Preparing Buffers from Weak Base and Strong Acid

Introduction

A weak base B can be converted partly to its conjugate acid BH⁺ by adding a measured amount of strong acid. The mixture of B and BH⁺ can buffer pH. The calculation mirrors partial neutralisation of a weak acid, but the strong reagent consumes B rather than HA. Keep the stoichiometric mole ledger separate from the later acid-base equilibrium calculation.

Core explanation

The rapid stoichiometric reaction is B + H⁺ → BH⁺. If b0 moles of B are initially present and h moles of strong acid are added with 0 < h < b0, then nB = b0−h and nBH = h, provided no BH⁺ was initially present. The buffer ratio is base/conjugate acid = nB/nBH. In the ordinary concentration approximation, pH ≈ pKa(BH⁺) + log(nB/nBH).

Using the correct pKa is crucial. BH⁺ is the acid member of the conjugate pair. If a question supplies Kb for B at 25 °C, one can calculate Ka(BH⁺) = Kw/Kb and pKa(BH⁺) = pKw−pKb. Directly inserting pKb into pH = pKa + log ratio gives a wrong result. Temperature matters because pKw is not universally 14.

At half-protonation, h = b0/2, so B and BH⁺ have equal moles and pH ≈ pKa(BH⁺). At h = b0, essentially all B has been protonated and the solution is no longer a useful B/BH⁺ buffer. Its pH is then controlled by the weak conjugate acid BH⁺ and any other species. If h exceeds b0, excess strong acid has an even more direct effect. A Henderson–Hasselbalch expression with zero B would be mathematically undefined and chemically inappropriate.

If the starting solution already contains BH⁺, include it: nBH,final = nBH,initial + h, while nB,final = nB,initial−h. As with weak-acid buffers, both species share one final volume, so volume cancels from their ratio but matters for actual concentrations and capacity. This lets a chemist adjust an existing buffer using a strong-acid dose while accounting for the original conjugate acid.

An ammonia/ammonium pair is a standard example. NH₃ accepts H⁺ to form NH₄⁺. Adding a controlled sub-equivalent of strong acid to ammonia creates NH₃ and NH₄⁺ in one solution. The pair can consume later added acid through NH₃ and added base through NH₄⁺. If the pH target is close to the ammonium pKa, comparable amounts work well; a different target calls for a different ratio.

Strong acid first reacts with B on the stoichiometric scale, so the added H⁺ is not simply left free at its initial concentration. A common error is to calculate pH from the strong-acid moles as if no weak base were present. Another is to use the inverted ratio BH⁺/B in the plus-log formula. A sensible direction check prevents it: adding more acid converts B into BH⁺, so pH should decrease through the buffer region.

For a high-accuracy solution, activity coefficients and base hydrolysis can be incorporated in a full mass/charge-balance model. These refinements do not alter the order of reasoning: neutralisation changes the analytical composition first, then equilibrium establishes the final pH. At extreme dilution, water autoionisation may also require explicit inclusion.

Step-by-step reasoning

1. Convert B and strong-acid amounts to moles. 2. Apply B + H⁺ → BH⁺ and calculate the remaining pair. 3. Confirm neither B nor strong acid is exhausted in the wrong direction. 4. Convert Kb to pKa(BH⁺) if needed, using the stated temperature. 5. Use pH ≈ pKa + log(nB/nBH) when both forms remain.

Visual explanation

Draw a row of B circles and convert a chosen number to BH⁺ circles as acid is added. The ratio of unconverted to converted circles appears in the logarithmic pH formula.

Real-world analogy

A batch of empty seats is partly occupied. The remaining empty seats can receive more arrivals, while occupied seats represent the paired form; the balance depends on how many arrivals have already occurred.

Real-world example

An ammonia buffer can be prepared by protonating only part of an ammonia solution with a strong acid. The resulting ammonia and ammonium inventories determine its approximate pH and its ability to absorb later additions.

Why?

Why does adding strong acid lower pH even though much of its H⁺ is consumed? It changes the B/BH⁺ ratio toward more conjugate acid, and that equilibrium ratio sets a lower pH.

Common misconception

“Use the pKb of NH₃ directly as the pKa in the Henderson–Hasselbalch equation.” The equation uses the pKa of NH₄⁺; pKa and pKb are related through pKw at the specified temperature.

Worked example

Begin with 0.0200 mol NH₃ and add 0.0080 mol HCl. The strong acid protonates NH₃, leaving 0.0120 mol NH₃ and forming 0.0080 mol NH₄⁺. If pKa(NH₄⁺) is supplied as 9.25 for the conditions, pH ≈ 9.25 + log(0.0120/0.0080) = 9.25 + 0.176 = 9.43. The answer exceeds pKa because unprotonated base remains more abundant than conjugate acid.

Quick check

1. What pair remains after adding less than one equivalent of strong acid to a weak base B? Answer: Unreacted B and the newly formed conjugate acid BH⁺, provided the strong acid is fully consumed.

Exam focus

Use B/BH⁺, not its inverse, in the plus-log formula. Convert pKb to the conjugate acid's pKa only after checking temperature.

Advanced insight

A real buffer made from weak base and strong acid also contains the strong acid's counterion. It contributes to ionic strength and charge balance even though it is a spectator in the principal B/BH⁺ proton-transfer equilibrium.

Summary

Partial protonation converts h moles of an initial b0 moles of B into BH⁺, leaving b0−h moles of B. When both remain, pH is estimated from their ratio and pKa(BH⁺). At full or excess strong-acid addition, the buffer-ratio approximation fails and a new equilibrium problem must be solved.

Practice questions

1. If 0.030 mol B receives 0.010 mol strong acid, what pair amounts remain? Answer: 0.020 mol B and 0.010 mol BH⁺. 2. What is pH relative to pKa when B and BH⁺ are equimolar? Answer: Approximately equal to pKa(BH⁺). 3. At 25 °C, what is pKa(BH⁺) if pKb(B) = 4.75? Answer: Approximately 14.00−4.75 = 9.25. 4. What happens if more strong acid is added than the initial B moles? Answer: B is fully protonated and excess strong acid remains, so a simple B/BH⁺ buffer equation is invalid.