The most consequential buffer mistake often occurs before any logarithm: treating the initial acid and base amounts as though they survived an added strong acid unchanged. A second mistake is subtler: assuming two solutions with the same pH have the same capacity to resist a disturbance.
Both disappear when the problem is divided into two questions: what reacts first, and what equilibrium remains afterward? Buffers and acid–base equilibria appear in the AAMC content outline. The examples here are original applications, with explicitly idealized chemistry.
Know what the ratio does and does not describe
For a weak acid HA and its conjugate base A⁻:
pH = pKa + log₁₀(aA⁻/aHA).
The activity form follows from the acid-dissociation equilibrium. Replacing activities with concentrations gives the familiar Henderson–Hasselbalch expression. In routine buffer calculations, we additionally approximate the equilibrium concentrations by the amounts left after neutralization, divided by their common final volume.
That last approximation needs checking. Both buffer components must remain appreciable, and their further equilibration must change their amounts only slightly. At extreme dilution, near depletion of one component, or when activity effects matter, a fuller equilibrium calculation may be needed. OpenStax explains buffer reactions and finite capacity.
A ratio is not an inventory. Ten millimoles of each component and one millimole of each have the same ratio. They do not have the same reserve against added acid.
Worked example: update millimoles before calculating pH
A fictional monoprotic acid has pKa = 4.80 at 25°C. A 100.0 mL solution initially contains 10.0 mmol HA and 10.0 mmol A⁻. Assume ideal behavior, negligible volume changes beyond the stated added liquid, and no competing reactions.
Its initial pH is approximately 4.80. Add 2.00 mL of 1.00 M HCl, or 2.00 mmol H⁺. Strong acid reacts with the conjugate base:
A⁻ + H⁺ → HA.
| Amount | HA | A⁻ |
|---|---|---|
| Before addition | 10.0 mmol | 10.0 mmol |
| Neutralization change | +2.00 mmol | −2.00 mmol |
| After neutralization | 12.0 mmol | 8.00 mmol |
Both species remain. Therefore:
pH ≈ 4.80 + log₁₀(8.00/12.0) = 4.62.
The common final volume, 102.0 mL, cancels in the concentration ratio. It does not disappear from chemistry generally: volume will matter if you need a concentration, a hydrolysis calculation, or the concentration of excess strong acid.
The result also passes a directional check. Added acid consumes A⁻, decreases the base-to-acid ratio and lowers pH. Using 12/8 would predict a pH increase and reveal an inverted ratio.
Same initial pH, very different outcome
Now start with only 1.00 mmol HA and 1.00 mmol A⁻ in 100.0 mL. Its initial pH is again approximately 4.80. Add the same 2.00 mmol HCl in 2.00 mL.
The first 1.00 mmol H⁺ consumes all the available A⁻. The remaining 1.00 mmol is excess strong acid. The post-reaction inventory contains approximately 2.00 mmol HA plus that excess H⁺.
The dominant hydrogen-ion concentration is:
[H⁺] ≈ 0.00100 mol / 0.1020 L = 0.00980 M, giving pH ≈ 2.01.
Weak-acid dissociation adds a small correction; its contribution is suppressed by the existing H⁺. With Ka = 10⁻⁴·⁸⁰, the extra H⁺ is about 0.000032 M, small relative to 0.00980 M. The approximation is therefore adequate at the precision used here.
A negative conjugate-base amount in a buffer table is not a strange new equilibrium. It signals that neutralization exhausted that reactant and the calculation must change regions. Likewise, log(0) is not a physical prediction of infinitely low pH; it indicates that the assumed buffer approximation is no longer appropriate.
A titration has several chemical regions
Consider a separate sample: 100.0 mL of 0.100 M HA, titrated with 0.100 M NaOH at 25°C. There are initially 10.0 mmol HA.
Before any base is added, calculate weak-acid dissociation rather than assuming an established equal-component buffer. After 50.0 mL NaOH, 5.00 mmol HA has become A⁻ and 5.00 mmol HA remains. At this half-equivalence point, pH ≈ pKa = 4.80.
At 100.0 mL NaOH, the stoichiometric equivalence point has been reached. The solution contains 10.0 mmol A⁻ in 200.0 mL: approximately 0.0500 M conjugate base. Equivalence means equal reacting amounts according to stoichiometry; it does not mean neutral pH for every titration. OpenStax’s titration treatment distinguishes these regions.
Here A⁻ hydrolyzes water. With Kw = 1.00 × 10⁻¹⁴:
- Kb = Kw/Ka = 6.31 × 10⁻¹⁰.
- [OH⁻] ≈ √(KbC) = 5.62 × 10⁻⁶ M.
- pOH ≈ 5.25 and pH ≈ 8.75.
The approximation is consistent because the hydrolyzed fraction is tiny compared with 0.0500 M. Beyond equivalence, excess strong base becomes the dominant calculation. These are different models for different remaining inventories.
Transfer exercise: addition versus dilution
Return to the original 10.0 mmol HA/10.0 mmol A⁻ buffer. In a fresh sample, add 4.00 mmol NaOH, with both components remaining in solution. What is its approximate pH?
OH⁻ converts HA to A⁻. The new amounts are 6.00 mmol HA and 14.0 mmol A⁻, so:
pH ≈ 4.80 + log₁₀(14.0/6.00) = 5.17.
The answer 4.43 reverses the ratio. The answer 4.80 assumes a buffer prevents all pH change. A buffer resists change within its capacity; it does not lock pH to one value.
Now imagine a fresh original buffer diluted tenfold with water, without losing any solute. Its approximate pH remains near 4.80 because the ratio remains one. But equal-volume portions of the diluted and original buffers contain different numbers of buffer molecules: the diluted portion has less capacity.
The entire enlarged solution still contains the original total millimoles. Saying dilution automatically removes nine-tenths of the whole sample’s stoichiometric reserve would confuse concentration with total amount. At sufficiently extreme dilution, the simple pH approximation itself also becomes less reliable.
Use an inventory check on your next passage
Before choosing an equation, label the added reagent, write its dominant reaction, and calculate the remaining amounts. Then decide whether you have a buffer, a weak acid or base alone, or excess strong acid or base.
Finally, check direction and scale. A small addition to a substantial buffer should not produce the same result as overwhelming a nearly exhausted component. The equation should describe the chemistry your inventory actually leaves behind.