Buffer Solutions — How They Resist pH Change
Add one drop of hydrochloric acid to a glass of pure water and the pH crashes. Add the same drop to blood, and the pH barely moves. That difference is the whole idea of a buffer, and it is the reason your body, the soil in a field and half the reactions in a laboratory do not go out of control. This article explains why a buffer holds its pH — not just the formula you plug numbers into — and then works through three fully computed examples.
What a buffer actually is
A buffer solution contains two species that are chemically related but do opposite jobs:
- An acidic buffer — a weak acid plus its conjugate base, usually supplied as a salt. Example: acetic acid CH₃COOH with sodium acetate CH₃COONa.
- A basic buffer — a weak base plus its conjugate acid. Example: ammonia NH₃ with ammonium chloride NH₄Cl.
The key word is weak. A weak acid stays mostly un-ionised in solution, so a large reservoir of intact HA molecules sits there. The salt supplies a large reservoir of A⁻ ions. The solution therefore holds a big stock of a proton donor and a big stock of a proton acceptor at the same time — something a strong acid can never do, because a strong acid is fully ionised and has no undissociated molecules left in reserve.
The mechanism — where the added H⁺ and OH⁻ go
Consider the acetic acid / acetate buffer. Two reactions are waiting:
Alkali added: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
Added H⁺ is captured by the acetate ions and converted into more un-ionised acetic acid. Added OH⁻ is neutralised by the acetic acid molecules and converted into more acetate. In both cases the added ion is removed from solution almost completely, and the only thing that changes is the ratio of the two reservoirs. Since pH depends on the logarithm of that ratio, a modest change in the ratio moves the pH only slightly.
Seen through Le Chatelier's principle it is the same story. The equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts left when H⁺ is added and right when H⁺ is removed, and both reservoirs are large enough to absorb the shift.
The working formula
Here pKa = −log Ka for the weak acid, and the square brackets are the concentrations left after mixing. Because both concentrations are divided by the same total volume, the volumes cancel — you may put moles straight into the ratio. For a basic buffer the matching form is pOH = pKb + log([conjugate acid]/[base]), and then pH = 14 − pOH at 25 °C.
For acetic acid, Ka = 1.8 × 10⁻⁵, so pKa = −log(1.8 × 10⁻⁵) = 5 − 0.2553 = 4.74. Some textbooks quote pKa = 4.76, which comes from the slightly different value Ka = 1.75 × 10⁻⁵. Either is acceptable in an exam — just use the value given in your own question paper and stay consistent.
Worked example 1 — the buffer versus plain water
1.00 L of buffer contains 0.10 mol CH₃COOH and 0.10 mol CH₃COONa.
Starting pH: ratio = 0.10/0.10 = 1, log 1 = 0, so pH = 4.74 + 0 = 4.74.
Add 0.01 mol HCl. The H⁺ eats 0.01 mol of acetate and makes 0.01 mol
of acid:
acetate = 0.10 − 0.01 = 0.09 mol; acid = 0.10 + 0.01 = 0.11 mol
ratio = 0.09/0.11 = 0.8182, log 0.8182 = −0.0872
pH = 4.74 − 0.087 = 4.65 — a fall of only 0.09 units.
The same 0.01 mol HCl in 1.00 L of pure water: [H⁺] = 0.01 M, pH = −log(0.01) = 2.00. The pH has fallen from 7.00 to 2.00 — a change of 5.00 units, more than fifty times bigger.
Worked example 2 — adding alkali to the same buffer
Add 0.01 mol NaOH to the original 0.10 / 0.10 buffer. The OH⁻ eats acid and makes base:
acid = 0.10 − 0.01 = 0.09 mol; acetate = 0.10 + 0.01 = 0.11 mol
ratio = 0.11/0.09 = 1.2222, log 1.2222 = +0.0872
pH = 4.74 + 0.087 = 4.83.
Notice the symmetry: acid pushed the pH down 0.09 units, alkali pushed it up 0.09 units. A buffer sitting exactly at its pKa resists both directions equally well, and that is the condition of maximum buffer capacity.
Worked example 3 — a basic buffer
A solution is 0.20 M in NH₃ and 0.30 M in NH₄Cl. For ammonia Kb = 1.8 × 10⁻⁵, so pKb = 4.74 and the conjugate acid NH₄⁺ has pKa = 14 − 4.74 = 9.26.
pH = 9.26 + log(0.20 / 0.30) = 9.26 + log(0.6667) = 9.26 − 0.1761 = 9.08.
Cross-check by the pOH route: pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.30/0.20) = 4.74 + 0.1761 = 4.92, so pH = 14 − 4.92 = 9.08. The two routes agree, which is exactly the check you should run in an exam when you are unsure which form of the equation to use.
Buffer range and buffer capacity — two different ideas
Students often merge these. They are not the same thing.
- Buffer range is where a buffer works: roughly pKa ± 1. Beyond a 10 : 1 ratio in either direction, one reservoir is nearly exhausted and the pH starts moving fast. So an acetate buffer is useful from about pH 3.7 to 5.7, and you would not choose it to hold pH 9.
- Buffer capacity is how much acid or alkali it can absorb before failing. It depends on the total concentration. A 1.0 M / 1.0 M acetate buffer and a 0.01 M / 0.01 M acetate buffer have the same starting pH (both ratios are 1), but the concentrated one can soak up a hundred times more added acid.
This is also why diluting a buffer with water leaves the pH almost unchanged — the ratio is untouched — while quietly destroying its capacity.
Common mistakes that cost marks
- Using a strong acid and its salt. HCl + NaCl is not a buffer. There are no undissociated HCl molecules to neutralise added OH⁻.
- Inverting the ratio. In the pH form, the conjugate base is on top. Put the acid on top and every answer is wrong by twice the log term.
- Forgetting to update both species. Adding acid does not only reduce the base — it increases the acid by the same amount. Both numbers change.
- Mixing pKa and pKb. For NH₃ the value 4.74 is pKb, not pKa. Convert with pKa + pKb = 14 (at 25 °C) before using the pH form.
- Claiming the pH never changes. It does change — just slightly. Write "resists change" or "changes only slightly", never "remains constant".
- Ignoring the limits. If you add more strong acid than there is conjugate base, the buffer is destroyed and you must go back to a plain strong-acid calculation.
Where buffers appear in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 12 | Define a buffer, explain the mechanism with equations, calculate pH from given moles |
| Class 12 practicals & viva | Why a buffer is used to fix pH in a salt-analysis or titration procedure |
| NEET / JEE | Henderson–Hasselbalch numericals, choosing the right buffer for a target pH |
| IIT-JAM / CUET-PG | pH after adding a measured amount of strong acid or base; buffer range reasoning |
| GATE / CSIR-NET | Buffer capacity, titration-curve interpretation, biological buffer systems |
Always read the question carefully: "calculate the pH of the buffer" and "calculate the pH after adding 10 mL of 0.1 M HCl" are two different amounts of work.
Check your buffer working in seconds. The Buffer (Henderson–Hasselbalch) calculator takes pKa and the two concentrations and returns the pH, so you can confirm every step above and test what happens as you shift the ratio.
Open the Buffer (Henderson–Hasselbalch) Calculator →Struggling with the ionic-equilibrium chapter as a whole? ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and online across India — details at abcchemistry.in. If one-to-one help at home suits you better and you are in Delhi, Noida or Gurgaon, see delhihometutor.com.