Part 4: Buffers: action, calculation and preparation

Explain which component removes added acid or alkali, calculate the resulting ratio and design a buffer of a chosen pH.

A buffer contains two useful reservoirs

A buffer resists large pH changes when small quantities of acid or alkali are added. An acidic buffer contains appreciable amounts of a weak acid HA and its conjugate base A⁻, usually supplied by a soluble salt. A basic buffer contains a weak base B and its conjugate acid BH⁺, such as NH₃ and NH₄Cl.

When acid is added to an HA/A⁻ buffer, A⁻ + H⁺ → HA removes most of the added H⁺. When alkali is added, HA + OH⁻ → A⁻ + H₂O removes most of the added OH⁻. Each reaction converts one buffer component into the other; large initial amounts mean a small addition changes their ratio only slightly.

The equilibrium explanation and the removal equations should agree. Removing H⁺ with OH⁻ allows more HA to dissociate. Do not say that the salt “releases H⁺” to oppose added acid, or that a buffer keeps pH exactly constant. If an addition consumes essentially all of one component, effective buffering fails.

For NH₃/NH₄⁺, added H⁺ reacts with NH₃ to make NH₄⁺; added OH⁻ reacts with NH₄⁺ to give NH₃ and water. You can calculate its pH with the acid dissociation constant of NH₄⁺, provided you put NH₄⁺ in the acid denominator and NH₃ in the conjugate-base numerator.

Rearrange Ka instead of using the pure-acid shortcut

From Ka = [H⁺][A⁻]/[HA], [H⁺] = Ka[HA]/[A⁻]. The equivalent logarithmic form is pH = pKa + log₁₀([A⁻]/[HA]). These use the equilibrium buffer component concentrations, approximated by amounts after mixing or neutralisation when those reservoirs are large.

Mix 30.0 cm³ of 0.200 mol dm⁻³ HA with 45.0 cm³ of 0.200 mol dm⁻³ NaA. Their amounts are 0.00600 and 0.00900 mol respectively. Since they share a final volume, [HA]/[A⁻] = n(HA)/n(A⁻) = 2/3. For Ka = 1.80 × 10⁻⁵, [H⁺] = 1.20 × 10⁻⁵ mol dm⁻³ and pH = 4.92.

If a solid salt is used, find its amount from mass divided by the molar mass of the entire formula, including waters of crystallisation if present. An ion concentration can determine the amount of salt needed, but the mass of an ion alone is not the mass weighed. This distinction was relevant to Pearson 9CH0/03 June 2023 Q7(b)(ii).

Simple dilution leaves the HA/A⁻ ratio approximately unchanged, so pH changes little within the buffer approximation. It reduces both concentrations and the capacity of a given volume of buffer to absorb further acid or alkali. At extreme dilution the approximations and neglect of water cease to hold.

Worked example: react first, then equilibrate

A buffer is made from 50.0 cm³ of 0.200 mol dm⁻³ HA and 30.0 cm³ of 0.200 mol dm⁻³ NaOH. Initial n(HA) = 0.0100 mol and n(OH⁻) = 0.00600 mol. The essentially complete reaction HA + OH⁻ → A⁻ + H₂O leaves 0.00400 mol HA and forms 0.00600 mol A⁻.

Use these remaining amounts, not 0.0100 mol HA: [H⁺] = 1.80 × 10⁻⁵ × 0.00400/0.00600 = 1.20 × 10⁻⁵ mol dm⁻³, giving pH 4.92. Both buffer components are present, which validates the choice of model. If OH⁻ had equalled or exceeded initial HA, this would not be a two-component buffer.

Now add 0.00100 mol NaOH to that buffer. The acid amount falls to 0.00300 mol and the base amount rises to 0.00700 mol. [H⁺] = 1.80 × 10⁻⁵ × 0.00300/0.00700 = 7.71 × 10⁻⁶ mol dm⁻³; pH = 5.11. The pH rises by about 0.19, demonstrating resistance rather than perfect constancy.

For addition of strong acid, reverse the amount changes: A⁻ decreases and HA increases. First compare added moles with buffer capacity. If a component is exhausted, calculate excess strong acid or alkali, or the appropriate remaining weak-species equilibrium, instead of forcing negative numbers into the buffer equation.

Design a specified pH and a practical preparation

For a target pH, first choose a conjugate pair with pKa reasonably close to that target so that both components are appreciable. Rearranging gives [A⁻]/[HA] = 10^(pH − pKa) = Ka/10⁻ᵖᴴ. The ratio determines pH; an additional concentration or total-amount requirement determines capacity and fixes a unique recipe.

For Ka = 1.80 × 10⁻⁵ and target pH 5.10, [A⁻]/[HA] = 2.266. To make 250.0 cm³ with total formal buffer concentration [HA] + [A⁻] = 0.200 mol dm⁻³, solve [HA](1 + 2.266) = 0.200. Thus [HA] = 0.0612 and [A⁻] = 0.1388 mol dm⁻³.

Required amounts are 0.01531 mol HA and 0.03469 mol A⁻. With separate 1.00 mol dm⁻³ stock solutions, measure 15.31 cm³ acid and 34.69 cm³ salt solution using suitable calibrated apparatus, transfer quantitatively to a 250.0 cm³ volumetric flask, dilute to the mark and mix. Verify pH with a calibrated meter at the intended temperature; real activity effects can shift a measured value from an ideal concentration calculation.

An alternative is to start with 0.0500 mol HA and neutralise 0.03469 mol using a standard strong base, leaving the same 0.01531:0.03469 ratio, before dilution to 250.0 cm³. Avoid adjusting a buffer with large unrecorded additions, which can change its intended capacity.

Carbonic acid and hydrogencarbonate in blood

The carbonic acid–hydrogencarbonate equilibrium is H₂CO₃ ⇌ H⁺ + HCO₃⁻. Hydrogencarbonate removes added H⁺ by forming H₂CO₃; carbonic acid reacts with added OH⁻ to form HCO₃⁻ and water. These processes limit changes in hydrogen-ion concentration.

Carbonic acid is also linked to dissolved carbon dioxide and water: CO₂ + H₂O ⇌ H₂CO₃. Gas exchange and regulation of hydrogencarbonate make blood an open regulated system, so it should not be treated as just a sealed school buffer flask. Other buffer systems also contribute. The chemistry required here is the role of the H₂CO₃/HCO₃⁻ pair, not a medical interpretation of a blood-pH result.

The buffer region on a weak acid–strong base titration curve is the same chemistry in a changing mixture: increasing added OH⁻ gradually replaces HA by A⁻. Near half-equivalence, comparable amounts provide resistance to additions in either direction, and pH ≈ pKa gives a useful experimental reference point.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources