Acid Dissociation and Ka

A single number tells you whether an acid will sting your tongue or barely register. That number is KaK_a, the acid dissociation constant. It converts a vague label like "weak acid" into a precise, predictive quantity. Drop a crystal of acetic acid and a crystal of trichloroacetic acid into separate beakers of water. Use the same concentration for both. The second solution will be measurably more acidic. This is not because chemists say so. It is because trichloroacetic acid's KaK_a is roughly 10,000 times larger. Everything about how an acid behaves traces back to this equilibrium constant. This includes its role in a titration, a buffer, a natural water, or a drug's absorption in the gut. This matters because analytical chemistry is fundamentally about controlling and predicting chemical environments. pH governs an enormous share of chemistry. This includes metal ion speciation, indicator color changes, enzyme activity, and solubility of precipitates. It also shapes every titration curve you will ever plot. Bronsted-Lowry acid-base theory — proton donors and proton acceptors — gives the conceptual language. KaK_a gives the numbers. You must be able to write the equilibrium expression for a proton-transfer reaction. You must also know how to extract KaK_a from a measured pH. Once you can do both, you have the single most reusable tool in solution equilibrium chemistry. The big picture is that acid strength is not a binary property. It is a continuum measured on a logarithmic scale, pKa=−log⁡Ka\text{p}K_a = -\log K_a. This scale spans from strongly negative values to values above 40. Strongly negative values mean essentially complete dissociation, like HCl. Values above 40 mean essentially no proton loss, like methane. Every Bronsted acid has a conjugate base. The strength of the acid is mathematically locked to the strength of its conjugate base. This link runs through water's autoprotolysis constant, KwK_w. This linkage — the Ka-Kb relationship — means that knowing one member of a conjugate pair tells you everything about the other. This theme becomes central once you start building buffers and interpreting titration curves. This article builds the equilibrium-constant machinery for weak acid and weak base dissociation from the ground up. It covers the mass-action expressions. It also covers the algebra connecting KaK_a to measured pH and concentration. It further covers the reciprocal relationship between KaK_a and KbK_b. It also covers the practical skill of estimating relative acid strength from molecular structure. Three fully worked problems will anchor the theory in numbers you could reproduce on a bench. These include a foundational KaK_a calculation. They also include an applied problem using a real environmental buffer system. Finally, they include a multi-step problem requiring you to back out an unknown concentration from a measured pH.

Key takeaways
  • Ka=[H3O+][A−]/[HA]K_a = [\text{H}_3\text{O}^+][\text{A}^-]/[\text{HA}] is the equilibrium constant for a Bronsted acid transferring a proton to water.
  • pKa=−log⁡10Ka\text{p}K_a = -\log_{10} K_a; smaller pKaK_a means a stronger acid (more dissociation at a given concentration).
  • Every conjugate acid-base pair obeys Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0\times10^{-14} at 25 °C.
  • The 5% approximation lets you skip the quadratic formula when Ca/Ka≳400C_a/K_a \gtrsim 400; otherwise solve exactly.
  • Acid strength tracks bond polarity, electronegativity, resonance stabilization of A−\text{A}^-, and inductive effects of nearby groups.
  • Percent dissociation increases as concentration decreases, even though KaK_a itself never changes with concentration.
  • Key idea: KaK_a is a fixed number that quantifies how far a proton-transfer equilibrium sits toward products, and it lets you calculate — rather than guess — the pH, buffering behavior, and titration shape of any weak acid or base system.

Where this lives today

Every buffer used in a biochemistry lab — Tris, HEPES, phosphate, acetate — is selected by matching its pKaK_a to the target pH, because a buffer works best within about one pH unit of its pKaK_a; this selection criterion is a direct, everyday application of the KaK_a framework built here and is formalized quantitatively once you reach Buffers and the Henderson-Hasselbalch Equation. Pharmaceutical chemists use pKaK_a to predict how much of an ionizable drug will be in its neutral, membrane-permeable form versus its charged form at stomach pH (~1.5) versus blood pH (7.4), directly governing oral bioavailability and the design of prodrugs. Environmental chemists model the buffering capacity of natural waters — lakes, oceans, soil pore water — using exactly the carbonate Ka1/Ka2K_{a1}/K_{a2} system worked in Example 2, which underlies concerns about ocean acidification as rising atmospheric CO2\text{CO}_2 shifts the carbonate equilibrium and lowers seawater pH. Analytical instrumentation, from potentiometric titrators to ion-selective electrodes, is calibrated and interpreted using KaK_a values to predict the shape and equivalence-point behavior of titration curves before any experiment is run — a topic connecting directly to how pH of Weak Acid and Weak Base Solutions is calculated in practice. Structural chemists and pharmacologists routinely estimate unknown pKaK_a values from Hammett-type substituent constants, extending Ostwald's original 19th-century structure-reactivity correlations into modern quantitative structure-activity relationship (QSAR) software used in drug design. Finally, every conjugate acid-base pair you will encounter — ammonium/ammonia in nitrogen cycling, bisulfate/sulfate in atmospheric chemistry, imidazole side chains in protein histidines — is governed by the same KaK_a-KbK_b reciprocity derived here, a relationship that becomes the working tool for Conjugate Pairs and the Ka-Kb Relationship.

Ask the tutor: How do I decide, before doing any algebra, whether a given weak-acid problem needs the full quadratic or can use the square-root shortcut?

The physical picture

A Bronsted-Lowry acid is a proton donor; a Bronsted-Lowry base is a proton acceptor. In water, a weak acid HA does not fully ionize. Instead it reaches a dynamic equilibrium:

HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)\text{HA(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{A}^-\text{(aq)}

At any instant, some HA molecules are donating protons to water while, simultaneously, some H3O+\text{H}_3\text{O}^+ ions are handing protons back to A−\text{A}^- ions — the reverse reaction, in which A−\text{A}^- acts as a Bronsted base and H3O+\text{H}_3\text{O}^+ as its conjugate acid. At equilibrium the forward and reverse rates are equal, not the concentrations. This is the same equilibrium framework used for any reaction quotient: a ratio of product activities to reactant activities that becomes constant at a given temperature. The magnitude of KaK_a tells you where that dynamic balance sits. A large KaK_a (small or negative pKaK_a) means the equilibrium lies far to the right — most HA molecules have given up their proton. A small KaK_a (large pKaK_a) means most HA molecules remain intact. Because water is both the solvent and a reactant/product in huge excess, its concentration is treated as constant and folded into the equilibrium constant, so [H2O][\text{H}_2\text{O}] never appears in the KaK_a expression — the same convention used for KwK_w itself. Acid strength is a molecular-structure story as much as a thermodynamic one. Three structural factors dominate: bond polarity and electronegativity (an O–H bond next to a strongly electron-withdrawing group is more easily broken heterolytically), resonance delocalization of the resulting anion (a carboxylate spreads negative charge over two oxygens, making acetic acid millions of times stronger than ethanol), and inductive effects of nearby substituents (each additional chlorine on acetic acid's methyl group pulls electron density away from the O–H bond, dramatically raising KaK_a).

Figure
Proton-transfer equilibrium between acetic acid and water, showing the conjugate acid–base pairs
CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻ acid 1 (donor) base 2 (acceptor) acid 2 (conjugate) base 1 (conjugate) Ka = [H₃O⁺][CH₃COO⁻] / [CH₃COOH] = 1.8 × 10⁻⁵ pKa = −log₁₀(1.8 × 10⁻⁵) = 4.74 Water's concentration (≈ 55.5 M) is constant and absorbed into Ka.

The Ka-Kb relationship, foreshadowed here and developed fully elsewhere, arises because the conjugate base of any weak acid is itself a weak base toward water, and the product of the two equilibrium constants must reproduce KwK_w — the autoprotolysis constant you already know governs neutral water's own [H3O+][\text{H}_3\text{O}^+] and [OH−][\text{OH}^-].

Figure
pKa scale placing common acids and their conjugate bases
pKa −7 −1.7 4.74 9.24 15.7 HCl H₃O⁺ CH₃COOH NH₄⁺ H₂O strong acid ← ————————————————————————— → weak acid Increasing pKa: weaker acid, stronger conjugate base, equilibrium shifts left.

The core ideas

For the generic weak acid HA in water:

Ka=[H3O+][A−][HA]K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

where: [H3O+][\text{H}_3\text{O}^+], [A−][\text{A}^-], and [HA][\text{HA}] are equilibrium molar concentrations (strictly, activities — but concentrations are used at low ionic strength), and KaK_a is dimensionless (each term is implicitly divided by the standard-state concentration 1 M1\ \text{M}).

Reading: KaK_a is the ratio of "dissociated" to "intact" acid at equilibrium, weighted by how much free proton is floating around. A bigger KaK_a means more product-side character at equilibrium — a stronger acid. For the conjugate base A−\text{A}^- acting as a Bronsted base toward water:

Kb=[HA][OH−][A−]K_b = \dfrac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}

where: KbK_b is the base dissociation (hydrolysis) constant of the conjugate base A−\text{A}^-.

Reading: KbK_b measures how strongly the conjugate base pulls a proton away from water to regenerate HA and release hydroxide. The p-scale compresses these constants logarithmically, exactly as p H compresses [H3O+][\text{H}_3\text{O}^+]:

pKa=−log⁡10Ka,pKb=−log⁡10Kb\text{p}K_a = -\log_{10} K_a, \qquad \text{p}K_b = -\log_{10} K_b

where: the negative log turns a small, awkward number like 1.8×10−51.8\times10^{-5} into a manageable one, 4.744.74.

Reading: lower pKaK_a = stronger acid; lower pKbK_b = stronger base. A change of one pKaK_a unit is a tenfold change in KaK_a. The linkage between an acid and its conjugate base is fixed by water's own equilibrium:

Ka×Kb=Kw=1.0×10−14 (25 ∘C)K_a \times K_b = K_w = 1.0 \times 10^{-14} \ (25\ ^\circ\text{C})

where: KwK_w is the autoprotolysis constant of water, itself equal to [H3O+][OH−][\text{H}_3\text{O}^+][\text{OH}^-] in any aqueous solution at that temperature.

Reading: the stronger the acid, the weaker its conjugate base must be, and vice versa — you cannot independently choose both. Equivalently in p-notation:

pKa+pKb=pKw=14.00 (25 ∘C)\text{p}K_a + \text{p}K_b = \text{p}K_w = 14.00 \ (25\ ^\circ\text{C})

For the ICE-table treatment of a weak acid at nominal (formal) concentration CaC_a, letting x=[H3O+]x = [\text{H}_3\text{O}^+] produced by dissociation (ignoring water's own tiny autoprotolysis contribution):

Ka=x2Ca−xK_a = \dfrac{x^2}{C_a - x}

where: xx is simultaneously [H3O+][\text{H}_3\text{O}^+] and [A−][\text{A}^-] generated, and Ca−xC_a - x is the equilibrium concentration of remaining HA.

Reading: this is a single quadratic in xx; when x≪Cax \ll C_a (weak acid, not too dilute) it simplifies to Ka≈x2/CaK_a \approx x^2/C_a, giving the widely used approximation x≈KaCax \approx \sqrt{K_a C_a}.

Derivation AFrom the ICE table to the quadratic-in-x formula, with the 5% test

Goal. Derive the exact expression for [H3O+][\text{H}_3\text{O}^+] in a solution of a monoprotic weak acid at formal concentration CaC_a, and state precisely when the simplified square-root shortcut is valid.

Step 1 — Set up the equilibrium and the ICE table. Start from the dissociation reaction and assign initial, change, and equilibrium concentrations:

HA+H2O⇌H3O++A−\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-

Initial: [HA]0=Ca[\text{HA}]_0 = C_a, [H3O+]0≈0[\text{H}_3\text{O}^+]_0 \approx 0, [A−]0=0[\text{A}^-]_0 = 0 (ignoring water autoprotolysis, which contributes only 10−710^{-7} M and is negligible whenever CaKa≫KwC_a K_a \gg K_w). Change: −x-x, +x+x, +x+x. Equilibrium: Ca−xC_a - x, xx, xx.

Step 2 — Substitute into the mass-action expression.

Ka=[H3O+][A−][HA]=x⋅xCa−x=x2Ca−xK_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} = \dfrac{x \cdot x}{C_a - x} = \dfrac{x^2}{C_a-x}

Step 3 — Rearrange into standard quadratic form. Multiply both sides by (Ca−x)(C_a - x):

x2=KaCa−Kaxx^2 = K_a C_a - K_a x
x2+Kax−KaCa=0x^2 + K_a x - K_a C_a = 0

This is a quadratic ax2+bx+c=0ax^2+bx+c=0 with a=1a=1, b=Kab=K_a, c=−KaCac=-K_a C_a, solved by the quadratic formula, keeping only the physically meaningful positive root:

x=−Ka+Ka2+4KaCa2x = \dfrac{-K_a + \sqrt{K_a^2 + 4K_aC_a}}{2}

Step 4 — Identify the approximation and its validity condition. If x≪Cax \ll C_a, then Ca−x≈CaC_a - x \approx C_a, collapsing Step 2 to Ka≈x2/CaK_a \approx x^2/C_a, so

x≈KaCax \approx \sqrt{K_a C_a}

The common rule of thumb — valid when the approximation introduces less than 5% error in xx — is

xCa≤0.05⟺CaKa≳400\dfrac{x}{C_a} \le 0.05 \quad \Longleftrightarrow \quad \dfrac{C_a}{K_a} \gtrsim 400

This threshold comes from substituting x=0.05 Cax = 0.05\,C_a back into the exact quadratic and solving for the boundary ratio Ca/KaC_a/K_a; when concentration is at least ~400 times KaK_a, the linear (square-root) approximation is safe, and otherwise the full quadratic (Step 3) must be used.

Go deeper · Derivation B · 1 minDeriving Ka·Kb = Kw for a conjugate acid–base pair
Derivation BDeriving Ka·Kb = Kw for a conjugate acid–base pair

Goal. Show algebraically why the product of a weak acid's KaK_a and its conjugate base's KbK_b must equal KwK_w, independent of which acid you pick.

Step 1 — Write both mass-action expressions. For the acid dissociation of HA:

Ka=[H3O+][A−][HA]K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

For the base hydrolysis of its conjugate base A−\text{A}^-:

Kb=[HA][OH−][A−]K_b = \dfrac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}

Step 2 — Multiply the two expressions together.

Ka×Kb=[H3O+][A−][HA]×[HA][OH−][A−]K_a \times K_b = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \times \dfrac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]}

Step 3 — Cancel the common factors. Both [HA][\text{HA}] and [A−][\text{A}^-] appear once in a numerator and once in a denominator, so they cancel exactly, leaving:

Ka×Kb=[H3O+][OH−]K_a \times K_b = [\text{H}_3\text{O}^+][\text{OH}^-]

Step 4 — Recognize the remaining product as Kw. By definition, the autoprotolysis constant of water is Kw=[H3O+][OH−]K_w = [\text{H}_3\text{O}^+][\text{OH}^-] in any aqueous solution at a given temperature (this is why KwK_w is universal — it does not depend on which acid or base is dissolved). Therefore:

Ka×Kb=Kw=1.0×10−14 (25 ∘C)K_a \times K_b = K_w = 1.0\times10^{-14}\ (25\ ^\circ\text{C})

This derivation used no approximation — it is an exact consequence of the two equilibrium definitions and the identity Kw=[H3O+][OH−]K_w = [\text{H}_3\text{O}^+][\text{OH}^-], so it holds for every conjugate pair, from HCl/Cl⁻ to NH4+\text{NH}_4^+/NH3\text{NH}_3.

Worked examples

Example 1 (Foundational) — Finding Ka from measured pH. A 0.150 M aqueous solution of a monoprotic weak acid HX is measured with a calibrated pH electrode and gives pH = 2.68 at 25 °C. Find KaK_a and pKaK_a. Step 1 — Get [H3O+][\text{H}_3\text{O}^+] from pH.

[H3O+]=10−pH=10−2.68=2.09×10−3 M[\text{H}_3\text{O}^+] = 10^{-\text{pH}} = 10^{-2.68} = 2.09\times10^{-3}\ \text{M}

(keeping 3 sig figs, consistent with pH given to 2 decimal places, i.e., 2 sig figs in concentration — we carry an extra guard digit through the calculation). Step 2 — Build the ICE table at equilibrium. Since x=[H3O+]=[X−]=2.09×10−3x=[\text{H}_3\text{O}^+]=[\text{X}^-]=2.09\times10^{-3} M (1:1 stoichiometry, and autoionization of water is negligible here since x≫10−7x \gg 10^{-7} M):

[HX]eq=Ca−x=0.150−0.00209=0.1479 M[\text{HX}]_{eq} = C_a - x = 0.150 - 0.00209 = 0.1479\ \text{M}

Step 3 — Substitute into the Ka expression.

Ka=x2Ca−x=(2.09×10−3)20.1479=4.37×10−60.1479=2.96×10−5K_a = \dfrac{x^2}{C_a-x} = \dfrac{(2.09\times10^{-3})^2}{0.1479} = \dfrac{4.37\times10^{-6}}{0.1479} = 2.96\times10^{-5}

Step 4 — Convert to pKa.

pKa=−log⁡10(2.96×10−5)=4.53\text{p}K_a = -\log_{10}(2.96\times10^{-5}) = 4.53

Sanity check via the 5% rule. Here Ca/Ka=0.150/(2.96×10−5)≈5,070≫400C_a/K_a = 0.150/(2.96\times10^{-5}) \approx 5{,}070 \gg 400, so the square-root shortcut would have been valid too: x≈(2.96×10−5)(0.150)=2.11×10−3x\approx\sqrt{(2.96\times10^{-5})(0.150)} = 2.11\times10^{-3} M, matching the exact value to within rounding — a useful cross-check that the exact and approximate methods agree here, confirming internal consistency of the arithmetic.

Example 2 (Applied) — Carbonic acid buffering in a lake, using Ka·Kb = Kw. Carbonic acid, H2CO3\text{H}_2\text{CO}_3, has a first dissociation constant Ka1=4.3×10−7K_{a1} = 4.3\times10^{-7} at 25 °C (this is the effective Ka1K_{a1} used in aquatic chemistry, lumping dissolved CO2\text{CO}_2 and true H2CO3\text{H}_2\text{CO}_3 together). A limnologist wants the KbK_b of the conjugate base, bicarbonate ion HCO3−\text{HCO}_3^-, to model how a lake resists pH change as bicarbonate reacts with incoming acid rain. Step 1 — Identify the conjugate pair.

H2CO3 (acid)⇌H++HCO3− (conjugate base)\text{H}_2\text{CO}_3 \ (\text{acid}) \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \ (\text{conjugate base})

Step 2 — Apply the Ka·Kb = Kw relationship.

Kb(HCO3−)=KwKa1=1.0×10−144.3×10−7=2.3×10−8K_b(\text{HCO}_3^-) = \dfrac{K_w}{K_{a1}} = \dfrac{1.0\times10^{-14}}{4.3\times10^{-7}} = 2.3\times10^{-8}

Step 3 — Interpret physically. Kb=2.3×10−8K_b = 2.3\times10^{-8} is small — bicarbonate is a very weak base — but it is not zero, and this is precisely why bicarbonate can neutralize small additions of strong acid (like H+\text{H}^+ from dissolved atmospheric SO2\text{SO}_2 or NOx\text{NO}_x) without a large pH swing, the chemical basis of natural water buffering. Step 4 — Compute pKb and cross-check with pKa1.

pKa1=−log⁡(4.3×10−7)=6.37,pKb=−log⁡(2.3×10−8)=7.64\text{p}K_{a1} = -\log(4.3\times10^{-7}) = 6.37, \qquad \text{p}K_b = -\log(2.3\times10^{-8}) = 7.64
pKa1+pKb=6.37+7.64=14.01≈14.00 ✓\text{p}K_{a1}+\text{p}K_b = 6.37+7.64 = 14.01 \approx 14.00\ \checkmark

The 0.01 discrepancy is rounding error from carrying only 2 sig figs through logarithms — well within expected precision, confirming the relationship.

Example 3 (Exam/competition level) — Back-calculating an unknown concentration from measured pH, exact quadratic required. A student prepares a solution of monochloroacetic acid, ClCH2COOH\text{ClCH}_2\text{COOH} (Ka=1.4×10−3K_a = 1.4\times10^{-3}), by dissolving an unknown mass in water to make 250.0 mL of solution. The measured pH is 1.96. What was the formal (analytical) concentration CaC_a, and is the 5% approximation valid? Step 1 — Get [H3O+][\text{H}_3\text{O}^+].

x=[H3O+]=10−1.96=1.096×10−2 Mx = [\text{H}_3\text{O}^+] = 10^{-1.96} = 1.096\times10^{-2}\ \text{M}

Step 2 — Attempt the simple approximation first, to see if it's even usable. Rearranged from Ka≈x2/CaK_a \approx x^2/C_a:

Ca≈x2Ka=(1.096×10−2)21.4×10−3=1.202×10−41.4×10−3=0.0858 MC_a \approx \dfrac{x^2}{K_a} = \dfrac{(1.096\times10^{-2})^2}{1.4\times10^{-3}} = \dfrac{1.202\times10^{-4}}{1.4\times10^{-3}} = 0.0858\ \text{M}

Check the ratio: Ca/Ka=0.0858/(1.4×10−3)=61.3C_a/K_a = 0.0858/(1.4\times10^{-3}) = 61.3, which is far below 400 — the approximation is not valid here, since KaK_a is relatively large (monochloroacetic acid is a "medium-strength" weak acid due to the electron-withdrawing chlorine). We must redo this with the exact quadratic. Step 3 — Use the exact mass-action expression, solving for Ca instead of x this time. Since xx is now the known quantity (from pH) and CaC_a is unknown, rearrange

Ka=x2Ca−xK_a = \dfrac{x^2}{C_a - x}

directly for CaC_a:

Ca−x=x2KaC_a - x = \dfrac{x^2}{K_a}
Ca=x+x2Ka=1.096×10−2+(1.096×10−2)21.4×10−3C_a = x + \dfrac{x^2}{K_a} = 1.096\times10^{-2} + \dfrac{(1.096\times10^{-2})^2}{1.4\times10^{-3}}
Ca=0.01096+0.0858=0.0968 MC_a = 0.01096 + 0.0858 = 0.0968\ \text{M}

Step 4 — Compare the two answers and quantify the error of the naive approximation.

percent error=0.0968−0.08580.0968×100%=11.4%\text{percent error} = \dfrac{0.0968-0.0858}{0.0968}\times100\% = 11.4\%

This confirms the earlier flag: skipping the quadratic step here would have introduced an 11% error in CaC_a — well above the acceptable 5% threshold — precisely because monochloroacetic acid's KaK_a is large enough that a non-trivial fraction of the acid dissociates. Step 5 — Back out the mass, given molar mass 94.50 g/mol and 250.0 mL volume.

moles=Ca×V=0.0968 M×0.2500 L=0.02420 mol\text{moles} = C_a \times V = 0.0968\ \text{M} \times 0.2500\ \text{L} = 0.02420\ \text{mol}
mass=0.02420 mol×94.50 g/mol=2.29 g\text{mass} = 0.02420\ \text{mol} \times 94.50\ \text{g/mol} = 2.29\ \text{g}

Limiting cases, scope & common traps

SituationWhat happensWhy
Ca/Ka≳400C_a/K_a \gtrsim 400x≈KaCax\approx\sqrt{K_aC_a} is accurate to within 5%HA barely dissociates; Ca−x≈CaC_a - x \approx C_a
Ca/Ka≲400C_a/K_a \lesssim 400Must solve the full quadraticDissociation is large enough that xx is not negligible vs. CaC_a
Extremely dilute weak acid (Ca→0C_a \to 0)Water's own autoprotolysis contributes comparably to acid dissociation; simple ICE table failsCannot ignore [H3O+][\text{H}_3\text{O}^+] from KwK_w itself
Very strong acid limit (Ka→∞K_a \to \infty)Dissociation is essentially complete; [H3O+]≈Ca[\text{H}_3\text{O}^+]\approx C_aThe equilibrium lies so far right that HA effectively vanishes
Polyprotic acids (e.g. H2CO3\text{H}_2\text{CO}_3, H3PO4\text{H}_3\text{PO}_4)Each proton loss has its own Ka1,Ka2,…K_{a1}, K_{a2}, \dots, usually with Ka1≫Ka2≫Ka3K_{a1}\gg K_{a2}\gg K_{a3}Successive protons are removed from an increasingly negatively charged, harder-to-deprotonate species
Conjugate base of a strong acid (e.g. Cl⁻)Essentially zero basicity; does not hydrolyzeKb=Kw/KaK_b = K_w/K_a is vanishingly small when KaK_a is huge

The chemist behind it

Svante Arrhenius is usually credited with the first quantitative theory of acids and bases (1884), defining acids as substances producing H+\text{H}^+ in water — a theory limited to aqueous solvents but revolutionary for treating dissociation as a genuine equilibrium governed by a constant. Wilhelm Ostwald, working alongside Arrhenius in the emerging discipline of physical chemistry, systematically measured dissociation constants for dozens of organic acids in the 1880s, establishing the experimental methodology — conductivity measurements correlated against concentration — that let chemists tabulate KaK_a values and connect them to molecular structure for the first time. It was this body of careful, comparative measurement across a series of substituted acetic and benzoic acids that first revealed the systematic relationship between electron-withdrawing substituents and increased acid strength, a structural insight still taught exactly the same way today. The theory was generalized decades later. In 1923, Johannes Nicolaus Bronsted in Denmark and Thomas Martin Lowry in England independently proposed the proton-transfer definition of acids and bases used throughout this article — an acid is any proton donor, a base any proton acceptor, freeing acid-base chemistry from the requirement of water as solvent and unifying the treatment of NH4+\text{NH}_4^+, HCO3−\text{HCO}_3^-, and countless organic and inorganic species under one equilibrium framework.

Origins · 1 minJohannes Nicolaus Bronsted and how this was discovered
JN
Johannes Nicolaus Bronsted
1879–1947 · Danish physical chemist, University of Copenhagen

Bronsted trained originally in chemical engineering before turning to physical chemistry, where his careful thermodynamic measurements of ionic reactions led him to recognize that "acid" and "base" were relational terms — properties of a pair, not of a molecule in isolation. His 1923 paper proposing the proton-donor/proton-acceptor definition (published the same year as Lowry's independent, similar proposal) reframed acid-base chemistry as fundamentally about equilibrium and conjugate relationships, the exact structure used in the KaK_a/KbK_b formalism taught today. He also did foundational work connecting reaction rates to acid-base catalysis, an early bridge between equilibrium thermodynamics and kinetics.

History
- 1884 — Svante Arrhenius proposes that acids dissociate in water to release H+\text{H}^+, treating dissociation as an equilibrium with a characteristic degree of ionization.
- 1880s — Wilhelm Ostwald systematically tabulates dissociation constants for organic acids using conductivity, establishing KaK_a as a comparative, structure-linked quantity.
- 1909 — Søren Sørensen introduces the pH scale, giving chemists a practical logarithmic measure to pair with KaK_a and pKaK_a.
- 1923 — Bronsted and Lowry independently generalize acid-base theory to proton transfer, unifying aqueous and non-aqueous acid-base equilibria.
- 1923 onward — G. N. Lewis proposes the still more general electron-pair donor/acceptor definition, broadening acid-base theory beyond proton transfer entirely, while KaK_a remains the standard quantitative tool for Bronsted acids specifically.
Why doesn't water's concentration appear in the Ka expression?

Water is both the solvent and, formally, a reactant in HA+H2O⇌H3O++A−\text{HA}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{A}^-. In dilute aqueous solution, water is present in such vast excess (about 55.5 M) that its concentration barely changes during the reaction — losing a few millimoles of water to protonation is a negligible perturbation. By convention, [H2O][\text{H}_2\text{O}] is treated as constant and absorbed into the equilibrium constant itself, exactly the same convention applied when defining Kw=[H3O+][OH−]K_w = [\text{H}_3\text{O}^+][\text{OH}^-] without a separate [H2O]2[\text{H}_2\text{O}]^2 term. This is why KaK_a expressions only ever show the solute species.

Is a larger Ka always a "more dangerous" or more corrosive acid?

Not necessarily — KaK_a measures the position of equilibrium, not concentration or total acidic material present. A very dilute solution of a strong acid (small pKaK_a, essentially complete dissociation) can have a lower [H3O+][\text{H}_3\text{O}^+] than a concentrated solution of a weaker acid. Corrosiveness and hazard depend on both KaK_a and concentration together, plus other factors like oxidizing power (concentrated sulfuric versus concentrated acetic acid behave very differently despite comparable [H+][\text{H}^+] at high concentration, because sulfuric acid is also a strong dehydrating and oxidizing agent).

How is Ka actually measured in the lab?

The most common approach is potentiometric titration: a calibrated pH electrode records pH continuously as standard base is added to a weak acid solution. At the half-equivalence point, exactly half the acid has been converted to its conjugate base, so [HA]=[A−][\text{HA}]=[\text{A}^-] and the Ka expression collapses to Ka=[H3O+]K_a = [\text{H}_3\text{O}^+], meaning the measured pH at that point directly gives pKaK_a. Spectrophotometric methods are used when the acid and its conjugate base have distinct absorbance spectra, tracking the ratio of the two forms directly across a pH range. Conductivity measurements, historically Ostwald's method, remain useful for simple systems.

Why do polyprotic acids have Ka1 ≫ Ka2 ≫ Ka3?

Each successive proton is removed from a species that is already more negatively charged than the last. Removing H+\text{H}^+ from neutral H3PO4\text{H}_3\text{PO}_4 is comparatively easy; removing a second H+\text{H}^+ from the resulting H2PO4−\text{H}_2\text{PO}_4^- requires pulling a proton away from an anion that already attracts it electrostatically, and removing a third from the doubly-charged HPO42−\text{HPO}_4^{2-} is harder still. Typically each successive KaK_a drops by a factor of 10410^4–10510^5, which is why polyprotic titration curves show well-separated, distinguishable equivalence points.

Can Ka be greater than 1, and what does that mean?

Yes — strong acids like HCl, HNO3\text{HNO}_3, and H2SO4\text{H}_2\text{SO}_4 (first proton) have KaK_a values far greater than 1 (often reported as very large numbers or simply treated as "complete dissociation" because the reverse reaction is negligible). A Ka>1K_a>1 (negative pKaK_a) simply means the equilibrium lies overwhelmingly toward products in water — so far, in fact, that these acids are usually assumed to dissociate 100% for practical pH calculations, and their true KaK_a values are difficult to measure precisely by aqueous methods because there's essentially no remaining HA to detect.

Does temperature affect Ka?

Yes. Like any equilibrium constant, KaK_a depends on temperature through the van't Hoff relationship, since acid dissociation has an associated enthalpy change. For acetic acid, KaK_a changes only mildly with temperature near room temperature (the dissociation enthalpy is small), but for water's own autoprotolysis, KwK_w increases noticeably with temperature (it roughly doubles from 25 °C to 37 °C), which is why physiological pH calculations at body temperature must use a KwK_w appropriate to 37 °C rather than the standard 25 °C value.

What's the difference between Ka and "acid strength" as used loosely in everyday speech?

In casual usage, "strong" sometimes just means "concentrated" or "corrosive," but in chemistry, acid strength specifically means the extent of dissociation in water, quantified by KaK_a. A concentrated solution of a weak acid (large CaC_a, small KaK_a) can have a lower pH — more free H3O+\text{H}_3\text{O}^+ — than a dilute solution of a strong acid, yet the weak acid is still, by definition, the "weaker" acid because its KaK_a is smaller. Always distinguish the equilibrium property (KaK_a) from the bulk solution property (pH, which depends on both KaK_a and concentration).

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