Response of a Buffer to Strong Base

Introduction

A buffer must defend against change in both directions. When strong base such as sodium hydroxide is added, it is the weak-acid member of the pair that does the work, donating protons to destroy the hydroxide ions. The calculation mirrors the treatment of added acid, with a mole ledger followed by a revised pH, but it also reveals a hard ceiling: once the weak acid has been used up, the buffer offers no further protection and the pH climbs steeply.

Core explanation

The neutralisation. Added hydroxide reacts with the weak acid:

OH⁻ + HA → A⁻ + H₂O

The equilibrium constant is Ka/Kw. For acetic acid this is 1.74 × 10⁻⁵ / 1.0 × 10⁻¹⁴ ≈ 1.7 × 10⁹, so the reaction is complete for all practical purposes. Each mole of OH⁻ removes one mole of HA and creates one mole of A⁻. The equilibrium [OH⁻] left behind is tiny, which is exactly why the pH changes so little.

The ledger. In moles: HA decreases by n(OH⁻) added, A⁻ increases by the same amount. The revised pH follows from:

pH = pKa + log(n(A⁻) / n(HA))

For 1.00 L of buffer containing 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻ (pH 4.76), adding 0.020 mol NaOH gives HA = 0.080 mol and A⁻ = 0.120 mol, so pH = 4.76 + log(1.5) = 4.76 + 0.18 = 4.94. The same amount of NaOH in 1.00 L of water would give pOH 1.70 and pH 12.30.

Capacity limits. The buffer can neutralise at most as many moles of OH⁻ as it contains HA. Three regimes follow:

- n(OH⁻) well below n(HA): use the Henderson–Hasselbalch ratio. - n(OH⁻) equal to n(HA): all HA has become A⁻; the solution is a solution of the weak base A⁻, and its pH follows from Kb = Kw/Ka. - n(OH⁻) greater than n(HA): excess OH⁻ remains, and it dominates the pH because the extra OH⁻ from A⁻ hydrolysis is negligible by comparison.

Asymmetric protection. A buffer's resistance to base depends on n(HA), while its resistance to acid depends on n(A⁻). A buffer with a ratio of 3:1 in favour of A⁻ copes well with acid but poorly with base. Designers therefore match the ratio not only to the target pH but also to the direction of the expected challenge.

Volume effects. Adding base as a solution increases the volume, but because HA and A⁻ share that volume, the ratio is unaffected. Volume only matters in the excess regime, where the concentration of leftover OH⁻ must be calculated from the total volume.

Formulae

n(HA)after = n(HA)before − n(OH⁻)added; n(A⁻)after = n(A⁻)before + n(OH⁻)added. Excess regime: [OH⁻] = (n(OH⁻)added − n(HA)before) / Vtotal; pH = 14.00 − pOH at 25 °C.

Step-by-step reasoning

1. Convert the buffer components and the added base to moles. 2. Compare n(OH⁻) with n(HA) to decide which regime applies. 3. If HA is in excess, update the ledger and use Henderson–Hasselbalch. 4. If OH⁻ is in excess, find the leftover moles and divide by the total volume. 5. Convert pOH to pH and compare with the starting value.

Visual explanation

Imagine a bar chart with two columns, HA and A⁻. Each dose of OH⁻ shortens the HA column and lengthens the A⁻ column by the same amount. The pH moves slowly while both columns are tall, then leaps upwards as the HA column shrinks to zero.

Real-world analogy

A buffer facing base is like a sponge soaking up spilled water. It absorbs spill after spill with no visible puddle, until it is saturated; after that, every further drop spreads straight across the table.

Real-world example

Swimming-pool water is buffered by hydrogencarbonate. When alkaline chemicals are added to adjust it, the hydrogencarbonate acts as the proton donor and moderates the pH rise. Pools with low total alkalinity show large, erratic pH swings because their capacity is small.

Why?

Why does the pH jump so sharply at the capacity limit? Before it, OH⁻ is destroyed and only the log ratio changes. After it, OH⁻ accumulates freely, and even a millimole per litre of free hydroxide corresponds to pH 11, far above any weak-acid buffer range.

Common misconception

Many learners think a buffer "absorbs" base indefinitely. In fact its protection is strictly stoichiometric: once the moles of added OH⁻ equal the moles of HA originally present, the buffer has no proton donor left and behaves like unbuffered solution.

Worked example

Question: A 250 cm³ buffer contains 0.0250 mol CH₃COOH and 0.0150 mol CH₃COO⁻ (pKa 4.76). Find the pH (a) initially, (b) after adding 0.0100 mol solid NaOH, (c) after adding 0.0300 mol solid NaOH instead. Assume no volume change.

Reasoning: (a) pH = 4.76 + log(0.0150/0.0250) = 4.76 − 0.22 = 4.54. (b) HA = 0.0150 mol, A⁻ = 0.0250 mol; pH = 4.76 + 0.22 = 4.98. (c) The 0.0300 mol OH⁻ exceeds the 0.0250 mol HA, leaving 0.0050 mol OH⁻ in 0.250 dm³: [OH⁻] = 0.020 mol dm⁻³, pOH = 1.70, pH = 12.30.

Answer: 4.54, 4.98 and 12.30; the third addition exceeds the capacity.

Quick check

1. Which component of an acetic acid/acetate buffer limits how much sodium hydroxide can be neutralised? Answer: The acetic acid (HA), because each mole of OH⁻ consumes one mole of HA.

Exam focus

Show the comparison of n(OH⁻) with n(HA) explicitly before choosing a method. Examiners often set a second addition that exceeds the capacity; applying Henderson–Hasselbalch with a negative amount of HA is a classic error that earns no credit.

Advanced insight

At exactly the capacity limit the solution contains only A⁻. For 0.100 mol dm⁻³ acetate, Kb = 5.7 × 10⁻¹⁰, so [OH⁻] = √(Kb × c) ≈ 7.6 × 10⁻⁶ mol dm⁻³ and pH ≈ 8.9. This is the equivalence point of a weak-acid titration, linking buffer exhaustion to titration curves.

Summary

Strong base added to a buffer is consumed by the weak-acid component in a complete neutralisation, OH⁻ + HA → A⁻ + H₂O. A mole ledger gives the new ratio, and the pH rises only slightly. The protection ends when the added OH⁻ equals the moles of HA; beyond that point excess hydroxide sets the pH, which rises steeply.

Practice questions

1. Write the equation for the reaction of NaOH with an NH₄⁺/NH₃ buffer. Answer: OH⁻ + NH₄⁺ → NH₃ + H₂O, which goes essentially to completion. 2. A 1.00 L buffer has 0.080 mol HA and 0.120 mol A⁻ (pKa 4.76). Calculate the pH after adding 0.040 mol NaOH. Answer: HA = 0.040 mol, A⁻ = 0.160 mol; pH = 4.76 + log 4 = 4.76 + 0.60 = 5.36. 3. Why does a buffer rich in A⁻ resist added base poorly? Answer: Resistance to base depends on the amount of HA available to donate protons; a buffer with little HA runs out quickly. 4. A 100 cm³ buffer contains 0.0040 mol HA. Calculate the pH after 0.0050 mol NaOH is added, ignoring volume change. Answer: Excess OH⁻ = 0.0010 mol in 0.100 dm³, so [OH⁻] = 0.010 mol dm⁻³, pOH = 2.00 and pH = 12.00.