Buffer Calculations A Level Chemistry Guide

This guide explains how to calculate the pH of buffer solutions for A level chemistry, covering the key equations, worked examples, and common pitfalls. You will learn how to use the Henderson–Hasselbalch equation and the Ka expression to find the pH of acidic and basic buffers, and how to handle dilution and addition of strong acids or bases.

Formula

The pH of an acidic buffer is calculated using the Henderson–Hasselbalch equation:

pH = pKa + log10([salt] / [acid])

where pKa = –log10(Ka), [salt] is the concentration of the conjugate base (usually from a salt like sodium ethanoate), and [acid] is the concentration of the weak acid. All concentrations are in mol dm⁻³. The equation is derived from the Ka expression: Ka = [H⁺][A⁻] / [HA], which rearranges to [H⁺] = Ka × [HA] / [A⁻]. Taking negative logarithms gives the form above.

Worked Example

Using the default values from the calculator: acid concentration = 0.10 mol dm⁻³, salt concentration = 0.20 mol dm⁻³, Ka = 1.74 × 10⁻⁵ mol dm⁻³.

1. First calculate pKa: pKa = –log10(1.74 × 10⁻⁵) ≈ 4.76
2. Calculate the ratio [salt]/[acid] = 0.20 / 0.10 = 2.0
3. log10(2.0) ≈ 0.301
4. pH = 4.76 + 0.301 = 5.06

The calculator gives pH = –log10(1.74e-5 × 0.10 / 0.20) = –log10(8.7e-6) ≈ 5.06, matching the step-by-step result.

How Buffer Calculations Work

A buffer resists pH change when small amounts of acid or base are added. For an acidic buffer (weak acid HA and its salt NaA), the equilibrium is HA ⇌ H⁺ + A⁻. Adding H⁺ shifts the equilibrium left, consuming the added acid; adding OH⁻ is neutralised by H⁺, which is replenished by HA dissociating. The calculation assumes that the concentrations of HA and A⁻ are large enough that the shift is negligible, so you can use the initial concentrations directly in the Henderson–Hasselbalch equation.

Choosing the Right Buffer System

For a buffer to be effective, the desired pH should be close to the pKa of the weak acid (within about 1 pH unit). For example, to buffer at pH 5, choose an acid with pKa near 5, such as ethanoic acid (pKa ≈ 4.76). The ratio [salt]/[acid] then adjusts the exact pH. If the required pH is far from any available pKa, the buffer capacity will be low and the pH will change significantly upon addition of small amounts of acid or base.

Common Mistakes and Misconceptions

• Using concentrations instead of moles when the buffer is diluted: the ratio [salt]/[acid] stays the same because both concentrations change equally, so pH is independent of dilution.
• Forgetting that the salt concentration must be the concentration of the conjugate base, not the total salt added if the salt contains more than one ionisable group.
• Confusing pKa with pH: pKa is a constant for a given acid at a given temperature, while pH depends on the ratio.
• Ignoring the change in volume when adding a strong acid or base to the buffer – you must calculate new concentrations after the addition.

Buffers After Adding Acid or Base

When a strong acid (e.g., HCl) is added to an acidic buffer, the H⁺ reacts with A⁻ to form HA. The new concentrations are: [HA] = initial [HA] + moles of H⁺ added / total volume; [A⁻] = initial [A⁻] – moles of H⁺ added / total volume. Then plug these into the Henderson–Hasselbalch equation. For a strong base (e.g., NaOH), the OH⁻ reacts with HA to form A⁻ and water, so [HA] decreases and [A⁻] increases. Always use the total final volume after mixing.

Basic Buffers and the pOH Approach

For a basic buffer (weak base B and its salt BH⁺), the same logic applies using Kb. The equation is pOH = pKb + log10([salt] / [base]), then pH = 14 – pOH (at 25 °C). For example, an ammonia buffer with [NH₃] = 0.10 M and [NH₄Cl] = 0.20 M, Kb = 1.8 × 10⁻⁵, gives pKb = 4.74, pOH = 4.74 + log10(2) = 5.04, so pH = 8.96. The calculator above is for acidic buffers; for basic buffers you would use the analogous equation with Kb.

Limitations of the Henderson–Hasselbalch Equation

The equation assumes that the acid is weak (so dissociation is negligible) and that the salt fully dissociates. It works well when the concentrations are not too dilute (above about 0.01 M) and when the ratio [salt]/[acid] is between 0.1 and 10. For very dilute buffers or extreme ratios, the approximation fails and you must solve the full equilibrium expression. Also, the equation is temperature-dependent because Ka changes with temperature, though the change is often small over typical lab temperatures.

Frequently Asked Questions

Why does dilution not change the pH of a buffer?

Dilution reduces both [acid] and [salt] by the same factor, so the ratio [salt]/[acid] remains constant. Since pH depends only on that ratio (plus pKa), the pH stays the same. This is true as long as the buffer is not diluted so much that the assumptions of the Henderson–Hasselbalch equation break down.

What is the difference between buffer capacity and buffer pH?

Buffer pH is the actual pH of the solution, determined by pKa and the ratio of salt to acid. Buffer capacity is the amount of strong acid or base that can be added before the pH changes significantly. Capacity is highest when [salt] = [acid] (ratio 1:1) and increases with higher absolute concentrations.

Do I use moles or concentrations in the Henderson–Hasselbalch equation?

You can use either moles or concentrations because the volume cancels out in the ratio. However, when adding a strong acid or base, you must work in moles first to account for the reaction, then convert to concentrations using the final total volume.

How do I calculate the pH of a buffer made from a weak base and its salt?

Use the analogous equation with Kb: pOH = pKb + log10([salt]/[base]), then pH = 14 – pOH (at 25 °C). Alternatively, convert Kb to Ka for the conjugate acid using Ka × Kb = Kw, then use the Henderson–Hasselbalch equation with the conjugate acid concentration.

Why does the Henderson–Hasselbalch equation give negative pH for very strong acids?

The equation is only valid for weak acids and bases. For strong acids, dissociation is essentially complete, so the assumption that [HA] ≈ initial concentration fails. In those cases, you must calculate pH directly from the concentration of H⁺ released by full dissociation.

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in c_acid | Acid concentration | mol dm^-3 | 0.10 | 0
in c_salt | Salt concentration | mol dm^-3 | 0.20 | 0
in ka | Acid dissociation constant Ka | mol dm^-3 | 1.74e-5 | 0
out pH | pH of buffer | - | -log10(ka * c_acid / c_salt) | 2