Ka and Kb Relation through Kw
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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Grade 12Grade 12 Chemistry
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Ka and Kb Relation through Kw explained
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
- = Acid dissociation constant
- = Base dissociation constant of the conjugate
- = Ion product of water
Missing one of these? Work it out first, then come back
- Acid dissociation constant — pKa from Acid Dissociation Constant, pH of a Weak Acid from Ka
- Base dissociation constant of the conjugate — pKb from Base Dissociation Constant, pKa from Acid Dissociation Constant