“Weak” is not a measurement. Ethanoic acid and hydrocyanic acid are both weak, but one is about a thousand times better at letting go of its proton than the other. To rank them properly you need a number, and since a weak acid in water is just an equilibrium, that number is an equilibrium constant.
📘 What you need to know
For a weak acid, HA ↔ H+ + A−, the acid dissociation constant is Ka = [H+][A−] ÷ [HA].
For a weak base, B + H2O ↔ BH+ + OH−, the base dissociation constant is Kb = [BH+][OH−] ÷ [B].
Water never appears in either expression, for the same reason it vanished from Kw.
Bigger Ka = stronger acid. Bigger Kb = stronger base.
pKa = −log10Ka, so bigger pKa means a weaker acid — the opposite way round.
For a conjugate pair: Ka × Kb = Kw, and taking logs, pKa + pKb = 14.00 at 298 K.
Most weak acids have pKa between about 3 and 11.
Writing the expressions
Nothing new is going on here. It is the ordinary equilibrium law applied to a dissociation, with the water folded into the constant just as it was for Kw.
So for propanoic acid you would write Ka = [CH3CH2COO−][H+] ÷ [CH3CH2COOH], and for methylamine Kb = [CH3NH3+][OH−] ÷ [CH3NH2].
The commonest error in this whole topic is leaving [H2O] on the bottom of a Kb expression. Water is the solvent, it is present in enormous excess, and its concentration barely moves — so it has already been absorbed into the constant. Leave it out.
Big number, small number
Ka compares products to reactants. A weak acid barely dissociates, so the top of that fraction is tiny and the bottom is large. That is why Ka values come out as awkward things like 1.74 × 10−5.
Comparing 1.74 × 10−5 with 6.2 × 10−10 in your head is annoying, so we take a negative log and turn them into friendly numbers between roughly 3 and 11. That is pKa.
Converting both ways
pKa = −log10Ka and Ka = 10−pKa
pKb = −log10Kb and Kb = 10−pKb
Watch the direction. Because of the minus sign, the ranking flips. A largeKa means a strong acid, but a large pKa means a weak one. Ethanoic acid (pKa 4.76) is a much stronger acid than the ammonium ion (pKa 9.25).
Spot HCO3− in both columns. It is the conjugate base of carbonic acid and an acid in its own right — exactly what being amphiprotic means.
Acid
Ka
pKa
Strength
Methanoic acid, HCOOH
1.77 × 10−4
3.75
strongest of these
Benzoic acid, C6H5COOH
6.46 × 10−5
4.19
next
Ethanoic acid, CH3COOH
1.74 × 10−5
4.76
middle
Carbonic acid, H2CO3
4.30 × 10−7
6.37
weaker
Ammonium ion, NH4+
5.60 × 10−10
9.25
very weak
Hydrogencarbonate, HCO3−
4.80 × 10−11
10.32
weakest of these
The link between a pair
Take any weak acid and its conjugate base and multiply their two constants together. Something remarkable happens.
Cancel [A−] top and bottom, then [HA] top and bottom, and only [H+][OH−] is left. That is a two-line derivation worth being able to reproduce.
This is the algebra behind something you learned back on the conjugate pairs page. If Ka is large, then Kb must be small, because their product is stuck at 10−14. Strong acid, weak conjugate base, proved rather than just asserted.
Worked examples
WORKED EXAMPLE
Write the Ka expression for hydrocyanic acid, HCN, and the Kb expression for ethylamine, C2H5NH2.
Step 1: write each equilibrium firstHCN(aq) ↔ H+(aq) + CN−(aq)C2H5NH2(aq) + H2O(l) ↔ C2H5NH3+(aq) + OH−(aq)Step 2: products on top, reactants on the bottomStep 3: leave water outKa = [H+][CN−] ÷ [HCN]Kb = [C2H5NH3+][OH−] ÷ [C2H5NH2]Both written with no [H2O] termwrite the equation first, every time — the expression then writes itself
WORKED EXAMPLE
Ethanoic acid has Ka = 1.74 × 10−5. Find pKa, then find Kb and pKb for the ethanoate ion.
Step 1: take the negative logpKa = −log10(1.74 × 10−5) = 4.76Step 2: use the pair relationship for pKbpKb = 14.00 − 4.76 = 9.24Step 3: convert back for KbKb = 10−9.24 = 5.75 × 10−10Step 4: check with the other route1.00 × 10−14 ÷ 1.74 × 10−5 = 5.75 × 10−10 — agreespKa = 4.76, Kb = 5.75 × 10−10, pKb = 9.24the tiny K₋ confirms ethanoate is a very weak base
WORKED EXAMPLE
Acid X has pKa 3.20 and acid Y has Ka = 2.0 × 10−8. Which is the stronger acid, and which has the stronger conjugate base?
Step 1: put both on the same scaleY: pKa = −log10(2.0 × 10−8) = 7.70Step 2: compare, remembering the flip
Lower pKa means stronger acid, and 3.20 is lower than 7.70.
Step 3: apply the pair rule for the bases
Stronger acid gives weaker conjugate base, so the weaker acid Y has the stronger one.
X is the stronger acid; Y has the stronger conjugate basealways convert to the same units before comparing — mixing Kₐ with pKₐ causes chaos
💡 Exam tip
Write the equilibrium equation before the expression. It stops you inventing the wrong conjugate.
Never put [H2O] in a Ka or Kb expression.
Say it out loud once: big Ka, strong acid; big pKa, weak acid. The reversal is tested constantly.
The relationships KaKb = Kw and pKa + pKb = 14 only apply to a conjugate pair, not to any random acid and base.
Both constants are equilibrium constants, so they change with temperature but not with concentration.
Look up values in the data booklet rather than memorising them, but do memorise the shape of the ladder.
⚠ Common mix-up
Thinking a bigger pKa means a stronger acid. The minus sign in the log reverses everything.
Including water in the expression. It has been absorbed into the constant already.
Pairing Ka of one acid with Kb of an unrelated base. The product rule needs a genuine conjugate pair.
Assuming Ka changes when you dilute. Concentrations shift, the constant does not.
Writing a Kb expression with H+ in it. A base equilibrium produces OH−, not H+.
Confusing Ka with degree of dissociation. The constant is fixed; the fraction that dissociates depends on concentration too.
Up next: Solving Acid–Base Dissociation Problems. You now have the constants. Next we turn them into pH values, using two approximations that make the algebra collapse into a square root.
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