solubility.