Preparing Buffers by Partial Neutralisation

Introduction

A buffer need not be made by mixing separate acid and salt bottles. A weak acid HA can be partly neutralised by a measured amount of strong base, creating A⁻ while leaving some HA. The word “partly” is essential: if the strong base consumes all HA, the intended conjugate pair is gone. Solve the neutralisation in moles before applying an equilibrium formula.

Core explanation

The stoichiometric reaction is HA + OH⁻ → A⁻ + H₂O. If n0 moles of HA are present and b moles of strong base are added with 0 < b < n0, the post-reaction inventory is nHA = n0−b and nA = b, assuming no initial A⁻. Both components then occupy the same final volume, so their concentration ratio equals b/(n0−b). In the standard concentration approximation, pH ≈ pKa + log[b/(n0−b)].

At b = n0/2, half the acid has become A⁻ and half remains HA. Their amounts are equal and pH ≈ pKa. This is the half-neutralisation point. It is distinct from the equivalence point b = n0, where all starting HA has been converted to A⁻ and pH is determined largely by conjugate-base hydrolysis, not by an HA/A⁻ buffer ratio.

Volume changes need careful handling. If acid solution and base solution are mixed, total volume affects the final concentrations and buffer capacity. However, the HA/A⁻ ratio can be calculated from final moles directly because both species share the same volume. If a problem also asks for component concentrations or for capacity, divide by the actual final volume. Ignoring volume is acceptable for the ratio, not for every quantity.

If the acid initially contains some A⁻, as in a partially prepared buffer, add the initial A⁻ moles to the A⁻ inventory: nA,final = nA,initial + b, while nHA,final = nHA,initial−b. A pH calculation based only on newly formed A⁻ would undercount the base component. Likewise, if strong acid impurities are present, neutralisation accounting must include them first.

The direction of the logarithm is a common error. More added base produces more A⁻ and less HA, so pH should rise as b increases through the buffer region. If an algebraic expression predicts the opposite, the ratio was inverted. The limits also provide a check: b → 0 gives almost no A⁻ and Henderson–Hasselbalch becomes unreliable at the extreme; b → n0 gives almost no HA and the equation again breaks down. The useful range lies between these extremes.

For accurate work at high ionic strength, pH is tied to activities and the apparent pKa can depend on the medium. The mole-ledger chemistry remains the correct first step, while a refined equilibrium calculation replaces the simple concentration ratio. Similarly, if the weak acid is polyprotic, strong base may pass through more than one neutralisation stage and the correct conjugate pair must be chosen from the current stoichiometric region.

Buffer preparation is therefore a two-layer calculation. Stoichiometry determines what remains after a nearly complete strong-base reaction. Equilibrium determines the pH of the remaining weak acid/conjugate base mixture. Reversing those layers—calculating pH from the initial acid before accounting for added OH⁻—answers a different question.

Step-by-step reasoning

1. Calculate initial HA and added OH⁻ moles from concentration × volume. 2. Apply HA + OH⁻ → A⁻ + H₂O to a mole ledger. 3. Confirm both HA and A⁻ remain and no OH⁻ is in excess. 4. Form nA/nHA and estimate pH using pKa + log ratio. 5. Use final volume for concentrations or capacity if requested.

Visual explanation

Draw a row of HA tokens. Convert b tokens to A⁻ when OH⁻ is added, leaving n0−b HA tokens. A balance scale under the row represents the final A⁻/HA ratio.

Real-world analogy

Partly exchanging one kind of voucher for another leaves two usable voucher types. The exchange count is solved first; only then can their ratio be used to describe the buffer's behaviour.

Real-world example

An ethanoic-acid buffer can be made by adding less than one equivalent of a strong base to ethanoic acid. The newly formed ethanoate and remaining acid then establish a pH near the acid's pKa when their amounts are comparable.

Why?

Why do moles come before pH? Strong hydroxide consumes HA almost completely on the stoichiometric scale, changing the pair amounts that set the later weak-acid equilibrium.

Common misconception

“At one equivalent of added base, pH equals pKa.” Equality of HA and A⁻ occurs at half an equivalent; one full equivalent leaves mostly A⁻ and requires a different pH calculation.

Worked example

Mix 50.0 mL of 0.200 M HA with 20.0 mL of 0.200 M NaOH. Initially nHA = 0.0100 mol and nOH = 0.00400 mol. After reaction, nHA = 0.00600 mol and nA = 0.00400 mol. The final ratio is 2/3, so pH ≈ pKa + log(2/3) = pKa − 0.176. Final volume is 70.0 mL, but it cancels from the ratio; it matters if component molarities are requested.

Quick check

1. If 0.030 mol HA is partly neutralised by 0.010 mol OH⁻, how many moles of each buffer component remain? Answer: HA = 0.020 mol and A⁻ = 0.010 mol, assuming no initial A⁻ and complete strong-base neutralisation.

Exam focus

Make a mole table, identify half-equivalence correctly and test whether the strong base is in excess before using Henderson–Hasselbalch.

Advanced insight

In a polyprotic acid, partial neutralisation can create a pair such as H₂PO₄⁻/HPO₄²⁻ rather than a simple HA/A⁻ pair. The same mole-ledger principle applies, but there are multiple equivalence stages and pKa choices.

Summary

Adding less than one equivalent of strong base to HA converts part to A⁻ and leaves a buffer pair. The post-neutralisation moles are n0−b and b, so pH follows their ratio when both are appreciable. At half-equivalence pH is near pKa; at full equivalence the buffer equation no longer applies.

Practice questions

1. What fraction of HA is neutralised at the half-equivalence point? Answer: One half. 2. What is nA/nHA if 0.020 mol HA starts and 0.015 mol OH⁻ is added? Answer: 0.015/0.005 = 3. 3. Does the final volume cancel from the ratio of HA and A⁻ in one mixed solution? Answer: Yes, though it is needed for their separate concentrations and capacity. 4. Why is Henderson–Hasselbalch inappropriate after 0.025 mol OH⁻ is added to 0.020 mol HA? Answer: All HA is consumed and strong OH⁻ remains in excess, so the assumed buffer pair is absent.