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ChemistryMediumClass 11

Degree of Ionization — Ka, Kb, pKa

Ionic Equilibrium

6

JEE Qs

8%

Hard

75

min

Master the conditions for applying the 1-α ≈ 1 approximation and clearly distinguish between Ka/Kb for a weak acid/base and the Ka*Kb = Kw relationship for a conjugate pair to avoid common calculation errors.

🧮 Key Formulas

α = (moles ionized) / (initial moles)
Ka = [H⁺][A⁻] / [HA] (for weak acid HA)
Kb = [B⁺][OH⁻] / [BOH] (for weak base BOH)
Ka = Cα² / (1-α)
Kb = Cα² / (1-α)
If α << 1 (typically < 0.05), Ka ≈ Cα² => α = sqrt(Ka/C)
If α << 1 (typically < 0.05), Kb ≈ Cα² => α = sqrt(Kb/C)
[H⁺] = Cα = sqrt(Ka * C) (for weak monoprotic acid)
[OH⁻] = Cα = sqrt(Kb * C) (for weak monoprotic base)
pKa = -log(Ka)
pKb = -log(Kb)
Kw = Ka * Kb (for a conjugate acid-base pair)
pH = -log[H⁺]
pOH = -log[OH⁻]
pH + pOH = 14 (at 25°C)

✅ Key Points for JEE

  • 1The degree of ionization (α) for a weak electrolyte increases upon dilution, while Ka and Kb values are constant at a given temperature.
  • 2Larger Ka implies a stronger weak acid (lower pKa); larger Kb implies a stronger weak base (lower pKb).
  • 3The approximation 1-α ≈ 1 is valid if α < 0.05 (or 5%), which generally holds when C/Ka or C/Kb > 400-500. Always verify this assumption.
  • 4Ka and Kb values are critical for calculating the [H⁺], [OH⁻], pH, and pOH of weak acid and weak base solutions.
  • 5The product Ka * Kb = Kw only applies to a conjugate acid-base pair at a specific temperature (usually 10⁻¹⁴ at 25°C), not for any arbitrary acid and base.

⚠️ Common Mistakes

  • Applying Ostwald's Dilution Law or Ka/Kb concepts to strong acids or bases.
  • Incorrectly using the 1-α ≈ 1 approximation without checking if α is indeed small enough (e.g., when concentration is very low or Ka/Kb is relatively large).
  • Confusing the Ka and Kb values of an acid and its conjugate base with two unrelated acid and base dissociation constants.
  • Mistaking initial concentration 'C' for equilibrium concentration of the undissociated species in the denominator of Ka/Kb expressions when alpha is not negligible.

📝 Practice Questions

See all

Q74.If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10−x . The value of x is ______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]

2025·NumericalMedium

Q61. Ksp for Cr(OH)3 is 1.6 × 10−30 . What is the molar solubility of this salt in water? (1) 1.8×10−30 (2) 5√1.8 × 10−30 27 × 10−30 (3) 4√1.6×10−3027 (4) 2√1.6

2025·MCQEasy

Q58.The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3(PO3−4 )4 is given by relation : (1) Ksp 13 (2) Ksp 17 ( 9612 ) ( 6912 ) (3) Ksp 17 (4) Ksp 16 ( 8435 ) ( 5348 )

2025·MCQMedium

Q62.Which of the following happens when NH4OH is added gradually to the solution containing 1 M A2+ and 1MB3+ ions? Given : Ksp [A(OH)2] = 9 × 10−10 and Ksp [B(OH)3] = 27 × 10−18 at 298 K . (1) Both A(OH)2 and B(OH)3 do not show (2) A(OH)2 will precipitate before B(OH)3 precipitation with NH4OH (3) B(OH)3 will precipitate before A(OH)2 (4) A(OH)2 and B(OH)3 will precipitate together excess HCHO alkali−

2025·MCQMedium

Q68.pH of water is 7 at 25∘C. If water is heated to 80∘C., it's pH will : (1) Decrease (2) H+ concentration increases, OH− concentration decreases (3) Remains the same (4) Increase 2025 (23 Jan Shift 2) JEE Main Previous Year Paper

2025·MCQEasy

Q58.A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH pKa )? (1) 0 (2) log(1 + 2x) (3) log ( 1−xx ) (4) log ( 1−xx )

2025·MCQMedium

NCERT Chapters

  • Class 11 Chemistry Ch 7: Equilibrium