Chemistry • Year 12 • Module 5 • Lesson 14
Ka, Kb & Gibbs Free Energy
Apply Ka and Kb expressions, the Ka × Kb = Kw relationship, and ΔG° = −RT ln Keq to real-world Australian contexts including wine production, Great Barrier Reef chemistry, and aquaculture.
1. Interpret graph — Ka values and acid strength
The bar chart below shows the Ka values (on a log&sub10; scale) for five common weak acids discussed in this lesson. Use the graph to answer the sub-questions. 8 marks
Ka values at 25 °C. Data from standard thermodynamic tables. Note: x-axis shows log₁₀(Ka), so bars extending further right indicate smaller Ka (weaker acid).
1.1 Rank the five acids from strongest to weakest (largest Ka to smallest Ka). 2 marks
1.2 Estimate the Ka of ammonium (NH&sub4;¹+) from the graph and use Ka × Kb = Kw to calculate Kb for ammonia (NH&sub3;). Show working. 3 marks
1.3 Carbonic acid (H&sub2;CO&sub3;) is in equilibrium in Great Barrier Reef seawater. As atmospheric CO&sub2; dissolves, [H&sub2;CO&sub3;] increases. Using the Ka expression for carbonic acid and Le Chatelier’s Principle, explain the direction of the equilibrium shift and its impact on [H¹+] in the reef. 3 marks
2. Cause-and-effect chain — ammonia toxicity in aquaculture
CSIRO Marine & Atmospheric Research has studied ammonia (NH&sub3;) toxicity in Australian prawn aquaculture ponds. Ammonia is a weak base (Kb = 1.8 × 10−&sup5;). When water temperature increases, Kb increases and so does the proportion of un-ionised NH&sub3; relative to NH&sub4;¹+. Un-ionised NH&sub3; crosses gill membranes and is toxic to aquatic animals; NH&sub4;¹+ does not. Complete the cause-and-effect chain below. 5 marks
| Cause | → | Effect (fill in) |
|---|---|---|
| Atmospheric CO&sub2; levels rise; CO&sub2; dissolves in pond water | → | |
| Water temperature in the pond increases | → | |
| Kb for NH&sub3; increases with temperature | → | |
| Greater proportion of total ammonia exists as un-ionised NH&sub3; | → | |
| Overall outcome (so…): ____________________________________________________________________________________________________________ | ||
3. Data-table interpretation — ΔG° calculations at different temperatures
A student calculates ΔG° for an equilibrium reaction at five temperatures using ΔG° = −RT ln Keq. The results are shown below. 6 marks
| Temperature (°C) | T (K) | Keq | ln Keq | ΔG° (kJ mol−¹) |
|---|---|---|---|---|
| 25 | 298 | 150 | 5.01 | −12.4 |
| 100 | 373 | 45 | 3.81 | −11.8 |
| 200 | 473 | 8.2 | 2.10 | −8.3 |
| 350 | 623 | 1.05 | 0.049 | −0.25 |
| 400 | 673 | 0.62 | −0.478 | ? |
3.1 Calculate the missing ΔG° value at 400 °C and state whether the reaction is spontaneous under standard conditions at that temperature. Show full working. 2 marks
3.2 Describe the trend in ΔG° as temperature increases from 25 °C to 400 °C. What does this indicate about how the equilibrium position changes with temperature? 2 marks
3.3 At 350 °C, Keq = 1.05. Explain, without calculating, why ΔG° must be very close to zero at this temperature. 2 marks
4. Predict and justify — acetic acid in Australian wine
Acetic acid (CH&sub3;COOH) is the main acid in wine vinegar and is also present in Australian table wines. Its Ka = 1.8 × 10−&sup5;. A winemaker adds a small amount of sodium acetate (CH&sub3;COONa, a source of the conjugate base CH&sub3;COO−) to a batch of wine. 4 marks
4.1 Write the Ka expression for acetic acid. Then predict the direction of the equilibrium shift when CH&sub3;COO− is added, and justify your prediction using Le Chatelier’s Principle. 3 marks
4.2 Would the Ka of acetic acid change when CH&sub3;COO− is added? Justify your answer in one sentence. 1 mark
Q1.1 — Ranking by Ka (2 marks)
Strongest to weakest: HF > acetic acid > carbonic acid > HCN ≈ ammonium (NH&sub4;¹+). [1 mark for correct top 3 relative order; 1 mark for HCN and ammonium at similar low Ka with ammonium slightly larger]. Ka values: HF = 7.2 × 10−&sup4;; acetic acid = 1.8 × 10−&sup5;; carbonic acid = 4.3 × 10−&sup7;; ammonium = 5.6 × 10−¹&sup0;; HCN = 6.2 × 10−¹&sup0;.
Q1.2 — Ka(NH&sub4;¹+) and Kb(NH&sub3;) (3 marks)
From graph: Ka(NH&sub4;¹+) ≈ 5.6 × 10−¹&sup0; [1 mark]. Ka × Kb = Kw: Kb(NH&sub3;) = Kw / Ka(NH&sub4;¹+) = 1.0 × 10−¹&sup4; / 5.6 × 10−¹&sup0; [1 mark] = 1.8 × 10−&sup5; [1 mark]. This is the known Kb for ammonia at 25 °C, consistent with lesson data.
Q1.3 — Carbonic acid and GBR seawater (3 marks)
Equilibrium: H&sub2;CO&sub3;(aq) ⇌ H¹+(aq) + HCO&sub3;−(aq); Ka = [H¹+][HCO&sub3;−] / [H&sub2;CO&sub3;] [1 mark]. As CO&sub2; dissolves, [H&sub2;CO&sub3;] increases — this is equivalent to increasing a reactant concentration. By Le Chatelier’s Principle, the equilibrium shifts to the right (forward direction) to consume the excess H&sub2;CO&sub3; [1 mark]. This produces more H¹+ and HCO&sub3;−, so [H¹+] increases — ocean acidification — which reduces the pH of reef water and endangers calcifying organisms such as corals [1 mark].
Q2 — Cause-and-effect chain (5 marks, 1 per row + overall)
Row 1: CO&sub2; dissolves, forming H&sub2;CO&sub3; which dissociates to H¹+ and HCO&sub3;−, lowering pond pH (increasing [H¹+]).
Row 2: Higher temperature increases Kb for NH&sub3;, shifting the NH&sub3; / NH&sub4;¹+ equilibrium toward more un-ionised NH&sub3;.
Row 3: The equilibrium NH&sub3;(aq) + H&sub2;O ⇌ NH&sub4;¹+(aq) + OH−(aq) shifts right to a greater extent; but critically, the ratio [NH&sub3;] / total ammonia increases at higher T because Kb increases faster than the reverse reaction.
Row 4: Un-ionised NH&sub3; is lipid-soluble and crosses gill membranes, whereas charged NH&sub4;¹+ cannot — more NH&sub3; means greater gill exposure.
Overall: Rising water temperatures increase NH&sub3; toxicity risk in aquaculture ponds, requiring CSIRO-informed management interventions such as aeration and pH control to protect farmed prawns and fish.
Q3.1 — Missing ΔG° at 400 °C (2 marks)
ΔG° = −RT ln Keq = −(8.314)(673)(−0.478) [1 mark] = −(5595)(−0.478) = +2.67 kJ mol−¹ [1 mark]. ΔG° > 0 and Keq = 0.62 < 1, so the forward reaction is not spontaneous under standard conditions at 400 °C; reactants are favoured.
Q3.2 — Trend in ΔG° with temperature (2 marks)
ΔG° becomes less negative (more positive) as temperature increases from −12.4 kJ mol−¹ at 25 °C to +2.67 kJ mol−¹ at 400 °C [1 mark]. This indicates that the equilibrium shifts in favour of reactants as temperature increases — Keq decreases with rising temperature, so this is an exothermic reaction (Le Chatelier’s Principle: adding heat shifts equilibrium left for an exothermic process) [1 mark].
Q3.3 — Why ΔG° ≈ 0 when Keq = 1 (2 marks)
ΔG° = −RT ln Keq. When Keq = 1, ln(1) = 0, so ΔG° = −RT × 0 = 0 [1 mark]. A Keq of exactly 1.05 gives ln(1.05) = 0.049, which is nearly zero, confirming products and reactants are present in almost equal amounts under standard conditions and the reaction has no strong thermodynamic preference for either direction [1 mark].
Q4.1 — Ka expression and equilibrium shift (3 marks)
Ka = [H¹+][CH&sub3;COO−] / [CH&sub3;COOH] = 1.8 × 10−&sup5; [1 mark]. Adding CH&sub3;COO− increases the concentration of a product, so by Le Chatelier’s Principle the equilibrium shifts to the left (reverse direction) [1 mark] to partially oppose the increase — consuming H¹+ and CH&sub3;COO− and regenerating CH&sub3;COOH. This reduces [H¹+] and raises pH slightly [1 mark].
Q4.2 — Does Ka change? (1 mark)
No. Ka is an equilibrium constant; at fixed temperature it is constant regardless of concentration changes. Adding CH&sub3;COO− shifts the equilibrium position but does not alter Ka itself [1 mark].