General chemistry

Solutions, solubility, colligative properties

Describe solution composition, saturation and temperature-dependent solubility. Understand colligative effects through the number of dissolved particles and their ideal-solution limits.

1 Lower secondary 2 Upper secondary 3 Undergraduate 4 Advanced 5 Research

IntuitionIntuition: particles and change

Salt disappears when stirred into water, but it has not vanished: ions spread through the liquid. Adding enough solute eventually leaves undissolved solid at equilibrium.

Small blue solvent particles cross the dashed membrane; big red solute particles cannot. Solute stays on one side and crowds it — the origin of osmotic pressure. Raise the temperature to speed everything up.

SchoolSchool: models and rules

Definition: Key idea

A solution is a homogeneous mixture of solvent and solute. At a specified temperature and pressure, a saturated solution is in equilibrium with undissolved solute; solubility quantifies the composition at that condition.

For an ideal dilute solution, Raoult’s law gives solvent vapour pressure $p_A=x_Ap_A^\ast$ . Adding a nonvolatile solute lowers solvent mole fraction and vapour pressure. Real mixtures may deviate from ideality.

Four colligative effects
Effect	Ideal dilute law	Everyday example
Vapour-pressure lowering	p_A = x_A p_A*	sugar syrup evaporates slower
Boiling-point elevation	ΔT_b = iK_b m	pasta water boils slightly above 100 °C
Freezing-point depression	ΔT_f = iK_f m	salt melts ice on roads
Osmotic pressure	Π ≈ icRT	cells shrink or swell in salty/pure water

\Pi\approx i\,cRT,\qquad \Delta T_b=iK_bm,\qquad \Delta T_f=iK_fm

Example: Worked example

A solution contains 0.20 mol solute in 0.50 L. Find its concentration.

Solution

c = n/V = 0.20/0.50 = 0.40 mol L⁻¹. Use the final solution volume, not the solvent volume before mixing.

Example: Second example

A car radiator contains 1.0 kg water with 0.62 mol ethylene glycol (non-electrolyte). Estimate freezing-point depression using K_f = 1.86 °C kg mol⁻¹.

Solution

m = 0.62 mol / 1.0 kg = 0.62 mol kg⁻¹; i = 1 for a non-electrolyte; ΔT_f = 1 × 1.86 × 0.62 ≈ 1.2 °C, so the solution freezes near −1.2 °C.

For dilute solutions, boiling-point elevation and freezing-point depression are approximately proportional to solute molality: $\Delta T_b=iK_bm$ and $\Delta T_f=iK_fm$ . The factor $i$ reflects the effective number of dissolved particles.

For dilute solutions, colligative effects depend on the number of dissolved particles, not their identity: 1 mol of sugar and 1 mol of any other non-electrolyte give the same freezing-point depression. Electrolytes dissociate, so their effective i factor is higher.

UndergraduateDeeper view

At a deeper level, classify observations by the particles involved and by the quantities that remain invariant. Models describe charge, electron density or particle counting; choose the simplest model that accounts for the measured evidence.

Example: Calculation / application example

Calculate the osmotic pressure at 25 °C (298 K) of a 0.050 mol/L sucrose aqueous solution (non-electrolyte, i = 1). R = 0.0821 L·atm/(mol·K).

Solution

Π = i c R T = 1 × 0.050 mol/L × 0.0821 L·atm/(mol·K) × 298 K ≈ 1.22 atm (about 124 kPa).

References

Physical Chemistry · Peter AtkinsJulio de PaulaJames Keeler, 2023
Physical Chemistry · Ira N. Levine, 2009