Buffer Calculations A Level Chemistry AQA

This page explains how to do buffer calculations for AQA A-level Chemistry, covering the two main question types: finding the pH of a buffer made from a weak acid and its salt, and calculating the pH after adding a small amount of strong acid or base. You’ll get the exact formulas, worked examples, common mistakes, and exam-style tips.

Formula

The pH of an acidic buffer is calculated using the Henderson–Hasselbalch equation, which is derived from the Ka expression for a weak acid:

pH = pKa + log₁₀([A⁻]/[HA])

where:

• pH – the buffer’s pH (no unit)
• pKa = –log₁₀(Ka), where Ka is the acid dissociation constant (mol dm⁻³)
• [A⁻] – concentration of the salt (conjugate base) in mol dm⁻³
• [HA] – concentration of the weak acid in mol dm⁻³

For a buffer made by mixing a weak acid with its sodium or potassium salt, the salt is fully dissociated, so [A⁻] equals the salt concentration. The weak acid is only slightly dissociated, so [HA] is taken as the initial acid concentration.

Worked Example

Using the default values from the calculator: [HA] = 0.10 mol dm⁻³, [A⁻] = 0.10 mol dm⁻³, Ka = 1.7 × 10⁻⁵ mol dm⁻³.

1. Calculate pKa: pKa = –log₁₀(1.7 × 10⁻⁵) = 4.77
2. Calculate the log term: log₁₀([A⁻]/[HA]) = log₁₀(0.10/0.10) = log₁₀(1) = 0
3. Add them: pH = 4.77 + 0 = 4.77

The calculator gives pH = 4.77 (rounded to 2 decimal places). When the acid and salt concentrations are equal, the buffer pH equals pKa.

How Buffer Calculations Work in AQA

In AQA A-level Chemistry (specification 7405), buffer calculations appear in Physical Chemistry (3.1.12). You need to know two types:

• Making a buffer from a weak acid and its salt: you are given concentrations and Ka, and you find pH using the Henderson–Hasselbalch equation.
• Adding strong acid or alkali to a buffer: you first calculate the new concentrations of HA and A⁻ after the neutralisation reaction, then use the same equation.

For the second type, the strong acid reacts with the salt (A⁻) to form HA, decreasing [A⁻] and increasing [HA]. The strong alkali reacts with HA to form A⁻, increasing [A⁻] and decreasing [HA]. The change in moles is equal to the moles of strong acid or base added.

AQA exam questions often give the buffer made by mixing a weak acid with a solution of its salt, or by adding solid salt to the acid. You must convert all volumes to dm³ and use moles, not concentrations, when dealing with additions.

How to Choose the Right Approach

Read the question carefully. If it asks for the pH of a buffer already made, use the Henderson–Hasselbalch equation directly. If it asks what happens after adding acid or base, follow these steps:

1. Write the neutralisation equation (e.g., H⁺ + A⁻ → HA).
2. Calculate moles of H⁺ or OH⁻ added (concentration × volume in dm³).
3. Subtract from the relevant species: added H⁺ reduces moles of A⁻ and increases moles of HA; added OH⁻ reduces moles of HA and increases moles of A⁻.
4. Divide new moles by the total volume (in dm³) to get new concentrations.
5. Plug into pH = pKa + log₁₀([A⁻]/[HA]).

If the question gives masses of solid salt, convert to moles using Mr first. If it gives volumes of two solutions mixed, the total volume is the sum of the two volumes.

Common Mistakes and How to Avoid Them

• Forgetting to convert cm³ to dm³: divide by 1000. A common trap is using cm³ in the concentration formula.
• Using [HA] after addition incorrectly: when strong acid is added, [HA] increases, not decreases. Write the reaction to see which species is consumed.
• Ignoring the volume change: when two solutions are mixed, the total volume changes. Always use total volume when calculating new concentrations.
• Rounding too early: keep at least 3 significant figures in intermediate steps. Round only the final pH to 2 decimal places as AQA expects.
• Forgetting that strong acids and bases fully dissociate: the moles of H⁺ or OH⁻ added equals the moles of strong acid/base used.
• Using the wrong Ka: check the units – Ka is usually given in mol dm⁻³. If you are given pKa, convert to Ka using Ka = 10^(–pKa).

Limits of the Buffer Formula

The Henderson–Hasselbalch equation assumes that the weak acid is only slightly dissociated and that the salt is fully dissociated. This holds when the concentrations are reasonably high (above about 0.01 mol dm⁻³) and when the buffer pH is within about 1 unit of pKa. If the buffer is very dilute or the pH is far from pKa, the approximation breaks down because the dissociation of the acid becomes significant.

Also, the equation gives the pH of an ideal buffer. In reality, activity coefficients and temperature affect pH, but AQA ignores these at A-level. The equation works best for buffers with pH within ±1 of pKa; outside that, the buffer capacity is low and the pH changes rapidly with small additions.

Frequently Asked Questions

Do I need to memorise the Henderson–Hasselbalch equation for AQA?

Yes, it is not provided in the data booklet. You must know it and be able to derive it from the Ka expression. Practice writing it from Ka = [H⁺][A⁻]/[HA] by taking logs.

What is the difference between buffer capacity and buffer pH?

Buffer pH is the actual pH value, calculated from pKa and the ratio of salt to acid. Buffer capacity is how much strong acid or base the buffer can neutralise before the pH changes significantly. Capacity depends on the absolute concentrations, not just the ratio.

How do I calculate pH after adding NaOH to a buffer?

NaOH reacts with the weak acid HA: OH⁻ + HA → A⁻ + H₂O. Calculate moles of OH⁻ added, subtract from moles of HA, add to moles of A⁻, then divide by total volume and use the Henderson–Hasselbalch equation.

Why do we ignore the dissociation of water in buffer calculations?

In a buffer, the concentration of H⁺ from the weak acid is much higher than from water (which is 10⁻⁷ mol dm⁻³). The contribution from water is negligible, so it is ignored. This is valid for pH values between about 4 and 10.

Can a buffer be made from a weak base and its salt?

Yes, but AQA focuses on acidic buffers. For a basic buffer (e.g., ammonia and ammonium chloride), you use the Kb of the weak base and a similar equation: pOH = pKb + log₁₀([salt]/[base]), then pH = 14 – pOH.

What if the salt is not fully dissolved?

In AQA questions, salts are assumed to be fully soluble and fully dissociated. If a solubility limit is given, you would need to check that the salt concentration does not exceed it, but this is not tested at A-level.

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in c_ha | Concentration of weak acid (HA) | mol dm⁻³ | 0.10 | 0
in c_a | Concentration of salt (A⁻) | mol dm⁻³ | 0.10 | 0
in ka | Acid dissociation constant (Ka) | mol dm⁻³ | 1.7e-5 | 0
out ph | pH of buffer | - | -log10(ka * c_ha / c_a) | 2
out ratio | [A⁻]/[HA] ratio | - | c_a / c_ha | 3
out poh | pOH | - | 14 - (-log10(ka * c_ha / c_a)) | 2