Chemistry • Year 12 • Module 5 • Lesson 17
Solubility Product Ksp
Apply Ksp expressions and molar solubility calculations to real data and Australian contexts, including Jenolan Caves limestone dissolution and Sydney Water hardness treatment.
1. Ksp data table interpretation
The table below gives Ksp values for five sparingly soluble salts at 25°C. Use the data to answer the questions that follow. Show all working. 14 marks
| Salt | Dissolution equation | Formula type | Ksp (25°C) |
|---|---|---|---|
| AgCl | AgCl ⇌ Ag+ + Cl− | 1:1 | 1.8 × 10−10 |
| BaSO4 | BaSO4 ⇌ Ba2+ + SO42− | 1:1 | 1.1 × 10−10 |
| CaCO3 | CaCO3 ⇌ Ca2+ + CO32− | 1:1 | 3.3 × 10−9 |
| PbI2 | PbI2 ⇌ Pb2+ + 2I− | 1:2 | 9.8 × 10−9 |
| Mg(OH)2 | Mg(OH)2 ⇌ Mg2+ + 2OH− | 1:2 | 5.6 × 10−12 |
(a) Using the table data only, rank the three 1:1 salts (AgCl, BaSO4, CaCO3) from least soluble to most soluble. Justify why a direct Ksp comparison is valid here. 3 marks
(b) A student claims: “PbI2 has a larger Ksp than CaCO3, therefore PbI2 is more soluble.” Explain whether this comparison is valid. Calculate the molar solubility of both salts (show full working) and state which is actually more soluble. 5 marks
(c) Sydney Water adds Mg(OH)2 (milk of magnesia) to hard bore water to precipitate dissolved Ca2+ and reduce water hardness. Calculate the molar solubility of Mg(OH)2 in pure water at 25°C. Show all algebraic steps. 3 marks
(d) A lead paint chip found in an old NSW house is estimated to contain PbCO3 (Ksp = 7.4 × 10−14). PbCO3 is a 1:1 salt. Calculate its molar solubility. Explain how this relates to the risk of lead contamination in a child’s environment. 3 marks
2. Ksp versus temperature graph
The graph below shows how the Ksp of two salts — CaCO3 and AgCl — varies with temperature. Use the graph to answer the questions. 8 marks
(a) Describe the trend in Ksp for both salts as temperature increases from 10°C to 50°C. 2 marks
(b) Use the graph to estimate the molar solubility of CaCO3 at 30°C. Show the Ksp value you read from the graph and your calculation. 2 marks
(c) Jenolan Caves (NSW) are limestone cave systems formed by CaCO3 dissolution in groundwater. Using the trend in the graph, predict and explain what effect an increase in groundwater temperature (due to climate change) would have on the rate of cave formation at Jenolan. 2 marks
(d) At 30°C, read the Ksp of AgCl from the graph. Compare the molar solubility of AgCl with CaCO3 at this temperature. Explain whether the Ksp values alone are sufficient for this comparison. 2 marks
3. Cause-and-effect chain — acid rain and Jenolan Caves
Complete the cause-and-effect chain below. Each box describes a step in the process by which acidic rain accelerates limestone (CaCO3) dissolution at Jenolan Caves. Fill in the empty effect boxes (B, D, F) and the overall outcome (G). 4 marks
| Step | Event | Your response |
|---|---|---|
| A (given) | Rainwater absorbs CO2 from the atmosphere, forming carbonic acid H2CO3. | (no response needed) |
| B → | H2CO3 dissociates to release H+ ions in the groundwater. | The increased [H+] reacts with CO32− ions (a product of CaCO3 dissolution) to form _______. This shifts the dissolution equilibrium to the _______. |
| C (given) | The dissolution equilibrium of CaCO3 shifts to the right (Le Chatelier’s principle). | (no response needed) |
| D → | More CaCO3 dissolves. | In terms of Ksp, what happens to the ion product Q compared to Ksp as more solid dissolves? _______ |
| E (given) | Limestone walls of the cave system thin and erode over geological timescales. | (no response needed) |
| F → | Overall outcome: | Acid rain effectively lowers the effective Ksp conditions experienced by CaCO3 in the cave environment. Predict whether stalactites (calcium carbonate deposits) grow faster or slower under more acidic groundwater conditions and why. _______ |
Q1 — Ksp data table
(a) Same formula type (all 1:1) so direct Ksp comparison is valid. Ranking from least to most soluble: BaSO4 (Ksp = 1.1 × 10−10) < AgCl (Ksp = 1.8 × 10−10) < CaCO3 (Ksp = 3.3 × 10−9). Larger Ksp = more soluble for same formula type. (1 mark for correct ranking, 1 mark for noting same formula type, 1 mark for justification.)
(b) The comparison is NOT valid — CaCO3 is 1:1 and PbI2 is 1:2 (different formula types). Must calculate s for each.
CaCO3 (1:1): s = √(3.3 × 10−9) = 5.74 × 10−5 mol/L.
PbI2 (1:2): Ksp = 4s3; s = (9.8 × 10−9/4)1/3 = (2.45 × 10−9)1/3 = 1.35 × 10−3 mol/L.
Despite a smaller Ksp overall than it might appear, PbI2 is actually far more soluble than CaCO3 — s(PbI2) ≈ 24× larger. (1 mark validity statement, 2 marks calculation each, 1 mark correct conclusion.)
(c) Mg(OH)2 (1:2): Ksp = 4s3 = 5.6 × 10−12.
s3 = 1.4 × 10−12; s = (1.4 × 10−12)1/3 = 1.12 × 10−4 mol/L. (1 mark per step: expression, algebra, answer.)
(d) PbCO3 (1:1): s = √(7.4 × 10−14) = 2.72 × 10−7 mol/L. Very low solubility (~2.7 × 10−7 mol/L) but lead is highly toxic; even extremely small concentrations in soil or dust ingested by children pose neurological risk (NSW Health guidelines set limits in the µg/L range). A tiny Ksp does not mean zero dissolution. (1 mark calculation, 1 mark Ksp expression, 1 mark contextual link to health risk.)
Q2 — Graph interpretation
(a) Ksp increases for both salts as temperature rises from 10°C to 50°C — indicating that dissolution is an endothermic process; higher temperature favours the forward (dissolution) reaction, increasing ion concentrations at equilibrium.
(b) From the graph at 30°C, Ksp(CaCO3) ≈ 3.6 × 10−9. For 1:1 salt: s = √(3.6 × 10−9) ≈ 6.0 × 10−5 mol/L.
(c) Warmer groundwater would increase Ksp of CaCO3, so more limestone could dissolve — increasing the rate of cave enlargement / erosion. Stalactites and stalagmites could dissolve or grow more slowly as dissolution exceeds re-precipitation.
(d) From the graph at 30°C, Ksp(AgCl) ≈ 1.6 × 10−10. Both are 1:1 formula type so direct Ksp comparison is valid here. CaCO3 (larger Ksp) is more soluble than AgCl at 30°C. Ksp comparison alone is sufficient for same-formula-type salts.
Q3 — Cause-and-effect chain
B: H2CO3 reacts with CO32− to form H2O and CO2 (or HCO3−), removing a product of dissolution — by Le Chatelier's principle, the equilibrium shifts to the right (more dissolution).
D: As dissolution proceeds, [Ca2+] and [CO32−] increase until Q = Ksp at the new equilibrium. While Q < Ksp, more solid continues to dissolve.
F: Stalactites grow slower (or dissolve) under more acidic conditions. Acid lowers CO32− concentration by converting it to HCO3−/CO2, so Q < Ksp — the re-precipitation driving force is reduced. Stalactite growth (CaCO3 precipitation) requires Q > Ksp, which becomes harder to achieve in more acidic water.