GENERAL CHEMISTRY 2 • ACIDS AND BASES

Ka, Kb & Kw for Conjugates — Relate Ka, Kb, and Kw for conjugate pairs

Understanding how the acid and base dissociation constants of conjugate pairs are forever linked through the autoionization of water.

Historical Context & Motivation

The quantitative treatment of acid–base equilibria owes its modern form to more than a century of theoretical refinement. Early chemists recognized that acids and bases existed in opposing relationship, but placing that intuition on rigorous thermodynamic footing required a series of conceptual breakthroughs—from Arrhenius's ionic dissociation theory through the Brønsted–Lowry framework and ultimately to the recognition that conjugate acid–base pairs share a precise mathematical connection mediated by the solvent itself. Understanding this history clarifies why the relationship Ka × Kb = Kw is not merely a convenient formula but an inevitable consequence of how water participates in every proton-transfer event.

1884
Arrhenius Dissociation Theory
Svante Arrhenius proposes that acids produce H⁺ and bases produce OH⁻ in aqueous solution, laying the groundwork for quantifying acid and base strength through equilibrium constants.
1909
Sørensen Introduces pH
S. P. L. Sørensen defines the pH scale, creating a practical metric tied to the hydrogen-ion concentration and motivating a deeper understanding of water's autoionization.
1923
Brønsted–Lowry Acid–Base Theory
Johannes Brønsted and Thomas Lowry independently define acids as proton donors and bases as proton acceptors, formalizing the concept of conjugate acid–base pairs and enabling the Ka–Kb relationship.
1930s
Precise Measurement of Kw
Electrometric methods refine Kw to 1.0 × 10⁻¹⁴ at 25 °C, anchoring the quantitative link between conjugate-pair dissociation constants.
Modern Era
Computational & Thermodynamic Integration
Contemporary chemistry integrates the Ka·Kb = Kw relationship into buffer design, pharmaceutical formulation, and environmental water chemistry.

The central question that this lesson addresses is deceptively simple: if you know the acid dissociation constant of a weak acid, can you immediately determine the base dissociation constant of its conjugate base—and vice versa? The answer is yes, and the bridge between them is the ion-product constant of water, Kw. This relationship is indispensable for predicting solution pH, designing buffer systems, and understanding how molecular structure governs acid–base behavior.

Core Principles & Definitions

Before deriving the quantitative relationship, it is essential to anchor three equilibrium constants and the concept of conjugate pairing. In the Brønsted–Lowry framework, every acid–base reaction generates a conjugate pair: when an acid HA donates a proton to water, it produces its conjugate base A⁻. Conversely, when A⁻ accepts a proton from water, it reforms HA. The equilibrium constants for these two processes are not independent; they are connected through the autoionization equilibrium of the solvent.

1

Ka — Acid Dissociation Constant

Quantifies the extent to which a weak acid HA donates a proton to water: HA + H₂O ⇌ H₃O⁺ + A⁻. Larger Ka means a stronger acid.
2

Kb — Base Dissociation Constant

Quantifies the extent to which a base A⁻ accepts a proton from water: A⁻ + H₂O ⇌ HA + OH⁻. Larger Kb means a stronger base.
3

Kw — Ion-Product Constant of Water

Describes water's autoionization: 2 H₂O ⇌ H₃O⁺ + OH⁻. At 25 °C, Kw = 1.0 × 10⁻¹⁴. This constant is temperature-dependent.
4

Conjugate Pair Relationship

For any conjugate acid–base pair, the product Ka × Kb always equals Kw. A strong acid has a very weak conjugate base, and vice versa.
✦ KEY TAKEAWAY
Think of a conjugate acid–base pair like two sides of a see-saw anchored at Kw. When one side (the acid's Ka) goes up, the other side (the conjugate base's Kb) must come down by exactly the same factor, because the product of the two sides is locked at 1.0 × 10⁻¹⁴. The solvent—water—acts as the fulcrum, and its autoionization sets the height of that fulcrum.

Visual Explanation — The Conjugate Equilibrium Cycle

The diagram shows how the acid dissociation reaction (pink, left) and the base hydrolysis reaction (cyan, right) sum to give the water autoionization reaction (amber, center). Because the net reaction is the sum, the equilibrium constants multiply: Ka × Kb = Kw.

The diagram above encapsulates the central thermodynamic argument. When you write the Ka expression for HA and the Kb expression for A⁻ and then add the two chemical equations, the HA and A⁻ species cancel, leaving only the autoionization of water. Because adding balanced equations corresponds to multiplying their equilibrium constants, the product Ka × Kb must equal Kw. This is a direct application of Hess's Law applied to equilibrium constants rather than enthalpy changes.

Mathematical Framework

We now formalize the derivation. Consider a generic monoprotic weak acid HA and its conjugate base A⁻ in aqueous solution at 25 °C. Three equilibria are relevant, and we will show that only two are independent.

ACID DISSOCIATION
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq) Ka = [H₃O⁺][A⁻] / [HA]
Ka is the acid dissociation constant. Water, as a pure liquid, does not appear in the expression.
BASE HYDROLYSIS
A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq) Kb = [HA][OH⁻] / [A⁻]
Kb is the base dissociation constant for the conjugate base A⁻. Again, the activity of water is incorporated into the constant.
AUTOIONIZATION OF WATER
2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ (25 °C)
Kw is the ion-product constant of water. It is temperature-dependent: at 37 °C (body temperature), Kw ≈ 2.4 × 10⁻¹⁴.

Derivation by Summing Reactions

Add the acid dissociation reaction to the base hydrolysis reaction algebraically. On the left side you have HA + H₂O + A⁻ + H₂O; on the right, H₃O⁺ + A⁻ + HA + OH⁻. The species HA and A⁻ appear on both sides and cancel, yielding 2 H₂O ⇌ H₃O⁺ + OH⁻, which is precisely the autoionization equilibrium. By the multiplication rule for coupled equilibria (a consequence of Hess's Law applied to Gibbs energies and hence to equilibrium constants), the overall equilibrium constant for the net reaction is the product of the individual constants.

FUNDAMENTAL RELATIONSHIP
Ka × Kb = Kw
This holds for any conjugate acid–base pair in water. Taking the negative logarithm of both sides gives the equivalent logarithmic form: pKa + pKb = pKw = 14.00 at 25 °C.
📐 Logarithmic Form
Because pK = −log K, taking −log of both sides of Ka × Kb = Kw yields: pKa + pKb = 14.00 (at 25 °C). This additive form is often more convenient for quick mental estimation.

Inverse Strength Relationship — A Closer Look

The equation Ka × Kb = Kw carries a profound qualitative consequence: the stronger an acid, the weaker its conjugate base, and the stronger a base, the weaker its conjugate acid. This is not a vague tendency—it is an exact inverse proportionality. If Ka increases by a factor of 10, Kb must decrease by a factor of 10 to maintain the constant product. The table below illustrates this trend with familiar conjugate pairs.

Each data point lies on the line pKa + pKb = 14. Moving right along the x-axis (increasing pKa, weaker acid) forces the point upward (decreasing pKb, stronger conjugate base).
Selected conjugate pairs at 25 °C confirming Ka × Kb = 1.0 × 10⁻¹⁴
Conjugate AcidConjugate BaseKaKbKa × Kb
HFF⁻6.3 × 10⁻⁴1.6 × 10⁻¹¹1.0 × 10⁻¹⁴
CH₃COOHCH₃COO⁻1.8 × 10⁻⁵5.6 × 10⁻¹⁰1.0 × 10⁻¹⁴
NH₄⁺NH₃5.6 × 10⁻¹⁰1.8 × 10⁻⁵1.0 × 10⁻¹⁴
HCNCN⁻6.2 × 10⁻¹⁰1.6 × 10⁻⁵1.0 × 10⁻¹⁴
HCO₃⁻CO₃²⁻4.7 × 10⁻¹¹2.1 × 10⁻⁴1.0 × 10⁻¹⁴

Notice that for every pair in the table, the product Ka × Kb returns 1.0 × 10⁻¹⁴ regardless of whether the acid is relatively strong (HF) or extremely weak (HCO₃⁻ acting as an acid). This universality is the hallmark of a thermodynamic identity—it does not depend on molecular structure, concentration, or reaction mechanism, only on the temperature and the solvent.

Worked Example — Finding Kb from Ka

Suppose you are given that the Ka of nitrous acid (HNO₂) is 4.5 × 10⁻⁴ at 25 °C, and you need to determine the Kb of its conjugate base, the nitrite ion (NO₂⁻). We will also calculate pKb and verify the result using the pK relationship.

Calculating Kb for NO₂⁻ from Ka of HNO₂
1
Step 1 — Identify the Conjugate PairHNO₂ is the conjugate acid; NO₂⁻ is its conjugate base. The relationship Ka(HNO₂) × Kb(NO₂⁻) = Kw applies directly.
Conjugate pair: HNO₂ / NO₂⁻
2
Step 2 — Write the RelationshipRearrange the fundamental identity to isolate Kb: Kb = Kw / Ka.
Kb = Kw / Ka
3
Step 3 — Substitute ValuesKb = (1.0 × 10⁻¹⁴) / (4.5 × 10⁻⁴) = (1.0 / 4.5) × 10⁻¹⁴⁻⁽⁻⁴⁾ = 0.222 × 10⁻¹⁰.
Kb = 2.2 × 10⁻¹¹
4
Step 4 — Convert to pKbpKb = −log(2.2 × 10⁻¹¹) = −(log 2.2 + log 10⁻¹¹) = −(0.342 − 11) = 10.66.
pKb = 10.66
5
Step 5 — Verify with the pK SumpKa = −log(4.5 × 10⁻⁴) = 3.35. Check: pKa + pKb = 3.35 + 10.66 = 14.01, which rounds to 14.00, confirming consistency.
✓ pKa + pKb = 14.00
⚠️ Common Error
Students sometimes mistakenly add Ka and Kb to get Kw. Remember: the reactions are added, so the constants are multiplied. The logarithmic form (pKa + pKb = 14) is an addition, but that's because the 'p' operator converts multiplication into addition.

Strengths, Limitations & Common Misconceptions

Strengths and limitations of the Ka × Kb = Kw relationship
AspectStrength / Correct ViewLimitation / Misconception
UniversalityApplies to every conjugate pair in aqueous solution without exception.Only valid in water as solvent. In other solvents, the autoionization constant differs.
Temperature dependenceCorrectly predicts shifts in relative strength with temperature since Kw changes.Students often treat Kw as universally 1.0 × 10⁻¹⁴; it varies with T (e.g., 2.4 × 10⁻¹⁴ at 37 °C).
Logarithmic formpKa + pKb = 14 is easy to apply for mental estimation and quick checks.The '14' is only valid at 25 °C. At other temperatures, use the actual pKw.
Strong acid/base pairsExplains why conjugate bases of strong acids (e.g., Cl⁻) have negligible basicity.Ka for strong acids is not precisely measurable in water (leveling effect), so Kb of their conjugates is estimated indirectly.
Polyprotic speciesEach deprotonation step has its own Ka, and each conjugate base has a corresponding Kb linked through Kw.Must match the correct Ka with the correct Kb: Ka1 of H₂CO₃ pairs with Kb2 of CO₃²⁻, not Kb1.
✦ KEY TAKEAWAY — Polyprotic Pairing
For a diprotic acid H₂A, the pairing rule is: Ka1 (H₂A) pairs with Kb2 (A²⁻), and Ka2 (HA⁻) pairs with Kb1 (HA⁻). Think of it like numbering from opposite ends of a ladder—the first step up on the acid side matches the last step up on the base side. Each pair still satisfies Ka × Kb = Kw.

Connection to Thermodynamics & Non-Aqueous Systems

The relationship Ka × Kb = Kw is a specific instance of a deeper thermodynamic principle. The standard Gibbs energy change for the autoionization of water equals the sum of the Gibbs energy changes for the acid dissociation and base hydrolysis: ΔG°w = ΔG°a + ΔG°b. Since ΔG° = −RT ln K, the additivity of Gibbs energies translates directly into the multiplicativity of equilibrium constants. This Gibbs-energy perspective also explains why Kw depends on temperature: at higher temperatures, the endothermic autoionization of water becomes more favorable, Kw increases, and the pKa + pKb sum shifts below 14.

Aqueous vs. non-aqueous extensions of the conjugate pair relationship
FeatureAqueous System (This Lesson)Non-Aqueous / Advanced
Autoionization constantKw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °CSolvent autoionization constant (Ksolvent) differs. E.g., liquid NH₃: Ksolvent ≈ 10⁻³³
Conjugate pair relationshipKa × Kb = KwKa(solvent) × Kb(solvent) = Ksolvent — same form, different numerical anchor
Practical implicationpKa + pKb = 14.00pKa + pKb = pKsolvent, which can be 33 (NH₃) or 20 (DMSO)
Acidity scale rangepH 0–14 covers most aqueous chemistrySuperacid and superbase chemistry extends beyond the aqueous window via leveling effect considerations

As you advance into physical chemistry, biochemistry, and materials science, you will encounter systems where the solvent is not water. In each case, the structural logic remains identical: the acid dissociation and base association equilibria for a conjugate pair always sum to the solvent's autoionization reaction, locking their equilibrium constants to the solvent's own self-ionization constant. Mastering the aqueous case thoroughly prepares you for these extensions.

Practice Problems

PROBLEM 1 — CONCEPTUAL
The Ka of hydrocyanic acid (HCN) is 6.2 × 10⁻¹⁰. Without performing a calculation, predict whether the Kb of CN⁻ will be greater than, less than, or equal to 1.0 × 10⁻⁷. Justify your reasoning using the Ka × Kb = Kw relationship.
PROBLEM 2 — BASIC CALCULATION
The Ka of benzoic acid (C₆H₅COOH) is 6.3 × 10⁻⁵ at 25 °C. Calculate the Kb and pKb for the benzoate ion (C₆H₅COO⁻).
PROBLEM 3 — INTERMEDIATE
Methylamine (CH₃NH₂) has Kb = 4.4 × 10⁻⁴. Calculate the Ka of the methylammonium ion (CH₃NH₃⁺). Then determine the pH of a 0.10 M solution of CH₃NH₃Cl, assuming complete dissociation of the salt.
PROBLEM 4 — APPLIED
Phosphoric acid (H₃PO₄) has Ka2 = 6.2 × 10⁻⁸ for the equilibrium H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻. Calculate the Kb for HPO₄²⁻ acting as a base to regenerate H₂PO₄⁻. Explain which Ka is paired with which Kb and why.
PROBLEM 5 — CRITICAL THINKING
At 37 °C (body temperature), Kw = 2.4 × 10⁻¹⁴. The Ka of acetic acid at 37 °C is approximately 1.7 × 10⁻⁵. Calculate the Kb of acetate at 37 °C and comment on how the neutral pH and the pK sum change relative to 25 °C values.

Lesson Summary

Every conjugate acid–base pair in aqueous solution obeys the relationship Ka × Kb = Kw, which equals 1.0 × 10⁻¹⁴ at 25 °C. This identity arises because the acid dissociation and base hydrolysis reactions for a conjugate pair sum algebraically to the autoionization of water, and adding reactions means multiplying equilibrium constants. In logarithmic form, pKa + pKb = pKw = 14.00 at 25 °C.

The practical consequence is an exact inverse relationship between acid and base strength within a conjugate pair: a strong acid has an extremely weak conjugate base, and vice versa. For polyprotic acids, each Ka pairs with the Kb numbered from the opposite end. Remember that Kw is temperature-dependent, so the '14' in pKa + pKb = 14 applies only at 25 °C. This foundational relationship is your gateway to buffer design, salt hydrolysis calculations, and advanced equilibrium problems.

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