Test why buffers resist change but have finite capacity

Follow the material that accepts or donates a proton

A buffer contains appreciable amounts of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this fictional HA/A− mixture, A− accepts added acid to form more HA. Added strong base instead consumes HA and produces more A−. These conversions reduce the change in free hydronium compared with an otherwise comparable unbuffered solution.

The model begins with equal nominal amounts of HA and salt MA. The salt supplies A− and inert M+. Its assigned pKa is 5. The first slider sets the amount of each component, not their sum. A setting of 10 means 10 mmol HA plus 10 mmol MA, for 20 mmol total acid-derived material.

The second slider is signed: positive means added strong monoprotic acid HX; negative means added strong hydroxide MOH. Zero means neither. All cases are separate prepared solutions with final volume fixed at 1 L and temperature fixed at 25 °C. This is not a sequence of additions from a burette, and moving the slider does not accumulate earlier additions.

Calculate a moderate challenge by amounts first

Start with 10 mmol of each component and add 5 mmol acid equivalents. The nominal reaction ledger converts 5 mmol A− to HA, leaving 5 mmol A− and 15 mmol HA. The total remains 20 mmol. The acid’s X− remains as a spectator; it does not disappear when its proton is transferred.

The Henderson–Hasselbalch relation in the ideal concentration approximation is pH = pKa + log10([A−]/[HA]). When equilibrium changes are small relative to the post-reaction inventories, use the nominal amounts in the ratio because both share the same final volume. The estimate is 5 + log10(5/15) ≈ 4.523. The full model gives a slightly different value because it retains the subsequent equilibrium adjustment and water’s ions.

With a 5 mmol base challenge instead, the nominal ledger gives 15 mmol A− and 5 mmol HA. The estimate is 5 + log10(15/5) ≈ 5.477. In both cases pH changes; buffering does not mean a constant pH regardless of what is added.

The relation using actual equilibrium activities is distinct from the shortcut that substitutes nominal post-reaction amounts. This investigation displays the full equilibrium answer and labels the nominal-amount estimate separately.

Compare ratio with available amounts

At zero challenge, equal nominal amounts of HA and A− give pH close to pKa for every available amount setting. The full values are not exactly identical because the initial preparation also equilibrates. Compare the displayed baseline values before applying a challenge.

Now compare 5 mmol of each component with 25 mmol of each. Both start near pH 5. A 5 mmol acid challenge uses all the smaller mixture’s nominal A− inventory, but only one fifth of the larger mixture’s. For the larger mixture, the nominal ratio becomes 20/30 and the estimated pH is about 4.824. For the smaller mixture, the nominal A− amount becomes zero and that shortcut is no longer available. The full result behaves like a weak-acid solution, with a substantially larger pH change.

The larger inventory resists the same challenge more effectively. Initial pH alone cannot tell you this. Here “capacity” is explored through finite added amounts and resulting pH changes; the widget does not calculate a formal differential buffer-capacity coefficient or certify a practical operating range.

Know what exhaustion means

At each = 10 and acid challenge = 10 mmol, the stoichiometric ledger converts all nominal A− into HA. The equilibrium calculation still has some A− because HA partially ionizes. Nominal exhaustion does not mean an exactly zero equilibrium concentration.

At acid challenge = 20 mmol, the added acid exceeds the initial nominal A− by 10 mmol. Treating the leftover nominal A− as −10 mmol would be physically meaningless. The model does not pass a negative amount into a logarithm. It marks the shortcut unavailable and solves the complete balance instead. The corresponding large negative challenge produces a base-excess case.

A buffer also loses usefulness before a literal component is exhausted if the pH change exceeds what the application tolerates. A meaningful design decision therefore specifies a challenge size and acceptable pH range, rather than declaring all mixtures of a conjugate pair equally suitable.

Keep mass, charge and water consistent

The full calculation retains [HA] + [A−] = C, Ka = [H3O+][A−]/[HA], and [H3O+][OH−] = Kw in the ideal dilute approximation. It also requires [H3O+] + [M+] = [A−] + [OH−] + [X−]. The table makes both spectators visible so that the added charge is accounted for.

An 80-step numerical search solves these constraints, including when a nominal component is exhausted. The model assumes inert spectators, one acid proton, no activity corrections, no temperature change and no competing reactions. It does not model heat release, mixing time or laboratory preparation accuracy.

Transfer: a mixture initially has 20 mmol HA and 20 mmol A−. A 5 mmol strong-base challenge is applied at the same fixed final volume. Calculate the nominal amounts and estimate pH, then explain why simply averaging pH values is inappropriate.

Check the buffer transfer

The nominal ledger gives 15 mmol HA and 25 mmol A−. The estimate is pH = 5 + log10(25/15) ≈ 5.222. Proton-transfer amounts must be matched before evaluating the concentration ratio. pH is logarithmic and is not itself a conserved quantity that can be averaged during a reaction.

OpenStax buffers provides the reference conjugate-pair and buffer-response concepts. The widget uses assigned fictional parameters and original comparison cases.