Equilibrium: JEE Advanced PYQ, 2007–2026

Answer: (A)

Strong acid and strong base react completely, so find which one is in excess. HCl = 0.10 × 60 = 6.0 mmol. NaOH = 0.20 × 40 = 8.0 mmol. Excess OH⁻ = 2.0 mmol in a total volume of 100 mL, so [OH⁻] = 0.020 M. pOH = −log(2 × 10⁻²) = 2 − 0.30 = 1.70. pH = 14 − 1.70 = 12.30.

Answer: (A)

Principle: when solid Mg(OH)₂ is in equilibrium with solution, Ksp = [Mg²⁺][OH⁻]² holds. The buffer fixes [OH⁻], so [Mg²⁺] follows directly. pH = 11 gives pOH = 14 − 11 = 3, so [OH⁻] = 1.0 × 10⁻³ M. Then [Mg²⁺] = 4.0 × 10⁻¹²/(1.0 × 10⁻³)² = 4.0 × 10⁻⁶ M. Only 0.04% of the original 0.010 M Mg²⁺ remains dissolved.

Answer: (A)

For Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻, Ksp = s(2s)² = 4s³. So 4s³ = 3.2 × 10⁻¹¹, which gives s³ = 8 × 10⁻¹² and s = 2 × 10⁻⁴ M. Then [OH⁻] = 2s = 4 × 10⁻⁴ M. pOH = 4 − log 4 = 4 − 0.60 = 3.40, so pH = 14 − 3.40 = 10.60.

Answer: (C)

Write Kp = p(SO₃)²/[p(SO₂)²·p(O₂)] = n(SO₃)²·n(total)/[n(SO₂)²·n(O₂)·P]. At constant total pressure P, adding argon increases n(total), so Q becomes greater than Kp. The reaction therefore shifts backward, toward SO₂ and O₂, which have more moles of gas, until Q = Kp. Kp itself is unchanged because the temperature is constant. At constant volume, argon would not change the partial pressures, and there would be no shift.

Answer: (A)

Doubling the volume halves every concentration. Q = [CH₃OH]/([CO][H₂]²). Substituting the halved values gives Q = (½)/(½ × ¼) × K = 4K. Because Q > K, the system moves backward until Q returns to K. This is toward the side with more gas moles (3 mol vs 1 mol), which agrees with Le Chatelier's principle.

Answer: (A)

For a weak acid with small ionisation, [H⁺] = √(K_a·C). Originally [H⁺] = √(4 × 10⁻⁶ × 0.04) = 4 × 10⁻⁴ M, giving pH 3.40. After fourfold dilution C = 0.010 M, so [H⁺] = √(4 × 10⁻⁶ × 0.01) = 2 × 10⁻⁴ M, which is only half the original value. pH = −log(2 × 10⁻⁴) = 4 − 0.30 = 3.70. A strong acid would have given 4.00, because its [H⁺] is proportional to C rather than to √C.

Answer: (D)

CaCl₂ is a strong electrolyte that supplies a common ion, Cl⁻. Each formula unit gives two Cl⁻ ions, so [Cl⁻] = 2 × 0.010 = 0.020 M. For AgCl, Ksp = [Ag⁺][Cl⁻], and the solubility equals [Ag⁺]. So s = 1.8 × 10⁻¹⁰/0.020 = 9.0 × 10⁻⁹ mol L⁻¹, far less than the pure-water value of about 1.3 × 10⁻⁵ mol L⁻¹.

Answer: (D)

At such a low concentration, H⁺ from water cannot be neglected. Charge balance gives [H⁺] = [Cl⁻] + [OH⁻] = 10⁻⁸ + K_w/[H⁺]. This rearranges to x² − 10⁻⁸x − 10⁻¹⁴ = 0, where x = [H⁺]. Then x = [10⁻⁸ + √(10⁻¹⁶ + 4 × 10⁻¹⁴)]/2 = (0.1 + 2.0025) × 10⁻⁷/2 ≈ 1.051 × 10⁻⁷ M. So pH = 7 − 0.0216 ≈ 6.98. The pH of 8.00 ignores water. The pH of 7.00 ignores the acid. The pH of 6.96 double-counts water's contribution.

Answer: (C)

When the acid concentration is close to 10⁻⁷ M, the H⁺ from the ionisation of water cannot be neglected. Let [H⁺] = x. Charge balance gives x = [Cl⁻] + [OH⁻] = 10⁻⁸ + K_w/x. This rearranges to x² − 10⁻⁸x − 10⁻¹⁴ = 0, so x = (10⁻⁸ + √(10⁻¹⁶ + 4×10⁻¹⁴))/2 ≈ 1.05 × 10⁻⁷ M. Hence pH = −log(1.05 × 10⁻⁷) ≈ 6.98. The solution is still very slightly acidic.

Answer: (A)

Both salts precipitate once their ionic product exceeds Ksp. AgCl needs [Ag⁺] = 1.8×10⁻¹⁰/0.010 = 1.8×10⁻⁸ M, whereas Ag₂CrO₄ needs [Ag⁺] = √(Ksp/[CrO₄²⁻]) = √(1.1×10⁻¹⁰) ≈ 1.05×10⁻⁵ M. AgCl therefore precipitates first. When Ag₂CrO₄ is about to appear, [Ag⁺] = 1.05×10⁻⁵ M and the solution is still saturated with AgCl, so [Cl⁻] = Ksp(AgCl)/[Ag⁺] = 1.8×10⁻¹⁰/1.05×10⁻⁵ ≈ 1.7×10⁻⁵ M. This is only about 0.17% of the original chloride.