Buffer Action Revisited

Introduction

Earlier work introduced buffers as solutions that keep pH nearly constant. At this level we look more closely at the mechanism. A buffer does not stop added acid or base from reacting; it gives that acid or base something to react with, converting a strong species into a weak one. Seeing buffering as two nearly complete neutralisation reactions, followed by a small equilibrium adjustment, makes every later buffer calculation straightforward.

Core explanation

Two reserves. A buffer contains a weak acid HA and its conjugate base A⁻, for example ethanoic acid and ethanoate ions, or ammonium ions and ammonia. Each member of the pair has a job:

- Added strong base is removed by the acid reserve: HA + OH⁻ → A⁻ + H₂O - Added strong acid is removed by the base reserve: A⁻ + H₃O⁺ → HA + H₂O

In both cases a strong, pH-controlling species (OH⁻ or H₃O⁺) is replaced by a weak one (A⁻ or HA). The pair itself is not destroyed; one member is simply converted into the other.

These reactions go to completion. For ethanoic acid, Ka = 1.74 × 10⁻⁵ (pKa = 4.76). The constant for A⁻ + H₃O⁺ → HA + H₂O is 1/Ka ≈ 5.8 × 10⁴. The constant for HA + OH⁻ → A⁻ + H₂O is Ka/Kw ≈ 1.7 × 10⁹. Both are very large, so added strong acid or base is consumed almost entirely, provided the relevant reserve is not exhausted.

Why pH changes only a little. After the neutralisation, the solution is still a mixture of HA and A⁻, and its hydronium concentration is fixed by

[H₃O⁺] = Ka × [HA] ÷ [A⁻]

The pH depends on the ratio [HA]/[A⁻], and on it only logarithmically. Converting 10% of A⁻ into HA in an equimolar buffer changes the ratio from 1.00 to 1.22, which shifts the pH by only about 0.09 units. In pure water, the same amount of acid would change the hydronium concentration by many orders of magnitude.

The limits of the mechanism. Buffering works only while both reserves remain. If added acid exceeds the amount of A⁻, the excess H₃O⁺ has nothing to react with and the pH falls sharply. Similarly, excess base beyond the amount of HA is not buffered. The total amount of the conjugate pair therefore sets how much acid or base can be absorbed.

Accounting order. Every buffer problem involving added strong acid or base follows the same order: first a stoichiometry step (the complete reaction, done in moles), then an equilibrium step (using Ka and the new ratio). Mixing up the order, or trying to handle both at once with an ICE table in concentrations, is the commonest source of errors.

Formulae

pH = pKa + log(n(A⁻) ÷ n(HA)), valid when both are present in amounts much larger than the H₃O⁺ and OH⁻ produced by the water or the weak acid itself. Moles may replace concentrations because both species share the same volume.

Step-by-step reasoning

To find the pH of a buffer after adding strong acid:

1. Convert everything to moles: HA, A⁻ and the added H₃O⁺. 2. Subtract the added acid from A⁻ and add it to HA. 3. Check that some A⁻ remains; if not, calculate pH from the excess strong acid instead. 4. Substitute the new mole ratio into the pH expression. 5. Compare with the original pH to find the change.

Visual explanation

Imagine two columns representing the moles of HA and A⁻. Adding hydroxide moves a slice from the HA column to the A⁻ column; adding acid moves a slice the other way. The pH reading depends on the logarithm of the height ratio, so small transfers barely move the pointer until one column is almost empty.

Real-world analogy

A buffer is like a shop with both a till float and a cash safe. Customers paying in cash (added base) have their money absorbed by the till; customers wanting change (added acid) are served from the float. The balance of the shop hardly changes until the float or the till runs out.

Real-world example

Many medicines given by injection are formulated in dilute buffers such as citrate or phosphate. If the drug solution is slightly contaminated by acidic or basic impurities during manufacture or storage, the buffer absorbs them and keeps the pH in the range where the drug is stable and comfortable to inject.

Why?

Why does the conjugate base, rather than water, react with added hydronium ions? Water is an extremely weak base compared with ethanoate. The equilibrium A⁻ + H₃O⁺ ⇌ HA + H₂O lies far to the right, so the proton transfers to the stronger base, A⁻, and ends up locked in the weak acid.

Common misconception

"A buffer keeps the pH exactly constant." A buffer reduces a pH change; it does not eliminate it. The pH does shift, by an amount set by the change in the ratio, and a buffer fails completely once one reserve is used up.

Worked example

Question: 1.00 dm³ of buffer contains 0.100 mol ethanoic acid and 0.100 mol sodium ethanoate (pKa = 4.76). Find the pH after adding 0.0100 mol HCl, and compare with adding the same acid to 1.00 dm³ of pure water.

Reasoning: Stoichiometry: A⁻ falls to 0.0900 mol; HA rises to 0.110 mol. pH = 4.76 + log(0.0900 ÷ 0.110) = 4.76 − 0.087 = 4.67. In water, [H₃O⁺] = 0.0100 mol dm⁻³, so the pH falls from 7.00 to 2.00.

Answer: The buffer changes by only 0.09 units (4.76 to 4.67); water changes by 5 units.

Quick check

1. Which member of an ammonium/ammonia buffer removes added hydroxide ions, and what does it become? Answer: The ammonium ion, NH₄⁺, which is converted into ammonia and water.

Exam focus

Examiners reward equations for both buffering reactions, with correct species, and a clear two-step calculation: stoichiometry in moles, then Henderson–Hasselbalch. State that pH changes slightly, not that it stays constant, and mention that a buffer fails when one component is exhausted.

Advanced insight

A buffer also resists pH change on dilution, because diluting both HA and A⁻ equally leaves the ratio unchanged. However, at very low total concentrations the water autoionisation and the weak acid's own dissociation become comparable to the buffer components, the simple ratio equation fails and the pH drifts towards 7.

Summary

A buffer contains a weak acid and its conjugate base. The acid removes added OH⁻ and the base removes added H₃O⁺, in reactions with very large equilibrium constants. Because pH depends on the logarithm of the ratio of the pair, the pH changes only slightly until one reserve is exhausted. Always do stoichiometry in moles before the equilibrium step.

Practice questions

1. Write the equation for the reaction of an ethanoic acid/ethanoate buffer with added NaOH. Answer: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. 2. Calculate the equilibrium constant for NH₄⁺ + OH⁻ → NH₃ + H₂O, given pKa(NH₄⁺) = 9.25. Answer: K = Ka ÷ Kw = 10^(−9.25) ÷ 10^(−14.00) = 10^4.75 ≈ 5.6 × 10⁴, so the reaction is essentially complete. 3. A buffer contains 0.050 mol HA and 0.050 mol A⁻ (pKa = 4.76). Find the pH after adding 0.010 mol NaOH. Answer: HA = 0.040 mol, A⁻ = 0.060 mol; pH = 4.76 + log(1.5) = 4.76 + 0.18 = 4.94. 4. A buffer contains 0.020 mol A⁻. Explain what happens when 0.030 mol HCl is added. Answer: All the A⁻ is used up and 0.010 mol H₃O⁺ is left in excess, so buffering fails and the pH falls sharply, set by the excess strong acid.