23 public questions tagged with this topic.
For AgBr Ag+ + Br- , Ksp = [Ag+][Br-] = 5.0 × 10⁻¹³ , [Br-] ≈ 0.05 , [Ag+] = (5.0 × 10⁻¹³/0.05) = 1.0 × 10⁻¹¹ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For Ag₂CO₃ 2Ag+ + CO₃²⁻ , Ksp = [Ag+]²[CO₃²⁻] = 8.1 × 10⁻¹² , [CO₃²⁻] ≈ 0.01 , [Ag+]² = (8.1 × 10⁻¹²/0.01) = 8.1 × 10⁻¹⁰ , [Ag+] = 2.85 × 10⁻⁵ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
For AgIO₃ Ag+ + IO₃- , Ksp = [Ag+][IO₃-] = 3.0 × 10⁻⁸ . [IO₃-] = 0.02 + [Ag+] ≈ 0.02 M , [Ag+] = (3.0 × 10⁻⁸/0.02) = 1.5 × 10⁻⁶ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
For Al(OH)3 Al³⁺ + 3OH- , Ksp = [Al³⁺][OH-]³ = 1.3 × 10⁻³³ . Given [Al³⁺] = 0.01 , 0.01 [OH-]³ = 1.3 × 10⁻³³ , [OH-]³ = 1.3 × 10⁻³¹ , [OH-] = (1.3 × 10⁻³¹)¹/³ ≈ 5.07 × 10⁻¹¹ . pOH = -log(5.07 × 10⁻¹¹) ≈ 10.3 , pH = 14 - 10.3 = 3.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
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
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
