Buffers: explain pH resistance and its limits for the MCAT

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Two buffers can start at the same pH yet respond very differently to the same acid addition. Track the amounts before using the ratio.

About 16 minutes, including the worked examples and checks.

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Concept: a buffer needs both members of a weak conjugate pair
Pattern: the ratio sets working pH, while the amounts set resistance
Application: two recipes with the same starting ratio
Rule: neutralize first, then calculate the new ratio
Changed conditions: dilution and exhaustion are different
Connect the same account to a titration
Check your reasoning
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Concept: a buffer needs both members of a weak conjugate pair

A buffer limits the pH change caused by a modest addition of strong acid or base. A weak acid HA can consume added OH⁻; its conjugate base A⁻ can consume added H₃O⁺. Appreciable amounts of both give the solution a response in both directions. AAMC category 5A includes buffer systems and their effects on titration behavior.¹²

For an acid challenge, A⁻ + H₃O⁺ → HA + H₂O. For a base challenge, HA + OH⁻ → A⁻ + H₂O. Added strong acid therefore decreases A⁻ and increases HA by the same amount; strong base makes the opposite change. The acid or base is converted into a member of the weaker conjugate pair. The component ratio still changes, so “resists” does not mean “prevents.”

Pattern: the ratio sets working pH, while the amounts set resistance

Rearranging the acid equilibrium gives the Henderson–Hasselbalch relation: pH ≈ pK_a + log([A⁻]/[HA]) in the ideal concentration model. The ratio here is conjugate base over acid, and the logarithm is base ten. At equal equilibrium concentrations, pH ≈ pK_a; a tenfold base excess puts pH about one unit above pK_a.²

The equilibrium relation is not permission to insert any two numbers from a recipe. Using amounts calculated from preparation or neutralization assumes that the subsequent equilibrium changes are small relative to both components. Check that both partners remain appreciable. Within the same final vessel, their common volume cancels, so the concentration ratio can be replaced by n(A⁻)/n(HA).

For a target pH near 6, a chemically suitable pair with pK_a near 6 allows comparable acid and base amounts. A working range of roughly pK_a ± 1 corresponds to base/acid ratios from 0.1 to 10. This is a useful design range, not a sharp boundary where buffering suddenly switches off. Toward either end, the less abundant partner is more easily depleted.

Application: two recipes with the same starting ratio

The following recipes and acid challenge are illustrative. Use a hypothetical monoprotic HA with pK_a = 6.20 in ideal water at 25°C. Its sodium salt supplies A⁻, and sodium has negligible acid–base effect. Assume no side reactions.

Sample	Final volume	HA amount	A⁻ amount	Initial estimated pH
More concentrated	100.0 mL	4.00 mmol	8.00 mmol	6.50
Less concentrated	100.0 mL	0.400 mmol	0.800 mmol	6.50

Original equal-volume buffer comparison

Both ratios are two, giving pH ≈ 6.20 + log 2 = 6.50. The total concentrations differ tenfold: 0.120 M versus 0.0120 M when HA and A⁻ are added together. A pH measurement alone would not reveal that difference in inventory.

Rule: neutralize first, then calculate the new ratio

Add 1.00 mL of 0.200 M HCl to each sample. Each receives 0.200 mmol of strong acid. Use the reaction with A⁻ to update the amounts before applying the equilibrium relation. The final volume is 101.0 mL in this example, but it cancels from each within-sample ratio.

Sample	HA	A⁻	New base/acid ratio	Estimated pH
More concentrated	4.20 mmol	7.80 mmol	7.80/4.20 = 1.86	6.47
Less concentrated	0.600 mmol	0.600 mmol	0.600/0.600 = 1.00	6.20

Amounts after the acid challenge, before the small equilibrium adjustment

The larger buffer changes from about 6.50 to 6.47; the smaller one changes from about 6.50 to 6.20. In the smaller sample, the added acid consumed one-quarter of the initial A⁻. In the larger sample, it consumed only one-fortieth. The same initial pH and the same chemical pair therefore do not imply the same response to an acid challenge.

Original calculation: after the same 0.200 mmol acid addition to each 100 mL sample, pH falls by about 0.03 in the 12 mmol buffer and 0.30 in the 1.2 mmol buffer. The enlarged pH axis shows the changes; the tables supply the component amounts.

These calculations use millimoles consistently. Converting only one component to moles would introduce a factor-of-one-thousand error. They also use the amounts after neutralization: leaving the ratio at two would predict no change and would ignore where the added proton went.

The equilibrium approximation is appropriate here because both post-reaction components remain in the millimolar range, while the predicted hydronium is below 10⁻⁶ M. The remaining equilibrium shifts are tiny relative to the prepared components. If either component were extremely small, the full equilibrium calculation would be more reliable than substituting the preparation amounts.³

Changed conditions: dilution and exhaustion are different

Dilution leaves the component ratio nearly unchanged in this model, so it can leave the initial pH nearly unchanged. When comparing equal final sample volumes, a more dilute portion contains fewer buffer moles and shows a larger response to the same added acid amount. But adding water to an entire beaker does not remove that beaker’s buffer moles. Specify whether the challenge is a fixed amount, an amount per liter, or an addition to a fixed-size sample.⁴

The two recipes above compare fixed-size samples. Their different resistance follows from the amount available to react in each sample, rather than from dilution changing the acid’s pK_a. At very large dilutions, the assumptions about negligible equilibrium changes and water ionization can themselves fail.

For exhaustion, return to a fresh less-concentrated sample containing 0.800 mmol A⁻. An addition of 0.900 mmol strong acid exceeds that initial base amount. After the principal neutralization, 0.100 mmol strong acid remains, and the system is no longer an appreciable HA/A⁻ buffer pair. Calculate from the excess acid and the remaining equilibria; do not put a negative base amount into a logarithm. At exact stoichiometric depletion, the remaining weak acid has its own equilibrium and is not automatically neutral.²

Connect the same account to a titration

During titration of a weak acid with a strong base, partial neutralization creates its conjugate base while some acid remains. This produces a buffer region. At half-equivalence, approximately half the original acid remains and half has become conjugate base, so their ratio is one and pH ≈ pK_a. At equivalence, the principal species is instead the conjugate-base salt; hydrolysis determines its pH. Equivalence is a stoichiometric point, not a promise of pH 7.⁵

The same sequence handles all these cases: count the reacting amounts, perform any near-complete neutralization, identify what remains, and then choose its equilibrium model. The useful buffer question is not merely “What is its pH?” but “How much of each partner remains to respond to the next change?”

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A researcher wants a buffer near pH 6.0 in ideal water at 25°C. Each proposed recipe contains 0.0500 M of each listed solute in the final solution. All solutes are soluble and otherwise compatible with the experiment. Which preparation best supplies both appropriate conjugate partners near that pH?

A. Equimolar HCl and NaCl, using chloride as the proton-accepting component.

Review this choice. HCl is a strong acid in water, and chloride is a negligibly weak base. This is not an appreciable weak conjugate pair capable of buffering near pH 6.

B. Equimolar HX and NaX, where the weak acid HX has pKₐ = 3.0.

Review this choice. This is a conjugate buffer pair, but equal amounts place its pH near 3.0. Reaching pH 6 would require an extreme base/acid ratio, leaving little acid available against a base challenge.

C. Equimolar HY and NaY, where the weak acid HY has pKₐ = 6.0.

Correct. The weak acid and its conjugate base are both present. Their ratio of one places pH near the supplied pKₐ, 6.0, without making one component extremely scarce.

D. Equimolar NaOH and KCl, using potassium as the proton-donating component.

Review this choice. NaOH supplies strong base, while potassium is not an appreciable proton donor. These solutes do not supply the required weak conjugate pair.

The principle to carry forward

Choose both a conjugate relationship and a suitable pKₐ; two dissolved substances alone do not establish useful buffering at the target pH.

Check 2

A 100 mL buffer contains 3.00 mmol HA and 1.00 mmol A⁻, with pK_a = 7.00 at 25°C. Add 0.500 mmol NaOH in a small volume. Assume complete principal neutralization, no side reactions, and negligible later changes in component amounts. Using log(0.60) ≈ −0.22, what is the new pH?

A. 6.52

Review this choice. This uses the original ratio 1.00/3.00 and ignores neutralization. Added hydroxide consumes HA and produces A⁻.

B. 6.78

Correct. After neutralization, HA = 2.50 mmol and A⁻ = 1.50 mmol. Their ratio is 0.60, so pH ≈ 7.00 − 0.22 = 6.78. The common final volume cancels.

C. 7.22

Review this choice. This reverses the acid/base ratio. HA remains more abundant than A⁻, so pH should remain below the pKₐ, not above it.

D. 7.48

Review this choice. This is approximately the result of reversing the original ratio. It both ignores the amount changes and places the acid in the numerator.

The principle to carry forward

Neutralization changes both conjugate partners. Use their new amounts in the base-over-acid ratio.

Check 3

Two fresh 100 mL samples use the same HA/A⁻ pair with pK_a = 6.00. P contains 5.00 mmol of each partner; Q contains 0.500 mmol of each. Each receives 0.250 mmol HCl, with no side reactions and negligible volume change. Which prediction follows from the buffer model?

A. Both pH values decrease equally because their initial conjugate-partner ratios are identical.

Review this choice. Equal initial ratios give equal initial pH. They do not make a fixed added amount the same fraction of the two different inventories.

B. Both pH values remain at 6.00 because the strong acid is converted into weak acid.

Review this choice. Conversion protects against a much larger change, but it alters the component ratio. More HA and less A⁻ make the pH decrease.

C. P shows the smaller decrease because the addition consumes a smaller fraction of its A⁻.

Correct. P ends with 4.75/5.25 as its base/acid ratio, giving pH ≈ 5.96. Q ends with 0.250/0.750, giving pH ≈ 5.52. Both still contain both partners, but their fractional changes differ.

D. Q shows the smaller decrease because its initial buffer contains fewer dissolved particles.

Review this choice. The smaller inventory provides less A⁻ to consume the same acid amount. Its component ratio changes more, so its pH decreases more.

The principle to carry forward

For a fixed sample volume and acid challenge, compare available amounts as well as the initial ratio.