Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
In 1887, Svante Arrhenius at Uppsala University published his dissociation theory, defining acids as substances that produce H⁺ in water: HA(aq) ⇌ H⁺(aq) + A⁻(aq). He measured what he called the "dissociation constant" for acetic acid as 1.8 × 10⁻⁵ at 25°C — the same number we now write as Ka. A pharmacy student reads that hydrochloric acid has Ka >> 1 while acetic acid (vinegar) has Ka = 1.8 × 10⁻⁵. The student says: "Ka must be a different type of constant from Keq — it measures something about acids specifically that Keq can't."
Do you agree? Is Ka a fundamentally different type of constant from Keq, or is it the same concept Arrhenius applied to a specific type of reaction? Write your position with reasoning before reading on.
Know
- Ka is Keq for acid dissociation — the same concept with a specific context
- Strong acids have large Ka; weak acids have small Ka
- The Kw relationship: Ka × Kb = Kw for conjugate acid-base pairs
Understand
- Why aspirin solubility and stomach irritation can be explained using Ka and pH
- How to rank acid strength using Ka and why pKa is sometimes more convenient
- The quantitative relationship ΔG° = −RT ln Keq and what it means for spontaneity
Can Do
- Calculate Kb from Ka using Ka × Kb = Kw
- Rank acids and their conjugate bases by relative strength
- Apply Ka concepts to explain drug absorption and biological contexts
Module 5 — Key Formulas: Lesson 14
Misconceptions to Fix
Ka is not a new type of constant — it is Keq with a specific name for a specific category of reaction. Understanding this immediately makes Module 6 acid chemistry far less daunting.
When a weak acid HA dissociates in water: $\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)$
This is a reversible reaction in aqueous solution. The equilibrium expression is written exactly as you have been doing all module:
$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$
Water is the solvent (pure liquid) — excluded. All other species are aqueous — included. This is mathematically and conceptually identical to any other Keq you have written in L09–L13.
Ka is Keq. Nothing more. The subscript "a" simply tells you this particular Keq applies to an acid dissociation equilibrium.
The same logic applies to Kb (base dissociation constant) for a base B:
$$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$
Ka, Kb, Ksp, and Kw — all are just Keq applied to a specific type of equilibrium
Ka is Keq for HA(aq) ⇌ H⁺(aq) + A⁻(aq) — same expression, same exclusion rules (water = solvent, excluded). Ka = [H⁺][A⁻]/[HA]. Larger Ka means more dissociation at equilibrium (stronger acid). Kb is Keq for base dissociation: Kb = [BH⁺][OH⁻]/[B]. All of Ka, Kb, Ksp, and Kw are just Keq for specific equilibrium types.
Copy the Ka and Kb definitions and their expressions into your notes before the check below.
Which of the following correctly identifies Ka as a form of Keq?
We just saw that Ka and Kb are just Keq applied to acid and base dissociation equilibria — same expression, same exclusion rules. That raises a question: what does the numerical value of Ka actually tell you about how an acid behaves — and how do you use Ka to rank and classify acids? This card answers it → by showing how Ka magnitude distinguishes strong from weak acids and how to compare relative strengths.
Ka and Kb are the quantitative tools that distinguish strong acids from weak acids — and the logic is identical to using Keq magnitude to distinguish reactions that go to completion from those that reach a partial equilibrium.
| Acid/Base | Ka or Kb | Classification | Dissociation at equilibrium |
|---|---|---|---|
| HCl | >> 1 (e.g. 10⁷) | Strong acid | Essentially complete |
| Aspirin (HA) | 3.0 × 10⁻⁴ | Weak acid | Partial (~5% at 0.1 mol/L) |
| Acetic acid (CH₃COOH) | 1.8 × 10⁻⁵ | Weak acid | Partial (~1% at 0.1 mol/L) |
| NH₃ (ammonia) | 1.8 × 10⁻⁵ | Weak base | Partial |
| NaOH | >> 1 | Strong base | Essentially complete |
Comparing Ka values of weak acids: a larger Ka means more dissociation (stronger weak acid). Aspirin (Ka = 3.0 × 10⁻⁴) is more acidic than acetic acid (Ka = 1.8 × 10⁻⁵) — its equilibrium lies further to the right.
Strong acid: Ka >> 1 → essentially complete dissociation → no HA remaining at equilibrium. Weak acid: Ka << 1 → partial equilibrium → most HA stays undissociated. To rank weak acid strength: larger Ka = stronger weak acid (more H⁺ produced). Ka is not pH and not [H⁺] — it is the equilibrium constant for dissociation.
Write the strong vs weak acid Ka comparison and the ranking rule into your notes before the check below.
Three weak acids at 25°C: HCN Ka = 6.2 × 10⁻¹⁰; HF Ka = 6.8 × 10⁻⁴; HCOOH Ka = 1.8 × 10⁻⁴. Which ranking from strongest to weakest is correct?
We just saw how Ka magnitude distinguishes strong from weak acids and lets you rank relative acid strength. That raises a question: how does Ka as a Keq value directly explain a real medical phenomenon — why does aspirin irritate the stomach? This card answers it → by tracing how the Ka-driven equilibrium shift between stomach pH and cell pH leads to ion trapping inside stomach lining cells.
Aspirin's Ka of 3.0 × 10⁻⁴ is not just a number — it is the chemical explanation for one of the most commonly experienced drug side effects, and it has driven decades of pharmaceutical development to find safer analogues.
The Ion Trapping Mechanism
Aspirin equilibrium: $\text{HA}(aq) \rightleftharpoons \text{A}^-(aq) + \text{H}^+(aq)$, Ka = 3.0 × 10⁻⁴.
- Stomach (pH ≈ 1–2, high [H⁺]): High [H⁺] is a product of aspirin's dissociation. By LCP, the equilibrium shifts LEFT → aspirin remains mostly in the undissociated neutral form (HA).
- Neutral HA is lipid-soluble → can diffuse through the lipid bilayer of stomach lining cells (the cell membrane).
- Inside the cell (pH ≈ 7.4, low [H⁺]): Lower [H⁺] means the equilibrium shifts RIGHT → aspirin dissociates inside the cell → H⁺ and A⁻ ions released.
- Ion trapping: The charged ionic form (A⁻) cannot escape back through the lipid bilayer → it accumulates inside the cell, irritating the interior.
Ka determines how much dissociation occurs at each pH. A higher Ka would mean more dissociation in the stomach, more neutral form, more entry into cells, more irritation.
Aspirin ion-trapping mechanism: stomach (pH 1–2, high [H⁺]) → LCP shifts HA ⇌ A⁻ + H⁺ LEFT → neutral HA dominates → HA is lipid-soluble → diffuses into stomach lining cells → inside cell (pH 7.4, low [H⁺]) → equilibrium shifts RIGHT → A⁻ and H⁺ released → A⁻ is charged, cannot escape lipid bilayer → trapped → irritation.
Add the ion-trapping mechanism chain to your notes before the check below.
Why does aspirin (HA ⇌ A⁻ + H⁺) mostly remain in the neutral HA form inside the acidic stomach?
We just saw how aspirin's Ka drives the ion-trapping mechanism across the pH gradient between stomach and cell interior. That raises a question: Ka and Kb seem independent, but is there a mathematical relationship between them for a conjugate acid-base pair — and what does that imply about acid and base strength? This card answers it → by deriving Ka × Kb = Kw from first principles and applying it to calculate Kb from Ka for aspirin's conjugate base.
The product of Ka for a weak acid and Kb for its conjugate base is always Kw. This is not a coincidence — it is a direct consequence of equilibrium thermodynamics.
For any conjugate acid-base pair (HA and A⁻): $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 25°C.
Why this works (qualitative derivation):
- Acid dissociation: $K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]$
- Base dissociation of conjugate: $K_b = [\text{HA}][\text{OH}^-]/[\text{A}^-]$
- Multiplying: $K_a \times K_b = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \times \frac{[\text{HA}][\text{OH}^-]}{[\text{A}^-]} = [\text{H}^+][\text{OH}^-] = K_w$
The HA and A⁻ terms cancel exactly. The result is the Kw expression for water autoionisation.
Application: If Ka of a weak acid is known: $K_b = K_w / K_a = 1.0 \times 10^{-14} / K_a$
For aspirin: Ka = 3.0 × 10⁻⁴ → Kb(acetylsalicylate) = 1.0 × 10⁻¹⁴ / 3.0 × 10⁻⁴ = 3.33 × 10⁻¹¹
The conjugate base of aspirin is a very weak base (Kb << 1) — consistent with the rule: stronger acid → weaker conjugate base; weaker acid → stronger conjugate base.
Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C — applies to conjugate pairs only (HA and its A⁻). To find Kb: Kb = Kw / Ka. Stronger acid (larger Ka) → weaker conjugate base (smaller Kb). The derivation: multiplying Ka and Kb expressions cancels [HA] and [A⁻], leaving [H⁺][OH⁻] = Kw.
Write Ka × Kb = Kw, the conjugate-pair rule, and the Kb = Kw/Ka formula into your notes before the check below.
HF has Ka = 6.8 × 10⁻⁴ at 25°C. What is Kb for its conjugate base F⁻?
We just saw that Ka × Kb = Kw links conjugate pair strength via thermodynamics. That raises a question: we know ΔG° = −RT ln Keq qualitatively from L13, but how do you actually calculate ΔG° from Keq with real numbers — and what pitfalls exist? This card answers it → by demonstrating the calculation for the Haber process at 298 K and 500°C, including the Kelvin conversion and units checklist.
Acetic acid has Ka = 1.8 × 10⁻⁵. Put that into the equation van 't Hoff derived in 1886 and you get ΔG° = −RT ln(1.8 × 10⁻⁵) = +27 kJ/mol — meaning acetic acid's ionisation is thermodynamically uphill, explaining why it barely ionises. HCl has Ka >> 1, so ΔG° is large and negative — it ionises completely. The equation ΔG° = −RT ln Keq converts a number (Keq) into a thermodynamic energy statement about how strongly a reaction favours one direction.
$\Delta G^\circ = -RT \ln K_{eq}$, where R = 8.314 J mol⁻¹ K⁻¹ and T is in Kelvin.
Calculation example — Haber process at 298 K (Keq = 977):
$$\Delta G^\circ = -(8.314)(298)\ln(977) = -(2477.6)(6.884) = -17{,}060 \text{ J/mol} = -17.1 \text{ kJ/mol}$$
Negative ΔG° confirms the forward reaction is spontaneous under standard conditions at 25°C — consistent with Keq = 977 >> 1.
Same reaction at 500°C (Keq = 0.013):
$$\Delta G^\circ = -(8.314)(773)\ln(0.013) = -(6427)(-4.343) = +27{,}910 \text{ J/mol} = +27.9 \text{ kJ/mol}$$
Positive ΔG° confirms forward reaction is non-spontaneous under standard conditions at 500°C — consistent with Keq = 0.013 << 1.
ΔG° = −RT ln Keq calculation checklist: (1) convert T to Kelvin: T(K) = T(°C) + 273; (2) use R = 8.314 J mol⁻¹ K⁻¹; (3) result is in joules — divide by 1000 for kJ; (4) sign check: Keq > 1 → ln positive → ΔG° negative (spontaneous forward); Keq < 1 → ln negative → ΔG° positive (non-spontaneous). ΔG° ≠ ΔG; ΔG = 0 at equilibrium.
Write the ΔG° = −RT ln Keq checklist into your notes before the check below.
For the reaction A(g) ⇌ B(g) + C(g), Keq = 2.50 × 10⁻³ at 500 K. Which correctly calculates ΔG°?
Write the Ka expression for each of the following and identify the stronger acid. (a) HF(aq) ⇌ H⁺(aq) + F⁻(aq), Ka = 6.8 × 10⁻⁴. (b) HNO₂(aq) ⇌ H⁺(aq) + NO₂⁻(aq), Ka = 4.5 × 10⁻⁴. (c) Write the Kb expression for F⁻ and calculate Kb at 25°C. (d) Is F⁻ a stronger or weaker base than NO₂⁻?
Part (a): Ka for HF
$K_a = [\text{H}^+][\text{F}^-]/[\text{HF}]$. Water excluded (solvent). Powers all 1. Ka = 6.8 × 10⁻⁴.
Part (b): Ka for HNO₂, Compare Strengths
$K_a = [\text{H}^+][\text{NO}_2^-]/[\text{HNO}_2]$. Ka = 4.5 × 10⁻⁴.
Stronger acid: Ka(HF) = 6.8 × 10⁻⁴ > Ka(HNO₂) = 4.5 × 10⁻⁴ → HF is the stronger weak acid — its equilibrium lies further to the right → more H⁺ at equilibrium for the same initial concentration.
Part (c): Kb for F⁻
Conjugate base of HF is F⁻. Base dissociation: $\text{F}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HF}(aq) + \text{OH}^-(aq)$
$K_b = [\text{HF}][\text{OH}^-]/[\text{F}^-]$. Water excluded (pure liquid solvent).
$K_b = K_w/K_a = 1.0 \times 10^{-14} / 6.8 \times 10^{-4} = \mathbf{1.47 \times 10^{-11}}$
Part (d): Compare Kb of F⁻ and NO₂⁻
Kb(NO₂⁻) = Kw/Ka(HNO₂) = 1.0 × 10⁻¹⁴ / 4.5 × 10⁻⁴ = 2.22 × 10⁻¹¹
Kb(NO₂⁻) = 2.22 × 10⁻¹¹ > Kb(F⁻) = 1.47 × 10⁻¹¹ → NO₂⁻ is a stronger base than F⁻. Consistent with: stronger acid (HF) → weaker conjugate base (F⁻); weaker acid (HNO₂) → stronger conjugate base (NO₂⁻).
(a) Calculate ΔG° for $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ at 430°C, given Keq = 54.3. (b) Is the forward reaction spontaneous at 430°C? (c) At 25°C, Keq = 794. Calculate ΔG° at 25°C and explain why the result differs from part (a).
Part (a): ΔG° at 430°C
Convert: T = 430 + 273 = 703 K. $\ln(54.3) = 3.994$
$$\Delta G^\circ = -(8.314)(703)(3.994) = -(5844.7)(3.994) = -23{,}343 \text{ J/mol} = \mathbf{-23.3 \text{ kJ/mol}}$$
Part (b): Spontaneity
ΔG° = −23.3 kJ/mol (negative) → forward reaction is spontaneous under standard conditions at 430°C. Consistent with Keq = 54.3 > 1 (products favoured).
Part (c): ΔG° at 25°C
Convert: T = 25 + 273 = 298 K. $\ln(794) = 6.677$
$$\Delta G^\circ = -(8.314)(298)(6.677) = -(2477.6)(6.677) = -16{,}539 \text{ J/mol} = \mathbf{-16.5 \text{ kJ/mol}}$$
At 25°C, ΔG° = −16.5 kJ/mol; at 430°C, ΔG° = −23.3 kJ/mol. Despite Keq being larger at 25°C (794 vs 54.3), the product RT × ln Keq is larger at 430°C because the temperature factor (703 K) is much larger than at 25°C (298 K). ΔG° magnitude reflects both T and ln Keq.
Checking Your Understanding
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Define the acid dissociation constant Ka and explain its importance as a form of Keq.
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Identify one common misconception about Ka, Kb and Gibbs free energy. Explain why it is incorrect.
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Describe how the Ka × Kb = Kw relationship relates to the Ka of an acid being just a Keq, using a specific example from the lesson.
Applying Your Knowledge
Q1. Which of the following correctly identifies Ka as a form of Keq?
Q2. The Ka values at 25°C for three weak acids are: HCN Ka = 6.2 × 10⁻¹⁰; HF Ka = 6.8 × 10⁻⁴; HCOOH Ka = 1.8 × 10⁻⁴. Which ranking from strongest to weakest is correct?
Q3. For the reaction $\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$, Keq = 2.50 × 10⁻³ at 500 K. Which of the following correctly calculates ΔG°?
SAQ 1 (4 marks). Explain, using Ka as a form of Keq and Le Chatelier's Principle, why aspirin (HA ⇌ A⁻ + H⁺, Ka = 3.0 × 10⁻⁴) causes stomach irritation. Your answer should include: the Ka expression, what happens in the stomach (pH ≈ 1–2), what happens inside stomach cells (pH ≈ 7.4), and why the ionic form becomes trapped.
SAQ 2 (3 marks). For the conjugate pair CH₃COOH / CH₃COO⁻, Ka(CH₃COOH) = 1.8 × 10⁻⁵ at 25°C. (a) Write the Kb expression for CH₃COO⁻. (b) Calculate Kb for CH₃COO⁻ at 25°C. (c) Is CH₃COO⁻ a stronger or weaker base than the conjugate base of HF (Ka = 6.8 × 10⁻⁴)? Explain.
Reveal Answers
SAQ 1: Ka is the equilibrium constant for aspirin's dissociation HA(aq) ⇌ A⁻(aq) + H⁺(aq); Ka = [A⁻][H⁺]/[HA] = 3.0 × 10⁻⁴. In the stomach (pH 1–2, high [H⁺]): H⁺ is a product of the dissociation — by LCP, high [H⁺] shifts the equilibrium LEFT, favouring the undissociated neutral HA form. Neutral HA is lipid-soluble and diffuses through the stomach cell membrane lipid bilayer. Inside the cell (pH 7.4, low [H⁺]): the low [H⁺] shifts equilibrium RIGHT — aspirin dissociates back into A⁻ and H⁺. The charged A⁻ ion cannot pass back through the non-polar lipid bilayer (ion trapping) — it accumulates inside the cell, causing irritation.
SAQ 2: (a) Kb(CH₃COO⁻): CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq); Kb = [CH₃COOH][OH⁻]/[CH₃COO⁻]. (b) Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. (c) Kb(F⁻) = Kw/Ka(HF) = 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹. Kb(CH₃COO⁻) = 5.56 × 10⁻¹⁰ > Kb(F⁻) = 1.47 × 10⁻¹¹ → CH₃COO⁻ is a stronger base than F⁻. This is consistent with: acetic acid (weaker acid, smaller Ka) → stronger conjugate base; HF (stronger acid, larger Ka) → weaker conjugate base.
Complete these Ka and Kb relationship statements.
For a conjugate acid–base pair: Ka × Kb = at 25 °C.
A reaction with ΔG < 0 is under standard conditions.
The relationship between ΔG° and Keq: if Keq > 1, then ΔG° is .
Complete the Learn phase to unlock practice questions.
Ibuprofen (Ka ≈ 2.0 × 10⁻⁵) is a common anti-inflammatory drug. Like aspirin, it is a weak acid. (a) Write the Ka expression for ibuprofen's dissociation (HA ⇌ A⁻ + H⁺). (b) Compare the acidity of ibuprofen (Ka = 2.0 × 10⁻⁵) to aspirin (Ka = 3.0 × 10⁻⁴). Which would cause more stomach irritation at the same concentration, and why? Use Le Chatelier's Principle and Ka values in your answer. (c) A chemist calculates ΔG° = −RT ln Ka for ibuprofen's dissociation at 37°C (body temperature). Explain what this calculation would tell the chemist and whether the sign of ΔG° should be positive or negative. (d) Calculate Kb for the ibuprofen anion (A⁻) at 25°C. (6 marks)
Return to your Think First response about whether Ka is a different type of constant from Keq. Recall Arrhenius's 1887 dissociation theory at Uppsala University — Ka is simply Keq for the specific reaction HA ⇌ H⁺ + A⁻. Ka and Kb are both equilibrium constants — Ka for acid dissociation, Kb for base dissociation. For a conjugate pair, Ka × Kb = Kw (1.0 × 10⁻¹⁴ at 25°C). Were you right that Ka is the same concept as Keq?