Lesson 4 · 30 min

Forming solutions and solubility

“Like dissolves like” is a trade of intermolecular forces. Solid solubility changes with temperature, and gas solubility follows Henry’s law — opening a soda bottle is a Henry’s-law experiment.

In this lesson

Explain dissolving — and “like dissolves like” — as an energy account of interactions broken and made
Read solubility curves, classify solutions as saturated, unsaturated or supersaturated, and compute crystallization yields
Calculate gas solubility with Henry’s law and explain why it falls as the temperature rises
Predict how a solid’s solubility changes with temperature from the sign of its enthalpy of solution

Sugar vanishes into hot tea in seconds, yet oil and vinegar separate however hard you shake them. A soda straight from the fridge opens with a polite hiss, while one left in a hot car all afternoon sprays everywhere. Dissolving looks simple, but it is a trade in intermolecular forces — with temperature and pressure as two more hands on the deal.

Predict first

Two identical sodas were filled at 4 °C. One is still at 4 °C; the other has warmed to 40 °C in the sun. Opened at the same time, which one eventually loses more CO₂?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

The one at 4 °C The one at 40 °C The same: both end up losing almost all their CO₂ Neither loses gas unless shaken

Dissolving is an energy trade

Putting a solute into a solvent can be pictured in three steps:

pull the solute particles apart (absorbs heat);
open up gaps between solvent molecules (absorbs heat);
form new solute–solvent attractions (releases heat).

The three add up to the enthalpy of solution $\Delta_\text{sol}H$ . When solute–solvent attractions are comparable to the ones they replace, dissolving goes readily, helped along by the entropy gained on mixing. That is like dissolves like: polar or hydrogen-bonding substances (sugar, ethanol, salts) dissolve in water, while non-polar ones (oils, hexane) mix with each other but hardly with water — hexane dissolves only to about 10 mg/L. Iodine dissolves in water to only about 0.3 g/L, yet in hexane it quickly gives a violet solution.

How solid solubility changes with temperature

The largest mass of solute that 100 g of water can hold at a given temperature is its solubility. Points on the solubility curve are saturated solutions; below the curve, unsaturated; above it, supersaturated — unstable, so that a single seed crystal sets off crystallization of the excess.

The curve’s direction follows the enthalpy of solution: for $\Delta_\text{sol}H > 0$ (endothermic dissolving, like $\ce{KNO3}$ at +34.9 kJ/mol) warming shifts the equilibrium towards dissolving and the solubility climbs steeply; $\ce{NaCl}$ , at only +3.9 kJ/mol, has an almost flat curve; $\ce{Ca(OH)2}$ dissolves exothermically (−16.7 kJ/mol) and becomes less soluble on warming.

Measured solubility (g per 100 g water)

Solubility curves and cooling crystallization

KNO₃
NaNO₃
NH₄Cl
KCl
NaCl
Ca(OH)₂
measured points

Solution temperature · 60 °C Salt dissolved per 100 g water · 110 g Cool (or heat) to · 20 °C

Saturated: right on the curve

The beaker after cooling (schematic)

Solubility at this temperature: 110 g
Mass percent of the saturated solution: 52.4 %
Crystals deposited (per 100 g water) · 20 °C: 78.4 g
Enthalpy of solution ΔsolH (infinite dilution): +34.89 kJ/mol

$\ce{KNO3}$ Endothermic dissolving: warming raises the solubility

Cool a saturated KNO₃ solution from 60 °C to 20 °C: the state moves across, meets the curve and slides down it as the excess crystallizes. Switch to NaCl and almost nothing comes out; switch the axis to log to find Ca(OH)₂, whose solubility falls as it warms.

Worked example

Purifying potassium nitrate by cooling

100 g of water is saturated with KNO₃ at 60 °C and cooled to 20 °C. How much KNO₃ crystallizes, and what fraction is recovered? What about NaCl? (Solubilities: KNO₃ 110 g at 60 °C and 31.6 g at 20 °C; NaCl 37.3 g at 60 °C and 36.0 g at 20 °C.)

Gas solubility: Henry’s law

c = k_\mathrm{H}\,p \qquad k_\mathrm{H}(T) = k_\mathrm{H}^{\ominus}\exp\!\left[C\left(\frac{1}{T} - \frac{1}{298.15\ \mathrm{K}}\right)\right]

At fixed temperature, the dissolved amount is proportional to the gas’s partial pressure above the liquid; kH falls exponentially as the temperature rises.

$c$: dissolved gas concentration, mol/L
$p$: partial pressure of that gas, atm
$k_\mathrm{H}^{\ominus}$: Henry’s constant at 25 °C: CO₂ 0.034, O₂ 0.0013, N₂ 0.00061 mol L⁻¹ atm⁻¹
$C$: temperature coefficient: CO₂ 2400 K, O₂ 1700 K

Dissolving a gas in water is almost always exothermic (the molecules are captured and lose their freedom), so warming always lowers gas solubility. Soda is carbonated at about 3 atm of CO₂ and chilled: at 4 °C, $k_\mathrm{H} = 0.063$ mol L⁻¹ atm⁻¹, so about 0.19 mol/L dissolves — some 8 g of CO₂ per litre. Summer fish kills in ponds and rivers follow the same logic: the warmer the water, the less oxygen it can hold, just when fish need more.

Concept check

Why does KNO₃’s solubility soar with temperature while NaCl’s barely changes?

KNO₃ has a larger molar mass KNO₃ dissolves strongly endothermically (+34.9 kJ/mol), NaCl barely (+3.9 kJ/mol) NaCl decomposes in hot water KNO₃ is a covalent compound

Try the calculation

At 25 °C Henry’s constant for O₂ is $1.3\times10^{-3}$ mol L⁻¹ atm⁻¹; take the O₂ partial pressure in air as 0.21 atm. How many mg/L of oxygen does air-saturated water hold?

Solubility ∝ partial pressure · falls as temperature rises

Henry’s law: opening a bottle of soda

Partial pressure of the gas in the sealed bottle · 0.21 atm Temperature · 25 °C

Soda bottle (schematic)

0 °C
25 °C
40 °C
25 °C · Temperature

Henry’s constant kH · 25 °C: 0.0013 mol·L⁻¹·atm⁻¹
Equilibrium while sealed: 2.73 × 10⁻⁴ mol/L · 8.74 mg/L
Equilibrium once open (with air): 2.72 × 10⁻⁴ mol/L
Gas a 500 mL bottle eventually releases: 3.25 × 10⁻⁷ mol ≈ 8.0 × 10⁻⁶ L
Bunsen coefficient (“volumes”): 0.029

Switch to oxygen at 0.21 atm: drag the temperature and compare how much oxygen water holds at 0 °C and at 30 °C. Then compare the Henry’s constants of O₂ and N₂ — air dissolved in water is richer in oxygen than air itself.

Key takeaways

The enthalpy of solution balances the heat to break old interactions against the heat released by new ones; similar interactions mix readily (like dissolves like).
On the solubility curve a solution is saturated, below it unsaturated, above it supersaturated; salts that dissolve endothermically become more soluble when warm.
Cooling crystallization suits substances with steep curves (KNO₃); evaporation suits flat ones (NaCl).
Henry’s law c = kH·p: gas solubility is proportional to partial pressure and falls with rising temperature.

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Data and references

R. Sander, “Compilation of Henry’s law constants (version 4.0) for water as solvent”, Atmos. Chem. Phys. 15, 4399–4981 (2015).
W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016): aqueous solubility of inorganic compounds; enthalpy of solution of electrolytes.
人民教育出版社《化学》九年级下册，第九单元“溶液”（溶解度表）.
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 5A (ideal and ideal-dilute solutions).