Chemistry • Year 12 • Module 5 • Lesson 14

Ka, Kb & Gibbs Free Energy

Lock in the core vocabulary, equilibrium expressions for Ka and Kb, the Ka × Kb = Kw relationship, and the sign conventions for ΔG° = −RT ln Keq.

Build • Vocab & Recall

1. Term–definition match

Ten definitions are listed in shuffled order. In the right-hand column write the matching term from this list: acid dissociation constant (Ka), base dissociation constant (Kb), Kw, pKa, conjugate acid–base pair, Gibbs free energy (ΔG°), spontaneous reaction, weak acid, weak base, strong acid. 10 marks

#DefinitionMatching term
1.1The equilibrium constant for the partial ionisation of a weak acid: K = [H¹+][A−] / [HA]; water is excluded because it is the solvent.
1.2The equilibrium constant for the partial ionisation of a weak base in water: K = [BH¹+][OH−] / [B]; water is again excluded.
1.3The product Ka × Kb = 1.0 × 10−¹&sup4; at 25 °C, which is itself just Keq for the autoionisation of water: H&sub2;O ⇌ H¹+ + OH−.
1.4A dimensionless logarithmic scale defined as −log&sub10;(Ka); a smaller value indicates a stronger acid.
1.5An acid and its conjugate base (HA and A−), or a base and its conjugate acid (B and BH¹+), linked by the transfer of one proton.
1.6The thermodynamic potential (ΔG° = −RT ln Keq) that quantifies whether a reaction is product-favoured or reactant-favoured under standard conditions.
1.7A reaction for which ΔG° < 0 and Keq > 1; the forward reaction proceeds without an ongoing external energy input under standard conditions.
1.8An acid that partially dissociates in water, establishing an equilibrium; Ka << 1. Examples: acetic acid (Ka = 1.8 × 10−&sup5;), HF (Ka = 7.2 × 10−&sup4;).
1.9A base that partially ionises in water, establishing an equilibrium; Kb << 1. Example: ammonia (Kb = 1.8 × 10−&sup5;).
1.10An acid that essentially completely dissociates in dilute aqueous solution; Ka >> 1. Examples: HCl, HNO&sub3;, H&sub2;SO&sub4;.
Stuck? Revisit the Key Terms panel and formula panel in the lesson.

2. True or false — with correction

Circle T or F. If the statement is false, rewrite it correctly on the line below. 12 marks (1 T/F + 1 correction each)

2.1 Ka is a fundamentally different type of constant from Keq — it measures something about acids that Keq cannot.    T  /  F

2.2 Water is included in the Ka expression because it is a reactant in the proton-transfer equation.    T  /  F

2.3 A larger Ka value indicates a stronger weak acid — its equilibrium lies further to the right.    T  /  F

2.4 Ka × Kb = Kw applies to any acid paired with any base, not just a conjugate pair.    T  /  F

2.5 A negative ΔG° means a reaction will proceed quickly at room temperature.    T  /  F

2.6 If Keq < 1, then ΔG° is positive and the reverse reaction is favoured under standard conditions.    T  /  F

Stuck? Revisit Cards 1, 4 and 5 and the misconceptions box in the lesson.

3. Cloze — fill the blanks

Complete the paragraph using terms from the word bank below. Each term is used once. 10 marks

Word bank:   equilibrium constant  •  solvent  •  Ka  •  partial  •  larger  •  conjugate  •  Kw  •  Kelvin  •  negative  •  spontaneous

The acid dissociation constant,  _______________ , is simply the  _______________  written for the specific equilibrium in which a weak acid partially ionises in water. Water is excluded from the expression because it acts as the  _______________ . A weak acid undergoes  _______________  ionisation, meaning Ka << 1. A  _______________  Ka value indicates greater dissociation and therefore a stronger weak acid. For a  _______________  acid–base pair such as acetic acid (CH&sub3;COOH) and the acetate ion (CH&sub3;COO−), the product Ka × Kb equals  _______________  = 1.0 × 10−¹&sup4; at 25 °C. The relationship to Gibbs free energy is ΔG° = −RT ln Keq, where T must be expressed in  _______________ . When Keq > 1, ln Keq is positive, making ΔG°  _______________ , which confirms the forward reaction is  _______________  under standard conditions.

Stuck? Read the formula panel and Card 5 of the lesson carefully before completing this.

4. Build a concept map

Draw labelled arrows between the six terms below to show how they relate to each other. Each arrow must carry a linking phrase (e.g. “is a type of”, “determines”, “equals product of”). Aim for at least 6 labelled arrows. 6 marks

Supplied terms: KaKbKwKeqΔG°conjugate pair.

Keq
Ka
Kb
ΔG°
Kw
conjugate pair
Key relationships: Ka and Kb are both types of Keq; ΔG° = −RT ln Keq; Ka × Kb = Kw for a conjugate pair; stronger acid → weaker conjugate base (smaller Kb).

5. Function recall

Answer each in 1–2 sentences using precise lesson vocabulary. 8 marks (2 each)

5.1 What does a Ka value of 1.8 × 10−&sup5; tell you about the position of the equilibrium for acetic acid dissociation in water?

5.2 Explain what pKa is and why it is convenient when comparing Ka values that span many orders of magnitude (e.g. 10−&sup4; to 10−¹&sup0;).

5.3 Why is temperature always expressed in Kelvin, not Celsius, in the equation ΔG° = −RT ln Keq?

5.4 The conjugate base of carbonic acid (H&sub2;CO&sub3;, Ka = 4.3 × 10−&sup7;) is the bicarbonate ion (HCO&sub3;−). What does the relationship Ka × Kb = Kw tell you about the strength of HCO&sub3;− as a base?

Stuck? Re-read Cards 1, 2 and 4 and the key terms panel.
Answers — Do not peek before attempting

Q1 — Term–definition matches

1.1 acid dissociation constant (Ka) • 1.2 base dissociation constant (Kb) • 1.3 Kw • 1.4 pKa • 1.5 conjugate acid–base pair • 1.6 Gibbs free energy (ΔG°) • 1.7 spontaneous reaction • 1.8 weak acid • 1.9 weak base • 1.10 strong acid.

Q2 — True / false with corrections

2.1 False. Ka is not a different type of constant from Keq. Ka is simply Keq applied to the specific equilibrium of an acid dissociation (HA ⇌ H¹+ + A−). The expression, rules, and interpretation are identical.

2.2 False. Water is excluded from the Ka expression because it is the solvent (a pure liquid), not because it is not involved. The rule for writing Keq expressions is: exclude pure liquids and pure solids.

2.3 True.

2.4 False. Ka × Kb = Kw applies only to a conjugate acid–base pair — the specific acid HA and its conjugate base A− from the same dissociation equilibrium. Applying it to any random acid and base is incorrect.

2.5 False. A negative ΔG° means the forward reaction is thermodynamically spontaneous under standard conditions, but it says nothing about reaction rate. Some spontaneous reactions are extremely slow (e.g. diamond converting to graphite).

2.6 True.

Q3 — Cloze answers (in order)

Ka • equilibrium constant • solvent • partial • larger • conjugate • Kw • Kelvin • negative • spontaneous.

Q4 — Sample concept map (mark any 6 valid arrows)

Acceptable arrows include: Ka is a specific type of → Keq; Kb is a specific type of → Keq; ΔG° is related to by ΔG° = −RT ln → Keq; Ka × Kb equals for a → conjugate pair → Kw; conjugate pair has both → Ka and Kb; stronger acid (larger Ka) linked to → weaker conjugate base (smaller Kb). Award 1 mark per valid labelled arrow (minimum 6 required).

Q5.1 — Interpreting Ka = 1.8 × 10−&sup5;

Ka = 1.8 × 10−&sup5; << 1, so the equilibrium lies far to the left — at equilibrium, the vast majority of acetic acid molecules remain undissociated (as HA). Only a very small proportion (~1% at 0.1 mol/L) has ionised to form H¹+ and CH&sub3;COO−. Acetic acid is therefore a weak acid.

Q5.2 — pKa and its convenience

pKa = −log&sub10;(Ka). Because Ka values span many orders of magnitude (e.g. 10−&sup4; to 10−¹&sup0;), comparing raw Ka numbers is cumbersome. The logarithmic pKa scale compresses this range into small, easily comparable integers (pKa 4 vs pKa 10). A smaller pKa always means a stronger acid (larger Ka).

Q5.3 — Why Kelvin in ΔG° = −RT ln K

R = 8.314 J mol−¹ K−¹ is defined with temperature in Kelvin. Using Celsius would give a dimensionally inconsistent result (the zero of the Celsius scale is arbitrary, not an absolute zero). Using T = 25 °C instead of T = 298 K gives an answer approximately 12× too small, a common exam error.

Q5.4 — Kb of bicarbonate from Ka × Kb = Kw

Kb(HCO&sub3;−) = Kw / Ka(H&sub2;CO&sub3;) = 1.0 × 10−¹&sup4; / 4.3 × 10−&sup7; = 2.3 × 10−&sup8;. This Kb << 1 confirms HCO&sub3;− is a weak base. Consistent with the rule: carbonic acid is itself a weak acid (moderate Ka), so its conjugate base has a small but non-negligible Kb.