Construct buffer reaction inventories before equilibrium

AP scope: Initial buffer pH, qualitative conjugate conversion/resistance and capacity reasoning are required. Deriving the Henderson–Hasselbalch relation and calculating numerical pH changes after an acid/base addition are optional enrichment under current topic 8.9, not required AP assessment. The deeper solver/design comparisons below are labeled for that purpose; they do not create a new mastery requirement.

A buffer changes its components

The original nominal inventory has 10mmol HA and 20mmol A⁻, supplied as its sodium salt. Add a stipulated 5mmol strong acid. Keeping 10/20 conserves family total but misses the conversion A⁻+H₃O⁺→HA+H₂O. Build 15/15 instead, then inspect the full equilibrium.

You are constructing formal post-neutralization bookkeeping, not claiming a measured intermediate. Acid-base equilibria occur together; the reaction-first calculation is a useful way to organize conserved amounts.

Retain excess reagent without negative leftovers

When 20mmol acid is added, all 20mmol nominal A⁻ can be consumed, giving 30mmol HA and 0mmol A⁻. With 25mmol acid, the component leftovers are still 30/0, but 5mmol strong acid remains in the total inventory. A proposal 35/−5 does not represent physical species amounts.

Strong base converts HA to A⁻. A 15mmol base challenge exceeds the 10mmol initial HA, producing nominal 0/30 plus 5mmol excess base. Construct both directions and explain why the excess strong reagent cannot disappear from the charge balance.

The complete equilibrium calculation still has nonzero concentrations of both conjugate species. Nominal exhaustion does not mean an exactly zero equilibrium amount or an undefined physical pH.

Compare an identity with an approximation

Ka=h[A⁻]/[HA] rearranges to pH=pKa+log₁₀([A⁻]/[HA]) within the ideal concentration model. This is an equilibrium relationship. Using nominal leftovers in that ratio is an additional approximation.

When either nominal component is zero, the activity withholds that nominal-ratio estimate rather than taking a logarithm of zero or a negative number. The full mass, charge, water and acid-equilibrium equations remain solvable, including beyond nominal capacity.

For the accepted 15/15 packet, compare final volumes 1 L and 10 L. The nominal ratio is unchanged, so the shortcut gives the same pH. The full equilibrium correction changes with dilution. These are separate stipulated final states, not an animation of a physical dilution procedure.

Include every spectator

Let h and o denote hydronium and hydroxide concentrations. Initial NaA and added NaOH supply sodium; HCl supplies chloride. The complete calculation uses h+[Na⁺]=o+[A⁻]+[Cl⁻], ho=Kw, [HA]+[A⁻]=C and Ka=h[A⁻]/[HA]. C is the conserved family amount divided by stated final volume.

The solver uses the original conserved inventories and the strong challenge; it does not treat the nominal leftovers as exact final equilibrium species. Its target result is withheld until your independently constructed bookkeeping matches the specified conversion.

Capture a rejected ledger and an accepted exhaustion or excess case. Explain the difference between finite nominal neutralization capacity and empirical buffer performance. The practice changes the starting amounts and challenge, requiring you to retain excess acid.

Source and continuation

OpenStax Chemistry 2e, 14.6 supplies buffer reaction and equilibrium context. All packets, controls, questions and diagrams are original. The assigned pKa 5 and classroom 25°C Kw=10⁻¹⁴ are ideal concentration assumptions, not measured properties of an identified buffer. Activity corrections, gas exchange, multiple ionizations, heat and mixing kinetics remain outside the model.

Return to conjugate-salt construction to identify the reacting ion and its constant when only one conjugate component is initially supplied.