Separate water and solute-depletion approximations

Two shortcuts remove different information

The familiar square-root estimate for a weak acid, h≈√(KaC), makes two simplifications: it neglects water’s minority ion in the charge balance, and it replaces the remaining neutral acid concentration by the analytical total C. Those assumptions can fail separately.

Start with the packet Ka=10⁻⁷ and C=10⁻⁷mol/L. The square-root estimate gives h=10⁻⁷mol/L, apparently neutral. But if the parent acid is still C and the newly formed conjugate is also C, the proposal contains twice the supplied solute. Restore the minority water ion and inspect the charge ledger as well. A plausible pH hid two problems.

Turn off one simplification at a time. Then turn both off. Explain which balances each model keeps and which it only approximates. Capture the failed square-root construction and the water-inclusive result with your reasoning.

Use a species ledger, not a slogan about weak acids

For neutral acid HA, call the neutral parent concentration u=[HA] and charged conjugate y=[A⁻]. Let m=h=[H₃O⁺]. For a neutral base B, use u=[B], y=[BH⁺] and m=o=[OH⁻]. In either mirrored case, the complete ideal model requires:

• u+y=C: the parent and conjugate share one conserved solute inventory.
• m−Kw/m=y: the majority water ion, minority water ion and conjugate balance charge.
• my=Ku: the assigned acid or base equilibrium relates those species.

Here K means Ka in the acid packet and Kb in the base packet. The two mirrored packets are not conjugate partners. Their assigned constants are equal for comparison; real conjugate constants instead multiply to Kw.

The full-model ionization fraction is y/C. When water matters, m/C is not that fraction because m includes the minority-water contribution required by charge balance.

Repair the right approximation

The substantial-depletion packet has K=10⁻³ and C=10⁻⁴mol/L. The square-root proposal can assign more conjugate than total solute. Retaining depletion through the quadratic y²=K(C−y) repairs the major failure, while water is negligible at that scale.

The water-dominated packet has K=10⁻¹⁰ and C=10⁻⁶mol/L. Its small predicted ionization fraction may justify little parent depletion, but it does not justify ignoring water. Keeping water while approximating u≈C gives m≈√(KC+Kw). Compare that result with the complete model and explain why the square-root-only estimate fails.

Inspect the corresponding neutral-base packets. Calculate pH from hydronium, even though hydroxide is the majority ion being solved. The equivalent acid/base pictures exchange the majority and minority water ions, not the labels of every chemical species.

A declared acceptance rule

This activity requires three relative quantities to be at most 1%: majority-ion concentration error against the complete model, solute-budget residual divided by C, and charge residual divided by the candidate majority concentration. It also rejects a candidate conjugate exceeding the total inventory. These are teaching decision criteria, not experimental confidence intervals or universal chemistry rules.

The audit restores the minority water ion as Kw/m even when the selected derivation neglected it. That makes the omitted charge contribution inspectable. A model can retain a very small residual without being exact; quantify that distinction. Choosing the complete ideal model avoids these numerical simplifications but does not remove its physical assumptions.

Source and continuation

OpenStax Chemistry 2e, 14.3 supports the equilibrium expressions and distinction between strength and concentration-dependent ionization. The AP Chemistry CED 8.3 supplies the curricular contract. The numerical packets, diagrams, comparisons and practice are original synthetic material.

All cases use a stipulated ideal 25°C concentration model with Kw=10⁻¹⁴(mol/L)². No real compound identity, measurement, gas exchange, activity correction, added salt, multiple ionization or kinetics is supplied. This is separate from the preceding molal reference investigation. Return to strong-mixture inventory to compare complete dissociation with the distinct question of whether water can be neglected.

Let a weak acid react with a weak base

For HA + B ⇌ A⁻ + BH⁺, both reactants being weak in water does not imply that no proton transfer occurs. Combining the two conjugate acid equilibria gives K = Ka(HA)/Ka(BH⁺) under compatible activity conventions. The relative acid strengths, not the word “weak,” determine which side this equilibrium expression favors.

Original prediction: At the same solvent and temperature, fictional HA has pKa 5 and fictional BH⁺ has pKa 7. Construct the transfer direction and compare K with one. Then replace B by a different weak base whose conjugate acid has pKa 3, keeping HA unchanged. Does every weak-acid/weak-base mixture react quantitatively?

Compare the two proton-transfer equilibria

The first has K = 10⁻⁵/10⁻⁷ = 100, favoring the A⁻/BH⁺ side of the expression. The changed pair has K = 10⁻⁵/10⁻³ = 0.01 and favors HA/B. Neither value alone establishes complete conversion or a final pH; starting inventories, charge balance and water equilibria still matter. These are original assigned constants, not named-compound measurements.

OpenStax relative acid/base strengths, checked 31 October 2026, supports conjugate-strength reasoning. No mixing procedure is requested.