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α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
α = (Lambdam/Lambdam°) = (19/380) = 0.05 . Ka = (α² c/1 - α) = (0.05² × 0.02/1 - 0.05) = (0.0025 × 0.02/0.95) = 5.263 × 10⁻⁵ .
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Kw = Ka × Kb = 1.0 × 10⁻¹⁴ , Ka = (Kw/Kb) = (1.0 × 10⁻¹⁴/1.8 × 10⁻⁵) ≈ 5.56 × 10⁻¹⁰ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
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
pKa = -log Ka , so Ka = 10⁻pKa = 10⁻⁵.² ≈ 6.31 × 10⁻⁶ .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
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