The big idea: A weak acid only partially dissociates, so it sits at equilibrium with its ions:
HA ⇌ H⁺ + A⁻
That equilibrium has an equilibrium constant — and we give it a special name: the acid-dissociation constant, Ka. It tells you how far the acid dissociates, so it measures the strength of a weak acid.
- acid-dissociation constant (mol dm⁻³)
- equilibrium concentration of H⁺ (mol dm⁻³)
- equilibrium concentration of the conjugate base
- equilibrium concentration of the undissociated acid
Larger Ka = stronger weak acid: Ka is just Kc for the dissociation, so a bigger Ka means the equilibrium lies further right — more H⁺ released, so the stronger the weak acid.
The same idea for a weak base, B + H2O ⇌ BH⁺ + OH⁻, gives the base-dissociation constant Kb; a larger Kb means a stronger weak base.
Like Kc, neither Ka nor Kb appears in the data booklet — you write the expression yourself from the equation.
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Ka values span many orders of magnitude (e.g. 10⁻³ to 10⁻¹⁰), which is awkward to compare. So we take the negative logarithm — exactly as pH does for [H⁺].
- the lower the pK_{a}, the stronger the acid
- the lower the pK_{b}, the stronger the base
The lower the pKa, the stronger the acid: Because of the minus sign, the scale flips:
- larger Ka → smaller pKa → stronger acid - smaller Ka → larger pKa → weaker acid
So to rank weak acids, line up their pKa values: the lowest pKa is the strongest. (Same for bases: lowest pKb = strongest base.)
Worked example — convert between Ka and pKa
Ethanoic acid has Ka = 1.8 × 10⁻⁵ mol dm⁻³. (a) Calculate its pKa. (b) Another acid has pKa = 3.75 — calculate its Ka, and state which acid is stronger.
Solution
- (a) Formula first — pKa is the negative log of Ka:
- Substitute and evaluate:
- (b) Reverse it — take the inverse log to recover Ka:
- Evaluate:
- Compare: the second acid has the lower pKa (3.75 < 4.74), so it is the stronger acid.
Final answer
(a) pKa = 4.74. (b) Ka = 1.8 × 10⁻⁴ mol dm⁻³; the second acid (lower pKa) is stronger.
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An acid HA and its conjugate base A⁻ are linked. Multiply the acid's Ka by the conjugate base's Kb and the [HA] and [A⁻] terms cancel, leaving just the ionic product of water, Kw.
- dissociation constant of the acid
- dissociation constant of its conjugate base
- ionic product of water = 1.0 × 10⁻¹⁴ at 298 K
Two equivalent forms: The same relationship written two ways (at 298 K, where Kw = 1.0 × 10⁻¹⁴):
- in constants: Ka × Kb = Kw - in logs: pKa + pKb = 14
So a strong acid (small pKa) must have a weak conjugate base (large pKb) — they always add to 14.
Worked example — Kb of a conjugate base from Kw
Methanoic acid, HCOOH, has Ka = 6.2 × 10⁻⁵ mol dm⁻³ at 298 K. Calculate Kb for its conjugate base, the methanoate ion HCOO⁻, and find pKb.
Solution
- Formula first — for a conjugate pair, Ka × Kb = Kw, so rearrange for Kb:
- Substitute (Kw = 1.0 × 10⁻¹⁴ at 298 K):
- Work it out:
- Then pKb is its negative log:
- Check: pKa = −log(6.2 × 10⁻⁵) = 4.21, and 4.21 + 9.79 = 14
Final answer
Kb = 1.6 × 10⁻¹⁰ mol dm⁻³, pKb = 9.79 (and pKa + pKb = 14).
To find the pH of a weak acid you cannot use [H⁺] = c (that's only for a strong acid). Instead start from the Ka expression and make two approximations that lead to a tidy square-root formula.
- equilibrium hydrogen-ion concentration (mol dm⁻³)
- acid-dissociation constant
- initial (stated) concentration of the weak acid (mol dm⁻³)
The two assumptions (state them): Deriving [H⁺] = √(Kac) needs two approximations:
1. [H⁺] = [A⁻] — the acid is the only significant source of H⁺ (ignore the tiny amount from water).
2. [HA] ≈ c — dissociation is so small that the equilibrium acid concentration is unchanged from the initial value c.
These are valid when the acid is weak and not too dilute. State them to earn the reasoning marks.
Worked example — pH of a weak acid
Calculate the pH of a 0.10 mol dm⁻³ solution of a weak acid HX with Ka = 1.5 × 10⁻⁵ mol dm⁻³. State your assumptions.
Solution
- State the assumptions: [H⁺] = [X⁻] (acid is the only source of H⁺) and [HX] ≈ c = 0.10 (dissociation is negligible).
- Formula first — under those assumptions Ka = [H⁺]²/c, so:
- Substitute:
- Evaluate the concentration:
- Finally take the negative log for pH:
Final answer
pH = 2.91.
Worked example — pH of a weak base
Calculate the pH of a 0.20 mol dm⁻³ solution of a weak base B with Kb = 4.4 × 10⁻⁴ mol dm⁻³ at 298 K.
Solution
- Formula first — the base mirror gives [OH⁻] from Kb and c:
- Substitute:
- Get pOH:
- Convert to pH using pH + pOH = 14 (at 298 K):
Final answer
pH = 11.97.