Test a quantitative selectivity target against the second onset

Define what counts as selective

Two hypothetical1:1 solids share free anion X. M1X has Ksp10⁻⁹ and M2X has10⁻⁷. The first metal begins at .001 M. To remove at least99%, its remaining free concentration must be at most10⁻⁵ M, requiring X≥10⁻⁴ M once its solid phase exists.

For the second metal to remain wholly dissolved, X must not exceed Ksp2/M2initial. If M2 starts at .001 M, that threshold is also10⁻⁴ M. At the exact ideal boundary, the first target is met and the second just reaches saturation without a finite precipitated amount. There is no nonzero margin.

Vary the inventory, not the definition

At M2=.000500 M, its onset rises to2×10⁻⁴ M, leaving an ideal interval. At M2=.00200 M, onset falls to5×10⁻⁵ M, below the required first-metal threshold: no interval meets both criteria. The constants alone did not determine feasibility; the initial concentrations and removal target mattered.

The model controls free X through an external reservoir. It does not equate that value to added total reagent. Complexation, activity changes, kinetics and measurement bounds could alter a practical outcome. No separation procedure is provided.

Design against bounded concentration uncertainty

Keep the same fixed ideal constants K1=10⁻⁹ and K2=10⁻⁷, and require at least 99% removal of the first metal for every allowed initial concentration. Now M1initial is .0010±.0001 M and M2initial is .00050±.00005 M. These are independent feasible bounds, not confidence intervals. Activity coefficients remain fixed at one, and an independent reservoir is assumed to control the free X exactly; no uncertainty in X, constants or competing species is silently included.

For the first target, residual M1=K1/X must not exceed .01 M1initial. The smallest permitted first-metal inventory gives the most demanding lower bound: Xlower=10⁻⁹/(.01×.0009)=1.11111×10⁻⁴ M. For the second metal to avoid a finite precipitate, use its largest permitted inventory: Xupper=10⁻⁷/.00055=1.81818×10⁻⁴ M. A proposed independently controlled X=1.50×10⁻⁴ M lies strictly inside this robust interval. Predict the worst-case removal fraction and second-metal saturation product before checking: residual M1=6.66667×10⁻⁶ M, so even the .0009 M inventory has removal 99.2593%; the largest second-metal product is .00055×.00015=8.25×10⁻⁸, below K2.

Now change only the second-metal packet to .0010±.0001 M. Its worst-case onset falls to 10⁻⁷/.0011=9.09091×10⁻⁵ M, below the first target’s lower bound. There is no robust interval, even though nominal .001 M concentrations gave an exact equality-only boundary. An endpoint at saturation is not a positive-width practical margin. If the controlled X itself had a bounded error, its entire allowed interval would need to fit inside the feasible window; real activity or calibration uncertainty would require extending the calculation rather than treating this ideal result as a safe separation procedure.

Transfer and check

If the first-metal target is relaxed to90% removal, allowable remaining M1 is10⁻⁴ M and the required free X falls to10⁻⁵ M. For second-metal .001 M, compare that lower threshold with10⁻⁴ M onset. Explain why a wider ideal interval reflects a changed performance target rather than a magically more selective chemical constant.

Source and evidence boundary

Conceptual reference checked 30 October 2026: OpenStax Chemistry 2e. All cases, prose and diagrams are original. Fictional species and explicitly ideal or supplied-activity constraints are not measured solubilities, real assay performance or safe physical separation protocols. No material handling, mixing, filtration or reagent procedure is requested. Retain free versus total species, inventory constraints, boundary conditions and independent evidence limits.