10 public questions tagged with this topic.
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
For BOH B+ + OH- , Kb = (x²/0.01 - x) ≈ (x²/0.01) = 4.0 × 10⁻⁶ . Solving, x = [OH-] = sqrt4.0 × 10⁻⁸ = 2.0 × 10⁻⁴ . pOH = -log(2.0 × 10⁻⁴) ≈ 3.7 , pH = 14 - 3.7 = 10.3 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law
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
α = 0.02 , [OH-] = 0.05 × 0.02 = 1.0 × 10⁻³ , pOH = 3 , pH = 14 - 3 = 11 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
pH = 10.8 , pOH = 14 - 10.8 = 3.2 , [OH-] = 10⁻³.² ≈ 6.31 × 10⁻⁴ . [BOH] = 0.02 - 6.31 × 10⁻⁴ ≈ 0.0194 . Kb = ([OH-]²/[BOH]) = ((6.31 × 10⁻⁴)²/0.0194) ≈ 2.05 × 10⁻⁵ , but given Kb = 4.0 × 10⁻⁴ , ionization α = ([OH-]/[BOH]initial) = (6.31 × 10⁻⁴/0.02) = 0.03155 , % = 3.155%.
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
pOH = 14 - 10.5 = 3.5 , [OH-] = 10⁻³.⁵ ≈ 3.16 × 10⁻⁴ , α = ([OH-]/[BOH]) = (3.16 × 10⁻⁴/0.1) = 3.16 × 10⁻³ = 0.00316 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases
α = 0.005 , [OH-] = 0.02 × 0.005 = 1.0 × 10⁻⁴ , pOH = 4 , pH = 14 - 4 = 10 . Check: Kb = ((1.0 × 10⁻⁴)²/0.02) = 5.0 × 10⁻⁷ (discrepancy, but closest).
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
[OH-] = 0.1 × 0.005 = 5.0 × 10⁻⁴ , pOH = 3.3 , pH = 14 - 3.3 = 10.7 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
[OH-] = 0.1 × 0.00707 = 7.07 × 10⁻⁴ , pOH = 3.15 , pH = 14 - 3.15 = 10.85 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Acid-Base Theories - Arrhenius Bronsted-Lowry Lewis and Salts Hydrolysis
