Chemistry • Year 11 • Module 1 • Lesson 12
Solubility and Like-Dissolves-Like
Apply the “like dissolves like” principle and IMF reasoning to real data, solubility trends, and a real-world solvent-extraction scenario.
1. Interpret experimental data — solubility predictions
A student tested seven substances in both water and hexane. The table below records observations. 10 marks
| Substance | Structure / type | Observation in water | Observation in hexane |
|---|---|---|---|
| NaCl | Ionic solid | Dissolves completely — clear solution | |
| I2 | Non-polar covalent solid | Dissolves — brown/violet solution | |
| Ethanol (CH3CH2OH) | Polar molecule (–OH group) | ||
| Hexane (C6H14) | Non-polar molecule | Two distinct layers form | |
| BaSO4 | Ionic solid (high lattice energy) | Does not dissolve | |
| Glucose (C6H12O6) | Polar molecule (five –OH groups) | ||
| Candle wax (long-chain hydrocarbon) | Non-polar molecule |
1.1 Complete the empty cells in the table above, predicting the outcome in the relevant solvent. 7 marks (1 per empty cell)
1.2 BaSO4 is ionic yet insoluble in water. Explain this result using the concepts of lattice energy and hydration energy. 3 marks
2. Interpret graph — solubility of alcohols with increasing chain length
The graph below shows the solubility in water of four straight-chain primary alcohols at 25 °C. 7 marks
Figure 2. Solubility in water of straight-chain primary alcohols (C1–C4) at 25 °C. “Miscible” = mixes in all proportions. Illustrative data based on standard solubility tables.
2.1 Describe the trend in water solubility as the carbon chain length of the alcohol increases from C1 to C4. 2 marks
2.2 Using the principle of “like dissolves like” and IMF reasoning, explain why this trend occurs. In your answer, refer to both the –OH group and the hydrocarbon tail. 3 marks
2.3 Predict the approximate water solubility of pentanol (C5H11OH) relative to butanol, and justify your prediction. 2 marks
3. Compare water and hexane as solvents across five features
Complete the two-column table below. For each feature, write a concise description that contrasts the two solvents. 10 marks (1 per cell)
| Feature | Water (H2O) | Hexane (C6H14) |
|---|---|---|
| Polarity | ||
| Dominant IMF type | ||
| Type of solute it dissolves best | ||
| NaCl solubility | ||
| Real-world use as solvent |
4. Predict and justify — iodine extraction from seawater
Iodine (I2) occurs naturally dissolved in trace amounts in seawater. Chemists can extract it by adding hexane to the seawater in a separating funnel and shaking the mixture.
5 marks
4.1 Predict which layer (upper or lower) contains the iodine after the layers separate. Explain your reasoning using “like dissolves like” and the IMFs involved. 3 marks
4.2 Any dissolved NaCl remains entirely in the water layer throughout this extraction. Explain why, using IMF reasoning. 2 marks
Q1.1 — Solubility prediction table
NaCl in hexane: Does not dissolve (hexane has only dispersion forces; cannot form ion-dipole forces to overcome NaCl lattice energy). I2 in water: Barely dissolves / forms faint yellow solution (IMF mismatch: I2 dispersion forces cannot compensate for disrupting water’s H-bond network). Ethanol in water: Miscible (mixes in all proportions; –OH group forms H-bonds with water). Ethanol in hexane: Partially miscible / dissolves to some extent (the ethyl group is non-polar, giving some dispersion-force compatibility; –OH group reduces full miscibility). Hexane in hexane: Fully miscible (same substance — identical dispersion forces). BaSO4 in water: Does not dissolve / insoluble (very high lattice energy exceeds hydration energy). Glucose in water: Dissolves readily (five –OH groups form extensive H-bonds with water). Glucose in hexane: Does not dissolve (polar –OH groups incompatible with non-polar hexane). Candle wax in water: Does not dissolve (non-polar; incompatible with water H-bond network). Candle wax in hexane: Dissolves (both non-polar; dispersion–dispersion compatibility).
Q1.2 — BaSO4 insolubility (3 marks)
BaSO4 has a very high lattice energy due to the large charges involved: Ba2+ (2+ charge) and SO42− (2− charge) create strong electrostatic attractions in the crystal lattice [1 mark]. For dissolution to occur, the hydration energy (energy released when water molecules surround the separated Ba2+ and SO42− ions) must exceed the lattice energy [1 mark]. For BaSO4, the hydration energy is insufficient to overcome the very high lattice energy, so dissolution is thermodynamically unfavourable and BaSO4 remains insoluble [1 mark].
Q2.1 — Trend in alcohol solubility (2 marks)
As the carbon chain length increases from C1 (methanol, miscible) to C4 (butanol, ~73 g/L), the solubility in water decreases significantly [1 mark]. The decrease is dramatic: methanol and ethanol are fully miscible, propanol is highly soluble (~1000 g/L), but butanol drops sharply to ~73 g/L [1 mark for quantitative description of trend].
Q2.2 — IMF explanation of the trend (3 marks)
All primary alcohols possess a polar –OH group that can form hydrogen bonds with water, which drives solubility [1 mark]. However, as the hydrocarbon tail lengthens, the non-polar portion of the molecule becomes increasingly dominant [1 mark]. The longer the non-polar chain, the greater the disruption of water’s H-bond network needed to accommodate the molecule, without sufficient compensating H-bond energy released from the –OH group alone — so the overall dissolution becomes less favourable and solubility decreases [1 mark].
Q2.3 — Predict pentanol solubility (2 marks)
Pentanol would have a lower solubility in water than butanol (~73 g/L) — approximately 27 g/L [1 mark]. The longer five-carbon non-polar chain further increases the hydrophobic character of the molecule relative to the single –OH group, making the disruption of water’s H-bond network even less compensated, so water solubility continues to decrease [1 mark].
Q3 — Compare water and hexane
Polarity: Water — highly polar (δ− on O, δ+ on H; large permanent dipole). Hexane — non-polar (symmetric C–H bonds; no net dipole). Dominant IMF: Water — hydrogen bonding (strongest type between neutral molecules). Hexane — dispersion forces (London/van der Waals). Best solute type: Water — ionic and polar substances. Hexane — non-polar substances. NaCl solubility: Water — soluble (~360 g/L); ion-dipole forces overcome lattice energy. Hexane — insoluble; dispersion forces too weak to pull ions from the lattice. Real-world use: Water — biological solvent, industrial reactions, cleaning (ionic/polar substances). Hexane — extraction of non-polar lipids/fats, dry-cleaning, industrial degreasing.
Q4.1 — Iodine extraction layer (3 marks)
The iodine concentrates in the upper hexane layer [1 mark]. Iodine (I2) is non-polar and has only dispersion forces between molecules. Hexane is also non-polar with only dispersion forces — the IMFs are compatible, so I2 dissolves readily in hexane [1 mark]. Water’s strong H-bond network is disrupted by I2 for no compensating gain (I2–water dispersion is too weak) — so I2 migrates to the hexane layer. Hexane is less dense than water, so the hexane/I2 layer forms on top [1 mark].
Q4.2 — NaCl stays in water (2 marks)
NaCl is an ionic compound; Na+ and Cl− ions require ion-dipole forces with polar water molecules to remain in solution [1 mark]. Hexane is non-polar and can only offer dispersion forces, which are far too weak to stabilise the highly charged ions — so NaCl has no driving force to transfer to the hexane layer and remains entirely in the water layer [1 mark].