Buffer Capacity

Introduction

Two buffers can have exactly the same pH and yet behave very differently. One may hold its pH after a large dose of acid, while the other collapses after a few drops. The difference is buffer capacity , a quantitative measure of how much strong acid or base a buffer can absorb. Understanding capacity lets you choose not only the ratio of a buffer but also how concentrated it must be for a given job.

Core explanation

Definition. Buffer capacity β is the amount of strong base, per unit volume, needed to raise the pH by one unit:

β = dC(b) ÷ d(pH)

The same quantity describes resistance to acid, with a sign change. A large β means the pH moves only slightly for a given addition. Units are mol dm⁻³ per pH unit.

Formula for a single pair. For a weak acid/conjugate base pair of total concentration C,

β = 2.303 × (C × Ka[H₃O⁺] ÷ (Ka + [H₃O⁺])² + [H₃O⁺] + [OH⁻])

The first term is the contribution of the buffer pair. The [H₃O⁺] and [OH⁻] terms describe the "buffering" of water itself at very low or very high pH, where strong acid or base is already present in large amounts.

Proportional to total concentration. The buffer term is directly proportional to C. A 0.10 mol dm⁻³ buffer has ten times the capacity of a 0.010 mol dm⁻³ buffer with the same ratio, even though both have the same pH. This is the most important practical fact about capacity: pH comes from the ratio, capacity from the total.

Depends on ratio too. At fixed C, the buffer term equals 2.303C × α(HA) × α(A⁻). This product is greatest when both fractions equal 0.5, that is, at pH = pKa, giving β(max) = 2.303C/4 = 0.576C. Moving away from pKa lowers capacity; the next page examines this in detail.

Practical capacity. Often it is more useful to ask a finite question: how many moles of acid or base can be added before the pH shifts by a stated amount? This is found by stoichiometry followed by Henderson–Hasselbalch, solving for the unknown addition. Because β varies with pH, the finite answer is not exactly β × ΔpH for large shifts.

Absolute limit. No buffer can absorb more strong base than the moles of HA it contains, or more strong acid than the moles of A⁻. Beyond these limits the solution behaves like an unbuffered strong acid or base.

Formulae

β = 2.303(C·Ka[H₃O⁺]/(Ka + [H₃O⁺])² + [H₃O⁺] + [OH⁻]). At pH = pKa: β(max) ≈ 0.576C. Buffer term = 2.303C·α(HA)·α(A⁻).

Step-by-step reasoning

To find how much base raises a buffer's pH by a set amount:

1. Write the starting moles of HA and A⁻. 2. Let x be the moles of OH⁻ added; new amounts are HA − x and A⁻ + x. 3. Set pKa + log((A⁻ + x)/(HA − x)) equal to the new pH. 4. Rearrange the ratio equation and solve for x. 5. Check that x is less than the moles of HA available.

Visual explanation

Plot β against pH for a 0.10 mol dm⁻³ ethanoate buffer. It forms a bell-shaped peak centred on pH 4.76, of height 0.058 mol dm⁻³ per pH unit, falling towards zero about two units either side. At the extremes the curve rises again because of free H₃O⁺ at low pH and free OH⁻ at high pH.

Real-world analogy

Buffer capacity is like the size of a dam's reservoir. Two reservoirs can have the same water level (pH), but the larger one can absorb a much bigger flood (added acid or base) before the level changes significantly.

Real-world example

Swimming-pool maintenance distinguishes pH from "total alkalinity", which is mainly the hydrogencarbonate concentration. A pool with low total alkalinity has a poor buffer capacity, so its pH swings wildly when chemicals or rain are added, even if the starting pH is correct.

Why?

Why is capacity proportional to concentration? The number of moles of acid that can be neutralised equals the moles of A⁻, and the number of moles of base equals the moles of HA. Doubling C doubles both reserves, so it takes twice as much addition to cause the same change in ratio.

Common misconception

"A buffer with a lower pH is stronger." Buffer strength (capacity) has nothing to do with the pH value itself. It depends on total concentration and on how close the pH is to pKa.

Worked example

Question: 1.00 dm³ of buffer contains 0.0500 mol HA and 0.0500 mol A⁻ (pKa = 4.76). How much NaOH raises the pH to 5.76? Repeat for a buffer ten times more dilute.

Reasoning: Need (0.0500 + x)/(0.0500 − x) = 10. So 0.0500 + x = 0.500 − 10x, giving 11x = 0.450 and x = 0.0409 mol. For the dilute buffer, every amount scales by 0.1, so x = 0.00409 mol.

Answer: 0.041 mol for the concentrated buffer; only 0.0041 mol for the dilute one.

Quick check

1. What is the maximum buffer capacity of a 0.20 mol dm⁻³ buffer made from a single conjugate pair? Answer: β(max) = 0.576 × 0.20 ≈ 0.12 mol dm⁻³ per pH unit, reached at pH = pKa.

Exam focus

Distinguish clearly between buffer pH (set by the ratio) and buffer capacity (set mainly by total concentration). Be ready to calculate the amount of acid or base needed to shift pH by a given amount, and to state the absolute limit set by the moles of each reserve.

Advanced insight

Capacities are additive. A mixture of several buffer pairs, or a polyprotic acid, has a total β equal to the sum of the individual terms. Designers exploit this to build "universal" buffers from several acids with staggered pKa values, giving a fairly flat β over a wide pH range for broad-range titrations.

Summary

Buffer capacity β is the moles of strong base per dm³ needed to change pH by one unit. For a single pair it is proportional to the total concentration C and peaks at 0.576C when pH = pKa. Ratio sets pH; total concentration sets capacity. A buffer can never neutralise more acid than its moles of A⁻, or more base than its moles of HA.

Practice questions

1. Two ethanoate buffers both have pH 4.76, one at 0.50 mol dm⁻³ and one at 0.050 mol dm⁻³ total. Compare their capacities. Answer: Capacity is proportional to C, so the 0.50 mol dm⁻³ buffer has ten times the capacity. 2. Calculate β for a 0.10 mol dm⁻³ buffer at pH = pKa, ignoring the water terms. Answer: 2.303 × 0.10 × 0.5 × 0.5 = 0.058 mol dm⁻³ per pH unit. 3. A buffer contains 0.030 mol A⁻ and 0.010 mol HA. What is the most strong base it can absorb before buffering fails? Answer: 0.010 mol, the amount of HA present. 4. How many moles of HCl lower the pH of a buffer containing 0.100 mol each of HA and A⁻ by 0.30 units? Answer: Need (0.100 − x)/(0.100 + x) = 0.50, so 0.100 − x = 0.050 + 0.50x, giving x = 0.033 mol.