Weak Acid and Base Equilibria
Solve weak-acid and weak-base equilibria, check depletion and connect conjugate constants.
AP Chemistry Topic 8.3 · CED printed p. 148, 149 / PDF p. 155, 156. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Ka/Kb expressions
- ICE tables
- Logarithms
A weak acid reacts only partly with water. For HA + H2O ⇌ H3O+ + A−, . If only HA was added and water’s ion contribution is negligible, [HA] = C − x and both formed ion concentrations are x. Solve Ka = x2/(C − x). Most HA remains un-ionized in the usual weak-acid regime, so using C directly as hydronium would give the wrong pH. pKa = −log Ka; a smaller pKa means a larger Ka and a stronger acid under comparable conditions.
A weak base accepts a proton from water: B + H2O ⇌ HB+ + OH−. Its constant is Kb = [HB+][OH−]/[B], and pKb = −log Kb. Set the formed hydroxide concentration to x, solve Kb = x2/(C − x), calculate pOH, and then convert with that temperature’s pKw. Most B remains unprotonated in the usual weak-base regime. Do not use a weak-base hydroxide concentration directly as hydronium, and do not assume the original B concentration equals either formed ion concentration.
Percent ionization is the amount converted divided by the initial amount, multiplied by 100%. Under the simple single-solute assumptions it is (x/C) × 100%. If pH is measured, first convert it back to hydronium before computing an acid’s converted fraction. The initial concentration belongs in the denominator, not C − x. A small-x shortcut x ≈ √(KaC) or √(KbC) needs a fractional-change check. If that fails, solve the full equation. At extreme dilution, water’s contribution matters too. Dilution generally increases a weak acid/base’s fraction converted even while it decreases the concentration of the produced hydronium/hydroxide; the constant itself stays fixed at fixed T.
A conjugate pair differs by one proton. At a given temperature its constants satisfy KaKb = Kw and pKa + pKb = pKw. A stronger acid therefore has a weaker conjugate base. Use this to find the basicity of A− or acidity of HB+ in a salt solution. Apply it only to actual conjugate partners. Acid–base “strength” is about the equilibrium extent, not the rate of proton transfer or a solution’s analytical concentration.
Let's work through it
- A weak acid has C = 0.0500 M and Ka = 2.00 × 10−4. Write Ka = x2/(0.0500 − x).
- A trial shortcut gives x = 0.00316 M, or 6.32% of C. That exceeds a 5% check, so keep depletion in the equation.
- Solve x2 + 2.00 × 10−4x − 1.00 × 10−5 = 0. The positive root is 0.00306 M; pH = 2.514 and ionization is 6.13%.
- At 25 °C, [OH−] = Kw/x = 3.26 × 10−12 M, and Kb of A− = 5.00 × 10−11. Check mass balance [HA] + [A−] = C.
Worked weak-base example at 25 °C: C(B) = 0.0100 M and Kb = 1.00 × 10−5.
- Write B + H2O ⇌ HB+ + OH−. Let x be the concentration converted; [B] = 0.0100 − x and [HB+] ≈ [OH−] = x when the water contribution is negligible.
- Kb = x²/(0.0100 − x). A trial approximation gives x ≈ √(1.00 × 10−5 × 0.0100) = 3.16 × 10−4 M. The change is 3.16% of C, so the small-change estimate is reasonable.
- The approximate pOH is 3.50 and pH is 10.50. Most B remains unprotonated, while the produced HB+ and OH− are much less concentrated. The ion being calculated directly is hydroxide, not hydronium.
- The approximate converted fraction is 3.16%; Ka(HB+) = Kw/Kb = 1.00 × 10−9 and pKb = 5.00. If more precision is required, keep the full denominator and solve its quadratic.
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
Why is pH not −log C for a weak acid?
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
Why is pH not −log C for a weak acid?
Worked answer and reason
Only part of the acid ionizes, so hydronium is x, not the total analytical acid concentration.
The equilibrium divides the acid between HA and A−.
Retrieval 2
Which base belongs in KaKb = Kw for HA?
Worked answer and reason
Its conjugate base A−, which differs by one proton.
The two paired equilibria sum to water autoionization.
Catch the tempting mistake
- A weak acid can never be substantially ionized.
- At sufficiently low concentration its ionized fraction can be large; weak refers to Ka, not a universal fixed percentage.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 8 problem 3 · ap synthesis · multiple
- Unit 8 problem 4 · ap application · particle
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For the teacher
- Require students to state what x represents, test a neglected term, and distinguish fractional ionization from concentration.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 8.3.A
- Explain the relationship among pH, pOH, and concentrations of all species in a solution of a monoprotic weak acid or weak base.
- 8.3.A.1
- Weak acids react with water to produce hydronium ions. However, only a small percentage of molecules of a weak acid will ionize in this way. Thus, the concentration of H_3O^+ is much less than the initial concentration of the molecular acid, and the vast majority of the acid molecules remain un-ionized.
- 8.3.A.2
- A solution of a weak acid involves equilibrium between an un-ionized acid and its conjugate base. The equilibrium constant for this reaction is K_a, often reported as pK_a. The pH of a weak acid solution can be determined from the initial acid concentration and the pK_a. EQN: K_a = [H_3O^+][A^−]/[HA] EQN: pK_a = −log K_a
- 8.3.A.3
- Weak bases react with water to produce hydroxide ions in solution. However, ordinarily just a small percentage of the molecules of a weak base in solution will ionize in this way. Thus, the concentration of OH− in the solution does not equal the initial concentration of the base, and the vast majority of the base molecules remain un-ionized.
- 8.3.A.4
- A solution of a weak base involves equilibrium between an un-ionized base and its conjugate acid. The equilibrium constant for this reaction is K_b, often reported as pK_b. The pH of a weak base solution can be determined from the initial base concentration and the pK_b. EQN: K_b = [OH^−][HB^+]/[B] EQN: pK_b = −log K_b
- 8.3.A.5
- The percent ionization of a weak acid (or base) can be calculated from its pK_a (pK_b) and the initial concentration of the acid (base). The percent ionization can also be calculated from the initial concentration of the acid (base) and the equilibrium concentration of any of the species in the equilibrium expression.
- 8.3.A.6
- For any conjugate acid-base pair, the acid ionization constant and base ionization constant are related by K_w: EQN: K_w = K_a × K_b EQN: pK_w = pK_a + pK_b
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 · 8 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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ICE table
An ICE table organizes initial concentrations, stoichiometric changes, and equilibrium concentrations. Its change entries follow the balanced equation and do not imply equal changes for every species. For example, producing two units of a product for each reactant consumed requires twice the concentration-change entry for that product.
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Ion
An ion is an atom or group of atoms with a net electric charge. Ordinary ion formation changes electron count without changing the element's proton count. For example, a sodium atom losing one electron becomes a sodium ion with charge +1.
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Hydronium
Hydronium is the H₃O⁺ ion used in the introductory model of proton behavior in water. Aqueous H⁺ notation is shorthand for a solvated proton rather than a bare proton moving independently in the liquid. For example, a Brønsted acid can transfer a proton to water and form hydronium.
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pOH
pOH is the negative base-ten logarithm of hydroxide activity. At a specified temperature, pH plus pOH equals the corresponding pKw, which is approximately 14 at 25 °C. For example, increasing hydroxide activity tenfold lowers pOH by one unit.
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Percent ionization
Percent ionization is the fraction of an initially supplied acid or base that forms the relevant ions, multiplied by 100. It depends on concentration and conditions as well as the species' equilibrium constant. For example, a weak acid can have a larger ionized fraction after dilution even though its hydronium concentration decreases.
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Solute
A solute is a component dispersed in a solvent to form a solution. It need not be a solid, and a solution need not be liquid. For example, carbon dioxide is a gaseous solute in carbonated water.
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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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pKa
pKa is the negative base-ten logarithm of an acid's dissociation constant. A smaller pKa means a stronger acid under comparable conditions. For example, when an acid and its conjugate base have equal activities, their buffer's pH equals that acid's pKa.