A buffer doesn’t magically hold pH at one exact number. It works because a weak acid and its conjugate base can consume modest additions of strong base or strong acid. Understanding that chemistry makes the Henderson–Hasselbalch equation far easier to use—and much harder to misuse.
In this college-level chemistry tutorial, we’ll build a buffer from the molecular reaction, calculate its pH, then determine how that pH changes when a known amount of strong acid is added. You’ll also learn when to use moles instead of concentrations, how buffer capacity differs from pH, and why memorizing a formula is not enough.
What you should be able to do afterward
- Identify a valid weak acid/conjugate base buffer pair.
- Predict which component reacts when H3O+ or OH− is added.
- Calculate buffer pH using pH = pKa + log([A−]/[HA]).
- Do the strong-acid or strong-base stoichiometry before calculating the new pH.
- Explain when the Henderson–Hasselbalch shortcut becomes unreliable.
1. What is an acid–base buffer?
A buffer contains appreciable amounts of a weak acid HA and its conjugate base A−, or the analogous weak-base/conjugate-acid pair. A familiar example is acetic acid, CH3COOH, mixed with acetate, CH3COO− (often supplied by sodium acetate).
The relevant weak-acid equilibrium is HA + H2O ⇌ H3O+ + A−. Strong acid reacts primarily with the conjugate base: A− + H3O+ → HA + H2O. Strong base reacts primarily with the weak acid: HA + OH− → A− + H2O.
Both components must be present in useful amounts. A bottle labeled “acetic acid” is not automatically an effective buffer, and equal volumes of a strong acid and a strong base generally do not create a weak-acid buffer.
2. Where does Henderson–Hasselbalch come from?
For the weak-acid equilibrium, the acid dissociation constant is Ka = [H3O+][A−]/[HA] in a standard introductory concentration approximation. Rearrange for [H3O+] and take −log of both sides. The result is the Henderson–Hasselbalch relationship:
pH = pKa + log10([A−]/[HA]). Here, pKa = −log10Ka; [A−] means conjugate-base concentration and [HA] means weak-acid concentration.
This equation is an approximation that assumes equilibrium concentrations are sufficiently close to the starting buffer concentrations and that activity effects are reasonably represented by concentration ratios. It is typically most useful when both buffer components are present in substantial amounts.
3. Worked example: calculate the initial buffer pH
Problem: A 1.00 L buffer contains 0.200 mol sodium acetate and 0.100 mol acetic acid. Use pKa = 4.76 for acetic acid under the illustrative conditions. What is the pH? Assume ideal solution behavior.
Step 1: assign acid and base
The acid is CH3COOH (HA); the conjugate base is CH3COO− (A−). The acetate-to-acid mole ratio is 0.200/0.100 = 2.00. Because both species occupy the same final volume, that is also the concentration ratio.
Step 2: substitute
pH = 4.76 + log10(2.00) = 4.76 + 0.301 = 5.06. Since the solution contains more conjugate base than acid, its pH is above pKa. That’s a useful qualitative check on your numerical result.
4. Worked example: add strong acid to the buffer
Now add 0.0100 mol HCl to the original buffer and assume the volume change is negligible. Find the new pH.
Step 1: complete the neutralization bookkeeping
HCl supplies strong acid that reacts essentially completely with acetate. Each mole of H3O+ consumes one mole of acetate and generates one mole of acetic acid. We calculate the amounts after this stoichiometric step before using a buffer equation.
| Species | Before HCl | Change | After reaction |
|---|---|---|---|
| Acetate, A− | 0.200 mol | −0.0100 mol | 0.190 mol |
| Acetic acid, HA | 0.100 mol | +0.0100 mol | 0.110 mol |
Step 2: calculate the pH from the new ratio
pH = 4.76 + log10(0.190/0.110) = 4.997 ≈ 5.00. The initial pH was 5.06, so adding strong acid lowered the pH by only about 0.06 unit under these assumptions.
Notice that we did not insert the added HCl amount directly into the Henderson–Hasselbalch formula. First it neutralizes some acetate; only then do we use the remaining weak-acid/conjugate-base ratio.
5. What happens when you add strong base instead?
Suppose instead that the original buffer receives 0.0200 mol NaOH. Hydroxide consumes acetic acid, so HA changes from 0.100 to 0.0800 mol, and A− changes from 0.200 to 0.220 mol. Then pH = 4.76 + log10(0.220/0.0800) = 5.20. Adding strong base raises the buffer pH, but far less than it would raise the pH of an unbuffered solution.
6. Buffer pH versus buffer capacity
These are related but not interchangeable. A buffer’s pH is governed primarily by the conjugate-base/weak-acid ratio. Its capacity—how much acid or base it can neutralize before its pH changes substantially—also depends strongly on the total amounts of the pair.
| Buffer | [HA] | [A−] | Approximate initial pH | Relative capacity |
|---|---|---|---|---|
| Dilute pair | 0.010 M | 0.010 M | 4.76 | Lower |
| Concentrated pair | 0.100 M | 0.100 M | 4.76 | Higher |
A practical rule of thumb is that buffers often function well when the acid/base ratio lies roughly between 1:10 and 10:1, corresponding to pH within about one unit of pKa. That’s a useful design range, not a sharp boundary at which buffering suddenly disappears.
7. When should you NOT use the shortcut?
- One buffer component is depleted: after enough strong acid or base, the usual buffer approximation can fail completely.
- Very dilute solutions: water autoionization and weak-acid dissociation may no longer be negligible.
- Strongly nonideal or highly concentrated solutions: activity coefficients may matter; an activity-based treatment is more accurate.
- Wrong conjugate pair: two unrelated weak acids do not automatically form the simple HA/A− buffer used in this formula.
- Using the wrong direction in the log: the numerator is conjugate base; the denominator is weak acid.
8. College chemistry practice problems
Problem A: A buffer contains 0.300 mol acetate and 0.100 mol acetic acid. Use pKa = 4.76. What is the approximate initial pH?
Problem B: Add 0.0200 mol HCl to the buffer in Problem A. Assume the added volume is negligible and neither component is exhausted. What is the new pH?
Problem C: Two buffers have [HA] = [A−]. One contains 0.0100 M of each; the other contains 0.100 M of each. Which has greater capacity, and which has the higher approximate pH?
Problem D: A buffer initially contains 0.100 mol of HA and 0.100 mol of A−. What is wrong with using Henderson–Hasselbalch immediately after adding 0.130 mol of strong acid?
Worked answers and explanations
A: pH = 4.76 + log10(0.300/0.100) = 5.24. The base/acid ratio is 3, so pH is higher than pKa.
B: After neutralization, acetate = 0.280 mol and acetic acid = 0.120 mol. pH = 4.76 + log10(0.280/0.120) = 5.13.
C: Both have pH ≈ pKa = 4.76 by the same concentration-ratio approximation. The 0.100 M/0.100 M solution has greater buffering capacity because it contains more of the pair per liter.
D: The 0.130 mol strong acid would consume all 0.100 mol of A−, leaving excess strong acid. The usual buffer ratio is no longer valid; calculate residual strong-acid concentration accounting for the final volume and any necessary equilibria instead.
Always ask two questions in this order: What reacts completely first? What weak-acid equilibrium remains afterward?
Continue learning and explore
The underlying theory is covered in OpenStax Chemistry 2e: Buffers. You can also explore qualitative acid–base behavior through the University of Colorado PhET Acid–Base Solutions simulation. If you want stronger concentration and reaction setup skills first, work through the IntoChemistry stoichiometry tutorial.
Instructor note: These are idealized instructional examples; pKa, activity, temperature, ionic strength, and the calibration of a real pH meter all matter in laboratory measurements.