Buffers

What is a buffer solution?

A buffer is a system that minimises pH changes when small amounts of acid (H+) or base (OH–) are added. In H432 you focus on acidic buffers (weak acid + conjugate base)

How does a buffer minimise pH change?

It contains a weak acid (HA) and its conjugate base (A–). Added H+ is removed by A–: A– + H+ → HA. Added OH– is removed by HA: HA + OH– → A– + H2O. Because both partners are present in appreciable amounts, the ratio [A–]/[HA] changes only slightly, so pH changes only slightly.

Making a buffer: weak acid + its salt

Mix a weak acid (e.g., ethanoic acid, CH3COOH) with a soluble salt providing its conjugate base (e.g., sodium ethanoate, CH3COONa). The salt fully dissociates to give A–; together, HA and A– form the buffer.

Making a buffer: partial neutralisation (excess weak acid + strong alkali)

Add a limiting amount of strong alkali (e.g., NaOH) to an excess of a weak acid (HA). Reaction: HA + OH– → A– + H2O. You end with a mixture of unreacted HA (excess) and the newly formed A–: a buffer.

Roles in the buffer pair (who does what?)

• Weak acid (HA): removes added base via HA + OH– → A– + H2O.

• Conjugate base (A–): removes added acid via A– + H+ → HA.

• This dual removal keeps the [A–]/[HA] ratio—and therefore pH—nearly constant.

Buffer pH: key equations you need

• Acid dissociation: Ka = [H+][A–]/[HA].

• Rearrangement for buffers: [H+] ≈ Ka × ([HA]/[A–]).

• Or Henderson–Hasselbalch form: pH = pKa + log10([A–]/[HA]).

Using concentrations vs moles

• For a single solution, you can use concentrations.

• If you have volumes and amounts, use moles first (because both species are in the same total volume, the [A–]/[HA] ratio equals n(A–)/n(HA)).

Worked example (salt + acid)

• 0.200 mol dm–3 CH3COOH mixed with 0.100 mol dm–3 CH3COONa in the same volume. pKa(ethanoic acid) ≈ 4.76.

• Ratio [A–]/[HA] = 0.100/0.200 = 0.50.

• pH = 4.76 + log10(0.50) = 4.76 – 0.30 ≈ 4.46

Worked example (partial neutralisation)

• Mix 50.0 cm3 of 0.200 mol dm–3 CH3COOH with 25.0 cm3 of 0.200 mol dm–3 NaOH.

• Moles HA = 0.0500×0.200 = 0.0100 mol. Moles OH– = 0.0250×0.200 = 0.00500 mol.

• After reaction: HA left = 0.0100 – 0.00500 = 0.00500 mol; A– formed = 0.00500 mol.

• Ratio n(A–)/n(HA) = 1, so pH = pKa ≈ 4.76.

Approximations and when they hold

For weak acids/buffers, we assume [HA] and [A–] don’t change much upon equilibrium shift. This is valid when Ka is small and both components are present in reasonably high, similar amounts.

Effect of dilution on buffer pH

Diluting a buffer changes [HA] and [A–] by the same factor, so their ratio—and therefore pH—remains almost unchanged. Buffer capacity decreases, though, so it resists pH change less well.

Buffer capacity (qualitative only)

Capacity is greater when the absolute amounts of HA and A– are larger and when [A–] ≈ [HA]. It’s not a number you need to calculate in H432, but it helps explain why some buffers “hold” pH better.

Blood buffer system: components

The main blood buffer is carbonic acid/hydrogencarbonate: H2CO3/HCO3–, often represented as CO2(aq)/HCO3– because dissolved CO2 and H2CO3 interconvert rapidly.

Blood buffer: response to added acid

HCO3– removes H+: HCO3– + H+ → H2CO3 → CO2 + H2O (carbonic anhydrase catalyses the interconversion; CO2 is exhaled). This limits the fall in pH.

Blood buffer: response to added base

H2CO3 removes OH–: H2CO3 + OH– → HCO3– + H2O. The equilibrium shifts to make a little more H2CO3 from CO2 + H2O, buffering the rise in pH.

Blood pH and the ratio idea

pH ≈ 7.40 is maintained when the ratio [HCO3–]/[H2CO3] is about 20:1 at body temperature (pKa ≈ 6.1 for this system). The lungs (CO2 removal) and kidneys (HCO3– control) adjust this ratio.

A buffer is made so that [A–] = 3[HA]; pKa = 4.74. What is the pH?

pH = pKa + log10([A–]/[HA]) = 4.74 + log10(3) = 4.74 + 0.477 ≈ 5.22.

30.0 cm3 of 0.200 mol dm–3 ethanoic acid is partially neutralised with 12.0 cm3 of 0.250 mol dm–3 NaOH. Find the pH (pKa = 4.76).

• Moles HA = 0.0300 × 0.200 = 0.00600 mol.

• Moles OH– = 0.0120 × 0.250 = 0.00300 mol.

• After reaction: HA left = 0.00600 − 0.00300 = 0.00300 mol; A– formed = 0.00300 mol.

• Ratio n(A–)/n(HA) = 1, so pH = pKa = 4.76.

One sentence each: how does a buffer respond to added H+ and to added OH–?

• Added H+: the conjugate base A– removes H+ to form HA, limiting the fall in pH.

• Added OH–: the weak acid HA removes OH– to form A– and H2O, limiting the rise in pH