In this section, we review what buffers are and how they form, including why we care about buffers in the first place. For chemistry, we often look for solutions with unique properties, whether that be a specific compound, reaction, or observation made. In the case of buffers, these solutions resist changes in pH. This means that adding strong acids or strong bases to them does not impact the pH as much.
It is important to note, however, that buffers are not immune to changes in pH and do have a certain buffer capacity that we will talk about later.
Buffers Review
As we mentioned, buffers are special solutions that are resistant to pH changes when adding acids or bases to them. Buffers are formed in a very specific way: creating a solution of a weak acid and its conjugate base (or a weak base and its conjugate acid, but the former is much more common).
It is important that the acid you create a buffer with is weak because otherwise, the conjugate base would not be a significant base. For example, a mixture of HCl and NaCl would not be a buffer despite being a combination of an acid (HCl) and its conjugate base (Cl-).
You may be asking then why any weak acid isn't a buffer. At equilibrium, there is so much more acid than the conjugate base (assuming a low Ka) that the buffer effects are negligible. In order for a buffer to be effective, you must have comparable concentrations of acid and conjugate base. In fact, the maximum buffer, the point at which the buffer most effectively resists pH change, occurs when the concentration of acid is equal to the concentration of the conjugate base.
Solidify this concept by doing a few practice problems. For each of the pairs of compounds given, identify them as a pair that would form a buffer or not form a buffer:
- NaOH and Na+: - The answer to this question is no. Although NaOH and Na+ are a base-conjugate acid pair, remember that NaOH is a strong base. This means that Na+ is not a significant acid and will not form a buffer.
- CH3COOH and Ca(CH₃COO)₂: - The answer to this question is yes! When dissolved together, this pair will form a buffer. CH3COOH is a weak acid (acetic acid AKA vinegar) with a Ka=1.8 * 10^(-5). Ca(CH3COO)2 is calcium acetate, which will dissociate into Ca2+ (a spectator ion as far as the buffer is concerned), and two moles of CH3COO-, the conjugate base of CH3COOH! Because acetic acid is a weak acid, CH3COO- is a significant base, meaning that we will have a buffer.
- NH3 and NH4NO3: - This pair does form a buffer. NH3 is a weak base, and NH4+ is a significant acid (and its conjugate acid), meaning this pair forms a buffer. In this case, like Ca2+ in the previous example, the nitrate ion is simply a spectator.
- HI and I: - Like example one, this pair does not form a buffer. HI is a strong acid and cannot form buffers with its conjugate base I- because I- is not a significant base.
- KI and PbNO3: - It should be pretty easy to see that this pair does not form a buffer. There are no acids or bases involved. In fact, when you mix KI and PbNO3, you get the "golden rain" reaction, a precipitation reaction that forms PbI2 and KNO3. Take a look!
What Makes Buffers Cool: pH Resistance
Why do buffers have pH resistance, and what makes them so interesting and useful to study? Buffers have pH resistance because of the presence of an acid and a base that do not actively react together at equilibrium. This graphic shows what happens when an acid or a base is added to a buffer:
When a strong acid is added to a buffer, the conjugate base eats it up and forms HAn (An = anion). In the case of no buffer, the strong acid would completely dissociate into H+, increasing [H+] to a much higher degree. Similarly, if OH- from a strong base is added to a buffer, the HAn present in the solution reacts with it to form An- and H2O instead of letting it produce pure OH-. These two reactions lead to buffers being resistant to pH!
Vocabulary
The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.
| Term | Definition |
|---|---|
| buffer solution | A solution containing a large concentration of both members of a conjugate acid-base pair that resists changes in pH when small amounts of acid or base are added. |
| conjugate acid | The species formed when a base accepts a proton; the acid form in an acid-base conjugate pair. |
| conjugate acid-base pair | Two species that differ by one proton, where one is the acid form and the other is the base form of the same substance. |
| conjugate base | The species formed when an acid donates a proton; the base form in an acid-base conjugate pair. |
| pH stabilization | The ability of a buffer solution to maintain a relatively constant pH despite the addition of small amounts of acid or base. |
Frequently Asked Questions
What is a buffer solution and how does it work?
A buffer is a solution that contains a large concentration of both members of a conjugate acid–base pair (a weak acid HA and its salt A–, or a weak base B and its conjugate acid BH+). It stabilizes pH because when you add acid (H+), the conjugate base A– reacts with those protons (A– + H+ → HA), and when you add base (OH–), the conjugate acid HA donates a proton (HA + OH– → A– + H2O). The Henderson–Hasselbalch equation, pH = pKa + log([A–]/[HA]), predicts the buffer pH and shows why pH is tied to pKa and the ratio of conjugates. Buffer capacity (how much acid/base it can absorb) increases with total concentration; buffer range is typically pKa ± 1. On the AP exam, expect questions asking you to explain these neutralization (proton transfer) reactions and use Henderson–Hasselbalch (Topic 8.8). For a focused review, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why do buffers keep pH stable when you add acid or base?
Buffers keep pH stable because they contain a large amount of both members of a conjugate acid–base pair (a weak acid and its salt or a weak base and its salt). When you add acid (H+), the conjugate base in the buffer reacts with those H+ ions to form the weak acid, so free H+ rises only a little. When you add base (OH–), the conjugate acid donates H+ to neutralize OH–, forming water and the conjugate base. These proton-transfer (neutralization) reactions shift the acid/base ratio, not the strong change in [H+]. The Henderson–Hasselbalch equation (pH = pKa + log([base]/[acid]) explains why pH changes little if both concentrations are large and similar; buffer capacity and range depend on those concentrations and on pKa (useful ±1 pH unit around pKa). For AP review, see Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
What's the difference between a buffer and a regular solution?
A buffer is a special kind of solution that contains large concentrations of both members of a conjugate acid–base pair (for example, CH3COOH and CH3COO–). Because of that, when you add a small amount of acid (H+) the conjugate base reacts with it, and when you add base (OH–) the conjugate acid reacts with it. Those neutralization/proton-transfer reactions minimize changes in pH—that’s buffer action (described by the Henderson–Hasselbalch equation). A regular solution lacks a large amount of a conjugate pair, so adding acid or base causes a much larger pH change. Key ideas for AP (LO 8.8.A): buffer capacity (how much acid/base it can absorb) depends on concentrations, and buffer range depends on pKa. For a focused review, check the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and more Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8). For practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).
How does a conjugate acid-base pair make a buffer work?
A buffer works because it contains a conjugate acid–base pair (a weak acid and its salt or a weak base and its salt) in fairly large, comparable amounts. If you add acid (H+), the conjugate base (A–) reacts: A– + H+ → HA, removing most added H+ so pH changes little. If you add base (OH–), the conjugate acid donates a proton: HA + OH– → A– + H2O, neutralizing OH–. Those proton-transfer (neutralization) reactions are why a buffer “stabilizes” pH; the pair soaks up added acid or base. The Henderson–Hasselbalch equation, pH = pKa + log([A–]/[HA]), quantifies how the ratio controls pH and shows why buffer capacity is highest when pH ≈ pKa (CED 8.8.A.1). For more examples and practice aligned to the AP exam, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I'm confused about how buffers neutralize acids and bases - can someone explain?
A buffer works because it contains a lot of both members of a conjugate acid–base pair (for example, HA and A−). When you add a strong acid (H3O+), the conjugate base reacts: A− + H3O+ → HA + H2O. When you add a strong base (OH−), the conjugate acid reacts: HA + OH− → A− + H2O. Those neutralization (proton-transfer) reactions remove most added H+ or OH− so the pH only shifts a little. How well it resists change depends on buffer capacity (how many moles of HA and A− are present) and buffer range (≈ pKa ± 1). You can estimate pH with the Henderson–Hasselbalch equation: pH = pKa + log([A−]/[HA]). This is exactly what AP LO 8.8.A expects you to explain. For a quick review, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and grab extra practice from the AP Chemistry practice set (https://library.fiveable.me/practice/ap-chemistry).
What happens when you add HCl to an acetate buffer?
When you add HCl (a strong acid) to an acetate buffer (acetic acid CH3COOH + its conjugate base CH3COO–), the H+ from HCl is consumed by the conjugate base: CH3COO–(aq) + H+(aq) → CH3COOH(aq) That means acetate is converted to more acetic acid, so the buffer’s [CH3COO–]/[CH3COOH] ratio falls and pH decreases only slightly. You can predict the new pH with the Henderson–Hasselbalch equation: pH = pKa + log([A–]/[HA]) (pKa of acetic acid ≈ 4.76). The smaller the amount of HCl relative to buffer concentrations, the smaller the pH change (buffer capacity). If you add more acid than the buffer’s capacity (run out of CH3COO–), pH will drop toward that of the strong acid. This is exactly the conjugate-base-reacts-with-added-acid idea in Topic 8.8 (see the Topic 8.8 study guide: https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD). For extra practice, check unit problems at (https://library.fiveable.me/practice/ap-chemistry).
Why do you need both the weak acid and its conjugate base for a buffer to work?
You need both the weak acid (HA) and its conjugate base (A–) because they do opposite “neutralizing” jobs that keep pH steady. When you add acid (H+), the conjugate base A– soaks it up: H+ + A– → HA. When you add base (OH–), the weak acid HA donates H+ to neutralize it: HA + OH– → A– + H2O. Having large concentrations of both members of the conjugate pair gives buffer capacity—the solution can absorb added H+ or OH– with only a small pH change. Mathematically, pH ≈ pKa + log([A–]/[HA]) (Henderson–Hasselbalch), so removing one component or making its concentration tiny makes the ratio change a lot and the buffer fails. For a quick AP-aligned review see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and extra practice (https://library.fiveable.me/practice/ap-chemistry).
How do I know which part of the buffer reacts with added acid vs added base?
Think of a buffer as a conjugate acid–base pair. The member that can donate a proton (the conjugate acid) neutralizes added base (OH−); the member that can accept a proton (the conjugate base) neutralizes added acid (H+). Example: in an acetic acid/acetate buffer (CH3CO2H/CH3CO2−) - added H+ + CH3CO2− → CH3CO2H (conjugate base reacts with added acid) - added OH− + CH3CO2H → CH3CO2− + H2O (conjugate acid reacts with added base) Quick checklist: (1) identify the acid and its conjugate base, (2) if you add acid (H+), the base form will react; (3) if you add base (OH−), the acid form will react. This is exactly the CED point (8.8.A.1). Use Henderson–Hasselbalch to predict pH changes from changes in amounts; buffer capacity is largest when [acid] ≈ [base]. For more examples and practice, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8).
What are some common examples of buffer systems we need to know for the AP exam?
Common buffers you should know for AP Chem (focus on conjugate acid–base pairs, pKa near desired pH, Henderson–Hasselbalch use): - Acetic acid / acetate (CH3CO2H / CH3CO2–). pKa ≈ 4.76 → good around pH 3.8–5.8. - Ammonium / ammonia (NH4+ / NH3). pKa (NH4+) ≈ 9.25 → good around pH 8.3–10.3. - Carbonic acid / bicarbonate (H2CO3 / HCO3–). Important for physiological/respiratory buffering near pH 6.1–8.1 (remember CO2⇌H2CO3). - Dihydrogen phosphate / hydrogen phosphate (H2PO4– / HPO4^2–). pKa2 ≈ 7.2 → good near physiological pH ~7.2. On the exam you’ll be asked to show how the conjugate base neutralizes added acid and the conjugate acid neutralizes added base (proton-transfer reactions), use Henderson–Hasselbalch to calculate pH, and discuss buffer capacity/range (±1 pH unit of pKa). For a quick topic review see the AP study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and practice problems (https://library.fiveable.me/practice/ap-chemistry).
Can a strong acid and strong base make a buffer solution?
Short answer: no—a mixture of a strong acid and a strong base does not make a buffer. Why: a buffer requires a large concentration of both members of a conjugate acid–base pair that can neutralize small additions of H+ or OH− (CED 8.8.A.1). Strong acids (like HCl) and strong bases (like NaOH) fully dissociate; when mixed they neutralize to form water and a salt whose conjugate acid or base is extremely weak. That salt + water solution won’t resist pH changes the way a weak acid + its salt (or a weak base + its salt) will. Use the Henderson–Hasselbalch idea: pH ≈ pKa + log([A−]/[HA])—you need a weak acid (with a pKa) and its conjugate base present. If you partially neutralize a weak acid with a strong base (or vice versa) you can create a buffer. For more on buffers and AP-style practice, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and Unit 8 resources (https://library.fiveable.me/ap-chemistry/unit-8).
Why is it important that buffer components are in large concentrations?
Buffers need relatively large concentrations of both members of the conjugate acid–base pair because buffer capacity (the amount of acid or base the buffer can neutralize with only a small pH change) depends on the absolute amounts, not just the ratio. The Henderson–Hasselbalch equation (pH = pKa + log([A–]/[HA])) tells you pH from the ratio, but the resistance to added H+ or OH– comes from how many moles of HA or A– are present. Larger [HA] and [A–] means more available reactant to neutralize additions, so pH changes less. Low concentrations still give the same nominal pH (if the ratio is unchanged) but have poor buffer capacity and will be driven toward the equivalence region quickly in a titration. This is why AP questions emphasize buffer capacity and neutralization reactions (CED 8.8.A). For a quick review, check the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
I don't understand the difference between a buffer and just mixing an acid with a base - what am I missing?
A buffer isn't just “an acid mixed with a base”—it’s a solution that contains a large concentration of both members of a conjugate acid–base pair (for example, CH3CO2H and CH3CO2–). The key differences: - Reaction mechanism: In a buffer the conjugate base (A–) reacts with added H+ and the conjugate acid (HA) reacts with added OH–, so added acid/base is neutralized by reversible proton-transfer, keeping pH nearly constant. - Composition matters: Buffers use a weak acid (or weak base) plus its salt so both forms are present in appreciable amounts. Simply mixing a strong acid and a strong base gives complete neutralization to water and a salt, so you lose that reversible pair and the pH can swing a lot. - Quantitative control: Buffer pH follows Henderson–Hasselbalch (pH = pKa + log([A–]/[HA])). Buffer capacity depends on absolute concentrations and works best within ±1 pKa of the weak acid (the buffer range). For AP exam connections, this addresses EK 8.8.A.1 and use of Henderson–Hasselbalch—review examples in the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD). For more practice, check Unit 8 overview (https://library.fiveable.me/ap-chemistry/unit-8) and thousands of practice problems (https://library.fiveable.me/practice/ap-chemistry).
How do you calculate the pH of a buffer solution?
Use the Henderson–Hasselbalch equation. For a buffer made from a weak acid HA and its conjugate base A− (or a weak base and its conjugate acid), pH ≈ pKa + log([A−]/[HA]). Steps: 1) Identify the conjugate pair and the acid’s Ka → find pKa = −log Ka. 2) Calculate molar concentrations (or moles/total volume) of conjugate base and acid in the final solution. 3) Substitute into Henderson–Hasselbalch and solve. Example: if you have 0.10 M HA and 0.15 M A− and pKa = 4.75, pH = 4.75 + log(0.15/0.10) = 4.92. Notes: Henderson–Hasselbalch is valid when both species are present in substantial amounts (buffer range ≈ pKa ±1). For large additions of strong acid/base, do strong-neutralization first (change moles), then use the equation. This aligns with CED keywords (pKa, conjugate pair, buffer capacity). See the Topic 8.8 study guide for examples (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).
What happens to a buffer when you keep adding more and more acid to it?
A buffer resists pH change at first because its conjugate base (A–) neutralizes added H+ (HA + H+ → HA). Using Henderson–Hasselbalch, pH ≈ pKa + log([A–]/[HA]), so adding acid converts A– → HA and slowly lowers pH as the ratio changes. Two things matter: buffer capacity (how much acid it can neutralize before the ratio shifts a lot) and buffer range (about pKa ±1). If you keep adding acid past the buffer capacity, you’ll run out of conjugate base; once A– is essentially consumed (near the equivalence point), pH will drop quickly and the solution behaves like the added strong acid (pH ≈ determined by excess H+). For AP exam framing: explain the neutralization reaction, relate pH change to Henderson–Hasselbalch, and identify when the buffer fails (conjugate base exhausted). For a refresher, see the Topic 8.8 study guide (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).
Why are buffers so important in biological systems like blood?
Buffers are crucial because they keep pH nearly constant when small amounts of acid or base are added. A buffer contains a conjugate acid–base pair (like H2CO3/HCO3− in blood). The conjugate base (A−) neutralizes added acid: A− + H+ → HA. The conjugate acid (HA) neutralizes added base: HA + OH− → A− + H2O. That’s the core CED idea (8.8.A.1). Buffer capacity depends on the absolute concentrations of HA and A− and buffer range is roughly pKa ± 1 (use Henderson–Hasselbalch to calculate pH). In biology this matters because enzymes and cellular processes need a narrow pH—human blood is ~7.40. The bicarbonate buffer (pKa ≈ 6.1 for H2CO3/HCO3−) plus respiratory and renal control lets the body resist pH shifts from CO2, metabolism, or diet. On the AP exam you might be asked to explain these proton-transfer neutralization reactions, compute pH changes with Henderson–Hasselbalch, or discuss buffer capacity (see the Topic 8.8 study guide for examples) (https://library.fiveable.me/ap-chemistry/unit-8/properties-buffers/study-guide/PlRbvlggdbKMOXSUWfmD). For more practice, check Unit 8 and the AP practice bank (https://library.fiveable.me/ap-chemistry/unit-8; https://library.fiveable.me/practice/ap-chemistry).