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Ionic equilibrium questions often give you Ka for an acid and then ask about its salt, or give Kb for a base and ask about its hydrochloride. The link between the two is one equation, Ka × Kb = Kw. Students usually remember it but use it in the wrong place: for an acid and a base that are not a conjugate pair, or at a temperature where Kw is not 10⁻¹⁴. This post shows where the equation comes from, what it lets you predict, and three calculations built on NCERT's own tables.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list acid-base equilibria "and ionization constants", ionisation of water, and "hydrolysis of salts and pH of their solutions" in Unit 6, Equilibrium. The JEE (Advanced) 2026 brochure lists Brønsted and Lewis acids and bases and hydrolysis of salts. NCERT Class 11 Chemistry, Unit 6 Equilibrium (2026-27 reprint), derives the relation in section 6.11.5. Check your own year's syllabus.
Where does Ka × Kb = Kw come from?
Take a weak acid HA and its conjugate base A⁻:
HA + H₂O ⇌ H₃O⁺ + A⁻, with Ka = [H₃O⁺][A⁻]/[HA]
A⁻ + H₂O ⇌ HA + OH⁻, with Kb = [HA][OH⁻]/[A⁻]
Add the two equations. HA and A⁻ cancel and you are left with 2H₂O ⇌ H₃O⁺ + OH⁻, whose constant is Kw. NCERT states the general rule: the equilibrium constant for a net reaction obtained by adding reactions "equals the product of the equilibrium constants for individual reactions". So Ka × Kb = Kw. Taking negative logs gives pKa + pKb = pKw = 14 at 298 K.
Two things follow. A strong acid has a very weak conjugate base, and the reverse: NCERT puts it as "weak acids have very strong conjugate bases", with NH₂⁻, O²⁻ and H⁻ as examples of very good proton acceptors. And the relation only links a pair that differs by one proton.
Worked example 1: rank four conjugate bases
NCERT's Table 6.6 gives Ka at 298 K for HF (3.5 × 10⁻⁴), formic acid (1.8 × 10⁻⁴), hypochlorous acid (3.0 × 10⁻⁸) and HCN (4.9 × 10⁻¹⁰). Kb = Kw/Ka with Kw = 1.0 × 10⁻¹⁴:
F⁻: 2.9 × 10⁻¹¹
HCOO⁻: 5.6 × 10⁻¹¹
OCl⁻: 3.3 × 10⁻⁷
CN⁻: 2.0 × 10⁻⁵
So base strength runs CN⁻ > OCl⁻ > HCOO⁻ > F⁻, the exact reverse of the acid order. You never needed a separate table of Kb values.
Worked example 2: two salts, same concentration, very different pH
Find the pH of 0.10 M NaCN and of 0.10 M NaF at 298 K. Both anions take a proton from water: A⁻ + H₂O ⇌ HA + OH⁻.
NaCN:
Kb(CN⁻) = 1.0 × 10⁻¹⁴ / 4.9 × 10⁻¹⁰ = 2.04 × 10⁻⁵.
[OH⁻] ≈ √(Kb × C) = √(2.04 × 10⁻⁶) = 1.43 × 10⁻³ M. That is 1.4% of 0.10 M, so the approximation holds.
pOH = 2.85, pH = 11.15.
NaF:
Kb(F⁻) = 1.0 × 10⁻¹⁴ / 3.5 × 10⁻⁴ = 2.9 × 10⁻¹¹.
[OH⁻] ≈ √(2.9 × 10⁻¹²) = 1.7 × 10⁻⁶ M.
pOH = 5.77, pH = 8.23.
The Ka values differ by a factor of 3.5 × 10⁻⁴ / 4.9 × 10⁻¹⁰ ≈ 7 × 10⁵, so the Kb values differ by the same factor the other way, and [OH⁻] by its square root, about 850. That is the three-unit gap in pH.
Worked example 3: salts of weak bases
NCERT's Table 6.7 gives Kb for pyridine as 1.77 × 10⁻⁹ and for aniline as 4.27 × 10⁻¹⁰. Which is more acidic, 0.10 M pyridinium chloride or 0.10 M anilinium chloride?
Ka(pyridinium) = 10⁻¹⁴ / 1.77 × 10⁻⁹ = 5.65 × 10⁻⁶, pKa 5.25.
Ka(anilinium) = 10⁻¹⁴ / 4.27 × 10⁻¹⁰ = 2.34 × 10⁻⁵, pKa 4.63.
[H⁺] ≈ √(Ka × C): 7.5 × 10⁻⁴ M for pyridinium (pH 3.12) and 1.53 × 10⁻³ M for anilinium (pH 2.82).
Aniline is the weaker base, so its conjugate acid is the stronger acid. If you need the pH of sodium acetate, the ammonia case or the 10⁻⁸ M HCl trap, our post on pH and buffers works them, and our guide to acid-base definitions shows how to spot conjugate pairs.
When does Ka × Kb = Kw not apply?
To an acid and a base from different pairs. Ka of HF times Kb of ammonia means nothing.
To the wrong step of a polyprotic acid. For H₂X, the base X²⁻ pairs with HX⁻, so Kb(X²⁻) = Kw/Ka₂, not Kw/Ka₁. HX⁻ acting as a base pairs with H₂X, so its Kb = Kw/Ka₁.
With 14 at another temperature. NCERT notes that "the value of Kw is temperature dependent as it is an equilibrium constant". pKa + pKb equals pKw at that temperature, which is 14 only at 298 K.
The adding-equations idea is the same one behind combining equilibrium constants in our post on the reaction quotient Q and K.
A practice routine
Derive Ka × Kb = Kw by adding equations, from memory, once a week.
From NCERT Table 6.6, rank the conjugate bases of all nine acids without a calculator, using only the exponents.
Redo example 2 for 0.10 M sodium hypochlorite.
Write Kb for carbonate and for hydrogen carbonate in terms of Ka₁ and Ka₂ of carbonic acid.
Mix these into your weekly test; Dunlosky and colleagues (2013) rated practice testing and spaced practice as high utility.
If you keep reaching for the wrong K, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, find the pH of 0.10 M sodium formate using only NCERT's Ka for formic acid.
