How do you read a weak-base titration curve?
You will be able to: Interpret a descending weak-base/strong-acid titration curve.
How do you read a weak-base titration curve?
When acid is added to ammonia, the pH falls. Around the middle of neutralization, NH₃ and NH₄⁺ coexist; at equivalence, ammonium makes the solution acidic.
A useful starting point: Why is a weak-acid equivalence point above pH 7? →
Words and symbols before equations
- Descending curve
- pH decreases as strong-acid titrant is added.
- Conjugate acid
- BH⁺ formed by protonating B.
- Equivalence salt
- Here the dissolved BH⁺ salt remaining after neutralization.
- pKa of BH⁺
- pKw−pKb of B at the same temperature.
What this picture assumes
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Initial analyte: 25.0 mL of 0.100 M solute. Titrant is 0.100 M. The pK control affects weak cases only. Additive volumes; exact monoprotic charge balance with water makes the curve continuous.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Equivalence. Added titrant 2.50 mmol; original analyte 2.50 mmol. Total volume=50.0 mL; pH=5.151. Half-equivalence is 12.5 mL; equivalence is 25.0 mL.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
At the start, weak-base equilibrium produces OH⁻. In the buffer region both B and BH⁺ are substantial, resisting a large pH change from a small titrant addition.
At half-equivalence [B]≈[BH⁺], so pOH≈pKb and pH≈pKa(BH⁺). The horizontal coordinate still measures delivered volume, even though the curve slopes downward.
At equivalence BH⁺ + H₂O ⇌ B + H₃O⁺ makes the pH lower than neutral. Use Ka=Kw/Kb and the combined volume to estimate it.
Beyond equivalence excess hydronium dominates. Comparing this curve with the weak-acid curve clarifies why the same word equivalence does not imply the same pH.
A worked example, step by step
Titrate 25.0 mL of 0.100 M B, Kb=1.0×10⁻⁵, with 0.100 M HCl at 25 °C. Estimate pH at half-equivalence and equivalence.
- Equivalence occurs at 25.0 mL; half-equivalence is 12.5 mL.
- pOH≈pKb=5.00 at half-equivalence, so pH≈9.00.
- At equivalence BH⁺ analytical concentration=2.50 mmol/50.0 mL=0.0500 M; Ka=10⁻⁹.
- [H₃O⁺]≈√(10⁻⁹×0.0500)=7.07×10⁻⁶ M, so pH≈5.15.
Do not apply the weak-acid rule pH=pKa using the base’s pKb value.
At weak-base equivalence, must excess strong acid be present to make the pH acidic?
Compare with an explanation
No. BH⁺ hydrolysis supplies hydronium even with exact stoichiometric matching.
Predict. Change one thing. Explain.
Select weak base + strong acid and pK=5 (pKb). Inspect 12.5 and 25 mL. Compare the descending curve with the weak-acid case at the same supplied constant.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Equivalence. Added titrant 2.50 mmol; original analyte 2.50 mmol. Total volume=50.0 mL; pH=5.151. Half-equivalence is 12.5 mL; equivalence is 25.0 mL.
Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Initial analyte: 25.0 mL of 0.100 M solute. Titrant is 0.100 M. The pK control affects weak cases only. Additive volumes; exact monoprotic charge balance with water makes the curve continuous.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using proton transfer, charge and atom conservation, a mole balance or the stated acid–base equilibrium. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA weak-base curve reaches equivalence at 30.0 mL. At 15.0 mL pH=9.50 at 25 °C. Estimate pKb and Kb; predict equivalence relative to neutrality.
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Compare with the answer and four-point rubric
- 1 point: 15.0 mL is half-equivalence.
- 1 point: pOH=14.00−9.50=4.50, hence pKb≈4.50.
- 1 point: Kb≈10⁻⁴⋅⁵=3.16×10⁻⁵.
- 1 point: Equivalence is below neutral because BH⁺ donates protons to water.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What is the curve’s direction here?
Downward as acid is added.
RECALL 2What does half-equivalence reveal?
pKb from pOH, or conjugate-acid pKa from pH.
RECALL 3Which species hydrolyzes at equivalence?
BH⁺.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you read a weak-base titration curve?
- Weak-base half-equivalence: pOH≈pKb; pH≈pKa(BH⁺).
- At equivalence use BH⁺ acid dissociation.
Remember: Do not apply the weak-acid rule pH=pKa using the base’s pKb value.
Conditions: Dilute ideal-solution concentration model at 25 °C, Kw=1.00×10⁻¹⁴. Concentrations are mol/L (M); displayed values are rounded. No household experiments are required. Initial analyte: 25.0 mL of 0.100 M solute. Titrant is 0.100 M. The pK control affects weak cases only. Additive volumes; exact monoprotic charge balance with water makes the curve continuous.
Refresh Kid · AP Chemistry Unit 8 · Objectives 8.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.5, objective 8.5.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 8: Acids and Bases, Topics 8.1–8.11. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Dilute ideal-solution concentrations approximate activities; numerical models use 25 °C and Kw=1.00×10⁻¹⁴ unless another pKw is supplied. pH need not be restricted to 0–14 in all real solutions. The optional 3D views show original schematic molecular geometry, not a measured trajectory or a reaction mechanism. Computation of a buffer’s pH change after adding acid/base, derivation of Henderson–Hasselbalch, concentrations of every species in a polyprotic titration, and solubility as a function of pH are excluded from assessed scope. Buffer response and pH-dependent solubility are taught qualitatively. Calculating the pH of a buffer formed by partial neutralization remains in Topic 8.4 scope.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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