Weak Acids and $K_a$Ka​

Spec Mapping — OCR H432 Module 5.1.3 — Acids, bases and buffers, sub-topic on weak acids: the definition of partial dissociation, the acid dissociation constant $K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]$Ka​=[H+][A−]/[HA] with units mol dm⁻³, the logarithmic form $\text{p}K_a = -\log_{10} K_a$pKa​=−log10​Ka​, comparing acid strengths using $K_a$Ka​ and $\text{p}K_a$pKa​, the relationship $K_a \cdot K_b = K_w$Ka​⋅Kb​=Kw​ for a conjugate pair, and the qualitative reasons why some acids are weak (refer to the official OCR H432 specification document for exact wording). This is the conceptual setup for lesson 5 (pH from $K_a$Ka​), lessons 7–8 (buffer derivation and calculation) and lesson 10 (indicator $\text{p}K_{In}$pKIn​ matching).

The leap from strong-acid pH (lesson 3) to weak-acid pH is the largest single conceptual jump in OCR Module 5.1.3. Strong-acid pH is a one-line calculation because the equilibrium lies fully to the right; weak-acid pH requires a genuine equilibrium expression because only a small fraction of HA molecules have dissociated at any instant. The acid dissociation constant $K_a$Ka​ is the workhorse for the rest of the module — it appears in every pH calculation for weak acids (lesson 5), every buffer derivation (lessons 7–8), every titration curve through the half-equivalence point (lesson 9), and every indicator selection (lesson 10). This lesson focuses on setting up $K_a$Ka​ correctly — writing the expression, getting the units right, understanding what "weak" means quantitatively, and comparing acid strengths via $\text{p}K_a$pKa​. The full quantitative calculation of pH from $K_a$Ka​ is reserved for lesson 5.

Key Equation: $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \quad \text{units: mol dm}^{-3}$Ka​=[HA][H+][A−]​units: mol dm−3 $\text{p}K_a = -\log_{10}(K_a) \qquad K_a = 10^{-\text{p}K_a}$pKa​=−log10​(Ka​)Ka​=10−pKa​ $K_a \cdot K_b = K_w \quad \text{(conjugate pair, 298 K)}$Ka​⋅Kb​=Kw​(conjugate pair, 298 K) A larger $K_a$Ka​ (smaller $\text{p}K_a$pKa​) means a stronger acid.

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What is a weak acid?

A weak acid is one that undergoes only partial dissociation in aqueous solution. For a generic weak monobasic acid HA:

$\text{HA(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{A}^-\text{(aq)}$HA(aq)⇌H+(aq)+A−(aq)

The double-headed arrow signals dynamic equilibrium. At any instant only a small fraction of HA molecules have shed their proton; the bulk remain intact. For a typical weak acid such as ethanoic acid at 0.1 mol dm⁻³, only about 1.3 % of molecules are dissociated — 98.7 % are still molecular CH₃COOH.

This is structurally different from the strong-acid story. For HCl, the operational reality is essentially complete dissociation (residual undissociated HCl below detection), and we write a single forward arrow because the reverse reaction is negligible. For a weak acid, the reverse reaction (recombination of H⁺ and A⁻) is comparable in rate to the forward reaction at equilibrium, and the population of HA is overwhelmingly the molecular form.

Common weak acids you must recognise and use at A-Level:

Acid	Formula	$K_a$Ka​ (298 K) / mol dm⁻³	$\text{p}K_a$pKa​
Hydrofluoric	HF	$5.6 \times 10^{-4}$5.6×10−4	3.25
Methanoic	HCOOH	$1.6 \times 10^{-4}$1.6×10−4	3.80
Benzoic	C₆H₅COOH	$6.3 \times 10^{-5}$6.3×10−5	4.20
Ethanoic	CH₃COOH	$1.74 \times 10^{-5}$1.74×10−5	4.76
Propanoic	CH₃CH₂COOH	$1.3 \times 10^{-5}$1.3×10−5	4.89
Carbonic ($K_{a1}$Ka1​)	H₂CO₃	$4.3 \times 10^{-7}$4.3×10−7	6.37
Ammonium	NH₄⁺	$5.6 \times 10^{-10}$5.6×10−10	9.25
Hydrogen cyanide	HCN	$4.9 \times 10^{-10}$4.9×10−10	9.31
Phenol	C₆H₅OH	$1.3 \times 10^{-10}$1.3×10−10	9.89

You are not required to memorise specific values, but you must be able to use them confidently in calculations and rank acid strengths from $\text{p}K_a$pKa​.

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The acid dissociation constant $K_a$Ka​

Applying the equilibrium law to the dissociation $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$HA⇌H++A−:

$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$Ka​=[HA][H+][A−]​

where every concentration is the equilibrium concentration in mol dm⁻³. The units of $K_a$Ka​ follow from dimensional analysis: $(\text{mol dm}^{-3})(\text{mol dm}^{-3})/(\text{mol dm}^{-3}) = \text{mol dm}^{-3}$(mol dm−3)(mol dm−3)/(mol dm−3)=mol dm−3. OCR mark schemes reserve a mark for the explicit unit; do not omit it.

Why water does not appear in the expression

The strictly Brønsted–Lowry equation involves water:

$\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}$HA(aq)+H2​O(l)⇌H3​O+(aq)+A−(aq)

The full $K_c$Kc​ would include $[\text{H}_2\text{O}]$[H2​O] in the denominator. But water is the solvent — its concentration ($\sim 55.5$∼55.5 mol dm⁻³) is essentially constant — so we absorb it into the equilibrium constant just as we did for $K_w$Kw​ in lesson 2:

$K_a = K_c \cdot [\text{H}_2\text{O}] = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$Ka​=Kc​⋅[H2​O]=[HA][H3​O+][A−]​=[HA][H+][A−]​

The shorthand H⁺ is interchangeable with H₃O⁺ in this expression — they refer to the same chemical species.

Magnitude of $K_a$Ka​

• $K_a \sim 10^{-2}$Ka​∼10−2 to $10^{-4}$10−4: a "stronger weak acid" (HF, HSO₄⁻).
• $K_a \sim 10^{-4}$Ka​∼10−4 to $10^{-6}$10−6: a typical weak acid (most carboxylic acids).
• $K_a \sim 10^{-7}$Ka​∼10−7 to $10^{-10}$10−10: a very weak acid (carbonic, hydrocyanic, phenol).

The wide range — eight orders of magnitude across A-Level examples — is precisely why we move to a logarithmic scale.

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pKa — the logarithmic scale

Define $\text{p}K_a = -\log_{10}(K_a)$pKa​=−log10​(Ka​), by analogy with pH. Then $K_a = 10^{-\text{p}K_a}$Ka​=10−pKa​. The minus sign converts the typically small fractional $K_a$Ka​ values into convenient small positive numbers.

Worked example 1 — pKa from Ka

Ethanoic acid has $K_a = 1.74 \times 10^{-5}$Ka​=1.74×10−5 mol dm⁻³. Calculate $\text{p}K_a$pKa​.

$\text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76 \quad (2 \text{ dp})$pKa​=−log10​(1.74×10−5)=4.76(2 dp)

Worked example 2 — Ka from pKa

Phenol has $\text{p}K_a = 9.89$pKa​=9.89. Calculate $K_a$Ka​.

$K_a = 10^{-9.89} = 1.29 \times 10^{-10} \text{ mol dm}^{-3}$Ka​=10−9.89=1.29×10−10 mol dm−3

The inverse relationship

A larger $K_a$Ka​ means a stronger acid (further dissociated) but a smaller $\text{p}K_a$pKa​. This catches students out at every exam — when ranking acids by strength, look at $K_a$Ka​ ↑ or $\text{p}K_a$pKa​ ↓.

Acid	$K_a$Ka​ / mol dm⁻³	$\text{p}K_a$pKa​	Strength (within "weak")
HF	$5.6 \times 10^{-4}$5.6×10−4	3.25	Strongest
HCOOH	$1.6 \times 10^{-4}$1.6×10−4	3.80	↓
C₆H₅COOH	$6.3 \times 10^{-5}$6.3×10−5	4.20	↓
CH₃COOH	$1.74 \times 10^{-5}$1.74×10−5	4.76	↓
HCN	$4.9 \times 10^{-10}$4.9×10−10	9.31	↓
C₆H₅OH	$1.3 \times 10^{-10}$1.3×10−10	9.89	Weakest

All of these acids are "weak" relative to HCl. Even HF, the strongest of them, has $K_a$Ka​ about $10^{10}$1010 times smaller than HCl's.

flowchart LR
    A["Strong acid HCl Ka ~ 10^7"] --> B["HF Ka ~ 10^-3.3 (strongest weak)"]
    B --> C["HCOOH Ka ~ 10^-3.8"]
    C --> D["CH3COOH Ka ~ 10^-4.8"]
    D --> E["HCN Ka ~ 10^-9.3"]
    E --> F["Phenol Ka ~ 10^-9.9 (weakest)"]
    F --> G["Pure H2O Ka effectively Kw = 10^-14"]

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Writing Ka expressions for specific acids

The Ka expression always has the form: $K_a = \frac{[\text{products}]}{[\text{reactant HA}]}$Ka​=[reactant HA][products]​, each concentration raised to its stoichiometric coefficient. Practise on the following examples — every weak-acid pH question begins with this step.

Ethanoic acid

$\text{CH}_3\text{COOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CH}_3\text{COO}^-\text{(aq)}$CH3​COOH(aq)⇌H+(aq)+CH3​COO−(aq) $K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]}$Ka​=[CH3​COOH][H+][CH3​COO−]​

Methanoic acid

$\text{HCOOH(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{HCOO}^-\text{(aq)}$HCOOH(aq)⇌H+(aq)+HCOO−(aq) $K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}$Ka​=[HCOOH][H+][HCOO−]​

Ammonium ion (acidic in water)

$\text{NH}_4^+\text{(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{NH}_3\text{(aq)}$NH4+​(aq)⇌H+(aq)+NH3​(aq) $K_a = \frac{[\text{H}^+][\text{NH}_3]}{[\text{NH}_4^+]}$Ka​=[NH4+​][H+][NH3​]​

Hydrogen cyanide

$\text{HCN(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{CN}^-\text{(aq)}$HCN(aq)⇌H+(aq)+CN−(aq) $K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]}$Ka​=[HCN][H+][CN−]​

Hydrofluoric acid

$\text{HF(aq)} \rightleftharpoons \text{H}^+\text{(aq)} + \text{F}^-\text{(aq)}$HF(aq)⇌H+(aq)+F−(aq) $K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]}$Ka​=[HF][H+][F−]​

In every case, H⁺ and the conjugate base appear in the numerator; the undissociated HA appears alone in the denominator. The conjugate base of a weak acid is always nucleophilic enough to act as a Brønsted base in its own right — F⁻, CN⁻, HCOO⁻, CH₃COO⁻ and C₆H₅COO⁻ all have small but measurable $K_b$Kb​ values, in contrast to Cl⁻ which is essentially inert.

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Ka, Kb and Kw — the conjugate-pair relationship

For every weak acid HA there is a corresponding conjugate base A⁻ that can act as a weak base in water:

$\text{A}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{HA(aq)} + \text{OH}^-\text{(aq)}$A−(aq)+H2​O(l)⇌HA(aq)+OH−(aq) $K_b = \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}$Kb​=[A−][HA][OH−]​

Multiply $K_a \cdot K_b$Ka​⋅Kb​:

$K_a \cdot K_b = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \cdot \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]} = [\text{H}^+][\text{OH}^-] = K_w$Ka​⋅Kb​=[HA][H+][A−]​⋅[A−][HA][OH−]​=[H+][OH−]=Kw​

So $K_a \cdot K_b = K_w$Ka​⋅Kb​=Kw​ at any given temperature for any conjugate acid-base pair. Equivalently $\text{p}K_a + \text{p}K_b = \text{p}K_w = 14.00$pKa​+pKb​=pKw​=14.00 at 298 K. This relationship is not always directly tested at OCR but it explains the inverse-strength rule: the conjugate base of a very weak acid (e.g. CN⁻, where $\text{p}K_a(\text{HCN}) = 9.31$pKa​(HCN)=9.31) is a moderately strong base ($\text{p}K_b = 4.69$pKb​=4.69). Conversely, Cl⁻ (conjugate base of HCl, with $\text{p}K_a \approx -7$pKa​≈−7) is essentially inert as a base.

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Why some acids are weak — the structural picture

The strength of an acid HA depends on four factors that govern how readily the H–A bond ionises:

1. Bond enthalpy of H–A. A weaker H–A bond means H⁺ is more easily released. HI ($D(\text{H-I}) = 295$D(H-I)=295 kJ mol⁻¹, strong acid) ionises more easily than HF ($D(\text{H-F}) = 565$D(H-F)=565 kJ mol⁻¹, weak acid).
2. Polarity of the H–A bond. A more polar bond places more positive charge on H, favouring ionisation. The electronegativity difference matters.
3. Stabilisation of A⁻. Resonance delocalisation of negative charge stabilises the conjugate base and shifts the equilibrium right. Carboxylate ions RCOO⁻ are resonance-stabilised between two equivalent C–O bonds, which is why carboxylic acids are far more acidic than alcohols ROH (no resonance stabilisation of RO⁻).
4. Solvation of H⁺ and A⁻. Aqueous solvation stabilises charged species; smaller, more concentrated charges are better solvated. F⁻ is heavily solvated (smaller than Cl⁻ but with the same charge), which partially explains why HF is weaker than HCl despite the higher H–F bond polarity.

Within the carboxylic acid family, the substituent on the α-carbon dramatically tunes acidity. Adding electron-withdrawing chlorines stabilises the carboxylate ion through induction and increases acidity:

Acid	$K_a$Ka​ / mol dm⁻³	$\text{p}K_a$pKa​	Inductive effect
CH₃COOH	$1.74 \times 10^{-5}$1.74×10−5	4.76	None
ClCH₂COOH	$1.36 \times 10^{-3}$1.36×10−3	2.87	One Cl pulls e⁻ density off COO⁻
Cl₂CHCOOH	$5.13 \times 10^{-2}$5.13×10−2	1.29	Two Cls amplify the effect
Cl₃CCOOH	$2.30 \times 10^{-1}$2.30×10−1	0.64	Three Cls — close to a strong acid

Each chlorine substitution drops $\text{p}K_a$pKa​ by 1–2 units. Trichloroethanoic acid is approaching strong-acid territory.

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Setting up a Ka calculation — the ICE table preview

The full quantitative treatment is in lesson 5, but the framework starts here. For a weak acid HA initially at concentration $c$c, let $x$x be the equilibrium concentration of H⁺:

	HA	H⁺	A⁻
Initial	$c$c	0	0
Change	$-x$−x	$+x$+x	$+x$+x
Equilibrium	$c - x$c−x	$x$x	$x$x

Two consequences:

• $[\text{H}^+] = [\text{A}^-] = x$[H+]=[A−]=x — each dissociation produces one H⁺ and one A⁻ together (provided water self-ionisation is negligible, which holds when pH < 6 or so).
• $[\text{HA}]_{\text{eq}} = c - x$[HA]eq​=c−x — the small amount $x$x has dissociated, so the molecular form is depleted by exactly that amount.

Substituting into $K_a$Ka​:

$K_a = \frac{x^2}{c - x}$Ka​=c−xx2​

This is a quadratic in $x$x, but at A-Level we make the standard approximation $c - x \approx c$c−x≈c (valid when ionisation is less than ~5 %), giving $x = \sqrt{K_a \cdot c}$x=Ka​⋅c​. The full treatment is in lesson 5.

Worked example 3 — Ka from pH

A 0.100 mol dm⁻³ solution of a weak monobasic acid has pH = 3.15 at 298 K. Calculate $K_a$Ka​ and $\text{p}K_a$pKa​.

$[\text{H}^+] = 10^{-3.15} = 7.08 \times 10^{-4} \text{ mol dm}^{-3}$[H+]=10−3.15=7.08×10−4 mol dm−3

Since $7.08 \times 10^{-4} \ll 0.100$7.08×10−4≪0.100, the approximation $[\text{HA}]_{\text{eq}} \approx c = 0.100$[HA]eq​≈c=0.100 holds. Also $[\text{A}^-] = [\text{H}^+]$[A−]=[H+] because every dissociation generates one of each.