Respect finite solid inventory when applying a solubility product

Try: the equality has a phase condition

A hypothetical solid AB dissolves into A+ and B−. The assigned dimensionless solubility product is Ksp = 0.04 at one fixed temperature. Start with 0.50 mol solid in 1 L water. With no initial ions, dissolving x mol gives both concentrations x mol/L. Saturation requires x² = 0.04, so x = 0.20 mol. Solid remaining is 0.30 mol. Construct that proposal and inspect the ledger.

The numbers are original ideal-model premises, not measured properties of a named salt. The simplification of activity by concentration is explicit, including at these comparatively large concentrations.

Predict: can a finite supply reach saturation?

Switch to only 0.05 mol solid in the same volume. Keeping x = 0.20 would create a negative solid amount. The software rejects the proposal rather than clipping it. Predict what happens when the entire available 0.05 mol dissolves.

Experiment: repair the endpoint

Set x = 0.05 mol in the limited-solid packet. Each ion is 0.05 mol/L, so Q = 0.0025, less than Ksp. There is no solid left to dissolve. This is an admissible final state, despite Q not equaling Ksp. At exactly 0.20 mol initially supplied, the all-dissolved endpoint lies on the saturation boundary instead.

Next select the common-ion packet: 0.30 mol A+ is already present with 0.30 mol nonreacting X− counterions. In 1 L, dissolving x mol gives Q = (0.30 + x)x. At x = 0.10, the ion concentrations are 0.40 and 0.10 mol/L; their product is 0.04. Less additional AB dissolves than in pure water, but the assigned Ksp has not changed.

In the 2 L packet the saturated concentrations are still 0.20 mol/L each, but 0.40 mol must dissolve. Concentration and amount are not interchangeable. Finally start with 0.30 mol of each ion and no solid in 1 L. A signed extent −0.10 precipitates 0.10 mol AB and leaves each ion at 0.20 mol/L. No initial solid is required in the model because it omits a nucleation barrier.

Understand: inventory and phase presence together

Let x be signed moles dissolved, nA,0 and nB,0 the initial ion amounts, nS,0 the initial solid amount and V the fixed solution volume. The proposal has nA = nA,0 + x, nB = nB,0 + x and nS = nS,0 − x. Every amount must be nonnegative. The sums nA + nS and nB + nS remain constant. The spectator carries the excess negative charge in the common-ion case.

Using standard concentration c° = 1 mol/L, Q = ([A+]/c°)([B−]/c°). A pure solid’s activity is one when that phase is present, so its amount does not enter this ion product. The criterion for an equilibrium endpoint is conditional: with solid remaining, Q = Ksp; with no solid, Q ≤ Ksp is allowed. Q > Ksp favors precipitation in this ideal model. This last criterion does not predict the waiting time for real nucleation.

Apply: another amount and volume

Complete the practice with Ksp = 0.01, only 0.02 mol solid and 0.50 L water. Compare the amount needed for saturation against the actual supply before selecting an endpoint. Explain why using the square root of Ksp alone is insufficient.

Continue: connect model and measured evidence

The earlier solubility-evidence activity distinguishes empirical plateaus, lower bounds and confounded comparisons. This activity adds ionic equilibrium algebra and finite phase inventory; neither substitutes for a controlled measurement.

AP Chemistry CED topics 7.11 and 7.12 address solubility products and common ions. OpenStax Chemistry 2e 15.1 develops dissolution, precipitation and ion-product comparisons. All displayed cases here use one ideal solid, fixed temperature/solution volume, no complexation or acid-base chemistry and no kinetic barrier. General phase equilibria, empirical validation and full Unit 7 evidence remain broader work.

Compare solubilities only after writing each ion ledger

For a separate ideal dilute fictional salt XY(s) ⇌ X⁺ + Y⁻ in pure water, molar solubility s gives Ksp = (s/c°)². For another fictional salt ZY₂(s) ⇌ Z²⁺ + 2Y⁻, the corresponding expression is Ksp = (s/c°)(2s/c°)² = 4(s/c°)³. Here c° = 1 mol/L. The same symbol s refers to dissolved formula-unit amount per solution volume, not every ion concentration.

Original comparison: XY has Ksp = 1.0×10⁻⁶ and ZY₂ has Ksp = 4.0×10⁻⁹ at the same stated temperature, with no common ions, competing reactions or activity corrections. Derive both molar solubilities before ranking them. Then reconstruct ZY₂’s Ksp from a supplied saturated free-Y⁻ concentration of 0.0020 mol/L and the stoichiometric charge balance.

Compare the two stoichiometries

Both have s = 0.0010 mol/L: the first is the square root of 10⁻⁶, while the second is the cube root of (4×10⁻⁹)/4. ZY₂ therefore has [Z²⁺] = 0.0010 and [Y⁻] = 0.0020 mol/L, whose normalized product is 4×10⁻⁹. A smaller Ksp alone did not establish smaller molar solubility. The inference from one ion requires that no other source, complexation or acid–base reaction alters its concentration.

The course framework connects Ksp greater than one with soluble salts. Under its simplified concentration convention, a large product is consistent with substantial dissolution; it does not imply unlimited solubility. For concentrated solutions, activity coefficients and solvent effects can invalidate a dilute calculation, and available solid can still be exhausted. Do not extrapolate s = √Ksp to enormous ideal concentrations or compare differently written dissolution equations as though their constants were identical. The conceptual bridge is supported by OpenStax precipitation and dissolution, checked 31 October 2026; all comparison values above are original.