Last updated: 18 August 2026
Many chemical and biological systems can function properly only within a fairly narrow pH range. Yet acids and bases may continuously enter these systems. A buffer provides a chemical way of absorbing small disturbances before they produce a large change in hydrogen-ion concentration.
The important word is resists. A buffer does not keep the pH perfectly constant. When acid or alkali is added, the pH still changes, but the change is much smaller than it would be in an unbuffered solution.
Buffer chemistry brings together ideas that you have already met in acid–base equilibria: weak acids and bases, conjugate acid–base pairs, Kₐ and Kᵦ, stoichiometry and logarithms. Once you understand what each component is doing, buffer calculations become a sequence of chemical reasoning rather than a formula to memorise.
Diagnostic Quiz
Try these five questions before reading the chapter. They check the acid–base, conjugate-pair and stoichiometric ideas that buffer chemistry builds upon.
- 5 quick prerequisite questions
- Focus on conjugate pairs, pH and neutralisation
- Every answer option is explained
What is a buffer solution?
A buffer solution is a solution that resists a significant change in pH when a small amount of acid or alkali is added.
For effective buffering, appreciable amounts of both members of a conjugate acid–base pair must be present. One member reacts with added H⁺ ions, while the other reacts with added OH⁻ ions.
| Type of buffer | Components | Example | Typical pH |
|---|---|---|---|
| Acidic buffer | Weak acid + its conjugate base | CH₃COOH and CH₃COO⁻ | Usually below 7 |
| Basic buffer | Weak base + its conjugate acid | NH₃ and NH₄⁺ | Usually above 7 |
The conjugate species is commonly supplied by a soluble salt. Sodium ethanoate, CH₃COONa, supplies CH₃COO⁻ ions, while ammonium chloride, NH₄Cl, supplies NH₄⁺ ions.
A weak acid on its own is not normally considered a complete buffer because there is not yet a substantial reservoir of its conjugate base available to remove added H⁺ ions.
Why does a strong acid and its salt not form a buffer?
A strong acid such as HCl is essentially completely ionised in aqueous solution. There is therefore no appreciable reservoir of undissociated HCl in equilibrium with Cl⁻ ions.
A mixture of HCl and NaCl therefore does not provide the weak-acid/conjugate-base pair required for effective buffer action.
However, this does not mean that a strong acid or strong base can never be used to prepare a buffer. Controlled partial neutralisation can generate the required conjugate pair, as we shall see later.
How an acidic buffer works
Consider a buffer containing ethanoic acid, CH₃COOH, and ethanoate ions, CH₃COO⁻. Ethanoic acid is weak and establishes the equilibrium:
The solution contains substantial amounts of both CH₃COOH and CH₃COO⁻. These act as two chemical reservoirs capable of responding to added acid or alkali.
When a small amount of acid is added
Added acid introduces extra H⁺ ions. The basic component, CH₃COO⁻, reacts with these ions:
Most of the added H⁺ is therefore converted into weakly ionised CH₃COOH rather than remaining free in solution.
The amount of CH₃COO⁻ falls slightly and the amount of CH₃COOH rises slightly. The pH does decrease, but only by a relatively small amount.
When a small amount of alkali is added
Added alkali introduces OH⁻ ions. These react with the acidic component of the buffer:
The strongly basic OH⁻ ions are converted into water. The amount of CH₃COOH decreases slightly while the amount of CH₃COO⁻ increases slightly.
The pH therefore rises, but again by much less than it would in an unbuffered solution.
Added H⁺: the conjugate base component removes the added acid.
Added OH⁻: the weak acid component removes the added alkali.
Buffer action converts one member of the conjugate pair into the other rather than leaving a large concentration of added H⁺ or OH⁻ in solution.
Connecting buffer action to equilibrium
The response can also be understood using Le Chatelier’s principle. For:
adding H⁺ favours the reverse reaction. Removing H⁺ by reaction with OH⁻ favours further ionisation of CH₃COOH.
In an examination explanation, however, it is usually clearer to write the direct reaction that consumes the added H⁺ or OH⁻ rather than merely saying that the equilibrium shifts.
How a basic buffer works
A basic buffer uses the same principle but contains a weak base and its conjugate acid. A common example is aqueous ammonia, NH₃, together with NH₄⁺ ions.
When acid is added
NH₃ acts as the basic member of the pair and removes added H⁺:
Most of the added H⁺ is therefore converted into the much weaker acid NH₄⁺.
When alkali is added
The conjugate acid NH₄⁺ reacts with OH⁻:
The added OH⁻ is converted into water, limiting the rise in pH.
| Buffer | Removes added H⁺ | Removes added OH⁻ |
|---|---|---|
| CH₃COOH / CH₃COO⁻ | CH₃COO⁻ | CH₃COOH |
| NH₃ / NH₄⁺ | NH₃ | NH₄⁺ |
The more basic member removes added H⁺. The more acidic member removes added OH⁻. This reasoning works for both acidic and basic buffers and avoids having to memorise separate mechanisms.
How buffer solutions can be prepared
A buffer can be prepared either by directly mixing the two members of a conjugate acid–base pair or by generating one member through partial neutralisation.
Directly mixing a conjugate pair
An acidic buffer can be made by mixing a weak acid with a soluble salt containing its conjugate base.
Example:
CH₃COOH + CH₃COONa
The acid provides CH₃COOH and the salt provides CH₃COO⁻.
Similarly, a basic buffer can be made by mixing a weak base with a salt containing its conjugate acid.
Example:
NH₃ + NH₄Cl
The weak base provides NH₃ and the salt provides NH₄⁺.
Partial neutralisation
Another method is to react only part of a weak acid with a strong base.
If there is insufficient OH⁻ to neutralise all the CH₃COOH, some CH₃COOH remains while some has been converted into CH₃COO⁻. The resulting mixture therefore contains both members of the conjugate pair.
In the same way, partial neutralisation of NH₃ with a strong acid forms NH₄⁺ while leaving some NH₃ unreacted.
The strong acid or strong base is not itself the buffering species. Its role is to convert part of the weak species into its conjugate partner.
Calculating the pH of an acidic buffer
The buffer equation can be obtained directly from the Kₐ expression. Represent a weak acid as HA:
The acid dissociation constant is:
Rearranging:
Taking −log₁₀ of both sides gives:
This is commonly called the Henderson–Hasselbalch equation. The important chemical meaning is that the buffer pH depends on the pKₐ of the weak acid and the ratio of conjugate base to weak acid.
Worked example: ethanoic acid and ethanoate ions
Calculate the pH of a buffer containing 0.100 mol dm⁻³ CH₃COOH and 0.200 mol dm⁻³ CH₃COO⁻.
Given:
First calculate pKₐ:
Substitute the concentration ratio:
Answer: pH = 5.05
The conjugate base is present at a higher concentration than the weak acid. The logarithmic term is therefore positive, making the pH greater than pKₐ.
Equal buffer components and the 1 : 1 condition
An especially important situation occurs when the weak acid and conjugate base are present in equal concentrations:
Their ratio is then 1:
Since log₁₀(1) = 0:
For a given total concentration of buffer components, this is the most balanced buffering condition. Comparable amounts of the acidic and basic members are available to deal with added OH⁻ and H⁺ respectively.
This does not mean that every 1 : 1 buffer has the same absolute buffer capacity. A concentrated buffer can neutralise considerably more added acid or base than a dilute buffer having exactly the same ratio.
Connecting buffer action to titration curves
The 1 : 1 condition becomes particularly useful when buffer chemistry is linked to acid–base titration curves. The volume at which equal amounts of the conjugate pair are present depends on whether the weak species is the analyte or the titrant.
When the weak species is the analyte
Suppose HA is placed in the conical flask and titrated with a strong base:
At the equivalence point, all the original HA has been converted into A⁻. At half the equivalence volume, exactly half has reacted.
If the initial amount of HA is n mol:
Therefore:
The 1 : 1 condition occurs at:
and hence:
This is the familiar half-equivalence point of a weak acid–strong base titration.
Likewise, when a weak base is titrated with a strong acid, equal amounts of B and BH⁺ occur at ½Veq, so:
When the weak species is the titrant
Now reverse the arrangement. Suppose a strong base is initially in the conical flask and the weak acid HA is added from the burette.
At equivalence, all the HA added so far has reacted with the original OH⁻ and has been converted into A⁻.
If n mol of HA was required to reach equivalence, approximately n mol of A⁻ has been formed. To obtain an equal amount of unreacted HA, a further n mol HA must now be added.
The total weak acid added is therefore 2n, so the 1 : 1 condition occurs at:
At this point:
and therefore:
| Arrangement | 1 : 1 conjugate-pair condition | Relationship |
|---|---|---|
| Weak species is the analyte | ½Veq | pH = pKₐ or pOH = pKᵦ |
| Weak species is the titrant | 2Veq | pH = pKₐ or pOH = pKᵦ |
Weak species in the flask: the 1 : 1 buffer condition occurs before equivalence, at ½Veq.
Weak species in the burette: the 1 : 1 buffer condition occurs after equivalence, at 2Veq.
Why can 2Veq be inconvenient experimentally?
A typical burette has a capacity of 50 cm³. Suppose the equivalence volume is 25.0 cm³.
Reaching the 1 : 1 buffer condition already requires the full burette capacity. If Veq were greater than 25.0 cm³, then 2Veq would exceed 50 cm³.
This is not chemically impossible. The burette could be refilled or the experiment could use different concentrations. It is simply less convenient experimentally.
The familiar half-equivalence relationship assumes that the weak acid or weak base is initially in the conical flask. If the weak species is instead the titrant, the 1 : 1 conjugate-pair condition occurs at 2Veq.
Calculating the pH of a basic buffer
For a weak base B and its conjugate acid BH⁺:
The Kᵦ expression is:
Rearranging:
Therefore:
At 298 K:
Worked example: ammonia and ammonium ions
Calculate the pH at 298 K of a buffer containing 0.200 mol dm⁻³ NH₃ and 0.100 mol dm⁻³ NH₄⁺.
Given:
First calculate pKᵦ:
Substitute:
Hence:
Answer: pH = 9.56
For an acidic buffer, use conjugate base ÷ weak acid. For a basic buffer written using pKᵦ, use conjugate acid ÷ weak base.
Calculating pH after adding strong acid or strong base
A buffer question may ask for the pH after a small amount of strong acid or strong base is added. This requires two different types of calculation.
- Stoichiometry first: determine how the added strong acid or base changes the amounts of the buffer components.
- Equilibrium second: use the new conjugate-pair ratio to calculate the pH.
Worked example: adding HCl to an acidic buffer
A buffer initially contains 10.0 mmol CH₃COOH and 10.0 mmol CH₃COO⁻. HCl containing 1.00 mmol H⁺ is added. Calculate the new pH.
Given:
H⁺ reacts with the conjugate base:
The reaction has a 1 : 1 mole ratio.
| Species | Before / mmol | Change / mmol | After / mmol |
|---|---|---|---|
| CH₃COO⁻ | 10.0 | −1.00 | 9.00 |
| H⁺ | 1.00 | −1.00 | 0 |
| CH₃COOH | 10.0 | +1.00 | 11.0 |
The solution still contains both members of the conjugate pair, so it remains a buffer.
Since both species occupy the same final volume, their common volume cancels and the mole ratio can be used directly:
Answer: pH = 4.66
Initially, equal amounts of acid and conjugate base were present, so the pH was 4.745. After addition of HCl, it falls only to about 4.66.
What if strong base is added?
Added OH⁻ reacts with the weak acid:
Calculate the moles of CH₃COOH consumed and CH₃COO⁻ formed before using the new ratio in the buffer equation.
Incorrect approach: put the amount of added HCl or NaOH directly into the Henderson–Hasselbalch equation.
Correct approach: complete the strong acid–base reaction first, then use the remaining buffer components in the equilibrium calculation.
When can moles be used instead of concentrations?
The acidic-buffer equation is formally written using concentrations:
If both components are in the same final solution of volume V:
The common volume cancels:
This is why mole ratios are particularly convenient after mixing solutions or carrying out a neutralisation reaction.
During neutralisation, moles determine how much reacts. Once the final mixture has formed, the buffer components share the same solution volume, so their mole ratio equals their concentration ratio.
Dilution and buffer capacity
For an acidic buffer:
If water is added, both concentrations decrease by the same dilution factor.
For example:
The ratio is unchanged, so under the usual ideal assumptions used in A-Level calculations, the buffer pH changes very little.
Does dilution have no effect on the buffer?
Dilution does reduce the amount of each buffer component present per unit volume. The diluted buffer therefore has less ability to absorb further additions of acid or base.
This ability is described as the buffer capacity.
Buffer pH: depends mainly on the ratio of the two buffer components.
Buffer capacity: depends on the amounts and concentrations of buffer components available to react with added acid or base.
Ratio and concentration are therefore different ideas
Consider two buffers that both contain equal concentrations of HA and A⁻ relative to one another:
| Buffer | [HA] | [A⁻] | Initial pH | Relative capacity |
|---|---|---|---|---|
| A | 0.500 mol dm⁻³ | 0.500 mol dm⁻³ | pKₐ | Greater |
| B | 0.0500 mol dm⁻³ | 0.0500 mol dm⁻³ | pKₐ | Smaller |
Both buffers initially have the same pH because their ratios are both 1 : 1. Buffer A, however, contains ten times as much of each component per unit volume and can therefore absorb more added acid or base.
What happens when a buffer is exhausted?
A buffer contains finite amounts of its acidic and basic components. It can only resist pH change while sufficient quantities of both remain.
If enough acid is added to consume essentially all of the conjugate base, further added H⁺ can no longer be removed effectively.
Similarly, if enough alkali is added to consume essentially all of the weak acid, further OH⁻ begins to accumulate.
A buffer resists small disturbances; it cannot neutralise an unlimited quantity of strong acid or strong base. Once one component has been depleted, a much larger pH change occurs.
The carbonate–hydrogencarbonate system in seawater
Natural waters contain interconnected acid–base equilibria involving dissolved carbon dioxide. These equilibria help seawater resist rapid changes in pH.
Carbon dioxide dissolves and can form carbonic acid:
Carbonic acid can dissociate:
Hydrogencarbonate can also dissociate:
The HCO₃⁻/CO₃²⁻ pair can therefore contribute to buffer action.
Responding to added acid
CO₃²⁻ removes added H⁺:
Responding to added alkali
HCO₃⁻ can react with OH⁻:
Why can more atmospheric CO₂ lower ocean pH?
When atmospheric CO₂ increases, more CO₂ can dissolve in seawater. The resulting equilibria can produce additional H⁺ ions:
Increasing [H⁺] lowers the pH. Some of the additional H⁺ is consumed by carbonate:
Buffering therefore reduces the magnitude of the pH change, but does not completely prevent a sustained change when increasing amounts of CO₂ continue to enter the system.
Common mistakes with buffer solutions
| Common mistake | Why it is wrong | Correct approach |
|---|---|---|
| Saying that a buffer keeps pH exactly constant | Adding acid or base changes the ratio of the buffer components | State that a buffer resists a significant change in pH |
| Calling a weak acid alone a complete buffer | There is not initially an appreciable reservoir of its conjugate base | Look for appreciable amounts of both members of the conjugate pair |
| Saying added H⁺ reacts with the weak acid in an acidic buffer | The weak acid is already the acidic member | Added H⁺ is removed by the conjugate base |
| Saying added OH⁻ reacts with the conjugate base | Both species are basic | Added OH⁻ is removed by the weak acid |
| Reversing [A⁻]/[HA] | This changes the sign of the logarithmic term | For an acidic buffer use conjugate base ÷ weak acid |
| Assuming the 1 : 1 condition always occurs at ½Veq | That relationship assumes the weak species is initially in the flask | Use ½Veq for weak analyte and 2Veq for weak titrant |
| Calling every 1 : 1 buffer the buffer with the greatest absolute capacity | Absolute capacity also depends on the total concentration of the components | Describe 1 : 1 as the most balanced condition for a given total concentration |
| Putting added HCl or NaOH straight into the buffer equation | The strong reagent reacts stoichiometrically first | Calculate the new buffer amounts before applying the equilibrium equation |
| Using the original buffer amounts after neutralisation | The strong acid or base has converted one buffer component into the other | Use the amounts remaining after reaction |
| Thinking dilution has no effect at all | The ratio may remain unchanged but the component concentrations decrease | Distinguish approximately unchanged pH from reduced buffer capacity |
| Saying an ocean buffer prevents any change in ocean pH | Buffer capacity is finite | State that buffering reduces, rather than eliminates, the pH change |
Buffer-solution checklist
Identify the buffer
- Look for a conjugate acid–base pair.
- Acidic buffer: weak acid + conjugate base.
- Basic buffer: weak base + conjugate acid.
- Check that appreciable amounts of both are present.
Explain the mechanism
- Identify the component that removes H⁺.
- Identify the component that removes OH⁻.
- Write the appropriate reaction.
- State that the pH change is relatively small.
Calculate buffer pH
- Calculate pKₐ or pKᵦ if required.
- Use the correct conjugate-pair ratio.
- At a 1 : 1 ratio, pH = pKₐ or pOH = pKᵦ.
- Use mole ratios when the common final volume cancels.
Handle titration and additions
- Weak analyte: 1 : 1 at ½Veq.
- Weak titrant: 1 : 1 at 2Veq.
- Complete strong acid–base stoichiometry first.
- Distinguish buffer pH from buffer capacity.
Mastery Quiz
Now apply the chapter to buffer composition, pH calculations, titration volumes, dilution and strong acid–base additions.
- 10 application-focused questions
- Includes calculations and titration reasoning
- Every answer option is explained
Frequently asked questions
Questions students often ask about buffer solutions
These questions connect the buffer equations to the equilibrium and titration reasoning that commonly causes confusion in A-Level Chemistry.
Why can't a strong acid and its salt make the usual type of buffer?
A strong acid such as HCl is essentially completely ionised, so there is no appreciable reservoir of undissociated weak acid in equilibrium with its conjugate base. A mixture such as HCl and NaCl therefore lacks the required weak-acid/conjugate-base pair.
Can a strong acid or strong base still be used to prepare a buffer?
Yes. A controlled amount of strong base can partially neutralise a weak acid, producing some conjugate base while leaving some weak acid unreacted. Similarly, a strong acid can partially neutralise a weak base to produce its conjugate acid.
Why does the pH of a buffer depend mainly on a ratio?
Rearranging the Kₐ expression gives [H⁺] = Kₐ[HA]/[A⁻]. The hydrogen-ion concentration therefore depends on the relative amounts of weak acid and conjugate base rather than simply on one concentration by itself.
Why is pH = pKₐ when [HA] = [A⁻]?
If [A⁻]/[HA] = 1, then log₁₀(1) = 0. The buffer equation therefore becomes pH = pKₐ.
Does pH = pKₐ mean the buffer has the greatest possible capacity?
A 1 : 1 ratio gives the most balanced ability to respond to additions of acid and alkali for a given total concentration.
Absolute buffer capacity also depends on how much buffer material is present. A concentrated 1 : 1 buffer can neutralise more acid or base than a dilute 1 : 1 buffer.
Why does the half-equivalence point give pH = pKₐ?
When a weak acid is initially in the flask and is titrated with a strong base, half-equivalence means that half the original HA has been converted into A⁻. Equal amounts of HA and A⁻ are therefore present, so pH = pKₐ.
Is the 1 : 1 buffer condition always at half the equivalence volume?
No. The ½Veq rule assumes the weak species is initially in the conical flask.
If the weak species is instead added from the burette to a strong species in the flask, equal amounts of the weak species and its conjugate partner occur at 2Veq.
Why can reaching 2Veq be inconvenient experimentally?
A typical burette holds about 50 cm³. If Veq is greater than 25 cm³, then 2Veq is greater than 50 cm³ and cannot be delivered in one normal burette filling.
The experiment is still possible by refilling the burette or adjusting concentrations, but the arrangement is less convenient.
Why must I do the neutralisation calculation before the buffer calculation?
Strong acids and bases react essentially completely with the appropriate buffer component. Their addition therefore changes the amounts of both members of the conjugate pair.
The Henderson–Hasselbalch equation must use the amounts remaining after this stoichiometric reaction has taken place.
Why can I often use moles instead of concentrations?
After mixing, both buffer components occupy the same final solution volume. When their concentrations are divided, the common volume cancels, leaving the ratio of their amounts in moles.
Does dilution leave buffer pH exactly unchanged?
In the ideal treatment used for most A-Level calculations, both buffer concentrations decrease by the same factor, so their ratio and calculated pH remain approximately unchanged.
The buffer capacity nevertheless decreases because less buffer material is present per unit volume.
What happens when one buffer component is completely used up?
Effective buffer action is lost. For example, once all the conjugate base has been consumed, further added H⁺ can accumulate directly in solution and produce a much larger fall in pH.
Why can ocean pH fall even though seawater contains buffer systems?
A buffer reduces the magnitude of a pH change but has finite capacity. Increasing dissolved CO₂ can lead to increased H⁺ production. Carbonate species consume part of this additional H⁺, but sustained CO₂ uptake can still lower the overall pH.
Buffer solutions show how acid–base equilibrium can stabilise pH without making it perfectly constant. A weak acid and its conjugate base, or a weak base and its conjugate acid, provide two chemical reservoirs that remove small additions of H⁺ and OH⁻. Their ratio controls the pH, while their amounts determine how much disturbance the buffer can absorb. The same reasoning connects directly to titration curves: the 1 : 1 conjugate-pair condition occurs at ½Veq when the weak species is the analyte but at 2Veq when it is the titrant. With that distinction secure, buffer questions become a logical sequence: identify the conjugate pair, complete any stoichiometric reaction first, and then use the resulting ratio to determine the pH.