Response of a Buffer to Strong Acid

Introduction

A buffer earns its name when something is added to it. Pour a small amount of hydrochloric acid into pure water and the pH can crash by several units; pour the same acid into a well-made buffer and the pH barely moves. To predict that small movement quantitatively you need a disciplined method: first let the added strong acid react completely with the buffer's base, keeping track in moles, and only then recalculate the pH from the new composition. This page builds that two-stage routine.

Core explanation

Two stages, not one. Adding strong acid to a buffer involves two very different processes. The first is a neutralisation that goes essentially to completion:

H₃O⁺ + A⁻ → HA + H₂O

Its equilibrium constant is 1/Ka, which for acetic acid (Ka = 1.74 × 10⁻⁵) is about 6 × 10⁴. Such a large constant means that, as long as A⁻ is in excess, virtually every added H₃O⁺ is converted into HA. The second stage is the ordinary weak-acid equilibrium HA ⇌ H⁺ + A⁻, which re-establishes itself around the new amounts and sets the pH.

Why moles, not concentrations. When a solution of acid is added, the volume changes. Amounts in moles are conserved through mixing, whereas concentrations are not. By working in moles you avoid having to recalculate every concentration after the addition. Because both HA and A⁻ share the same final volume, the ratio of their concentrations equals the ratio of their amounts:

pH = pKa + log(n(A⁻) / n(HA))

So the volume cancels in the Henderson–Hasselbalch expression, provided both species are in the same solution.

The mole ledger. Set up a table with rows for "before", "change" and "after", and columns for H₃O⁺ (added), A⁻ and HA. The strong acid is normally the limiting reagent. Its amount is subtracted from A⁻ and added to HA. The "after" row then feeds straight into the Henderson–Hasselbalch equation.

A benchmark calculation. Take 1.00 L of buffer containing 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻ (pH = pKa = 4.76). Add 0.010 mol HCl. After the ledger, A⁻ = 0.090 mol and HA = 0.110 mol:

pH = 4.76 + log(0.090/0.110) = 4.76 − 0.087 = 4.67

The pH falls by less than a tenth of a unit. The same 0.010 mol of HCl in 1.00 L of pure water would give [H₃O⁺] = 0.010 mol dm⁻³ and pH 2.00, a fall of five units from 7.00.

The condition for success. The method works only while A⁻ remains in clear excess. If the added acid approaches or exceeds the amount of A⁻, the buffer is being exhausted, and a different calculation is required.

Formulae

Neutralisation: n(A⁻)after = n(A⁻)before − n(H⁺)added; n(HA)after = n(HA)before + n(H⁺)added. Revised pH: pH = pKa + log(n(A⁻)after / n(HA)after).

Step-by-step reasoning

1. Convert every quantity to moles: n = c × V for each solution. 2. Write the neutralisation equation H₃O⁺ + A⁻ → HA + H₂O. 3. Build the ledger: subtract the added acid from A⁻, add it to HA. 4. Check that A⁻ is still present in a sensible amount. 5. Apply pH = pKa + log(n(A⁻)/n(HA)) using the "after" amounts. 6. Compare with the original pH to state the change.

Visual explanation

Picture a balance with A⁻ on one pan and HA on the other. Each added H₃O⁺ lifts one particle from the A⁻ pan and drops it onto the HA pan. Because the pans hold many particles, moving a few tilts the beam only slightly, and the pH needle, which reads the logarithm of the ratio, moves only a little.

Real-world analogy

A buffer resembles a car park with a barrier and a large queue of spare spaces. Each arriving car (an H₃O⁺ ion) simply takes a space (converts A⁻ into HA). The street outside stays clear until the spaces run out, and only then does traffic spill onto the road.

Real-world example

Acid-rain studies measure the buffering of lakes. Lakes on limestone contain dissolved hydrogencarbonate, which consumes incoming H₃O⁺ in exactly this stoichiometric way, so their pH changes little. Lakes on granite contain little base, and the same acid input lowers their pH sharply.

Why?

Why is the pH change so small? The pH depends on the logarithm of a ratio. Moving 0.010 mol between two reservoirs of about 0.1 mol alters the ratio from 1.00 to about 0.82, and the logarithm of 0.82 is only −0.087, so the pH barely shifts.

Common misconception

Students often add the H₃O⁺ from HCl to the equilibrium [H⁺] and then take −log. This is wrong: the strong acid does not stay as free H₃O⁺, because it is consumed almost completely by A⁻ before the weak-acid equilibrium re-establishes itself.

Worked example

Question: A 500 cm³ buffer contains 0.200 mol dm⁻³ NH₃ and 0.150 mol dm⁻³ NH₄Cl (pKa of NH₄⁺ = 9.25). Calculate the pH before and after adding 20.0 cm³ of 0.500 mol dm⁻³ HCl.

Reasoning: n(NH₃) = 0.200 × 0.500 = 0.100 mol; n(NH₄⁺) = 0.150 × 0.500 = 0.075 mol. Initial pH = 9.25 + log(0.100/0.075) = 9.25 + 0.125 = 9.37. Added acid: n(H⁺) = 0.500 × 0.0200 = 0.0100 mol. The base here is NH₃: H₃O⁺ + NH₃ → NH₄⁺ + H₂O. After: NH₃ = 0.090 mol, NH₄⁺ = 0.085 mol. pH = 9.25 + log(0.090/0.085) = 9.25 + 0.025 = 9.27.

Answer: The pH falls from 9.37 to 9.27, a change of only 0.10.

Quick check

1. In a buffer of HA and A⁻, which species is consumed when strong acid is added, and which increases? Answer: The conjugate base A⁻ is consumed and the weak acid HA increases by the same number of moles.

Exam focus

Examiners reward a clear ledger in moles, the correct neutralisation equation and the explicit statement that the ratio of amounts equals the ratio of concentrations. Always give the pH to two decimal places and state the direction of change; a pH rise after adding acid signals an error.

Advanced insight

The complete-reaction assumption is excellent while n(A⁻) after reaction is much larger than the equilibrium [H₃O⁺] times the volume. Near exhaustion the dissociation of HA itself becomes significant, and a full treatment combining the mass balance and charge balance is needed rather than the simple Henderson–Hasselbalch ratio.

Summary

When strong acid is added to a buffer, it first reacts completely with the conjugate base. A mole ledger tracks this: A⁻ falls and HA rises by the amount of acid added. The revised pH then follows from pKa + log(n(A⁻)/n(HA)), with the volume cancelling. The resulting change is usually a small fraction of a pH unit, whereas unbuffered water would change by several units.

Practice questions

1. A 1.00 L buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻ (pKa 4.76). Calculate the pH after adding 0.020 mol HCl. Answer: A⁻ = 0.080 mol, HA = 0.120 mol; pH = 4.76 + log(0.080/0.120) = 4.76 − 0.18 = 4.58. 2. Explain why moles rather than concentrations are used in the ledger. Answer: Amounts are conserved when solutions are mixed, whereas concentrations change with volume; both buffer species share one final volume, so the volume cancels in the ratio. 3. Write the equation for the reaction when HCl is added to an NH₃/NH₄⁺ buffer. Answer: H₃O⁺ + NH₃ → NH₄⁺ + H₂O, which goes essentially to completion. 4. A buffer contains 0.050 mol HA and 0.030 mol A⁻. What happens if 0.040 mol HCl is added? Answer: The acid exceeds the 0.030 mol of A⁻, so the buffer is exhausted; 0.010 mol H₃O⁺ remains in excess and the pH is set mainly by that excess strong acid.