Acid-Base Reactions and Buffers
Connect proton transfer to a mixed-solution species ledger before the buffer and titration chapter.
AP Chemistry Topic 8.4 · CED printed p. 150, 151 / PDF p. 157, 158. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Stoichiometry and limiting reagents
- Weak equilibria
- Conjugate constants
When acid and base are mixed, start with moles and the balanced proton-transfer equation. Strong acid plus strong base reacts essentially quantitatively: H3O+ + OH− → 2H2O. Compare their moles, subtract the limiting amount, and divide any excess by the total volume. Excess acid gives hydronium; excess base gives hydroxide. At exact stoichiometric equivalence, with nonreactive spectator ions and at 25 °C, the solution is neutral. Equal volumes alone do not guarantee equivalence.
Weak acid plus strong base also reacts essentially quantitatively in the usual AP case: HA + OH− → A− + H2O. If HA remains alongside the A− made, you have a buffer. At equivalence, mostly A− remains, and A− + H2O ⇌ HA + OH− makes the solution basic. Beyond equivalence, excess strong hydroxide controls pH. Select the model after bookkeeping, not before. Source clarification: the supplied CED’s printed 8.4.A.2 equation omits the plus sign between A− and H2O; the chemically complete equation is shown here.
Weak base plus strong acid follows the mirror logic: B + H3O+ → HB+ + H2O. Excess B leaves a B/HB+ buffer. At equivalence the conjugate acid HB+ reacts with water and makes the solution acidic. Beyond equivalence, excess hydronium dominates. Conjugate constants come from KaKb = Kw. The phrase “mostly HB+” permits the small amount of B regenerated by equilibrium; quantitative bookkeeping is the first stage, not a claim that subsequent equilibrium stops.
Weak acid plus weak base instead needs its own equilibrium: HA + B ⇌ A− + HB+. The reaction favors the side containing the weaker acid and weaker base. You may connect its K to Ka(HA)/Ka(HB+) by adding equilibrium equations, but do not assume every weak–weak mixture goes to completion. In all mixtures distinguish formal amounts after the fast reaction from the final equilibrium concentrations. Charge balance and atom balance remain useful checks.
Let's work through it
- At 25 °C, mix 20.0 mL 0.100 M B with 20.0 mL 0.100 M HCl. B + H3O+ → HB+ + H2O consumes 2.00 mmol of each.
- The formal HB+ concentration is 2.00 mmol/40.0 mL = 0.0500 M. There is no excess strong acid, but this does not imply neutrality.
- If Kb(B) = 1.00 × 10−5, Ka(HB+) = 1.00 × 10−9. Solve HB+ + H2O ⇌ B + H3O+. Approximate [H3O+] = √(1.00 × 10−9 × 0.0500) = 7.07 × 10−6 M.
- The depletion is only 0.0141%, and hydronium exceeds water’s neutral value substantially. pH ≈ 5.15. HB+ is the major base-related species; B and hydronium are minor.
Worked strong-acid/strong-base mixture at 25 °C.
- 20.0 mL 0.100 M HCl supplies 2.00 mmol hydronium. 30.0 mL 0.0500 M NaOH supplies 1.50 mmol hydroxide.
- H3O+ + OH− → 2H2O consumes the limiting 1.50 mmol of each, leaving 0.500 mmol hydronium.
- The final volume is 50.0 mL, so [H3O+] ≈ 0.0100 M and pH = 2.000. Spectator ions remain; strong acid is in excess.
- If the two initial amounts had instead been equal, only nonreactive spectators and water would remain and the ideal equivalence solution would be neutral at this temperature.
Worked weak-acid/strong-base regimes at 25 °C: separate 20.0 mL samples of 0.100 M HA, Ka = 1.00 × 10−5, receive 0.100 M NaOH.
- With 10.0 mL titrant, the initial 2.00 mmol HA reacts with 1.00 mmol OH− by HA + OH− → A− + H2O. There are 1.00 mmol HA and 1.00 mmol A− afterward, so this is a buffer with comparable forms. The hydronium is not the original formal acid concentration.
- With 20.0 mL titrant, 2.00 mmol OH− consumes the initial HA amount. Formal [A−] = 2.00 mmol/40.0 mL = 0.0500 M. A− + H2O ⇌ HA + OH− sets the pH; Kb = 1.00 × 10−9.
- At this equivalence composition, [OH−] ≈ √(1.00 × 10−9 × 0.0500) = 7.07 × 10−6 M; the fractional depletion is tiny and pH ≈ 8.85. No excess strong base is needed for a basic equivalence solution.
- With 30.0 mL titrant, 1.00 mmol OH− remains in 50.0 mL after reaction, so excess [OH−] = 0.0200 M and pH = 12.301. Identify the remaining species before choosing each model. These are separate current-composition calculations, not a calculation of pH change after adding acid/base to a prepared buffer.
Worked weak-acid/weak-base equilibrium model: HA + B ⇌ A− + HB+, Ka(HA) = 1.00 × 10−6 and Ka(HB+) = 1.00 × 10−4.
- Add HA dissociation to the reverse HB+ dissociation. The resulting proton-transfer K is 1.00 × 10−6/1.00 × 10−4 = 0.0100; the reactant side is favored.
- In a simplified ideal mixture initially 0.100 M in each reactant with negligible initial products, let x transfer. Then [HA] = [B] = 0.100 − x and [A−] = [HB+] = x.
- K = x²/(0.100 − x)² = 0.0100. The nonnegative concentration ratio is x/(0.100 − x) = 0.100, giving x = 0.00909 M.
- Each reactant remains about 0.0909 M and each product is about 0.00909 M in this proton-transfer model. Quantitative neutralization would be a poor approximation; only about 9.09% transfers. The model neglects water autoionization and subsidiary individual ionization changes; it illustrates why relative weak-pair strengths determine equilibrium extent.
Make the picture move
Interactive concept model
The species change across a titration curve.
Predict first. Visit 0, 12.5, 25.0, and 37.5 mL. Identify the species controlling pH at each point. Is equivalence necessarily pH 7?
HA + OH⁻ → A⁻ + H₂O; Kₐ = [H⁺][A⁻]/[HA]; K_w = [H⁺][OH⁻]
- Ideal weak-acid titration
- Labeled selected or reference state
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
At 12.5 mL, pH = 4.74. Half-equivalence: pH ≈ pKₐ. [HA] = 0.0333 M and [A⁻] = 0.0334 M.
Read the data and comparisons
| Added base (mL) | pH | [HA] (mol L⁻¹) | [A⁻] (mol L⁻¹) |
|---|---|---|---|
| 0.0 | 2.87 | 0.0987 | 0.00134 |
| 12.5 | 4.74 | 0.0333 | 0.0334 |
| 25.0 | 8.72 | 5.24 × 10^-6 | 0.05 |
| 37.5 | 12.30 | 1.10 × 10^-9 | 0.04 |
| 50.0 | 12.52 | 5.50 × 10^-10 | 0.0333 |
Macroscopic / data. The pH curve is a computed data representation, with a changing total liquid volume.
Particle view. HA, A⁻, H⁺, OH⁻, and spectator Na⁺ satisfy matter and charge conservation. Equivalence does not remove the hydrolysis equilibrium.
Symbolic view. Total acid-species concentration = 0.0667 M; [H⁺] = 1.82 × 10^-5 M; [OH⁻] = 5.50 × 10^-10 M.
Why it works. The initial 25.0 mL of 0.100 M HA contains 0.00250 mol of acid. Equivalence takes 25.0 mL of 0.100 M base regardless of pKₐ. At half-equivalence, comparable HA and A⁻ give pH approximately pKₐ. Within the buffer region, pH = pKₐ + log₁₀([A⁻]/[HA]) follows by rearranging Kₐ; estimating this ratio from neutralization requires appreciable amounts of both species. At equivalence A⁻ reacts with water, so a weak-acid titration is basic. Beyond equivalence excess OH⁻ dominates.
Conditions and limits. Monoprotic hypothetical HA and a fully dissociated strong base with Na⁺ as spectator; 25 °C, K_w = 1.0 × 10⁻¹⁴, ideal dilute activities approximated by concentrations divided by a 1 mol L⁻¹ standard state, additive volumes, no other equilibria. The curve solves acid mass balance, charge balance, Kₐ, and K_w together at every volume; it does not misuse Henderson–Hasselbalch at endpoints. The numerical solver is an implementation detail, not required AP work. pH is displayed to two decimal places.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
After HA reacts with strong hydroxide, what remaining species distinguish a buffer from equivalence and excess-base conditions?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
After HA reacts with strong hydroxide, what remaining species distinguish a buffer from equivalence and excess-base conditions?
Worked answer and reason
Appreciable HA and A− together identify a buffer; mostly A− at stoichiometric equivalence requires its weak-base equilibrium; excess OH− beyond equivalence determines pH from its remaining moles and total volume.
The neutralization bookkeeping identifies the remaining species before choosing the pH model.
Retrieval 2
What determines pH beyond weak-acid/strong-base equivalence?
Worked answer and reason
The concentration of excess strong hydroxide, using the total volume.
The excess strong reagent dominates the much weaker conjugate-base contribution.
Catch the tempting mistake
- No excess strong reagent means pH 7.
- A remaining conjugate acid or base can react with water and set an acidic or basic pH.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 8 problem 5 · ap application · symbolic
- Unit 8 problem 6 · ap synthesis · multiple
- Unit 8 problem 7 · ap synthesis · multiple
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For the teacher
- Have students draw a decision sequence: react moles, identify remaining major species, choose the equilibrium or excess-ion model.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 8.4.A
- Explain the relationship among the concentrations of major species in a mixture of weak and strong acids and bases.
- 8.4.A.1
- When a strong acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: H+(aq) + OH−(aq) → H2O(l). The pH of the resulting solution may be determined from the concentration of excess reagent.
- 8.4.A.2
- When a weak acid and a strong base are mixed, they react quantitatively in a reaction represented by the equation: HA(aq) + OH^−(aq) ⇌ A^−(aq) H_2O(l). If the weak acid is in excess, then a buffer solution is formed, and the pH can be determined from the Henderson-Hasselbalch (H−H) equation (see 8.9.A.1). If the strong base is in excess, then the pH can be determined from the moles of excess hydroxide ion and the total volume of solution. If they are equimolar, then the (slightly basic) pH can be determined from the equilibrium represented by the equation: A^−(aq) + H_2O(l) ⇌ HA(aq) + OH^−(aq).
- 8.4.A.3
- When a weak base and a strong acid are mixed, they will react quantitatively in a reaction represented by the equation: B(aq) + H3O+(aq) HB+(aq) + H2O(l). If the weak base is in excess, then a buffer solution is formed, and the pH can be determined from the H−H equation. If the strong acid is in excess, then the pH can be determined from the moles of excess hydronium ion and the total volume of solution. If they are equimolar, then the (slightly acidic) pH can be determined from the equilibrium represented by the equation: HB+(aq) + H2O(l) B(aq) + H3O+(aq).
- 8.4.A.4
- When a weak acid and a weak base are mixed, they will react to an equilibrium state whose reaction may be represented by the equation: HA(aq) + B(aq) A−(aq) + HB+(aq).
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §14.3, Relative Strengths of Acids and Bases — Conjugate equilibria, percent ionization, structure trends, and the incomplete second ionization of sulfuric acid.
Glossary and deeper reading · 10 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Stoichiometry
Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.
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Limiting reactant
The limiting reactant is consumed first under the specified complete-reaction model and limits the product amount. Identifying it requires comparing available amounts with the balanced equation's ratios. For example, abundant hydrogen cannot produce more water after the available oxygen has been consumed.
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Mole
A mole is an amount of substance containing exactly 6.02214076 × 10²³ specified entities. The entities can be atoms, molecules, ions, or formula units, so the counted entity must be named. For example, one mole of oxygen molecules contains two moles of oxygen atoms.
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Brønsted acid and base
A Brønsted acid donates a proton, and a Brønsted base accepts a proton. These roles describe a particular reaction, so some substances can act in either role. For example, water accepts a proton from hydrogen chloride but donates one to ammonia.
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Weak acid or base
A weak acid or base reacts only partially with water and establishes an equilibrium. Its equilibrium constant describes strength, while analytical concentration describes how much was prepared per volume. For example, acetic acid leaves both acid molecules and its conjugate base in water.
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Strong acid or base
A strong acid transfers protons essentially completely to water in the applicable aqueous-solution model. A dissolved strong hydroxide base dissociates into its constituent ions and supplies hydroxide; it does not have to react with water to generate that hydroxide. Strength is distinct from concentration and, for a solid base, from how much can dissolve. For example, dissolved sodium hydroxide supplies sodium and hydroxide ions even when the prepared solution is dilute.
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Buffer
A buffer contains appreciable amounts of a weak acid and its conjugate base, or the corresponding weak-base pair. Those partners consume small added amounts of strong base or acid, limiting the pH change within the buffer's capacity. For example, an acetic acid–acetate mixture can resist modest additions of acid or base.
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Conjugate acid–base pair
A conjugate acid–base pair consists of species differing by one proton. Losing a proton forms an acid's conjugate base, while gaining one forms a base's conjugate acid. For example, ammonium and ammonia form a conjugate acid–base pair.
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.
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Mixture
A mixture contains two or more substances whose proportions can vary. Its components retain their chemical identities until a chemical reaction changes them. For example, water and ethanol can form mixtures with many different compositions.