Buffer capacity

The formula

• β = Δn / (V · ΔpH)

Δn is the amount of strong acid or base added, in moles, V the volume of the buffer in cubic decimetres (litres) and ΔpH the change in pH that the addition caused, taken as a positive number. β is in mol per litre per pH unit: the moles a litre of the buffer absorbs for each unit the pH moves.

A worked example

Adding 0.010 mol of sodium hydroxide to 1 dm³ of a buffer moves its pH from 4.75 to 4.85, a change of 0.10: β = 0.010 / (1 · 0.10) = 0.10 mol/(dm³·pH). Turned round, a buffer with β = 0.10 can take 0.010 mol of base per litre for every tenth of a pH unit it is allowed to drift.

When a buffer is strongest

The capacity rises with the concentration of the buffer pair, and it is largest when the acid and its conjugate base are present in equal amounts, which is where the pH equals the pKa by the Henderson–Hasselbalch equation. A buffer a pH unit away from its pKa has one component ten times the other, and the smaller one is used up first; its capacity in that direction is a fraction of the capacity at the centre.

The formula as a ratio of finite changes

Written this way β is an average over the addition. The textbook definition is the derivative, dn / (V · dpH), and the two agree when the addition is small enough that the pH moves by a few tenths at most. A large addition that pushes the pH a whole unit reports a capacity averaged over a range in which the real value was falling.

Common mistakes

• Using millimoles or millilitres without saying so. Moles and litres give the standard unit.
• Reading β as a pH. It is a quantity of acid or base per litre per pH unit, and a buffer with β = 0.10 says nothing about which pH it holds.
• Measuring far from the pKa. The number is then the capacity of a nearly spent buffer.

Frequently asked questions