7 public questions tagged with this topic.
Δ Tb = Kb · m . 0.104 = Kb × 0.2 . Kb = (0.104/0.2) = 0.52 K kg mol⁻¹ .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Solubility - of Solids and Gases in Liquids Henry's Law
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/6.8 × 10⁻⁴) ≈ 1.47 × 10⁻¹¹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Let [NH₃] = C , α = 0.01 , [OH-] = C α = 0.01C . Kb = ((0.01C)²/C(1 - 0.01)) ≈ (0.0001C²/0.99C) = 1.01 × 10⁻⁴C = 1.8 × 10⁻⁵ , C = (1.8 × 10⁻⁵/1.01 × 10⁻⁴) ≈ 0.178 M .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
pH = 10 , pOH = 14 - 10 = 4 , [OH-] = 10⁻⁴ . For B + H₂O BH+ + OH- , Kb = ([OH-]²/[B]) = ((10⁻⁴)²/0.05) = (10⁻⁸/0.05) = 2.0 × 10⁻⁷ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (1.0 × 10⁻¹⁴/4.5 × 10⁻⁴) ≈ 2.22 × 10⁻¹¹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Kb = (Kw/Ka) = (1.0 × 10⁻¹⁴/6.8 × 10⁻⁴) ≈ 1.47 × 10⁻¹¹ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
pKb = -log Kb = -log(3.2 × 10⁻⁶) = 6 - log 3.2 ≈ 6 - 0.505 = 5.495 ≈ 5.5 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
