Solubility Product

Answer: B. Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻. The stoichiometric coefficient of OH⁻ is 2, so Ksp = [Mg²⁺]¹[OH⁻]² (NCERT Class 11 Chemistry Chapter 6, page 205).

Why A is wrong: A omits the stoichiometric power on [OH⁻]. Mg(OH)₂ produces 2 OH⁻ ions per formula unit, so [OH⁻] must be raised to the power 2 (trap: Ksp stoichiometric powers).

Why C is wrong: C raises [Mg²⁺] to power 2 instead of [OH⁻]. Only one Mg²⁺ ion is produced per formula unit; the exponent 2 belongs on [OH⁻].

Why D is wrong: D squares both ion concentrations. Only the ion with coefficient 2 (OH⁻) gets squared; Mg²⁺ has coefficient 1.

Answer: A. AgCl ⇌ Ag⁺ + Cl⁻ (1:1 salt). Ksp = s². So s = √(1.8 × 10⁻¹⁰) = √(18 × 10⁻¹¹) ≈ 1.34 × 10⁻⁵ mol/L.

Why B is wrong: B confuses Ksp with solubility directly. For AgCl, Ksp = s², so s = √Ksp, not Ksp itself (trap: forgetting the square-root step).

Why C is wrong: C divides Ksp by 2 instead of taking the square root. The 1:1 stoichiometry means Ksp = s², requiring a square root — not division.

Why D is wrong: D uses s = Ksp/10⁻⁵ or a miscalculation. The correct operation is s = √(1.8 × 10⁻¹⁰) ≈ 1.34 × 10⁻⁵.

Answer: C. C is correct. CaF₂ ⇌ Ca²⁺ + 2F⁻, so if s mol/L dissolves, [Ca²⁺] = s and [F⁻] = 2s. Ksp = s(2s)² = 4s³, so s = ∛(Ksp/4) = ∛(3.2 × 10⁻¹¹ / 4) = ∛(8.0 × 10⁻¹²) = 2.0 × 10⁻⁴ mol/L.

Why A is wrong: A is wrong because 5.66 × 10⁻⁶ mol/L is √Ksp — treating CaF₂ as a 1 : 1 salt with Ksp = s². CaF₂ releases two fluoride ions per formula unit, so Ksp = 4s³. (trap: Ksp stoichiometric powers)

Why B is wrong: B is wrong because 3.2 × 10⁻¹¹ mol/L is Ksp itself, reported as the solubility. Ksp is a product of ion concentrations raised to powers; the solubility has to be extracted from it through 4s³.

Why D is wrong: D is wrong because 1.6 × 10⁻⁴ mol/L comes from cubing the 2 instead of squaring it: writing (2s)³ = 8s³ and taking ∛(Ksp/8). Only the fluoride concentration is squared, because the 2 in 2F⁻ is its coefficient in the equilibrium; calcium enters to the first power.

Answer: D. For AB₂: Ksp = [A²⁺][B⁻]². Dimensions: (mol/L)(mol/L)² = mol³/L³. Equilibrium constants involving concentrations carry units.

Why A is wrong: A gives mol²/L², which would apply to a 1:1 salt (Ksp = [A⁺][B⁻] = (mol/L)²). AB₂ has three concentration factors, not two.

Why B is wrong: B gives mol/L, which is the unit of solubility (s), not of Ksp. The product of three concentration terms cannot reduce to mol/L.

Why C is wrong: C claims Ksp is dimensionless. Only the thermodynamic equilibrium constant (written in terms of activities) is dimensionless. Ksp expressed in molar concentrations carries units.

Answer: C. Q (2.5 × 10⁻⁴) > Ksp (1.7 × 10⁻⁵). When Q > Ksp, the solution is supersaturated and precipitation occurs until Q decreases to equal Ksp.

Why A is wrong: A applies when Q < Ksp (unsaturated solution). Here Q > Ksp, so the reverse is true — excess ions must leave solution as precipitate.

Why B is wrong: B applies only when Q = Ksp exactly. Q here is roughly 15 times larger than Ksp, so the solution is well beyond saturation.

Why D is wrong: D is incorrect. Q and Ksp are directly compared to predict dissolution or precipitation — this comparison is the core utility of the solubility product concept.

Answer: B. Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻. [Ag⁺] = 2s, [CrO₄²⁻] = s. Ksp = (2s)²·s = 4s²·s = 4s³.

Why A is wrong: A forgets both stoichiometric factors. Writing Ksp = s·s·s = s³ ignores that [Ag⁺] = 2s, not s (trap: Ksp stoichiometric powers).

Why C is wrong: C uses (2s)·s² = 2s³, reversing which ion gets the coefficient. Ag₂CrO₄ produces 2 Ag⁺ ions — the power goes on [Ag⁺], giving (2s)², not 2s.

Why D is wrong: D is wrong: 27s³ is not the Ksp of any salt — a 1 : 3 salt such as AB₃ gives s(3s)³ = 27s⁴. For Ag₂CrO₄, (2s)²·s = 4s³.

Answer: D. BaSO₄ ⇌ Ba²⁺ + SO₄²⁻. With 0.01 M SO₄²⁻ from Na₂SO₄ (common-ion effect), [SO₄²⁻] ≈ 0.01 (s is negligible compared to 0.01). Ksp = s × 0.01. s = 1.1 × 10⁻¹⁰ / 0.01 = 1.1 × 10⁻⁸ mol/L. Solubility drops dramatically due to the common ion.

Why A is wrong: A gives the solubility in pure water (s = √Ksp = √(1.1 × 10⁻¹⁰) ≈ 1.05 × 10⁻⁵), ignoring the common-ion effect entirely. Na₂SO₄ supplies extra SO₄²⁻, suppressing BaSO₄ dissolution.

Why B is wrong: B divides Ksp by 0.02 instead of 0.01, perhaps doubling the SO₄²⁻ concentration. Na₂SO₄ is fully dissociated but gives only one SO₄²⁻ per formula unit; 0.01 M Na₂SO₄ → 0.01 M SO₄²⁻.

Why C is wrong: C confuses Ksp with solubility. Ksp is 1.1 × 10⁻¹⁰, but solubility in the presence of 0.01 M SO₄²⁻ is Ksp/0.01 = 1.1 × 10⁻⁸.

Answer: A. M₃(PO₄)₂ ⇌ 3M²⁺ + 2PO₄³⁻. [M²⁺] = 3s, [PO₄³⁻] = 2s. Ksp = (3s)³·(2s)² = 27s³ · 4s² = 108s⁵.

Why B is wrong: B uses Ksp = 3s·2s·s³ or some partial coefficient application. The correct expression requires FULL powers: (3s)³ = 27s³ and (2s)² = 4s² (trap: Ksp stoichiometric powers).

Why C is wrong: C uses 3² instead of 3³ in the coefficient: 9 × 4 = 36. The powers must match the stoichiometry, (3s)³ × (2s)² = 27 × 4 s⁵ = 108s⁵.

Why D is wrong: D uses (3s)·(2s)² · s² = 3s·4s²·s² = 12s⁵, raising [M²⁺] to power 1 instead of 3. The dissolution produces 3 M²⁺ ions, demanding the cube.