Buffer Ratio and Henderson–Hasselbalch

Introduction

A buffer contains a weak acid and its conjugate base in comparable amounts. Its pH depends on the ratio of the two, which makes buffer calculations quick once the ratio is known. The difficulty in multi-step problems is that the ratio is often hidden: solutions are mixed, a strong base partly neutralises the acid, or acid is added later. This page shows how to do the mole accounting first and only then apply the Henderson–Hasselbalch equation.

Core explanation

Origin of the equation. For HA ⇌ H⁺ + A⁻, rearranging Ka = [H⁺][A⁻]/[HA] and taking −log of both sides gives

pH = pKa + log₁₀([A⁻]/[HA])

Why moles work. Both [A⁻] and [HA] are in the same solution, so they share the same volume. The ratio of concentrations equals the ratio of moles, and the total volume cancels. This is why you can work in moles throughout and ignore volumes at the final step.

Reactions come first. A strong base added to a weak acid reacts essentially completely:

HA + OH⁻ → A⁻ + H₂O

Likewise, a strong acid added to a buffer converts A⁻ into HA. These reactions go to completion before equilibrium is considered, so treat them like limiting-reagent problems: find which is in excess and what remains.

Example. Mix 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid (pKa = 4.76) with 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide.

- Moles HA = 0.0500 × 0.200 = 0.0100 mol; moles OH⁻ = 0.0250 × 0.100 = 0.00250 mol. - OH⁻ is limiting. After reaction: HA = 0.00750 mol, A⁻ = 0.00250 mol. - pH = 4.76 + log(0.00250 / 0.00750) = 4.76 − 0.48 = 4.28.

Special points. When exactly half the acid has been neutralised, [A⁻] = [HA] and pH = pKa. This half-neutralisation point is a common way to read pKa from a titration curve.

Validity. The equation assumes that the equilibrium amounts of HA and A⁻ are essentially the amounts put in, which holds when both are reasonably large compared with [H⁺] and the ratio lies between about 0.1 and 10. Outside that range the buffer is weak and a full equilibrium treatment is safer.

Formulae

pH = pKa + log₁₀(n(A⁻) / n(HA)); moles = concentration × volume (in dm³); at half-neutralisation pH = pKa.

Step-by-step reasoning

1. Convert every volume and concentration into moles. 2. Write any strong acid or strong base reaction and let it go to completion. 3. Record the moles of HA and A⁻ remaining. 4. Substitute the mole ratio into the Henderson–Hasselbalch equation. 5. Check the ratio lies between 0.1 and 10 before trusting the result.

Visual explanation

Imagine a seesaw with HA on one side and A⁻ on the other, pivoted at pKa. Adding alkali moves weight from the HA side to the A⁻ side and tips the pH up; adding acid moves weight back. Because the seesaw is logarithmic, small transfers barely tilt it when both sides are heavily loaded.

Real-world analogy

A buffer behaves like a large water tank with inflow and outflow valves. Small splashes in or out barely change the level because the tank is big. The level changes only noticeably if you add or remove an amount comparable with what is already there, which is the equivalent of exceeding buffer capacity.

Real-world example

Blood is held near pH 7.4 largely by the carbonic acid–hydrogencarbonate system. The ratio of hydrogencarbonate to dissolved carbon dioxide is roughly 20 : 1, and the lungs and kidneys adjust each component to keep that ratio, and so the pH, within narrow limits.

Why?

Why does the pH of a buffer hardly change on dilution? Dilution lowers both [HA] and [A⁻] by the same factor, so their ratio, and therefore the pH, stays the same. Only the buffer capacity falls.

Common misconception

"Use the concentrations of the original solutions in the equation." The original concentrations ignore the neutralisation reaction and the mixing. Only the amounts present after reaction belong in the ratio.

Worked example

Question: A buffer contains 0.0200 mol ethanoic acid and 0.0200 mol sodium ethanoate. 0.00100 mol hydrochloric acid is added. Find the pH before and after.

Reasoning: Before: ratio = 1, pH = pKa = 4.76. The strong acid converts A⁻ to HA: A⁻ = 0.0190 mol, HA = 0.0210 mol. After: pH = 4.76 + log(0.0190 / 0.0210) = 4.76 − 0.04 = 4.72.

Answer: pH falls from 4.76 to 4.72, a change of only 0.04.

Quick check

1. A buffer contains equal amounts of a weak acid and its conjugate base. What is its pH in terms of pKa? Answer: pH = pKa, because log(1) = 0 in the Henderson–Hasselbalch equation.

Exam focus

Write the neutralisation equation and the moles before and after reaction as a small table. Many marks are lost by skipping this step and using starting concentrations. State that the volume cancels because both species are in the same solution.

Advanced insight

Buffer capacity is greatest when [A⁻] = [HA], and it is proportional to the total buffer concentration. For precise work, activity coefficients differ between charged A⁻ and neutral HA, so the true pH of a buffer shifts slightly with ionic strength, which is why standard buffers are prepared to specified recipes.

Summary

Handle strong acid or base reactions first, in moles. Then use pH = pKa + log(n(A⁻)/n(HA)); volume cancels. At half-neutralisation pH = pKa. The equation is reliable when the ratio lies between about 0.1 and 10, and a buffer's pH changes little on dilution or on small additions.

Practice questions

1. Calculate the pH of a solution containing 0.10 mol NH₃ and 0.050 mol NH₄Cl, given pKa(NH₄⁺) = 9.25. Answer: pH = 9.25 + log(0.10 / 0.050) = 9.25 + 0.30 = 9.55. 2. What ratio of ethanoate to ethanoic acid gives pH 5.00? Answer: log ratio = 5.00 − 4.76 = 0.24, so the ratio is 10^0.24 ≈ 1.7. 3. 0.0300 mol ethanoic acid is mixed with 0.0100 mol NaOH. Find the pH. Answer: HA = 0.0200 mol, A⁻ = 0.0100 mol; pH = 4.76 + log(0.5) = 4.76 − 0.30 = 4.46. 4. A buffer is diluted tenfold with water. What happens to its pH and its capacity? Answer: The pH stays essentially the same because the ratio is unchanged, but the capacity falls to about one tenth.