Year 12 Chemistry Module 5 ⏱ ~35 min Lesson 14 of 18

Ka and Kb as Equilibrium Constants

Ka and Kb are not new constants to memorise. They are Keq written for a dissociation reaction, which means every rule you already have for Keq applies to them unchanged.

Today's hook, Acetic acid has Ka = 1.8 × 10⁻⁵ and hydrochloric acid has Ka around 10⁷. Those two numbers say everything about which one dissociates almost completely, and they are read exactly the same way as any other equilibrium constant.

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Get oriented

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First, Before You Read

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
Kb is Keq for base ionisation, written the same way
Strong acids have large Ka; weak acids have small Ka

Understand

Why the same Keq rules (products over reactants, pure liquids and solids excluded) apply unchanged to Ka, Kb and Ksp
What the magnitude of Ka or Kb tells you about how far a dissociation actually goes
The quantitative relationship ΔG° = −RT ln Keq and what it means for spontaneity

Can Do

Write the Keq expression for an acid dissociation, a base ionisation and a dissolving ionic solid
Rank weak acids by strength from their Ka values and justify the ranking
Explain what a very large or very small Keq implies about the position of equilibrium

Module 5, Key Formulas: Lesson 14

Ka = Keq for: HA(aq) ⇌ H⁺(aq) + A⁻(aq)

Ka = [H⁺][A⁻] / [HA] water excluded as solvent

Kb = Keq for: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)

Kb = [BH⁺][OH⁻] / [B] H₂O excluded as pure liquid solvent

Strong acid: Ka >> 1; Weak acid: Ka << 1; Larger Ka → stronger acid

Ksp = Keq for: M_xA_y(s) ⇌ xM^y+(aq) + yA^x−(aq) solid excluded

ΔG° = −RT ln Keq R = 8.314 J mol⁻¹ K⁻¹; T in Kelvin

Positive ΔG° → Keq < 1 → reactants favoured; Negative ΔG° → Keq > 1 → products favoured

Syllabus boundary

MODULE 5 BOUNDARY: this lesson uses Ka and Kb only as names for Keq applied to a dissociation. pH and pKa calculations, Kw and the conjugate-pair relationship, Brønsted-Lowry proton transfer, ICE tables for weak acids and buffers all belong to Module 6.

Key Terms, scan these before reading

Acid dissociation constant (Ka)

The equilibrium constant for the partial ionisation of a weak acid: Ka = [H⁺][A⁻]/[HA].

Base dissociation constant (Kb)

The equilibrium constant for the partial ionisation of a weak base: Kb = [BH⁺][OH⁻]/[B].

Solubility product (Ksp)

The equilibrium constant for a sparingly soluble ionic solid dissolving; the solid itself is excluded.

Gibbs free energy (ΔG°)

ΔG° = −RT ln Keq; negative ΔG° means spontaneous reaction and Keq > 1.

Extent of dissociation

How far a dissociation proceeds before equilibrium; read from the magnitude of Ka or Kb, not from concentration alone.

Weak acid behaviour

Weak acids partially dissociate; Ka quantifies the extent of dissociation at equilibrium.

Misconceptions to Fix

✗ 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.

Fill the blanks+4 XP

Complete these K_a and K_b relationship statements.

A larger K_a value indicates a acid (more dissociation).

In the expression for K_sp of a dissolving ionic solid, the is left out, exactly as it is in any other K_eq expression.

A reaction with ΔG < 0 is under standard conditions.

The relationship between ΔG° and K_eq: if K_eq > 1, then ΔG° is .

Ka is Keq for a dissociation

Ka Is Just Keq for an Acid Dissociation, Nothing New

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:

$$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$

Think First Answer

The student was wrong. Ka is NOT a different type of constant, it is Keq applied to the specific equilibrium of an acid dissociation. The only new element is the naming convention. The expression, the rules (exclude solids and pure liquids), and the interpretation (magnitude tells you equilibrium position) are identical to everything in L09.

Common Error

Students write water in the Ka expression: $K_a = [\text{H}^+][\text{A}^-][\text{H}_2\text{O}]/[\text{HA}]$, wrong. Water is the solvent (pure liquid) and is excluded from all equilibrium expressions, including Ka and Kb, exactly as it is excluded from any other Keq.

Ka and Kb mind map as a tree with Keq as root branching to Ka, Kb, Ksp, Kw

Ka, Kb, Ksp, and Kw, all are just Keq applied to a specific type of equilibrium

Exam Tip

When explaining equilibrium shifts, always state the direction (left or right) and justify using Le Chatelier's Principle language, simply stating the direction alone will not earn full marks.

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?

Interactive, Ka Expression Builder

Use the builder. Work through the five acids, then answer for HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq): which is the correct K_a expression?

What the size of Ka and Kb tells you

Strong vs Weak Acids and Bases: Ka and Kb as Discriminators

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
Hydrofluoric acid (HF)	6.8 × 10⁻⁴	Weak acid	Partial (a few per cent 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 (a stronger weak acid). Hydrofluoric acid (Ka = 6.8 × 10⁻⁴) is a stronger weak acid than acetic acid (Ka = 1.8 × 10⁻⁵), because its equilibrium lies further to the right.

Must Know

A larger Ka means a stronger acid and more dissociation at equilibrium. To rank acids by strength, rank them by Ka value: largest Ka is the strongest acid. This is exactly the same logic you already use to rank any reactions by the magnitude of Keq, which is the whole point of this lesson.

Common Error

Confusing Ka with pH or with [H⁺]. Ka is not pH and not the concentration of H⁺, it is the equilibrium constant for the dissociation. Ka determines how much dissociation occurs; pH depends on both Ka and the concentration of the acid.

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.

Strong vs weak acids: a strong acid (e.g. HCl) dissociates essentially completely (Ka ≫ 1, ~100%), leaving no HA; a weak acid (e.g. CH₃COOH) only partially dissociates (Ka ≪ 1, ~1% at 0.1 mol/L, Ka = 1.8×10⁻⁵), so most stays as HA. Other strong acids: H₂SO₄, HNO₃, HBr, HI, HClO₄; other weak acids: aspirin, HF, H₂CO₃, lactic acid.

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?

The same idea for a dissolving solid

Keq for the Dissociation of Ionic Solutions

We just saw that Ka and Kb are Keq written for a dissociation, and that their size tells you how far that dissociation goes. That raises a question: the syllabus asks you to use Keq for different types of reaction, so what does the same expression look like when the thing dissociating is not a molecule in solution but a solid? This card answers it → by writing Keq for a sparingly soluble ionic solid and showing that the exclusion rules you already know do all the work.

Drop a little silver chloride into water and almost none of it dissolves. That "almost none" is not a vague statement, it is a very small equilibrium constant, and it is written by exactly the same recipe as every other Keq in this module.

A sparingly soluble ionic solid sitting in contact with its saturated solution is at equilibrium:

$$\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$$

Write Keq the way you always do, products over reactants. Then apply the exclusion rule from IQ3: a pure solid has no meaningful concentration, so it never appears. What is left is

$$K_{sp} = [\text{Ag}^+][\text{Cl}^-]$$

and that constant gets its own name, the solubility product, for the same reason Ka got one: it is Keq for a particular kind of reaction, not a new kind of constant.

Stoichiometric coefficients become powers here, exactly as they do everywhere else. For lead(II) iodide,

$$\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \qquad K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2$$

Type of reaction	Equilibrium	Name given to Keq	What is excluded
Acid dissociation	HA(aq) ⇌ H⁺(aq) + A⁻(aq)	K_a	Water, as the solvent
Base ionisation	B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)	K_b	Water, as a pure liquid
Ionic solid dissolving	AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)	K_sp	The undissolved solid

Read the magnitudes the same way too. K_sp for AgCl is about 1.8 × 10⁻¹⁰, which is tiny, so the equilibrium sits far to the left and very little dissolves. That is the same reading you gave a small Ka a moment ago.

Must Know

The syllabus asks you to use Keq for different types of chemical reactions. The examiner is testing one skill three times: write products over reactants, raise each to its coefficient, and leave out pure solids and pure liquids. Ka, Kb and Ksp are the three names that skill wears.

Common Error

Writing K_sp = [Ag⁺][Cl⁻] / [AgCl]. The undissolved solid is not in the expression at all. Adding more solid to a saturated solution does not change K_sp and does not dissolve any more of it.

Ksp is Keq for an ionic solid dissolving: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) gives Ksp = [Ag⁺][Cl⁻], with the solid excluded because it is a pure solid. Coefficients become powers: PbI₂(s) ⇌ Pb²⁺ + 2I⁻ gives Ksp = [Pb²⁺][I⁻]². A small Ksp means very little dissolves.

Pause, copy the two Ksp expressions and the exclusion rule into your notes before the check below.

What is the correct Ksp expression for Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq)?

Where this goes next: Ksp is introduced here only as another name for Keq. Calculating solubility from Ksp, comparing salts fairly, and predicting whether a precipitate forms are the work of L17 and L18 in IQ4.

Extension: Gibbs free energy and Keq

Beyond the syllabus. Applying Keq to acid and base dissociation is core. The quantitative link $\Delta G^\circ = -RT \ln K_{eq}$ is extension for this syllabus — read it for insight, but exam marks come from writing Ka and Kb expressions and using them.

Deep dive, Extension ΔG° = −RT ln Keq: Quantitative Introduction

We just saw that Ka, Kb and Ksp are all Keq wearing different names for different reaction types. 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.

Must Know (Calculation Checklist)

(1) Always convert T to Kelvin: T(K) = T(°C) + 273. (2) Use R = 8.314 J mol⁻¹ K⁻¹, answer in joules; divide by 1000 to convert to kJ. (3) Check the sign: if Keq > 1, ln Keq > 0, ΔG° is negative; if Keq < 1, ln Keq < 0, ΔG° is positive.

Common Error

Using T = 25 (Celsius) instead of T = 298 (Kelvin), gives a result approximately 12× too small. Also: ΔG° ≠ ΔG. ΔG = 0 at equilibrium for any system; ΔG° is a fixed property of the reaction at a given temperature and equals zero only when Keq = 1.

Module Boundary Reminder

Ka, Kb and Ksp are presented here as naming conventions for Keq only. Weak-acid pH calculations, pKa, the Ka and Kb relationship through Kw, conjugate-pair reasoning, buffers and titration curves all belong to Module 6.

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°?

Deep dive, Extension. Applying Keq to acid and base dissociation, and Ka/Kb calculations, are core. Quantitative use of ΔG° = −RT ln Keq is Extension for this syllabus and does not affect your core score.

Worked examples and activities

Worked Examples

Worked Example 1 Apply/Analyse

Write the Keq 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 Keq expression for the base ionisation NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). (d) Write the Keq expression for BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) and state what its very small value (about 1.1 × 10⁻¹⁰) tells you.

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 ammonia

Products over reactants: $K_b = [\text{NH}_4^+][\text{OH}^-]/[\text{NH}_3]$.

Water is the solvent and is a pure liquid, so it is excluded, exactly as it was in the Ka expressions above. Nothing new is being applied here.

Part (d): Ksp for barium sulfate

$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$. The undissolved BaSO₄ is a pure solid, so it does not appear.

A value of about 1.1 × 10⁻¹⁰ is extremely small, so the equilibrium lies far to the left and only a tiny amount dissolves. This is the same reading you gave a small Ka: the size of the constant tells you how far the dissociation goes.

Answer: (a) Ka = [H⁺][F⁻]/[HF]. (b) Ka = [H⁺][NO₂⁻]/[HNO₂]; HF is the stronger weak acid (larger Ka). (c) Kb = [NH₄⁺][OH⁻]/[NH₃], water excluded. (d) Ksp = [Ba²⁺][SO₄²⁻], solid excluded; a value near 1.1 × 10⁻¹⁰ means very little dissolves. ✓

Worked Example 2 Analyse/Evaluate

(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.

Answer: (a) ΔG° = −23.3 kJ/mol. (b) Spontaneous, ΔG° negative, Keq > 1. (c) ΔG° = −16.5 kJ/mol at 25°C, less negative than at 430°C because the smaller T (298 K vs 703 K) gives a smaller RT × ln Keq product despite the larger Keq value. ✓

Activities

🔎 Activity 1, Spot + Fix

Checking Your Understanding

Answer the questions below to check your understanding of the key concepts from this lesson.

Define the acid dissociation constant Ka and explain its importance as a form of Keq.
Identify one common misconception about Ka, Kb and Gibbs free energy. Explain why it is incorrect.
Write the Keq expression for a dissolving ionic solid of your choice, state what is excluded and why, and explain what a very small value of that constant tells you.

🔬 Activity 2, Apply + Analyse

Applying Your Knowledge

Use what you have learned to solve the problems below. Show your reasoning clearly.

Question A

Explain why Ka, Kb and Ksp are described as the same constant applied to different types of reaction rather than as three separate constants. Use one worked expression of each to support your answer.

Question B

Compare and contrast Ka and Keq using the specific example of acetic acid dissociation.

Question C

Predict what would happen to the Ka of acetic acid (1.8 × 10⁻⁵ at 25°C) if the temperature were raised above 25°C. Explain your reasoning using ΔG° = −RT ln Keq, and state which factors could not change Ka no matter how much you changed them.

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Short Answer

(4 marks) 1. Methanoic acid dissociates as HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq) with Ka = 1.8 × 10⁻⁴ at 25°C, and hydrocyanic acid as HCN(aq) ⇌ H⁺(aq) + CN⁻(aq) with Ka = 6.2 × 10⁻¹⁰. (a) Write the Ka expression for each and state what is excluded from it and why. (b) Identify the stronger acid and justify your choice using the magnitude of Ka. (c) Explain what the value 6.2 × 10⁻¹⁰ tells you about the position of the HCN equilibrium.

(3 marks) 2. Silver chromate dissolves according to Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq). (a) Write the Keq expression for this dissociation and name the constant. (b) Explain why the solid does not appear in the expression, and why the coefficient 2 does appear. (c) A student says adding more solid Ag₂CrO₄ to the saturated solution will increase the value of the constant. Explain why that is wrong.

Calculation practice

Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.

ApplyBand 4(4 marks) A 0.100 mol/L solution of a monoprotic weak acid HA has a pH of 2.87 at 25 °C. Calculate the acid dissociation constant Ka for HA, stating any assumption you make about the extent of dissociation.

ApplyBand 5(4 marks) Ethanoic acid has Ka = 1.8 × 10⁻⁵ at 25 °C. Calculate the base dissociation constant Kb for the ethanoate ion at the same temperature, and explain what the relative sizes of Ka and Kb indicate about the strength of this conjugate pair.

ApplyBand 3(1 mark) At 25 °C a solution has [OH⁻] = 2.5 × 10⁻³ mol/L. Calculate [H⁺] in this solution.

Show all answers

Short Answer Model Answers

SA1 (a) Ka(HCOOH) = [H⁺][HCOO⁻]/[HCOOH] and Ka(HCN) = [H⁺][CN⁻]/[HCN] [1]. Water is excluded from both because it is the solvent and a pure liquid, so its concentration is effectively constant, which is the same exclusion rule used for every Keq expression [1]. (b) Methanoic acid is the stronger acid because its Ka (1.8 × 10⁻⁴) is far larger than that of HCN (6.2 × 10⁻¹⁰), and a larger equilibrium constant means the dissociation proceeds further before equilibrium is reached [1]. (c) A Ka of 6.2 × 10⁻¹⁰ is extremely small, so the HCN equilibrium lies far to the left: at equilibrium the solution is almost entirely undissociated HCN with very little H⁺ and CN⁻ [1].

SA2 (a) Ksp = [Ag⁺]²[CrO₄²⁻]; the constant is called the solubility product [1]. (b) Ag₂CrO₄ is a pure solid, and a pure solid has no meaningful concentration to vary, so it is excluded exactly as pure liquids are; the coefficient 2 appears as a power on [Ag⁺] because stoichiometric coefficients always become powers in a Keq expression [1]. (c) Adding more solid does not change the constant. Ksp is fixed at a given temperature, and the solid does not appear in the expression at all, so the extra solid simply sits undissolved; only a change in temperature can change the value of Ksp [1].

✓

Retrieve and reflect

Check what actually stuck

Extended Response

A chemistry student claims: "Ka, Kb and Ksp are three different constants that happen to look similar, so they each need their own rules." Evaluate this claim. In your response: (a) write the equilibrium and the Keq expression for one acid dissociation, one base ionisation and one dissolving ionic solid of your choice; (b) identify what is excluded from each expression and state the single rule that covers all three cases; (c) explain what the magnitude of each constant tells you about the position of that equilibrium, using a specific value; and (d) use ΔG° = −RT ln Keq to explain why a very small constant corresponds to a positive ΔG°. (8 marks)

How did your thinking change?

Return to your Think First response about whether Ka is a different type of constant from Keq. Arrhenius's 1887 dissociation theory gave the acid dissociation constant its name, but the lesson has shown it is simply Keq for the reaction HA ⇌ H⁺ + A⁻. The same is true of Kb for a base ionisation and Ksp for a dissolving ionic solid: one rule, products over reactants with coefficients as powers and pure solids and liquids excluded, wearing three different names. Were you right that Ka is the same concept as Keq?

I have completed this lesson

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