Aspirin (acetylsalicylic acid) has Ka = 3.0 × 10⁻⁴ — a number that tells you exactly why it irritates your stomach lining and why taking it with a full glass of water helps.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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 applied to a specific type of reaction? Write your position with reasoning before reading on.
Wrong: A negative ΔG means a reaction will happen quickly.
Right: A negative ΔG means a reaction is thermodynamically spontaneous — it can occur without external energy input. It says nothing about reaction rate. Thermodynamic spontaneity and kinetic rate are independent concepts. Some spontaneous reactions are extremely slow at room temperature.
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 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.
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.
Aspirin equilibrium: $\text{HA}(aq) \rightleftharpoons \text{A}^-(aq) + \text{H}^+(aq)$, Ka = 3.0 × 10⁻⁴.
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.
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):
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.
The equation ΔG° = −RT ln Keq is the quantitative version of everything you have known qualitatively since L02 — it converts a number (Keq) into a thermodynamic energy statement (how strongly the 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.
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⁻?
$K_a = [\text{H}^+][\text{F}^-]/[\text{HF}]$. Water excluded (solvent). Powers all 1. Ka = 6.8 × 10⁻⁴.
$K_a = [\text{H}^+][\text{NO}_2^-]/[\text{HNO}_2]$. 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.
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).
$K_b = K_w/K_a = 1.0 \times 10^{-14} / 6.8 \times 10^{-4} = \mathbf{1.47 \times 10^{-11}}$
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).
Convert: T = 430 + 273 = 703 K. $\ln(54.3) = 3.994$
$$\Delta G^\circ = -(8.314)(703)(3.994) = -(5844.7)(3.994) = -23{,}343 \text{ J/mol} = \mathbf{-23.3 \text{ kJ/mol}}$$
ΔG° = −23.3 kJ/mol (negative) → forward reaction is spontaneous under standard conditions at 430°C. Consistent with Keq = 54.3 > 1 (products favoured).
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.
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). A large Ka means a strong acid; a small Ka means a weak acid. The relationship to Gibbs free energy is ΔG° = −RT ln K. When K > 1, ΔG° is negative and the forward reaction is spontaneous under standard conditions.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Q1. Which of the following correctly identifies Ka as a form of Keq?
Q1. Select the option that 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°?
Q3. For the reaction $\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)$, Keq = 2.50 × 10⁻³ at 500 K. Select the option that correctly calculates ΔG°?
Lesson 14 complete — Ka, Kb & Gibbs Free Energy