The Relationship Ka × Kb = Kw

Data books list Ka values for dozens of acids, but they rarely give Kb values for anions like ethanoate, fluoride or cyanide. They don’t need to. For any conjugate acid–base pair, there’s a simple rule that lets you calculate one from the other:

Ka × Kb = Kw

At 25 °C, Kw = 1.0 × 10⁻¹⁴, so in logarithmic form:

pKa + pKb = 14.00

This short equation is one of the most useful tools in acid–base chemistry. It explains why some salts are acidic and others basic, it links acid strength to base strength, and it saves you from ever having to memorise two tables.

Where the relationship comes from

Take a weak acid HA and its conjugate base A⁻. Write both equilibria in water.

The acid ionises:

HA ⇌ H⁺ + A⁻ Ka = [H⁺][A⁻] ÷ [HA]

The conjugate base reacts with water:

A⁻ + H₂O ⇌ HA + OH⁻ Kb = [HA][OH⁻] ÷ [A⁻]

Now multiply the two expressions together:

Ka × Kb = ([H⁺][A⁻] ÷ [HA]) × ([HA][OH⁻] ÷ [A⁻])

The [A⁻] and [HA] terms cancel, leaving:

Ka × Kb = [H⁺][OH⁻] = Kw

That’s all there is to it. Adding the two chemical equations gives H₂O ⇌ H⁺ + OH⁻, the self-ionisation of water. When you add equations, you multiply their equilibrium constants.

What it tells you: the see-saw

Because the product is fixed, if Ka is large, Kb must be small, and vice versa:

• Strong acid → negligible conjugate base. HCl has an enormous Ka, so Cl⁻ has a vanishingly small Kb. Chloride ions don’t make water alkaline.
• Weak acid → weak (but real) conjugate base. Ethanoic acid (Ka 1.8 × 10⁻⁵) has a conjugate base with Kb 5.6 × 10⁻¹⁰. Small, but enough to raise pH noticeably.
• Very weak acid → fairly strong conjugate base. HCN has Ka 6.2 × 10⁻¹⁰, so cyanide has Kb 1.6 × 10⁻⁵, almost exactly as strong a base as ammonia.

Acid	Ka	pKa	Conjugate base	Kb	pKb
HF	6.8 × 10⁻⁴	3.17	F⁻	1.5 × 10⁻¹¹	10.83
CH₃COOH	1.8 × 10⁻⁵	4.74	CH₃COO⁻	5.6 × 10⁻¹⁰	9.26
NH₄⁺	5.6 × 10⁻¹⁰	9.25	NH₃	1.8 × 10⁻⁵	4.75
HCN	6.2 × 10⁻¹⁰	9.21	CN⁻	1.6 × 10⁻⁵	4.79
HCO₃⁻	4.7 × 10⁻¹¹	10.33	CO₃²⁻	2.1 × 10⁻⁴	3.67

Notice that every row adds up to 14.00 in the pKa and pKb columns (within rounding).

Worked example 1: Kb of fluoride

Ka of HF is 6.8 × 10⁻⁴. Find Kb of F⁻.

Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 6.8 × 10⁻⁴ = 1.5 × 10⁻¹¹

Or in log form: pKa = 3.17, so pKb = 14.00 − 3.17 = 10.83.

Worked example 2: Ka of the ammonium ion

Kb of ammonia is 1.8 × 10⁻⁵. What is Ka of NH₄⁺, and what is the pH of 0.10 mol/dm³ ammonium chloride?

• Ka(NH₄⁺) = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰
• NH₄⁺ is a weak acid, so [H⁺] ≈ √(Ka × c) = √(5.6 × 10⁻¹⁰ × 0.10) = 7.5 × 10⁻⁶ mol/dm³
• pH = 5.13

The chloride ion contributes nothing (it’s the conjugate base of a strong acid), so the solution is mildly acidic because of the ammonium ion alone. This is why ammonium-based fertilisers gradually make soil more acidic.

Worked example 3: sodium cyanide

Calculate the pH of 0.10 mol/dm³ NaCN. Ka of HCN = 6.2 × 10⁻¹⁰.

• Kb(CN⁻) = 1.0 × 10⁻¹⁴ ÷ 6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵
• [OH⁻] ≈ √(1.6 × 10⁻⁵ × 0.10) = 1.27 × 10⁻³ mol/dm³
• pOH = 2.90, so pH = 11.10

A “salt” solution with a pH above 11 surprises many students. It happens because hydrocyanic acid is so weak that its conjugate base is correspondingly strong. (Cyanide salts are highly toxic and are not used in school labs; this example is for calculation only.)

Worked example 4: which salt is more basic?

Which is more alkaline at the same concentration: sodium fluoride or sodium ethanoate?

Compare Kb values, or compare the Ka values of the parent acids:

• HF: pKa 3.17 → F⁻ pKb 10.83
• CH₃COOH: pKa 4.74 → CH₃COO⁻ pKb 9.26

Ethanoate has the lower pKb, so it’s the stronger base. Sodium ethanoate is more alkaline. The weaker the parent acid, the more basic its salt.

Worked example 5: working with pKb directly

Methylamine has pKb = 3.36. What is the pKa of the methylammonium ion, CH₃NH₃⁺?

pKa = 14.00 − 3.36 = 10.64

Many modern textbooks and databases quote base strength this way, as the pKa of the conjugate acid (sometimes written pKaH). A higher pKaH means a stronger base. It’s one scale for everything, which is why organic chemists prefer it.

Predicting the pH of any salt

The relationship turns salt pH into a quick decision:

Salt made from	Example	Solution
Strong acid + strong base	NaCl, KNO₃	Neutral (pH 7)
Strong acid + weak base	NH₄Cl	Acidic (cation is a weak acid)
Weak acid + strong base	CH₃COONa, Na₂CO₃	Basic (anion is a weak base)
Weak acid + weak base	CH₃COONH₄	Depends: compare Ka of cation with Kb of anion

For the last case: ammonium ethanoate is almost exactly neutral, because Ka(NH₄⁺) = 5.6 × 10⁻¹⁰ and Kb(CH₃COO⁻) = 5.6 × 10⁻¹⁰ happen to be equal. Ammonium carbonate is basic, because Kb(CO₃²⁻) is far bigger than Ka(NH₄⁺).

A real-world example: why fertilisers change soil pH

Farmers see Ka × Kb = Kw at work every season. Ammonium sulfate and ammonium nitrate are popular nitrogen fertilisers. Their anions (sulfate and nitrate) come from strong acids, so they have essentially no basic character. Their cation, NH₄⁺, is the conjugate acid of ammonia, a weak base, so it has a small but real Ka of about 5.6 × 10⁻¹⁰. Year after year, that mild acidity adds up, and soil bacteria that convert ammonium into nitrate release even more H⁺. Left unchecked, the soil slowly acidifies and crops suffer.

The fix uses the same logic in reverse. Farmers spread ground limestone, calcium carbonate. Carbonate is the conjugate base of HCO₃⁻, a very weak acid, so carbonate has a relatively large Kb and neutralises the excess acid. Knowing one constant for each conjugate pair lets an agronomist predict both the problem and the cure.

Practice questions

1. Ka of benzoic acid is 6.3 × 10⁻⁵. Find Kb and pKb for the benzoate ion.
2. The conjugate acid of pyridine has pKa 5.23. What is Kb for pyridine?
3. Without calculating, rank these 0.1 mol/dm³ solutions from most acidic to most alkaline: NaF, NH₄Cl, NaCl, NaCN.

Answers:

1. Kb = 1.0 × 10⁻¹⁴ ÷ 6.3 × 10⁻⁵ = 1.6 × 10⁻¹⁰; pKb = 9.80.
2. pKb = 14.00 − 5.23 = 8.77, so Kb = 10⁻⁸·⁷⁷ = 1.7 × 10⁻⁹.
3. NH₄Cl (acidic) < NaCl (neutral) < NaF (very slightly basic) < NaCN (clearly basic).

Temperature: Kw isn’t always 10⁻¹⁴

The rule is really Ka × Kb = Kw, and Kw changes with temperature (5.5 × 10⁻¹⁴ at 50 °C, for example). At temperatures other than 25 °C, use the correct Kw and don’t assume pKa + pKb = 14. For most exam questions, though, 25 °C is assumed.

Common mistakes

• Applying it to unrelated acids and bases. Ka × Kb = Kw only works for a conjugate pair, such as NH₄⁺/NH₃, never for, say, HF and NH₃.
• Multiplying pKa and pKb. In log form you add: pKa + pKb = pKw.
• Forgetting polyprotic subtleties. Carbonate’s Kb pairs with the Ka of HCO₃⁻ (Ka2 of carbonic acid), not Ka1.

Key takeaways

• For a conjugate pair, Ka × Kb = Kw, so pKa + pKb = 14.00 at 25 °C.
• It follows from adding the acid and base equilibria, which gives water’s self-ionisation.
• The weaker the acid, the stronger its conjugate base, and the more alkaline its salts.
• Use it to find Kb for anions and Ka for cations like NH₄⁺, then calculate salt pH as in pH of weak bases.
• Revise the basics in Ka and pKa, or check pH values with the pH calculator.