Buffer Calculations A Level Chemistry OCR

This page explains how to work through buffer calculations for OCR A Level Chemistry (A and B specifications). You enter the concentrations and volumes of the weak acid, its salt, or the weak base and its salt, and the calculator returns the pH of the buffer solution using the Henderson–Hasselbalch equation. It also covers the key steps for calculating pH after adding small amounts of strong acid or strong base, which is a common exam requirement.

Formula

The pH of an acidic buffer is given by the Henderson–Hasselbalch equation:

pH = pKa + log₁₀([salt] / [acid])

where:

• pKa = –log₁₀(Ka), unitless (Ka has units mol dm⁻³)
• [salt] = concentration of the conjugate base (e.g., sodium ethanoate) in mol dm⁻³
• [acid] = concentration of the weak acid (e.g., ethanoic acid) in mol dm⁻³

For a basic buffer (weak base + its salt), the equivalent form is:

pOH = pKb + log₁₀([salt] / [base]) then pH = 14 – pOH (at 25 °C).

The equation works because the weak acid is mostly undissociated, so its concentration stays close to the initial value, and the salt provides a reservoir of the conjugate base.

Worked Example

Using the default values from the calculator: Ka = 1.74 × 10⁻⁵ mol dm⁻³, [acid] = 0.10 mol dm⁻³, [salt] = 0.10 mol dm⁻³.

1. Calculate pKa: pKa = –log₁₀(1.74 × 10⁻⁵) = 4.76
2. Calculate the ratio [salt]/[acid] = 0.10 / 0.10 = 1.00
3. log₁₀(1.00) = 0
4. pH = 4.76 + 0 = 4.76

The calculator gives pH = 4.76, matching the step-by-step result. When the acid and salt concentrations are equal, the buffer pH equals the pKa.

How Buffer Calculations Work in OCR Exams

OCR A Level Chemistry (both the A and B specifications) expects you to be able to:

• Define a buffer solution as a system that resists changes in pH when small amounts of acid or base are added.
• Explain how an acidic buffer (weak acid + its sodium/potassium salt) works in terms of equilibrium shifts.
• Calculate the pH of a buffer from given Ka, acid concentration, and salt concentration.
• Calculate the pH after adding a small volume of strong acid or strong base to a buffer.
• Calculate the mass of salt needed to make a buffer of a target pH.

Questions often give you the Ka value, the concentration of the weak acid, and the concentration of the salt, and ask for the pH. Sometimes you are given the pH and asked to find the ratio of salt to acid, or the mass of salt to add.

How to Choose the Correct Equation

First identify whether the buffer is acidic (weak acid + salt of that acid) or basic (weak base + salt of that base).

• Acidic buffer (e.g., ethanoic acid + sodium ethanoate): use pH = pKa + log([salt]/[acid]).
• Basic buffer (e.g., ammonia + ammonium chloride): use pOH = pKb + log([salt]/[base]), then pH = 14 – pOH.

If you are given Kb for the weak base, convert to pKb first: pKb = –log₁₀(Kb). If you are given Ka for the conjugate acid of the base (e.g., NH₄⁺), then use the acidic buffer equation with the ammonium ion as the acid and ammonia as the salt (conjugate base).

Remember that for a buffer made by mixing a weak acid and its salt, the concentrations used in the equation are the equilibrium concentrations, but because the acid is weak, you can assume they are the same as the initial concentrations. This assumption is valid when the acid is weak (Ka < 10⁻³) and the concentration is not extremely dilute.

Common Mistakes and How to Avoid Them

Students frequently lose marks on buffer calculations due to these errors:

• Using the wrong pKa/pKb: Check whether the question gives Ka for the acid or Kb for the base. For basic buffers, use pKb, not pKa.
• Forgetting to convert volumes: When mixing solutions, concentrations change. Use the total volume after mixing to find the new concentrations of acid and salt.
• Ignoring the reaction with added strong acid/base: When a small amount of strong acid (H⁺) is added to an acidic buffer, it reacts with the conjugate base (salt) to form the weak acid. So [salt] decreases and [acid] increases by the same amount. For strong base (OH⁻), [acid] decreases and [salt] increases.
• Rounding too early: Keep at least 3 significant figures in intermediate steps. Round only the final pH to 2 decimal places as OCR typically does.
• Assuming pH = pKa: This is only true when [salt] = [acid]. If the ratio is 10:1, pH = pKa + 1; if 1:10, pH = pKa – 1.

Another common misconception is that adding a strong acid to a buffer changes the pH dramatically. In fact, the buffer resists change, but the pH does shift slightly. The calculation accounts for the stoichiometric reaction first, then recalculates the new ratio.

Step-by-Step for Adding Strong Acid or Base

When a question asks for the pH after adding a small volume of strong acid (e.g., HCl) or strong base (e.g., NaOH) to a buffer, follow these steps:

1. Calculate the moles of weak acid and salt present in the buffer before addition.
2. Calculate the moles of H⁺ or OH⁻ added (concentration × volume in dm³).
3. For H⁺ addition: subtract from salt moles, add to acid moles. For OH⁻ addition: subtract from acid moles, add to salt moles.
4. Divide the new moles by the total volume (buffer volume + added volume) to get new concentrations.
5. Plug into the Henderson–Hasselbalch equation to find the new pH.

Example: 100 cm³ of a buffer containing 0.10 mol dm⁻³ ethanoic acid and 0.10 mol dm⁻³ sodium ethanoate (Ka = 1.74 × 10⁻⁵). Add 1.0 cm³ of 1.0 mol dm⁻³ HCl.

• Initial moles acid = 0.10 × 0.100 = 0.010 mol; salt = 0.010 mol.
• Moles H⁺ added = 1.0 × 0.001 = 0.001 mol.
• New acid moles = 0.010 + 0.001 = 0.011 mol; new salt moles = 0.010 – 0.001 = 0.009 mol.
• Total volume = 0.101 dm³; [acid] = 0.109 mol dm⁻³; [salt] = 0.089 mol dm⁻³.
• pH = 4.76 + log(0.089/0.109) = 4.76 – 0.088 = 4.67.

The pH drops only slightly from 4.76 to 4.67, demonstrating the buffer action.

Limits of the Henderson–Hasselbalch Equation

The equation is an approximation. It assumes that the weak acid is so weak that its dissociation is negligible, and that the salt dissociates completely. It works well when:

• The acid is weak (Ka < 10⁻³).
• The concentrations are not extremely dilute (above ~10⁻³ mol dm⁻³).
• The ratio [salt]/[acid] is between 0.1 and 10 (pH within ±1 of pKa).

For very dilute buffers or very strong acids, the approximation fails and you would need to solve the full equilibrium expression. At A Level, you are expected to use the Henderson–Hasselbalch form, but you should be aware of these limitations.

Frequently Asked Questions

Do I need to use concentrations or moles in the Henderson–Hasselbalch equation? Use concentrations (mol dm⁻³). However, if the acid and salt are in the same solution (same total volume), the volume cancels, so you can use moles directly. If you mix two different solutions, you must calculate the new concentrations using the total volume.

What if the buffer is made from a weak base and its salt? Use the analogous equation pOH = pKb + log([salt]/[base]), then convert to pH using pH = 14 – pOH at 25 °C. For example, ammonia (NH₃) with ammonium chloride (NH₄Cl): pKb for ammonia is 4.75, so if [salt] = [base], pOH = 4.75 and pH = 9.25.

How do I find the mass of salt needed to make a buffer of a specific pH? Rearrange the equation to solve for [salt]: [salt] = [acid] × 10^(pH – pKa). Then multiply by the volume (in dm³) to get moles, and by the molar mass to get mass in grams. For example, to make 250 cm³ of ethanoic acid buffer at pH 5.00 with [acid] = 0.10 mol dm⁻³ and pKa = 4.76: [salt] = 0.10 × 10^(0.24) = 0.174 mol dm⁻³; moles = 0.174 × 0.250 = 0.0435 mol; mass of sodium ethanoate (Mr = 82.0) = 3.57 g.

Why does adding a small amount of strong acid not change the pH much? The added H⁺ ions react with the conjugate base (salt) to form the weak acid, which is only slightly dissociated. This removes the added H⁺ from solution, so the pH change is small. The buffer works best when the concentrations of acid and salt are comparable and not too low.

What is the difference between a buffer and a neutralisation? A buffer is a mixture of a weak acid and its conjugate base (or weak base and conjugate acid) that resists pH change. Neutralisation is a reaction between an acid and a base that produces a salt and water. A buffer is not a neutralisation product; it is a specific equilibrium system.

Does the temperature affect buffer pH? Yes. The value of Ka (and pKa) changes with temperature, so the pH of a buffer changes with temperature. The Henderson–Hasselbalch equation uses the Ka at a given temperature. At A Level, you usually work at 25 °C, where Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶.

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in ka | Acid dissociation constant (Ka) | mol dm⁻³ | 1.74e-5 | 0
in ca | Concentration of weak acid | mol dm⁻³ | 0.10 | 0
in cs | Concentration of salt (conjugate base) | mol dm⁻³ | 0.10 | 0
out ph | pH of buffer | - | -log10(ka) + log10(cs / ca) | 2
out ratio | Salt : Acid ratio | - | cs / ca | 2