Ka and Kb Relation through Kw

Ka Kb=KwK_a \, K_b = K_w

Worked example: Acetate Kb from acetic acid Ka 1.8e-5 → 5.5556e-10 — press Try an example to run it live, then adjust anything.

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Ka and Kb Relation through Kw explained

KaKbKw

Add the dissociation of an acid to the hydrolysis of its conjugate base and the two half-reactions sum to the autoionisation of water — so their equilibrium constants multiply to Kw. That single line means you never need to look up both numbers: acetic acid's Ka of 1.8 × 10⁻⁵ instantly gives acetate a Kb of 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰, an extremely feeble base, exactly as expected from a strong-ish weak acid.

The relation encodes chemistry's most useful see-saw: the stronger the acid, the weaker its conjugate base, and the product is fixed. It also explains why sodium acetate solutions are mildly alkaline while sodium chloride solutions are neutral — chloride's parent HCl is so strong that its conjugate Kb is vanishingly small. Kw itself is temperature-dependent (1.0 × 10⁻¹⁴ at 25 °C but about 5.5 × 10⁻¹⁴ at 50 °C), so the pairing of tabulated Ka and Kb values only balances at the temperature they were measured.

Ka and Kb Relation through Kw formula

Ka Kb=KwK_a \, K_b = K_w
Where
  • KaK_a= Acid dissociation constant
  • KbK_b= Base dissociation constant of the conjugate
  • KwK_w= Ion product of water

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