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Henry’s law: p = KH · x . x = (4/80) = 0.05 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Ksp = [A+]²[B²⁻] = (2S)² S = 4S³ = 1.0 × 10⁻⁹ , S³ = 2.5 × 10⁻¹⁰ , S ≈ 6.3 × 10⁻⁴ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For PbCl₂ Pb²⁺ + 2Cl⁻ , Ksp = [Pb²⁺][Cl⁻]² = S (2S)² = 4S³ = 1.6 × 10⁻⁵ . Solving, S³ = 4 × 10⁻⁶ , S = (4 × 10⁻⁶)¹/³ ≈ 1.6 × 10⁻² .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For Mg(OH)2 Mg²⁺ + 2OH- , Ksp = [Mg²⁺][OH-]² = S (2S)² = 4S³ = 1.8 × 10⁻¹¹ . Solving, S³ = (1.8 × 10⁻¹¹/4) = 4.5 × 10⁻¹² , S ≈ 1.65 × 10⁻⁴ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
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