Build strong-acid/base inventories and test the water approximation

Count formula units before taking a logarithm

A synthetic record contains .10mmol HCl and .05mmol dissolved Ca(OH)₂ in a stated final volume of 20 mL. Complete dissociation supplies two hydroxides per calcium hydroxide formula unit. The base therefore supplies .10mmol of hydroxide equivalents, matching the acid equivalents.

Start with the wrong coefficient 1. The candidate calculation predicts an acid excess, but the calcium spectator carries two positive charges. Inspect the charge residual, then repair the coefficient. Next deliberately choose a wrong volume while retaining the correct coefficient. The candidate pH remains7 at equivalence, yet the construction still fails the stated volume. A numerically plausible answer does not establish every step.

Amount, concentration and ionic charge

Analytical amounts describe the supplied dissolved inventories, not the amount of HCl molecules or free hydroxide remaining after equilibrium. Subtract equivalents before dividing by final volume. With nH mmol of HCl, nCa mmol of calcium hydroxide and final volume V mL, the analytical excess is Δ=(nH−2nCa)/V in mol/L: the factors of 1000 in mmol and mL cancel.

The spectator concentrations are [Cl⁻]=nH/V and [Ca²⁺]=nCa/V. Define h=[H₃O⁺] and o=[OH⁻]. Electrical neutrality requires h+2[Ca²⁺]=o+[Cl⁻], so h−o=Δ. The model also requires ho=Kw. Neither equation says that one equilibrium water ion vanishes.

A wrong final volume can preserve charge balance because all concentrations share the same wrong denominator. That is why the activity checks the given volume separately from the charge equation and pH.

When water changes the answer

For the acid-excess record, (.12−2×.05)/20=.001mol/L. The water-inclusive solution is extremely close to h=.001mol/L, so the dominant-ion approximation is useful. Diluting the same inventory to 1000 mL reduces the analytical excess fiftyfold. The number of excess equivalents is unchanged; the concentration and pH change.

Now inspect the very dilute HCl record. Its analytical excess is 10⁻⁸mol/L. Ignoring water would give pH 8, apparently basic despite adding acid to otherwise pure water in this closed ideal model. Solving both equations instead gives h≈1.05124922×10⁻⁷mol/L and pH≈6.97829. The water contribution matters even though the acid is fully dissociated. The paired very dilute base case reverses the ion predominance.

The activity compares the approximate majority-ion concentration |Δ| with the water-inclusive majority concentration. It accepts an approximation only after the inventory passes and its relative concentration error is at most 1%. This is a declared teaching criterion, not a measurement uncertainty. At zero excess, the dominant-excess logarithm is undefined; the water-inclusive solution remains well defined.

Optional derivation

Substituting o=Kw/h into h−o=Δ gives h²−Δh−Kw=0. For numerical stability, calculate the majority concentration as (|Δ|+√(Δ²+4Kw))/2 and obtain the minority by division into Kw. This avoids cancellation when the acid is the minority. The model’s stable algebra is exact for its declared equations; its physical assumptions are still approximations.

Capture one rejected inventory and one approximation decision. Explain which independent evidence rejects the former and quantify the latter’s error. The practice changes the amounts and volume; do not memorize a displayed pH.

Source and boundaries

OpenStax Chemistry 2e, 14.2 supplies the classroom water-ion and logarithmic relationships. The AP Chemistry CED, 8.2.A requires concentration calculations with strong acid/base stoichiometry. The numerical packets, derivation, diagrams and tasks here are original, not copied source experiments.

This lesson deliberately uses the familiar 25°C concentration approximation Kw=10⁻¹⁴ (mol/L)². It does not silently reuse the previous temperature-dependent molal reference. Stated amounts are already dissolved; this is not a calcium hydroxide solubility or preparation model. Activity coefficients, carbon dioxide exchange, precipitation, complexes, salt hydrolysis and mixing heat are not modeled. Ultra-dilute examples expose an approximation boundary; they are not predictions for uncharacterized laboratory water.

Do not assign every acid the same proton count

Sulfuric acid has an essentially complete first proton donation in the usual aqueous treatment, giving H₃O⁺ and HSO₄⁻. The second donation, HSO₄⁻ ⇌ H⁺ + SO₄²⁻ in shorthand, has its own finite equilibrium and depends on concentration and conditions. Neither a blanket monoprotic h = C nor a blanket h = 2C is a universal sulfuric-acid calculation.

Prediction: A record stipulates 0.010 mol of already-dissolved H₂SO₄ formula units but supplies no final volume or second-dissociation information. Identify the first-donation amount under the stated complete-first-step approximation, then explain why final hydronium concentration is still unidentified.

Compare amount and equilibrium

The first donation supplies approximately 0.010 mol of proton equivalents in that bookkeeping step. The second donation and water equilibrium can change the final hydronium amount, and concentration also needs final volume. It is not justified to report either 0.010 mol/L or 0.020 mol/L from the supplied amount alone.

Likewise, a formula such as Ca(OH)₂ gives two hydroxides per dissolved formula unit, but does not prove every amount of solid added dissolves. Dissociation of a known dissolved inventory and a solid’s solubility are separate questions. The model above deliberately starts after that distinction has been resolved. OpenStax polyprotic acids, checked 31 October 2026, supports the staged sulfuric-acid distinction.