Chemistry • Year 12 • Module 6 • Lesson 7
Conjugate Pairs, Amphiprotic Substances & Water's Role
Apply conjugate pair rules, amphiprotic equations, and Kw to real data, Australian contexts, and novel scenarios.
1. Interpret Kw data across temperature
The table below shows Kw values for pure water at four temperatures. Use the data to answer the sub-questions. 8 marks
| Temperature (°C) | Kw | Neutral [H3O+] = [OH−] (mol L−1) | Neutral pH |
|---|---|---|---|
| 10 | 2.9 × 10−15 | 5.4 × 10−8 | 7.27 |
| 25 | 1.0 × 10−14 | 1.0 × 10−7 | 7.00 |
| 37 | 2.4 × 10−14 | 1.55 × 10−7 | 6.81 |
| 60 | 9.6 × 10−14 | 3.10 × 10−7 | 6.51 |
Data: CRC Handbook of Chemistry and Physics, 97th edition.
1.1 Describe the trend in Kw as temperature increases from 10°C to 60°C. 2 marks
1.2 Using the data, calculate the value of pKw at 37°C and show that pH + pOH ≠ 14 at this temperature. 2 marks
1.3 A blood sample taken at Westmead Hospital has pH 7.35 at 37°C. Using the table data, determine whether this blood is acidic, basic or neutral at body temperature. Justify your answer. 2 marks
1.4 Explain why AGL steam turbine operators must consider that neutral pH is not 7 when working with steam at high temperatures, and predict whether a reading of pH 6.6 at 60°C indicates an acidic, neutral or basic steam condensate. 2 marks
2. Interpret graph — ocean acidification and HCO3− speciation on the Great Barrier Reef
The figure below shows how the proportions of three carbonate species (CO2(aq), HCO3−, CO32−) in seawater change as pH decreases due to ocean acidification. 9 marks
Figure 2.1. Carbonate speciation in seawater as a function of pH. Adapted from Zeebe & Wolf-Gladrow (2001), CO2 in Seawater: Equilibrium, Kinetics, Isotopes. Current mean surface pH of Great Barrier Reef seawater shown at 8.2 (AIMS, 2023).
2.1 At the current reef pH of approximately 8.2, which carbonate species is most abundant? Estimate its proportion from the graph. 2 marks
2.2 Using the graph, describe what happens to the proportion of CO32− as ocean pH decreases from 8.5 to 7.5. Explain what this means for reef-building corals, which require CO32− to deposit CaCO3. 3 marks
2.3 HCO3− is described as amphiprotic. Using the context of this graph, write the two equations (one where HCO3− acts as acid, one where it acts as base) that interconvert the species shown. Identify the conjugate pair in each equation. 4 marks
3. Cause-and-effect chain — blood buffer at Westmead Hospital
The bicarbonate buffer in human blood uses the CO2/HCO3− system to resist pH change. When blood pH drops (acidosis), HCO3− acts as a base to absorb the excess H+. Complete the cause-and-effect chain below. 5 marks
4. Predict and justify — Kw scenario
A researcher dissolves 0.025 mol of NaOH in enough water to make 500 mL of solution at 25°C. They then heat the solution to 60°C (Kw = 9.6 × 10−14). Assume NaOH is fully dissociated at both temperatures and its concentration does not change significantly on heating. 4 marks
4.1 Calculate [OH−] in the solution at 25°C. Then use Kw to find [H3O+] and classify the solution.
4.2 At 60°C the Kw increases to 9.6 × 10−14. Predict whether the solution is still basic at 60°C, and justify by calculating [H3O+] and comparing to the neutral [H3O+] at 60°C.
Q1.1 — Kw trend
Kw increases consistently as temperature increases from 10°C to 60°C (2.9 × 10−15 to 9.6 × 10−14, approximately a 33-fold increase). This reflects the endothermic nature of water autoionisation: higher temperature drives the equilibrium to the right. [1 mark for direction of trend; 1 mark for reasoning using endothermic / Le Chatelier.]
Q1.2 — pKw at 37°C
pKw = −log(2.4 × 10−14) = 13.62. Therefore pH + pOH = 13.62 at 37°C, not 14. (pH + pOH = 14 applies only at 25°C.) [1 mark pKw calculation; 1 mark explicit statement that pH + pOH ≠ 14.]
Q1.3 — Blood pH 7.35 at 37°C
From the table, neutral pH at 37°C = 6.81. Blood pH 7.35 > 6.81 (neutral pH at 37°C), so [H3O+] < [OH−]. The blood is basic (alkaline) at 37°C, which corresponds to healthy arterial blood. [1 mark for comparing 7.35 to 6.81; 1 mark for correct classification as basic.]
Q1.4 — Steam turbines at AGL
At high temperatures Kw is much larger than at 25°C, so neutral pH is well below 7. Using pH 7 as a reference for acidity/neutrality at high temperatures would give incorrect corrosion-risk assessments. At 60°C, neutral pH = −log(√(9.6 × 10−14)) = −log(3.10 × 10−7) ≈ 6.51. A reading of pH 6.6 > 6.51 (neutral pH), so the condensate is basic (very slightly alkaline) at 60°C, not acidic. [1 mark for reasoning about T-dependent neutral pH; 1 mark for correct classification.]
Q2.1 — Most abundant species at pH 8.2
HCO3− is by far the most abundant. From the graph at pH 8.2, HCO3− comprises approximately 95–97% of the carbonate system. [1 mark for identifying HCO3−; 1 mark for a reasonable estimate 90–99%.]
Q2.2 — CO32− as pH drops
As pH decreases from 8.5 to 7.5, the proportion of CO32− falls steeply, from roughly 15% at pH 8.5 to near 0% at pH 7.5 [1]. Because reef-building corals require CO32− to precipitate CaCO3 for their skeletons, a lower pH means less carbonate ion is available, reducing calcification rates and threatening structural integrity of the reef [1]. This is the chemical basis of ocean acidification impacts on the Great Barrier Reef [1 for linking both consequences].
Q2.3 — Amphiprotic equations for HCO3−
As acid (proton donor): HCO3−(aq) ⇌ H+(aq) + CO32−(aq). Conjugate pair: HCO3− (acid) / CO32− (conjugate base). [1 equation + 1 conjugate pair = 2 marks.]
As base (proton acceptor): HCO3−(aq) + H+(aq) ⇌ H2CO3(aq). Conjugate pair: H2CO3 (conjugate acid) / HCO3− (base). [1 equation + 1 conjugate pair = 2 marks.]
Q3 — Blood buffer chain
Effect 1: HCO3−(aq) + H+(aq) → H2CO3(aq). [1]
Effect 2: [H+] decreases as excess H+ is consumed by HCO3−. [1]
Effect 3: Blood pH increases (rises toward 7.35–7.45 normal range), counteracting the acidosis. [1]
Effect 4: The conjugate acid formed is H2CO3. HCO3− acted as a base (proton acceptor). [1]
Overall: Because HCO3− is amphiprotic, it can neutralise both acid (acting as a base) and base (acting as an acid) depending on the situation, making it an ideal pH buffer for blood. [1]
Q4.1 — NaOH at 25°C
n(NaOH) = 0.025 mol; V = 0.500 L; [OH−] = 0.025/0.500 = 0.050 mol L−1. [H3O+] = Kw/[OH−] = 1.0 × 10−14 / 0.050 = 2.0 × 10−13 mol L−1. Since [OH−] >> [H3O+], the solution is basic. [1 for [H3O+]; 1 for classification.]
Q4.2 — At 60°C
[OH−] is still ≈ 0.050 mol L−1 (NaOH fully dissociated). [H3O+] = 9.6 × 10−14 / 0.050 = 1.92 × 10−12 mol L−1. Neutral [H3O+] at 60°C = √(9.6 × 10−14) = 3.10 × 10−7 mol L−1. Since [H3O+] = 1.92 × 10−12 << 3.10 × 10−7 (neutral), the solution is still strongly basic at 60°C. [1 for [H3O+] at 60°C; 1 for comparison to neutral value; 1 for correct classification.]