“Like dissolves like” needs named forces
Lesson 2.9.1 used “like dissolves like” to predict whether something dissolves. AP topic 3.10 goes further: name the interactions between solute particles, between solvent molecules, and between solute and solvent, and compare their strengths. “Polar dissolves in polar” on its own earns no credit.
Dissolving: break two kinds of attraction, form one
Think of dissolving in three steps:
- Separate the solute particles from each other (requires energy).
- Make room between solvent molecules, breaking some of the attractions between them (requires energy).
- Form new attractions between solute particles and solvent molecules (releases energy).
When the new solute–solvent interactions溶质–溶剂作用力 are comparable in strength to the ones broken, the solute dissolves, and two liquids are miscible互溶. When the new interactions are much weaker, solute particles and solvent molecules stay with their own kind: a solid stays undissolved, and two liquids form separate layers (they are immiscible).
Ionic compounds: soluble in water, not in nonpolar solvents
When NaCl dissolves in water, each Na⁺ is surrounded by the partially negative O atoms of water molecules, and each Cl⁻ by the partially positive H atoms (the particle diagrams of U3.8). These ion–dipole interactions离子–偶极作用 are strong enough to replace the attractions between the ions and the hydrogen bonds between water molecules.
In carbon tetrachloride () or hexane, the solvent molecules are nonpolar. They attract the ions only through weak ion–induced dipole interactions, which come nowhere near making up for the energy needed to pull the ionic crystal apart, so NaCl is practically insoluble.
Molecular compounds: compare the hydrogen-bonding part with the nonpolar part
- Nonpolar molecules (such as and pentane, ) attract each other only through London dispersion forces, and they attract hexane or the same way, with similar strength, so they dissolve or are miscible. In water, they would break the fairly strong hydrogen bonds between water molecules and get only weak dipole–induced dipole and dispersion attractions in return, so they barely dissolve.
- Molecules with –OH groups can hydrogen-bond with water. The more –OH groups and the shorter the nonpolar carbon chain, the more soluble in water.
| Molecule | Condensed formula | –OH groups | Carbon atoms | In water |
|---|---|---|---|---|
| Pentane | CH₃CH₂CH₂CH₂CH₃ | 0 | 5 | Practically insoluble |
| 1-Butanol | CH₃CH₂CH₂CH₂OH | 1 | 4 | Partly soluble |
| 1,3-Propanediol | HOCH₂CH₂CH₂OH | 2 | 3 | Miscible |
So the solubility in water increases in the order pentane < 1-butanol < 1,3-propanediol. In hexane the order reverses: the larger the nonpolar part and the fewer the groups that can hydrogen-bond, the closer the solute–hexane dispersion forces come to the attractions that must be broken, and the more soluble the compound.
Choose a solute and a solvent to see the attractions broken, the attractions formed, and what happens in the test tube:
Strengths are only schematic. When layers form, the less dense liquid is on top: hydrocarbons and alcohols are less dense than water; hexane is 0.66 g/cm³ and carbon tetrachloride 1.59 g/cm³ (20 °C).
Extraction: density decides the top layer, forces decide where the solute goes
is only slightly soluble in water, giving a pale yellow-brown solution. Add hexane to the iodine solution, shake, and let it stand:
- Hexane and water are immiscible (for the same reason as pentane and water), so two layers form.
- The dispersion forces between and hexane are comparable to the attractions broken, so most of the moves into the hexane layer, which turns violet.
- Hexane (0.66 g/cm³) is less dense than water (1.00 g/cm³), so the hexane layer is on top.
With (1.59 g/cm³) instead, the violet layer is on the bottom. Using one solvent to pull a solute out of another solvent it doesn’t mix with is called extraction萃取. Which layer is on top depends only on density; which layer the solute ends up in depends on intermolecular forces. Keep the two questions apart.
Solubility and temperature: crystallization and percent recovered
The solubility of most solids increases with temperature (lesson 1.7.5). Cooling a solution that is saturated at a high temperature makes the solute that no longer fits crystallize out, a common way to purify a solid.
Some solute always stays dissolved after cooling, so the percent recovered is always below 100%. The more the solubility changes with temperature, the higher it is. Drag the two temperatures to try it:
- Potassium nitrate KNO₃
- Ammonium chloride NH₄Cl
- Potassium chloride KCl
- Sodium chloride NaCl
Calcium hydroxide (vertical scale magnified)
Calcium hydroxide Ca(OH)₂
Crystallization by cooling
Example 1: Ranking solubility with reasons
Rank pentane, 1-butanol, and 1,3-propanediol in order of increasing solubility in water, and justify the order in terms of intermolecular forces.
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List the key structural features.
Substance –OH C atoms e⁻ Pentane 0 5 42 1-Butanol 1 4 42 1,3-Propanediol 2 3 42 All three have the same number of electrons and similar dispersion forces, so the difference comes from hydrogen bonding with water.
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What breaks and what forms. To enter water, a solute must break some water–water hydrogen bonds; the more hydrogen bonds it forms with water, the better it makes up for that energy.
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The effect of the nonpolar part. A carbon chain wedged between water molecules breaks hydrogen bonds but offers only weak attractions in return; the longer the chain and the fewer the –OH groups, the less soluble the compound.
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Conclusion. Pentane < 1-butanol < 1,3-propanediol. Pentane can’t hydrogen-bond with water and is nearly insoluble; both –OH groups of 1,3-propanediol hydrogen-bond with water, and it mixes with water in all proportions.
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When writing the justification. Base it on the ability to hydrogen-bond with water, and note that the electron counts are equal, so dispersion forces aren’t the difference; “smaller molecules dissolve better” earns no credit.
Example 2: Percent recovered on cooling
A saturated solution of is made with 50.0 g of water at 60 °C and cooled to 20 °C. The solubility of is 110 g per 100 g of water at 60 °C and 31.6 g per 100 g of water at 20 °C. Find the mass of crystals formed and the percent recovered.
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Solubility is per 100 g of water, so scale to 50.0 g first. m₆₀ and m₂₀ are the masses of (in g) that 50.0 g of water holds at 60 °C and 20 °C:
55.0 g dissolved at 60 °C; at 20 °C, 50.0 g of water can hold only 15.8 g.
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Crystals formed.
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Percent recovered. The crystals as a percentage of the originally dissolved:
The other 15.8 g stays in solution.
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Common mistakes. Dividing 39.2 by 15.8 (it should be divided by the 55.0 g originally dissolved), or forgetting to scale the solubilities to 50.0 g of water.
- Sodium chloride, NaCl
The ions form strong ion–dipole interactions with water molecules.
- Iodine, I₂
A nonpolar molecule; its dispersion forces match those of hexane.
- Glucose, C₆H₁₂O₆ (5 –OH groups)
It can form many hydrogen bonds with water.
- Octane, C₈H₁₈
Only dispersion forces, like hexane.
- Methanol, CH₃OH
Its –OH group hydrogen-bonds with water, and its nonpolar part is tiny.
- Naphthalene, C₁₀H₈
A nonpolar molecule.
- Potassium nitrate, KNO₃
An ionic compound; it forms ion–dipole interactions in water.
- 1,3-Propanediol, HOCH₂CH₂CH₂OH
Both –OH groups can hydrogen-bond with water.
- Carbon tetrachloride, CCl₄
A nonpolar molecule: miscible with hexane, forms a layer with water.
- Pentane, C₅H₁₂
Only dispersion forces.
- Ammonia, NH₃
NH₃ can hydrogen-bond with water.
Common mistakes
Mistake 1: “NaCl dissolves in water because water is polar.” That names no interaction. Write instead: Na⁺ and Cl⁻ form ion–dipole interactions with water molecules that are strong enough to replace the attractions between the ions and the hydrogen bonds between water molecules.
Mistake 2: “Oil doesn’t dissolve in water because water repels it.” Water molecules and nonpolar molecules do attract each other (dipole–induced dipole and dispersion forces); the attraction is just too weak to make up for breaking the hydrogen bonds between water molecules.
Mistake 3: “After an extraction, the layer with the solute is always on the bottom.” Which layer is on top depends only on the densities of the two liquids: a hexane layer floats, a layer sinks. Which layer the solute goes into depends on intermolecular forces, not density.
Quick check
NaCl dissolves readily in water but is practically insoluble in . Which explanation is best?
Whether a solute dissolves depends on whether the solute–solvent interactions can replace the attractions broken. A confuses density with solubility; B and C are not the reason.
Which substance is most soluble in water, and why?
The ability to hydrogen-bond with water decides it. A and D ignore how weakly pentane attracts water; B is wrong because the C–H bonds of the chain can't hydrogen-bond with water.
Hexane (density 0.66 g/cm³) is added to an aqueous iodine solution, and the mixture is shaken and left to stand. Which description is correct?
Hexane and water don't mix; I₂ moves into the hexane, where the forces match; hexane is lighter, so it's on top. B uses the solute's density instead of the layer's; C is wrong because I₂ and hexane can't hydrogen-bond.
Which pair of liquids is miscible?
Hexane and octane both have only dispersion forces, of similar strength. In each of the other pairs, one liquid hydrogen-bonds and the other has only dispersion forces.
A KCl solution saturated at 80 °C (51.1 g of KCl in 100 g of water) is cooled to 20 °C, where the solubility is 34.0 g per 100 g of water. What is the percent recovered?
Crystals formed: 51.1 − 34.0 = 17.1 g, so the percent recovered is 17.1 ÷ 51.1 × 100% = 33.5%. A is the fraction left in solution; B is the mass of crystals, not a percentage; D divides by the wrong mass (17.1 ÷ 34.0).
is far more soluble in hexane than in water. Which explanation is best?
Compare the attractions broken with the attractions formed. A is wrong because neither I₂ nor hexane can hydrogen-bond; C confuses density with solubility.
AP-style free response
An aqueous solution contains small amounts of dissolved and NaCl. A student pours it into a separatory funnel, adds hexane (density 0.66 g/cm³), shakes the funnel, and lets it stand. Two layers form.
(a) Which layer is on top? Explain why hexane and water are immiscible.
(b) Which layer contains most of the ? Justify your answer using intermolecular forces.
(c) Which layer contains the NaCl? Justify your answer using the interactions between particles.
(d) The student also purifies by cooling: a solution saturated at 60 °C (55.2 g of in 100 g of water) is cooled to 20 °C, where the solubility is 37.2 g per 100 g of water. Calculate the mass of crystals formed and the percent recovered. Cooling from 60 °C to 20 °C recovers about 71% of ; explain why the two results differ so much.
Model answer and scoring
(a) The hexane layer is on top, because hexane is less dense than water. Hexane is nonpolar and has only dispersion forces; mixing it into water would break hydrogen bonds between water molecules and replace them only with weak dipole–induced dipole and dispersion attractions, so the liquids don’t mix.
(b) The hexane (top) layer, which turns violet. is nonpolar; the dispersion forces between and hexane are comparable to the – and hexane–hexane dispersion forces they replace. attracts water only through weak dipole–induced dipole interactions, which can’t make up for breaking the hydrogen bonds between water molecules.
(c) The water (bottom) layer. Na⁺ and Cl⁻ form strong ion–dipole interactions with water molecules; with hexane they form only weak ion–induced dipole interactions, which can’t make up for the attractions between the ions.
(d) Crystals formed: 55.2 g − 37.2 g = 18.0 g; percent recovered = 18.0 ÷ 55.2 × 100% = 32.6%. The solubility of changes much more with temperature (its solubility curve is steeper), so much more of it can no longer stay dissolved on cooling; the curve for is flatter, so most of it stays in solution.
Scoring: (a) hexane on top, explained by density, with hydrogen bonding versus dispersion forces explaining immiscibility; (b) hexane layer, comparing the attractions broken and formed; (c) water layer, ion–dipole interactions; (d) 18.0 g and 32.6%, explained by how much the solubility changes with temperature.
Summary
- Dissolving breaks the attractions between solute particles and between solvent molecules and forms solute–solvent attractions; a solute dissolves (or two liquids mix) only when the new attractions are comparable to the ones broken.
- Ionic compounds dissolve in water through ion–dipole interactions; in nonpolar solvents they form only weak ion–induced dipole interactions and barely dissolve.
- Molecular compounds: the more groups that can hydrogen-bond and the smaller the nonpolar part, the more soluble in water and the less soluble in hexane.
- Extraction: the solute moves into the layer whose forces match its own; which layer is on top depends only on density.
- Percent recovered on cooling = mass of crystals ÷ mass originally dissolved × 100%; it is higher when the solubility changes more with temperature.
AP tip: To explain solubility, name all three interactions (between solute particles, between solvent molecules, and between solute and solvent) and say whether the new ones can replace those broken. “Like dissolves like” or “polar dissolves polar” alone earns no credit.