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Grades 9–12 · Materials & Matter

How does a buffer resist a modest pH change?

A buffer limits a modest change in pH because it contains two chemical partners able to react with added acid or base. In this example, HA is a weak acid and A− is its conjugate base, the species formed when HA gives up a proton. Both partners must be available in useful amounts.

Added strong acid supplies H+. Much of it reacts with A− to make HA. Added strong base supplies OH−, which reacts with HA to form A− and water. The reagent therefore changes the balance between the partners rather than simply remaining as an equivalent amount of free strong acid or base.

The fictional initial solution contains 10 millimoles of HA and 10 millimoles of A− in 1 liter. Adding 2 millimoles of strong acid, with negligible added volume, converts 2 millimoles of A− into HA. The remaining amounts are 12 millimoles HA and 8 millimoles A−. The partner total remains 20 millimoles; the added protons change which form those units occupy.

Under suitable equilibrium approximations, pH ≈ pKa + log10([A−]/[HA]). Square brackets indicate concentrations. Because both partners share the same volume, their amount ratio can stand in for the concentration ratio. With the fictional pKa fixed at 5, the initial equal ratio gives pH 5. After the addition, pH ≈ 5 + log10(8/12), or about 4.82.

Each row contains twenty units. The highlighted two switch from A− to HA after accepting the added protons.

A buffer changes its partner balanceTwenty symbolic millimole units start as ten HA and ten A minus. Two acid units convert two A minus units to HA, ending at twelve and eight. The approximate pH moves from five to four point eight two.A reserve of two partnersBefore: 10 HA + 10 A−pH ≈ 5Add 2 mmol H+2 A− become 2 HAAfter: 12 HA + 8 A−pH ≈ 5 + log10(8/12)≈ 4.82Orange: HA · Blue: A−
The fictional buffer changes from 10 mmol HA and 10 mmol A− to 12 mmol HA and 8 mmol A− after 2 mmol acid. The stated approximation gives pH 4.82.

The pH has changed, so buffering does not mean holding an exact fixed value. It means the available acid–base pair has consumed much of the added strong reagent and limited the shift. The logarithm also explains why the ratio, rather than the simple difference between partner amounts, appears in this approximation.

Capacity is finite. Adding enough acid can use up the available A−; enough base can use up HA. Near depletion, approximating the equilibrium with remaining analytical amounts becomes less reliable. Once one partner is exhausted, this simple ratio expression is not used to report a pH. A fuller equilibrium calculation is required.

The drawing is an academic amount account, not a preparation procedure or a model of a particular body fluid. Its central mechanism is nevertheless clear: a reserve of both partners gives incoming acid or base something to react with. Resistance to change depends on that reserve, so a buffer can moderate a disturbance without being unlimited.

About this explanation

Starting knowledge: Recognize conjugate acid–base pairs, mole ratios, pKa and base-ten logarithms.

Sources

  • OpenStax — Buffers
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