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Ksp = [Hg₂²⁺][Cl-]² = 1.3 × 10⁻¹⁸ , [Cl-] ≈ 0.01 , [Hg₂²⁺] (0.01)² = 1.3 × 10⁻¹⁸ , [Hg₂²⁺] = (1.3 × 10⁻¹⁸/0.0001) = 1.3 × 10⁻¹⁴ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Kc = ([H₂][I₂]/[HI]²) = 0.04 . Since [H₂] = [I₂] = x , 0.04 = (x²/(0.2)²) = (x²/0.04) , x² = 0.0016 , x = 0.04 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 300 = 24.93 . Thus, Kc = Kp × RT = 0.25 × 24.93 ≈ 6.23 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
A Bronsted-Lowry acid donates a proton. HCO₃- can donate H+ to form CO₃²⁻ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Ka = ([H+][A-]/[HA]) = (x²/0.05 - x) ≈ (x²/0.05) = 1.0 × 10⁻⁵ , x² = 5.0 × 10⁻⁷ , x = 7.07 × 10⁻⁴ M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
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
Adding an inert gas at constant volume does not change the partial pressures or concentrations of the reactants and products, so the equilibrium position remains unaffected.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Kw = [H+][OH-] = 1.0 × 10⁻¹² , [H+] = sqrt1.0 × 10⁻¹² = 1.0 × 10⁻⁶ , pH = -log(1.0 × 10⁻⁶) = 6 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Initial: Q = ((1.5)³/(0.5)² (0.5)) = 27 > Kp = 9 , doubling pressure increases concentrations, Q increases further, shifts left.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Initial equilibrium: [A] = 0.2 M , [B] = 0.1 M , [C] = 0.4 M , Kc = ((0.4)²/(0.2)²(0.1)) = 4 . After adding 0.1 mol A , [A] = 0.3 M , reaction shifts right to restore equilibrium.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Total pressure = PA + PB + PC = 2 , PB = PC (stoichiometry), 1.6 + PB + PB = 2 , 2PB = 0.4 , PB = 0.2 atm . Check: Kp = (PB PC/PA²) = ((0.2)(0.2)/(1.6)²) = (0.04/2.56) ≈ 0.0156 (slightly off, but closest).
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
