Dilution of Buffers

Introduction

Take a buffer at pH 4.76 and add nine times its volume of pure water. Surprisingly, a pH meter still reads almost exactly 4.76. Dilution seems to have done nothing. Yet if you now add a drop of acid, the diluted buffer responds far more sharply than the original. Dilution preserves the ratio that sets the pH but reduces the amounts that provide protection. This page separates those two ideas and shows where the simple picture eventually breaks down.

Core explanation

The ratio survives dilution. The Henderson–Hasselbalch equation depends only on the ratio [A⁻]/[HA]. Diluting by a factor f divides both concentrations by f, so the ratio, and therefore the calculated pH, is unchanged:

pH = pKa + log([A⁻]/f ÷ [HA]/f) = pKa + log([A⁻]/[HA])

This is a key difference from solutions of a single strong or weak acid, whose pH rises on dilution.

Capacity does not survive. Buffer capacity is roughly proportional to the total concentration of the conjugate pair. A tenfold dilution leaves a tenth as many moles of HA and A⁻ in each litre, so a given amount of added acid or base alters the ratio ten times as much. The same challenge that shifted the original buffer by 0.01 pH units may shift the diluted buffer by nearly 0.1 units.

Small drift from activities. In reality the pH does change slightly. The thermodynamic Ka is defined in activities, and the activity coefficients of ions depend on ionic strength. On dilution the ionic strength falls, the activity coefficients rise towards 1, and the measured pH drifts. For acetate buffers the drift is a few hundredths of a unit upwards. For pairs involving doubly charged ions, such as H₂PO₄⁻/HPO₄²⁻, the drift is larger, often around 0.1 units for a tenfold dilution, because activity coefficients depend on the square of the charge.

Breakdown at extreme dilution. The Henderson–Hasselbalch approximation assumes that the ionisation of HA barely changes [HA] and [A⁻]. This holds when those concentrations far exceed [H₃O⁺]. At very high dilution this fails. For an acetate buffer with 1.0 × 10⁻⁴ mol dm⁻³ of each component, the equilibrium [H₃O⁺] is comparable to the component concentrations, and the true pH is noticeably higher than pKa. At still greater dilution the solution approaches the pH of water itself.

Practical message. Dilution is a convenient way to prepare working buffers from concentrated stocks without recalculating the ratio, but the working buffer must still be concentrated enough for the expected acid or base load.

Step-by-step reasoning

1. Note that dilution divides both [HA] and [A⁻] by the same factor. 2. Conclude that the ratio, and so the ideal pH, is unchanged. 3. Calculate the new concentrations to judge capacity. 4. Compare [HA] and [A⁻] with the expected [H₃O⁺] to test the approximation. 5. If they are not at least about 100 times larger, solve the full equilibrium.

Visual explanation

Draw a graph of pH against the logarithm of total buffer concentration. The line is flat across several orders of magnitude, then bends towards pH 7 at very low concentration. On a second graph, capacity falls as a straight line with the same slope as concentration.

Real-world analogy

Dilution is like shrinking a photograph. The proportions of the picture, the ratio, stay exactly the same, but the smaller print holds less detail. Shrink it far enough and the image dissolves into the blank page.

Real-world example

Laboratories often store buffers as tenfold concentrated stocks, such as "10× phosphate-buffered saline", and dilute them before use. The working solution has the intended pH, but a small pH adjustment is usually checked afterwards because of the activity drift.

Why?

Why does a single weak acid change pH on dilution but a buffer does not? For a lone weak acid, [H₃O⁺] ≈ √(Ka × c) depends on concentration. In a buffer, [H₃O⁺] = Ka × [HA]/[A⁻] depends only on a ratio.

Common misconception

"A diluted buffer is just as good because its pH is the same." The pH is the same, but the protection is not. Capacity scales with concentration, so a dilute buffer can be overwhelmed by small additions that a concentrated one would absorb easily.

Worked example

Question: Buffer P contains 0.100 mol dm⁻³ CH₃COOH and 0.100 mol dm⁻³ CH₃COO⁻. Buffer Q is P diluted tenfold. Calculate the pH of 1.00 L of each after adding 0.0010 mol HCl (pKa 4.76).

Reasoning: Both start at pH 4.76. For P: A⁻ = 0.099 mol, HA = 0.101 mol; pH = 4.76 + log(0.980) = 4.75. For Q: A⁻ = 0.0090 mol, HA = 0.0110 mol; pH = 4.76 + log(0.818) = 4.76 − 0.087 = 4.67.

Answer: P changes by 0.01 units and Q by 0.09 units: the same pH initially, but Q has one-tenth of the capacity.

Quick check

1. What happens to the pH and to the capacity of an ideal buffer when it is diluted fivefold? Answer: The pH stays essentially the same, but the capacity falls to about one-fifth.

Exam focus

State both halves of the answer: pH unchanged because the ratio is unchanged, capacity reduced because the amounts per litre are reduced. Supporting either claim with a short calculation earns full credit on most mark schemes.

Advanced insight

For the 1.0 × 10⁻⁴ mol dm⁻³ acetate buffer, solving Ka = x(1.0 × 10⁻⁴ + x)/(1.0 × 10⁻⁴ − x) gives x ≈ 1.3 × 10⁻⁵ mol dm⁻³ and pH ≈ 4.88, compared with the ideal 4.76. The ionisation of HA has shifted the ratio, showing why real buffers are rarely used below about 10⁻³ mol dm⁻³.

Summary

Diluting a buffer leaves the ratio [A⁻]/[HA] unchanged, so the pH stays nearly constant, apart from a small drift due to changing activity coefficients. Capacity, however, falls in proportion to concentration. At extreme dilution the ionisation of the weak acid and of water can no longer be ignored, and the pH moves away from the ideal value towards 7.

Practice questions

1. Explain, using the Henderson–Hasselbalch equation, why a buffer's pH is almost independent of dilution. Answer: pH = pKa + log([A⁻]/[HA]); dilution divides both concentrations by the same factor, so the ratio and hence the pH are unchanged. 2. A buffer absorbs 0.010 mol of acid per litre for a 0.10 pH change. Estimate the amount it absorbs after a fourfold dilution. Answer: About 0.0025 mol per litre, because capacity falls in proportion to concentration. 3. Why does dilution change the pH of an H₂PO₄⁻/HPO₄²⁻ buffer more than that of an acetate buffer? Answer: The doubly charged HPO₄²⁻ has an activity coefficient much more sensitive to ionic strength, so the apparent pKa shifts more as ionic strength falls. 4. State one sign that the simple buffer equation will fail for a very dilute buffer. Answer: The expected [H₃O⁺] is not negligible compared with [HA] and [A⁻], so the ionisation of HA significantly changes the ratio.