Practice questions covering buffer solutions, solubility product calculations, and the common ion effect in equilibrium systems.
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 Ag₂CO₃ 2Ag+ + CO₃²⁻ , [Ag+] = 0.1 + 2S ≈ 0.1 M , [CO₃²⁻] = S = 2.0 × 10⁻⁶ . Ksp = [Ag+]²[CO₃²⁻] = (0.1)² (2.0 × 10⁻⁶) = 2.0 × 10⁻⁸ .
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 Ag₂CrO₄ 2Ag+ + CrO₄²⁻ , [Ag+] = 2S = 2.6 × 10⁻⁴ , [CrO₄²⁻] = S = 1.3 × 10⁻⁴ . Ksp = (2.6 × 10⁻⁴)² (1.3 × 10⁻⁴) = 8.79 × 10⁻¹² .
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
For Fe(OH)3 Fe³⁺ + 3OH- , Ksp = [Fe³⁺][OH-]³ = 4.0 × 10⁻³⁸ . Given [Fe³⁺] = 1.0 × 10⁻¹⁰ , (1.0 × 10⁻¹⁰)[OH-]³ = 4.0 × 10⁻³⁸ , [OH-]³ = 4.0 × 10⁻²⁸ , [OH-] = (4.0 × 10⁻²⁸)¹/³ ≈ 1.59 × 10⁻⁹ . pOH = -log(1.59 × 10⁻⁹) ≈ 8.8 , pH = 14 - 8.8 = 5.2 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 350 ≈ 29.085 . Thus, Kc = Kp × RT = 4.0 × 29.085 ≈ 116.34 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For PbI₂ Pb²⁺ + 2I- , [Pb²⁺] = 1.0 × 10⁻⁴ , [I-] = 0.05 + 2 × 1.0 × 10⁻⁴ ≈ 0.05 M . Ksp = [Pb²⁺][I-]² = (1.0 × 10⁻⁴) (0.05)² = 2.5 × 10⁻⁷ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
CaF₂ Ca²⁺ + 2F⁻ , Ksp = [Ca²⁺][F⁻]² = S (2S)² = 4S³ = 1.0 × 10⁻⁶ . Solving, S³ = 2.5 × 10⁻⁷ , S ≈ 6.3 × 10⁻³ .
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 = 1.7 × 10⁻⁵ . Let solubility = S , [Cl-] = 0.1 + 2S ≈ 0.1 , S (0.1)² = 1.7 × 10⁻⁵ , S = (1.7 × 10⁻⁵/0.01) = 1.7 × 10⁻³ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For Ag₂SO₄ 2Ag+ + SO₄²⁻ , [Ag+] = 2 × 5.0 × 10⁻⁵ = 1.0 × 10⁻⁴ , [SO₄²⁻] = 0.1 + 5.0 × 10⁻⁵ ≈ 0.1 . Ksp = (1.0 × 10⁻⁴)² × 0.1 = 1.0 × 10⁻⁹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
For BaSO₄ Ba²⁺ + SO₄²⁻ , [Ba²⁺] = 2.0 × 10⁻⁹ , [SO₄²⁻] ≈ 0.05 , Ksp = (2.0 × 10⁻⁹) × 0.05 = 1.0 × 10⁻¹⁰ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect
