Buffer Capacity — How Much Acid a Buffer Can Absorb
Every buffer resists a change in pH — but only up to a point, and only over a limited pH range. Buffer capacity is the number that says how much. It is what separates a buffer that works in a real experiment from one that looks correct on paper and collapses at the first drop of acid. This guide gives the formula, four worked examples with the arithmetic shown, and the two conditions that make capacity a maximum.
The definition
Strictly, buffer capacity is the derivative β = dn ÷ d(pH), because the resistance changes as the buffer is consumed. For exam-sized additions the ratio above is the working definition, and it is what you will be asked to compute. The unit is mol L⁻¹ pH⁻¹: moles per litre per unit of pH. A large β means a strong buffer.
Two things control β, and both must be right:
- Total concentration. Capacity is roughly proportional to the total amount of the conjugate pair present. Double both concentrations and you double the capacity.
- The ratio of the pair. Capacity is highest when the acid and its conjugate base are present in equal amounts, that is when pH = pKa.
For a single conjugate pair of total concentration C, the theoretical maximum is
Worked example 1 — computing β
Question. One litre of buffer contains 0.100 mol acetic acid and 0.100 mol sodium acetate. Ka(CH₃COOH) = 1.8 × 10⁻⁵. Find the pH before and after adding 0.010 mol of HCl, then the buffer capacity.
Step 1 — pKa. pKa = −log(1.8 × 10⁻⁵) = 5 − 0.2553 = 4.74
Step 2 — initial pH from Henderson–Hasselbalch:
pH = pKa + log([salt] ÷ [acid]) = 4.74 + log(0.100 ÷ 0.100) = 4.74 + 0 =
4.74
Step 3 — after adding 0.010 mol HCl. The added H⁺ converts acetate into
acetic acid, mole for mole:
acetate: 0.100 − 0.010 = 0.090 mol
acetic acid: 0.100 + 0.010 = 0.110 mol
pH = 4.74 + log(0.090 ÷ 0.110) = 4.74 + log(0.8182) = 4.74 − 0.087 = 4.65
Step 4 — capacity.
β = 0.010 ÷ 0.087 = 0.115 mol L⁻¹ pH⁻¹
Cross-check by the formula. Total concentration C = 0.100 + 0.100 = 0.200 mol/L, and the buffer sits exactly at pH = pKa, so βmax = 0.576 × 0.200 = 0.115 mol L⁻¹ pH⁻¹ ✓ The two routes agree, which confirms both the arithmetic and the idea that capacity peaks at pH = pKa.
Worked example 2 — the same acid added to unbuffered water
The contrast is what makes the number meaningful.
Question. What happens when the same 0.010 mol of HCl is added to 1.00 L of pure water at pH 7.00?
[H⁺] = 0.010 mol/L (HCl is fully dissociated)
pH = −log(0.010) = 2.00
ΔpH = 7.00 − 2.00 = 5.00 units, against 0.087 units in the buffer. The buffer absorbed the same acid with about one-sixtieth of the pH shift.
Worked example 3 — overloading the buffer
Capacity is finite. Once the reserve of conjugate base is used up, there is nothing left to neutralise further acid and the pH falls off a cliff.
Question. Add 0.120 mol of HCl to the same 1.00 L buffer.
Only 0.100 mol of acetate is available, so all of it is consumed. The excess acid is
0.120 − 0.100 = 0.020 mol of HCl left over in 1.00 L.
[H⁺] ≈ 0.020 mol/L
pH = −log(0.020) = 2 − log 2 = 2 − 0.301 = 1.70
The pH crashed from 4.74 to 1.70. Beyond its capacity a buffer is not a weak buffer — it is no buffer at all. This is why the useful working range of a buffer is only about pKa ± 1, where the ratio of the pair stays between 1 : 10 and 10 : 1.
Worked example 4 — what dilution does
Question. The 1.00 L buffer of example 1 is diluted to 10.0 L. What happens to its pH and to its capacity?
pH: both concentrations fall by the same factor of 10, so the ratio [salt] ÷ [acid] is unchanged at 1, and pH stays at 4.74. This is the famous result that a buffer's pH is almost independent of dilution.
Capacity: the new total concentration is C = 0.200 ÷ 10 = 0.0200 mol/L, so
βmax = 0.576 × 0.0200 = 0.0115 mol L⁻¹ pH⁻¹ — one tenth of what
it was.
So dilution leaves the pH alone but destroys the capacity. A student who checks only the pH will conclude the diluted buffer is "the same", and be wrong in exactly the way that matters in the laboratory.
Choosing a buffer for a target pH
Pick a conjugate pair whose pKa is as close as possible to the pH you want, then make it as concentrated as the experiment allows. Standard pKa values at 25 °C:
| Buffer pair | pKa | Useful range (pKa ± 1) |
|---|---|---|
| HCOOH / HCOO⁻ (formate) | 3.75 | 2.8 – 4.8 |
| CH₃COOH / CH₃COO⁻ (acetate) | 4.74 | 3.7 – 5.7 |
| H₂PO₄⁻ / HPO₄²⁻ (phosphate, pKa2) | 7.20 | 6.2 – 8.2 |
| NH₄⁺ / NH₃ (ammonia) | 9.25 | 8.3 – 10.3 |
| HCO₃⁻ / CO₃²⁻ (carbonate, pKa2) | 10.33 | 9.3 – 11.3 |
Values shift a little with temperature and ionic strength, so for precise work measure the final pH with a meter rather than trusting the calculation alone.
Common mistakes that cost marks
- Confusing capacity with range. Range (pKa ± 1) says where a buffer works; capacity says how much it can take. A very dilute acetate buffer still has the same range and almost no capacity.
- Believing dilution weakens the pH. It is the other way round — dilution barely moves the pH and slashes the capacity.
- Working in concentrations when the volume changes. Adding a solution of acid changes the total volume. Work in moles through the neutralisation step, then divide by the final volume.
- Forgetting that added acid changes both terms. Acetate goes down by 0.010 and acetic acid goes up by 0.010. Changing only one gives the wrong pH.
- Using a buffer outside its range. An acetate buffer cannot hold pH 7.4; at that pH almost none of the pair is in the acid form.
- Dropping the sign of the log. log(0.8182) is negative, so the pH goes down when acid is added — which is physically obvious, and a useful check.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE Class 11–12 | Ionic equilibrium: buffer pH before and after adding acid or base |
| NEET / JEE | Henderson–Hasselbalch numericals, buffer range questions |
| IIT-JAM / CUET-PG | Buffer capacity definition and its maximum at pH = pKa |
| GATE / CSIR-NET | Titration curves, biological buffers, β as a derivative |
Get the pH at each stage in one step. Buffer capacity is calculated from two pH values, so the arithmetic hinges on the Henderson–Hasselbalch equation. The Buffer calculator gives you both — the pH before addition and the pH after the ratio has shifted — so you can divide with confidence.
Open the Henderson–Hasselbalch Buffer Calculator →Ionic equilibrium is one of the highest-yield chapters in Class 11–12 chemistry. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre plus online classes across India — abcchemistry.in.