Take timed practice tests on Chemical and Ionic Equilibrium for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.
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1. For the reaction N₂ + 3H₂ ⇌ 2NH₃, the expression for Kc is:
Explanation: Kc = [products]^stoich / [reactants]^stoich = [NH₃]²/([N₂][H₂]³). Pure solids and liquids are excluded.
2. The relation between Kp and Kc is:
Explanation: Kp = Kc(RT)^Δn where Δn = moles of gaseous products − moles of gaseous reactants. R = 0.0821 L·atm/mol·K.
3. For N₂ + 3H₂ ⇌ 2NH₃ (exothermic), increasing temperature will:
Explanation: Le Chatelier principle: increasing temperature for an exothermic reaction shifts equilibrium backward (to absorb heat), decreasing [NH₃] and increasing Keq decreases.
4. For N₂ + 3H₂ ⇌ 2NH₃ (Δn = 2−4 = −2), increasing pressure will:
Explanation: Increasing pressure shifts equilibrium toward fewer moles of gas (Δn
5. Adding a catalyst to an equilibrium reaction:
Explanation: A catalyst lowers activation energy for both forward and reverse reactions equally, reaching equilibrium faster but NOT changing K or equilibrium position.
6. If Q < Kc, the reaction will proceed:
Explanation: Q
7. For H₂ + I₂ ⇌ 2HI, if [H₂] = [I₂] = 0.1 M and [HI] = 0.8 M at equilibrium, Kc is:
Explanation: Kc = [HI]²/([H₂][I₂]) = (0.8)²/(0.1×0.1) = 0.64/0.01 = 64.
8. For PCl₅ ⇌ PCl₃ + Cl₂, if 1 mol PCl₅ is taken and α is the degree of dissociation, moles of each at equilibrium are:
Explanation: Starting with 1 mol PCl₅. At equilibrium: PCl₅ = 1−α, PCl₃ = α, Cl₂ = α. Total = 1+α moles.
9. For an endothermic reaction, increasing temperature:
Explanation: By van't Hoff equation: d(lnK)/dT = ΔH°/RT². For endothermic ΔH° > 0 → K increases with temperature.
10. The pH of a 0.01 M HCl solution (strong acid, fully dissociates) is:
Explanation: [H⁺] = 0.01 = 10⁻². pH = −log[H⁺] = −log(10⁻²) = 2.
11. If pOH = 3, then pH at 25°C is:
Explanation: At 25°C: pH + pOH = 14. pH = 14 − pOH = 14 − 3 = 11. The solution is basic.
12. For a weak acid HA with concentration C and Ka, the degree of dissociation α = √(Ka/C) when:
Explanation: For weak acid HA ⇌ H⁺ + A⁻: Ka = Cα²/(1−α) ≈ Cα² if α
13. A buffer is best prepared when:
Explanation: Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Maximum buffering capacity at pH = pKa where [acid] = [conjugate base]. This gives the flattest region on the buffer capacity vs pH curve.
14. The solubility product of BaSO₄ (Ksp = 1×10⁻¹⁰) in pure water. The molar solubility s is:
Explanation: BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. Ksp = s² = 10⁻¹⁰. s = √(10⁻¹⁰) = 10⁻⁵ M.
15. A buffer contains 0.1 M acetic acid (pKa = 4.74) and 0.1 M sodium acetate. The pH of the buffer is:
Explanation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = 4.74 + log(0.1/0.1) = 4.74 + 0 = 4.74.
16. Adding NaCl to a solution of AgCl (Ksp = 1.8×10⁻¹⁰) will:
Explanation: Common ion effect: added Cl⁻ from NaCl suppresses AgCl dissolution (shifts equilibrium left → less AgCl dissolves). Ksp remains constant.
17. An aqueous solution of CH₃COONa is:
Explanation: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. The acetate ion (weak acid anion) hydrolyses to produce OH⁻ → basic solution. pH > 7.
18. What is the pH of a solution prepared by mixing 100 mL of 0.1 M HCl with 100 mL of 0.1 M NaOH?
Explanation: Moles HCl = 0.01 mol, moles NaOH = 0.01 mol. They exactly neutralise: HCl + NaOH → NaCl + H₂O. Solution of NaCl in water is neutral → pH = 7.
19. The pH of a 0.1 M NH₃ solution (Kb = 1.8×10⁻⁵) is approximately:
Explanation: [OH⁻] = √(Kb×C) = √(1.8×10⁻⁵×0.1) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³. pOH = 2.87. pH = 14−2.87 ≈ 11.13.
20. A solution contains 0.001 M Ba²⁺ and 0.001 M SO₄²⁻. Ksp(BaSO₄) = 1×10⁻¹⁰. Will precipitation occur?
Explanation: Q = [Ba²⁺][SO₄²⁻] = 0.001 × 0.001 = 10⁻⁶. Ksp = 10⁻¹⁰. Since Q > Ksp, the solution is supersaturated and precipitation occurs.
21. An acid-base indicator is chosen such that its pKIn equals:
Explanation: The indicator should change colour at the equivalence point of the titration. Choose an indicator whose pKIn ≈ pH at equivalence point.
22. The ionic product of water Kw at 25°C is:
Explanation: Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C. This means at neutral pH, [H⁺] = [OH⁻] = 10⁻⁷ M, giving pH = 7.
23. For A + B ⇌ C + D, starting with 1 mol each of A and B in 1 L, at equilibrium [C] = 0.4 M. Kc is:
Explanation: ICE: [A]=[B]=1−0.4=0.6, [C]=[D]=0.4. Kc = (0.4)²/(0.6)² = 0.16/0.36 ≈ 0.44.
24. 50 mL of 0.1 M HCl is added to 50 mL of 0.05 M NaOH. The pH of the resulting solution is:
Explanation: Moles HCl = 5×10⁻³, moles NaOH = 2.5×10⁻³. Excess HCl = 2.5×10⁻³ mol in 100 mL → [H⁺] = 0.025 M. pH = −log(0.025) ≈ 1.6.
25. Ksp of Ag₂CrO₄ = 1.1×10⁻¹². Its molar solubility is:
Explanation: Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻. If s = solubility: [Ag⁺] = 2s, [CrO₄²⁻] = s. Ksp = (2s)²(s) = 4s³ = 1.1×10⁻¹². s³ = 2.75×10⁻¹³. s = (2.75×10⁻¹³)^(1/3) ≈ 6.5×10⁻⁵ M.
26. The pH of 0.01 M formic acid (HCOOH, Ka = 1.8×10⁻⁴) is approximately:
Explanation: [H⁺] = √(Ka × C) = √(1.8×10⁻⁴ × 0.01) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³. pH = −log(1.34×10⁻³) ≈ 2.87.
27. Ostwald dilution law states that for a weak electrolyte: α²/(1−α)V = Ka. For a very weak electrolyte (α << 1), this simplifies to:
Explanation: For α
28. Equal volumes of 0.2 M CH₃COOH (pKa = 4.74) and 0.2 M NaOH are mixed. The pH of the resulting solution is:
Explanation: Equal moles react completely, leaving 0.1 M CH₃COONa. For a salt of a weak acid and strong base, $pH=(pK_w+pK_a+\log C)/2=(14+4.74-1)/2= 8.87 $.
29. For the equilibrium N₂O₄(g) ⇌ 2NO₂(g), adding an inert gas at constant volume will:
Explanation: Adding inert gas at constant volume does not change the partial pressures of N₂O₄ or NO₂ (concentrations unchanged). No effect on equilibrium position.
30. For PCl₅ ⇌ PCl₃ + Cl₂ with degree of dissociation α at pressure P, Kp is:
Explanation: Mole fractions: PCl₅ = (1−α)/(1+α), PCl₃ = α/(1+α), Cl₂ = α/(1+α). Kp = [Pα/(1+α)]²/[P(1−α)/(1+α)] = Pα²/(1−α²).
31. In the titration of a strong acid with a strong base, which indicator is most appropriate?
Explanation: Strong acid−strong base titration: equivalence point at pH 7 with a sharp pH jump from ~3 to ~11. Both methyl orange and phenolphthalein fall within this jump. Either works.
32. The solubility of Mg(OH)₂ (Ksp = 1.2×10⁻¹¹) increases when:
Explanation: Adding acid decreases [OH⁻] → equilibrium Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻ shifts right → more dissolves. Ksp for Mg(OH)₂ depends on [Mg²⁺][OH⁻]² = 1.2×10⁻¹¹.
33. A solution contains 0.1 M each of Cl⁻ and I⁻. If AgNO₃ is gradually added, which precipitates first? Ksp(AgCl) = 1.8×10⁻¹⁰, Ksp(AgI) = 1×10⁻¹⁶.
Explanation: AgI has much lower Ksp → needs much less Ag⁺ to start precipitation. [Ag⁺] to start AgI precipitation = Ksp/[I⁻] = 10⁻¹⁵. To start AgCl: = 1.8×10⁻⁹. AgI precipitates first.
34. If K₁ and K₂ are equilibrium constants for A ⇌ B and B ⇌ C respectively, the equilibrium constant for A ⇌ C is:
Explanation: When reactions are added, equilibrium constants are multiplied: K(A⇌C) = K₁ × K₂. This follows from the fact that K is expressed as products of concentration terms with stoichiometric exponents.
35. At dynamic equilibrium, the rate of forward reaction is:
Explanation: At equilibrium, forward rate = reverse rate. Concentrations remain constant but the reaction is ongoing (dynamic).
36. For the reaction CO(g) + ½O₂(g) ⇌ CO₂(g), Δn =
Explanation: Δn = moles gaseous products − moles gaseous reactants = 1 − (1 + 1/2) = −1/2. So Kp = Kc(RT)^(−1/2).
37. In industrial synthesis of NH₃ (Haber process), a catalyst is used because:
Explanation: The iron catalyst in the Haber process lowers activation energy for both forward and reverse reactions, speeding up equilibrium attainment. K is unchanged.
38. 100 mL of 0.2 M HCl is mixed with 100 mL of 0.1 M Ba(OH)₂. The resulting pH is:
Explanation: HCl supplies 0.02 mol H⁺. The 0.01 mol Ba(OH)₂ supplies 0.02 mol OH⁻, so neutralisation is exact and the resulting solution has pH 7 at 25°C.
39. The solubility of CaF₂ (Ksp = 3.4×10⁻¹¹) in 0.1 M NaF solution (common ion effect) is:
Explanation: CaF₂ ⇌ Ca²⁺ + 2F⁻. With [F⁻] ≈ 0.1 M (from NaF): Ksp = s × (0.1)² → s = 3.4×10⁻¹¹/0.01 = 3.4×10⁻⁹ M. Much less than in pure water (s ≈ 2×10⁻⁴ M).
40. Ka of a weak acid HA is 4×10⁻⁵. The degree of dissociation in 0.01 M solution is approximately:
Explanation: α = √(Ka/C) = √(4×10⁻⁵/0.01) = √(4×10⁻³) = 0.0632 ≈ 6.3%. The approximation α
41. Among the following, which buffer has the highest capacity?
Explanation: Buffer capacity is proportional to the concentrations of both acid and conjugate base. Higher total concentration → higher capacity to resist pH changes. 1 M each gives maximum capacity here.