Ka, Kw and pKa

Lesson 1 developed pH for strong acids, where dissociation is essentially complete and [H⁺] equals the analytical concentration. Weak acids — ethanoic acid, hydrocyanic acid, the ammonium ion, every carboxylic acid in organic chemistry — dissociate only partially, and their pH cannot be read off the bottle label. To compare weak acids we need a thermodynamic measure of acid strength: the acid dissociation constant Ka and its logarithmic form pKa. This lesson introduces Ka, pKa, the analogous base constants Kb and pKb, the ionic product of water Kw, and the conjugate-pair relation Ka·Kb = Kw. By the end you will be able to compute pH from Ka and concentration, work Ka backwards from a measured pH, convert between Ka and pKa, and rank acid strengths from tabulated pKa values. The framework here underpins lesson 3 (weak-acid pH curves) and lesson 4 (the Henderson-Hasselbalch buffer equation).

Spec mapping (AQA 7405): This lesson maps to §3.1.12 (acids and bases — the equilibrium constants Ka, Kb, and Kw and the pH calculations they support). It builds on lesson 1 of this course (pH of strong acids and the definition pH = −log₁₀[H⁺]), prepares for lesson 3 (pH of weak monoprotic and polyprotic acids using the full quadratic treatment) and lesson 4 (buffer pH via the Henderson-Hasselbalch equation pH = pKa + log([A⁻]/[HA])), and connects synoptically to §3.1.6 (chemical equilibrium — Kc and the equilibrium law from which Ka is derived as a special case). Refer to the official AQA 7405 specification document for the exact wording of each statement.

Assessment objectives: AO1 recall items include the definitions of Ka, pKa, Kw, Kb, and pKb, and the relation Ka·Kb = Kw at 298 K (equivalently pKa + pKb = 14.00). AO2 calculation skills include computing Ka from a measured pH and analytical concentration, computing pH from Ka and concentration (with and without the small-x approximation), converting Ka to pKa and back, and applying Ka·Kb = Kw to find the Kb of a conjugate base from the Ka of its parent acid. AO3 reasoning includes comparing relative acid strengths from tabulated pKa values, rationalising pKa trends in organic chemistry (electron-withdrawing groups lower pKa; charge build-up on the conjugate base raises pKa for successive dissociations), and judging when the small-x approximation is acceptable (typical AQA tolerance: less than 5% dissociation).

---

Ka — The Acid Dissociation Constant

For a generic weak monoprotic acid HA dissociating in water:

HA(aq) ⇌ H⁺(aq) + A⁻(aq)

(Strictly the proton is solvated as H₃O⁺(aq); the AQA convention abbreviates this to H⁺(aq) and we will follow it throughout. Lesson 1 discussed the convention in detail.)

Applying the equilibrium law (the same framework as Kc in §3.1.6):

Ka = [H⁺][A⁻] / [HA]

where the square-bracket concentrations are the equilibrium concentrations in mol dm⁻³. Because [HA] appears in the denominator and the right-hand-side product in the numerator, Ka has units of mol dm⁻³ (the same units as a single concentration — the two numerator concentrations divided by one denominator concentration leave one unit of mol dm⁻³ overall). This is the AQA convention; you should always quote the units mol dm⁻³ unless the question explicitly asks for the dimensionless thermodynamic constant.

Ka depends on temperature but not on initial concentration. A stronger acid pushes the equilibrium further to the right at equilibrium, giving larger [H⁺][A⁻] for a given [HA], and so a larger Ka. A weaker acid has a smaller Ka. The range of Ka values in chemistry spans more than twenty orders of magnitude, which is why we routinely work with the logarithmic transform pKa instead.

Key Point: Ka is the equilibrium constant for acid dissociation specifically — the reaction HA ⇌ H⁺ + A⁻. It is not the equilibrium constant for the reverse reaction (protonation of A⁻), nor for the autoionisation of water, nor for any other process. Always check that the equilibrium expression matches the dissociation direction.

---

pKa — The Logarithmic Form

Because Ka values span many orders of magnitude, we work with the negative base-10 logarithm:

pKa = −log₁₀(Ka)

Equivalently, Ka = 10⁻ᵖᴷᵃ. The transformation compresses a range of Ka values from ~10⁻³⁰ to ~10⁺¹⁰ into a manageable scale roughly −10 to +30. The key intuition:

• Stronger acid → larger Ka → smaller (or more negative) pKa.
• Weaker acid → smaller Ka → larger (more positive) pKa.

A useful mental anchor: a pKa unit change of 1 corresponds to a tenfold change in Ka and therefore — at the same concentration — roughly a √10 ≈ 3.2-fold change in [H⁺] for a weak acid (the square root arises because [H⁺] ∝ √Ka in the small-x limit; see the worked examples below).

---

Table of pKa Values at 298 K

The following table lists pKa values for acids that appear repeatedly in AQA A-Level questions. All values are quoted at 298 K and refer to the dissociation in dilute aqueous solution.

Acid	Formula	Ka (mol dm⁻³)	pKa	Comment
Hydrochloric acid	HCl	~10⁷	≈ −7	Very strong; complete dissociation assumed
Sulfuric acid (1st)	H₂SO₄	~10³	≈ −3	Strong; first dissociation complete in dilute solution
Hydrogen sulfate ion	HSO₄⁻	1.0 × 10⁻²	2.0	The second dissociation of sulfuric acid — only partial
Phosphoric acid (1st)	H₃PO₄	7.08 × 10⁻³	2.15	First of three dissociations
Methanoic acid	HCOOH	1.78 × 10⁻⁴	3.75	Stronger than ethanoic — no electron-donating methyl
Ethanoic acid	CH₃COOH	1.74 × 10⁻⁵	4.76	The reference weak acid for A-Level work
Carbonic acid (1st)	H₂CO₃	4.45 × 10⁻⁷	6.35	First of two dissociations
Phosphoric acid (2nd)	H₂PO₄⁻	6.31 × 10⁻⁸	7.20	Buffering region for physiological systems
Hydrocyanic acid	HCN	6.17 × 10⁻¹⁰	9.21	A weak acid; lesson 3 example
Ammonium ion	NH₄⁺	5.62 × 10⁻¹⁰	9.25	Conjugate acid of ammonia
Hydrogen carbonate ion	HCO₃⁻	4.69 × 10⁻¹¹	10.33	Second dissociation of carbonic acid
Phenol	C₆H₅OH	1.00 × 10⁻¹⁰	10.00	The benchmark "acidic" phenol
Hydrogen phosphate ion	HPO₄²⁻	4.79 × 10⁻¹³	12.32	Third dissociation of phosphoric acid
Water	H₂O	1.8 × 10⁻¹⁶	15.7	Self-ionisation; see Kw treatment below

A few patterns worth noting:

• HCl and H₂SO₄ first-dissociation pKa values are approximate: for strong acids the equilibrium lies so far to the right that direct Ka measurement is experimentally challenging; values are inferred from thermodynamic cycles.
• Successive polyprotic dissociations are always weaker. H₃PO₄: pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.32 — each step ~5 pKa units weaker. Removing H⁺ from an already-anionic species is electrostatically unfavourable.
• NH₄⁺ (9.25) and HCN (9.21) have coincidentally close pKa values — useful sanity check, but they are unrelated chemistries.

---

Kb and pKb — The Analogous Base Constants

For a generic weak base B accepting a proton from water:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

The base equilibrium law (with [H₂O] absorbed into the constant, as it is in vast excess in dilute solution):

Kb = [BH⁺][OH⁻] / [B]

with units mol dm⁻³ on the same AQA convention as Ka. The logarithmic form is pKb = −log₁₀(Kb). Stronger base → larger Kb → smaller pKb. Ammonia, the reference weak base for A-Level work, has Kb = 1.78 × 10⁻⁵ mol dm⁻³ at 298 K, giving pKb(NH₃) = 4.75.

Note the close numerical similarity between pKb(NH₃) = 4.75 and pKa(CH₃COOH) = 4.76 — this is a celebrated coincidence in elementary acid-base chemistry, useful for sanity-checking calculations and for designing buffers that hold their conjugate-pair concentrations very close to 1:1.

---

Kw — The Ionic Product of Water

Pure water itself dissociates very slightly:

H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

The equilibrium constant for this autoionisation, with [H₂O] absorbed (because water is the solvent and effectively at constant concentration in dilute aqueous solutions), is the ionic product of water:

Kw = [H⁺][OH⁻]

At 298 K, the experimentally measured value is:

Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶

The units of Kw are mol² dm⁻⁶ because it is the product of two concentrations with no concentration in the denominator. Taking negative base-10 logarithms of both sides:

pKw = −log₁₀(Kw) = 14.00 at 298 K pH + pOH = 14.00 at 298 K

This last relation — pH + pOH = 14 — is one of the most-used identities in acid-base chemistry, and it is the basis of how we compute pH for solutions of strong bases (find [OH⁻], take pOH, subtract from 14).

Kw is strongly temperature-dependent. Because autoionisation is endothermic (it breaks an O—H bond), Le Chatelier's principle predicts that raising the temperature shifts the equilibrium to the right, increasing Kw. Approximate values:

Temperature	Kw (mol² dm⁻⁶)	pKw	pH of pure water
273 K (0 °C)	1.14 × 10⁻¹⁵	14.94	7.47
298 K (25 °C)	1.00 × 10⁻¹⁴	14.00	7.00
310 K (37 °C)	2.42 × 10⁻¹⁴	13.62	6.81
373 K (100 °C)	5.13 × 10⁻¹³	12.29	6.14

Body-temperature water is slightly acidic on the pH scale (pH ≈ 6.81), but it is still neutral in the technical sense — [H⁺] = [OH⁻] at every temperature for pure water. Neutrality is defined by [H⁺] = [OH⁻], not by pH = 7. An exam question that asks about pure water at any temperature other than 298 K is testing whether you understand this distinction.

---

The Conjugate-Pair Relation: Ka·Kb = Kw

The most important quantitative relationship in this lesson links Ka of a weak acid to Kb of its conjugate base. Take a weak acid HA and its conjugate base A⁻:

• Acid dissociation: HA ⇌ H⁺ + A⁻, with Ka = [H⁺][A⁻]/[HA]
• Conjugate base hydrolysis: A⁻ + H₂O ⇌ HA + OH⁻, with Kb = [HA][OH⁻]/[A⁻]

Multiplying the two expressions:

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

Therefore:

Ka × Kb = Kw (= 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K)

Taking negative base-10 logarithms:

pKa + pKb = pKw = 14.00 (at 298 K)

This is the conjugate-pair relation. Its practical use is enormous: if you have Ka of any weak acid, you can immediately deduce Kb of its conjugate base (and vice versa) without doing a separate measurement.

Worked Example: Kb of CN⁻ from Ka of HCN

Given Ka(HCN) = 6.17 × 10⁻¹⁰ mol dm⁻³ at 298 K, find Kb of the cyanide ion CN⁻.

Kb(CN⁻) = Kw / Ka(HCN) = (1.00 × 10⁻¹⁴) / (6.17 × 10⁻¹⁰) = 1.62 × 10⁻⁵ mol dm⁻³.

Equivalently, pKb(CN⁻) = 14.00 − pKa(HCN) = 14.00 − 9.21 = 4.79.

Cyanide is a moderately strong weak base — comparable in strength to ammonia — because its conjugate acid (HCN) is weak.

Worked Example: pKa of NH₄⁺ from pKb of NH₃

Given pKb(NH₃) = 4.75 at 298 K, find pKa of the ammonium ion NH₄⁺.

pKa(NH₄⁺) + pKb(NH₃) = 14.00 pKa(NH₄⁺) = 14.00 − 4.75 = 9.25

This matches the tabulated value of 9.25 quoted above.

Exam Tip: The conjugate-pair relation is the most common AO2 calculation in §3.1.12 questions. Any time you are given Ka and asked about the conjugate base, or given Kb and asked about the conjugate acid, the route is Ka·Kb = Kw. Memorise this relation and the logarithmic form pKa + pKb = 14 at 298 K.

---

Computing Ka from a Measured pH

The experimentally accessible quantity for a weak acid is pH, measured with a glass-electrode pH meter or estimated by indicator titration. Given the analytical concentration c and the measured pH, the route to Ka is:

1. Find [H⁺] from the pH: [H⁺] = 10⁻ᵖᴴ.
2. At equilibrium [A⁻] = [H⁺] (stoichiometry of HA → H⁺ + A⁻ — every proton released leaves one A⁻ behind).
3. [HA] at equilibrium = c − [H⁺] (whatever dissociated has been removed from the undissociated reservoir). For weakly dissociating acids the small-x approximation [HA] ≈ c is often valid (see below).
4. Compute Ka from Ka = [H⁺]²/[HA] (using [A⁻] = [H⁺] in the numerator).

Worked Example 1: Ka of an Unknown Weak Acid

A 0.100 mol dm⁻³ solution of a weak acid HA has measured pH = 2.88 at 298 K. Calculate Ka.

[H⁺] = 10⁻²·⁸⁸ = 1.32 × 10⁻³ mol dm⁻³ [A⁻] = [H⁺] = 1.32 × 10⁻³ mol dm⁻³ [HA]_eq = 0.100 − 1.32 × 10⁻³ = 0.09868 mol dm⁻³ (so [HA] ≈ 0.0987; the small-x approximation [HA] ≈ 0.100 is valid here because 1.32 × 10⁻³ / 0.100 = 1.32% — well under 5%)

Ka = [H⁺][A⁻]/[HA] = (1.32 × 10⁻³)² / 0.0987 = 1.74 × 10⁻⁶ / 0.0987 = 1.77 × 10⁻⁵ mol dm⁻³

So pKa = −log₁₀(1.77 × 10⁻⁵) = 4.75 — this is essentially ethanoic acid (pKa 4.76). The 0.01 discrepancy is rounding noise within the precision of the measured pH.

---

Computing pH from Ka and Concentration

The reverse direction — given Ka and the analytical concentration c, predict pH — is the more common AO2 problem. The standard procedure:

1. Write the ICE table (Initial, Change, Equilibrium) for HA ⇌ H⁺ + A⁻.
2. Apply the equilibrium law Ka = x²/(c − x), where x = [H⁺] at equilibrium.
3. Small-x approximation: if Ka is small and c is reasonably large (typically c/Ka > 100 — a quick rule of thumb), neglect x in the denominator: Ka ≈ x²/c, giving x ≈ √(Ka × c).
4. Take pH = −log₁₀(x) = −log₁₀(√(Ka × c)) = ½(pKa + p[HA]), where p[HA] = −log₁₀(c).

The approximation form gives the elegant relation:

pH ≈ ½ × (pKa − log₁₀(c)) for a weak monoprotic acid at concentration c.

Equivalently:

pH ≈ ½(pKa + p[HA])

where p[HA] = −log₁₀(c) is the "p-value" of the analytical concentration.

Worked Example 2: pH of 0.10 mol dm⁻³ Ethanoic Acid

Ka(CH₃COOH) = 1.74 × 10⁻⁵ mol dm⁻³; c = 0.10 mol dm⁻³.

Approximation: x = √(Ka × c) = √(1.74 × 10⁻⁵ × 0.10) = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³.

pH = −log₁₀(1.32 × 10⁻³) = 2.88.

Check the approximation: x/c = 1.32 × 10⁻³ / 0.10 = 1.32% — well under 5%, so the approximation is valid.

Alternatively, using the pH formula directly: pH = ½(pKa − log₁₀ c) = ½(4.76 − log₁₀(0.10)) = ½(4.76 − (−1.00)) = ½(5.76) = 2.88. Same answer, with less algebra.

Worked Example 3: pH of 0.05 mol dm⁻³ Hydrocyanic Acid

Ka(HCN) = 6.17 × 10⁻¹⁰; c = 0.05 mol dm⁻³.

x = √(6.17 × 10⁻¹⁰ × 0.05) = √(3.085 × 10⁻¹¹) = 5.55 × 10⁻⁶ mol dm⁻³.

pH = −log₁₀(5.55 × 10⁻⁶) = 5.26.

Approximation check: 5.55 × 10⁻⁶ / 0.05 = 1.1 × 10⁻⁴ = 0.011% — the small-x approximation is extremely safe for HCN at this concentration. Hydrocyanic acid is so weak that its dissociation is negligible compared with the analytical concentration.

Common student error: Treating HCN as if it were a strong acid (i.e. computing pH = −log₁₀(0.05) = 1.30). This would imply complete dissociation, which is wrong by five orders of magnitude. Always identify whether the acid is strong (complete) or weak (use Ka) before computing pH.

Worked Example 4: When the Small-x Approximation Fails

Consider 0.001 mol dm⁻³ chloroethanoic acid (ClCH₂COOH), with Ka = 1.36 × 10⁻³ mol dm⁻³.

Small-x estimate: x = √(1.36 × 10⁻³ × 0.001) = 1.17 × 10⁻³ mol dm⁻³. Check: x/c = 117% — physically impossible. The small-x form has broken down; use the full quadratic.

x² + Ka·x − Ka·c = 0; x = (−Ka + √(Ka² + 4·Ka·c))/2 = (−1.36 × 10⁻³ + √(7.29 × 10⁻⁶))/2 = (−1.36 × 10⁻³ + 2.70 × 10⁻³)/2 = 6.70 × 10⁻⁴ mol dm⁻³.

pH = −log₁₀(6.70 × 10⁻⁴) = 3.17. True dissociation x/c = 67%, far above the 5% threshold. The full quadratic is essential for moderately strong weak acids at dilute concentrations. Lesson 3 develops this in full; for most AQA questions at 0.10 mol dm⁻³, the small-x form is valid.

---

Polyprotic Acid Stepwise Dissociation

A polyprotic acid releases more than one proton, each with its own equilibrium constant Ka1, Ka2, Ka3, etc. Each successive dissociation is weaker than the previous, for the electrostatic reason given above: removing H⁺ from an already-anionic species is unfavourable.

Sulfuric Acid

H₂SO₄ → H⁺ + HSO₄⁻ (pKa1 ≈ −3; complete in dilute solution) HSO₄⁻ ⇌ H⁺ + SO₄²⁻ (pKa2 = 1.99; partial — a moderately weak acid)

For most A-Level purposes the second dissociation is neglected and [H⁺] ≈ 2 × c is used for dilute H₂SO₄ — though a rigorous treatment includes both steps because the second contributes detectably to pH above ~0.01 mol dm⁻³.

Phosphoric Acid

H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ (pKa1 = 2.15) H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (pKa2 = 7.20) HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (pKa3 = 12.32)

Each step is ~5 pKa units weaker than the previous, reflecting the electrostatic penalty of removing H⁺ from progressively more negative species. The second-step pKa (7.20) sits close to physiological pH (7.4) — the basis of the H₂PO₄⁻/HPO₄²⁻ buffer in intracellular fluid.

Carbonic Acid

H₂CO₃ ⇌ H⁺ + HCO₃⁻ (pKa1 = 6.35) HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (pKa2 = 10.33)