Explain why dissolution is a balance of competing effects

Do not inspect only one contribution

Dissolving an ionic solid separates ions from their lattice, reorganizes solvent molecules and establishes ion-solvent interactions. A schematic energy path can separate positive lattice-separation work from negative hydration contributions. The net enthalpy may be positive or negative. Dispersing ions also competes with changes in solvent organization, so dissolution entropy is not universally positive.

The lab assigns a deliberately simplified 1:1 salt and ΔS°=+50 J/(mol K). It computes ΔH° as lattice cost minus the positive magnitude of hydration release, then ΔG°=ΔH°−TΔS° at 298.15 K. These are fictional design variables, not a reference database for identifying salts.

Cancellation and sensitivity

With cost and release both 650 kJ/mol, net ΔH° is zero while ΔG°≈−14.908 kJ/mol. Raising the cost to 700 makes ΔH°=+50 and ΔG°≈+35.093. Lowering the cost to 600 gives ΔH°=−50 and ΔG°≈−64.908. Relatively modest changes in two large opposing terms can reverse the conclusion.

Before changing a control, predict the sign. Afterwards, explain which contribution dominates. A model that reproduces a net result while using incorrect large contributions may fail badly under a changed solvent or temperature. A correct-looking net number is not independent validation of every assumed microscopic term.

From a standard-state difference to saturation

For MX(s)⇌M⁺(aq)+X⁻(aq), pure solid has unit activity when it is present. Ksp is the product of the ion activities, and ΔG°diss=−RT ln Ksp. Actual direction depends on Q relative to Ksp. A positive ΔG° does not mean that absolutely no solid dissolves into initially pure water: when the ion product is very small, dissolution may still lower G until saturation or solid exhaustion.

Under ideal dilute conditions in pure water, with no common ion, set both ion concentrations to s and normalize by c°: Ksp=(s/c°)². This relation is not a license to accept an astronomically high concentration. Such a prediction violates the model’s dilute-activity assumptions and requires a better model.

Transfer

An experiment finds a cold solution after dissolution. Does that prove low solubility? No. Cooling suggests heat was absorbed under the apparatus conditions, but neither entropy nor the activity-dependent free energy has been determined. Identify the separate equilibrium measurement needed, and explain why a thermometer alone cannot establish Ksp.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.6 in the College Board course framework, effective Fall 2024, checked 30 October 2026. These pages do not reproduce restricted examination questions or establish AP authorization. The investigation is a bounded computational model, not a conducted laboratory.

Return to the preceding Unit 9 lesson when its reasoning is unfamiliar.

Continue to Require a mechanism before adding favorable reactions.