How can adding base to a weak acid create a buffer?
You will be able to: Track weak-acid neutralization before using an equilibrium relation.
How can adding base to a weak acid create a buffer?
Start with 10 mmol of a weak acid and add 4 mmol hydroxide. Four mmol of acid becomes its conjugate base, leaving both forms in the same solution.
A useful starting point: Which ions remain after strong acid and base are mixed? →
Words and symbols before equations
- Partial neutralization
- Strong base consumes only part of the initial weak acid.
- Buffer
- A solution with appreciable amounts of a weak conjugate pair that resists small acid/base additions.
- Stoichiometry first
- Determine reaction amounts before treating the remaining equilibrium.
- Base/acid ratio
- Amount or concentration of conjugate base divided by conjugate acid.
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.
- Before equivalence: conjugate pair forms. Added titrant 1.25 mmol; original analyte 2.50 mmol. Total volume=37.5 mL; pH=5.000. 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
The reaction HA + OH⁻ → A⁻ + H₂O is strongly favored. Before equivalence, added OH⁻ is consumed and creates A⁻ while decreasing HA.
With substantial HA and A⁻ remaining, the mixture is a buffer. The Henderson–Hasselbalch relation estimates pH from the ratio of their concentrations. Because both occupy the same volume, their mole ratio is also valid.
At equivalence no stoichiometric HA remains; conjugate-base hydrolysis controls pH. Beyond equivalence excess strong base dominates. Do not apply a logarithm to a zero or negative HA amount.
This calculation finds the pH of a buffer formed by partial neutralization, within Topic 8.4. Computing numerical pH changes after disturbing an existing buffer is excluded from assessed Topic 8.9 scope.
A worked example, step by step
A weak acid with pKa=5.00 initially has 10.0 mmol HA. Add 5.00 mmol NaOH. Estimate the pH after forming the buffer.
- React HA + OH⁻ → A⁻ + H₂O.
- Remaining HA=5.00 mmol; formed A⁻=5.00 mmol.
- Both forms share the same total volume, so [A⁻]/[HA]=1.
- pH≈5.00+log(1)=5.00; this is half-neutralization.
Use the amounts after the strong reaction, not the initial acid and the initially added hydroxide as though they were conjugate partners.
If all HA is neutralized, can the original buffer formula use nHA=0?
Compare with an explanation
No. Use conjugate-base hydrolysis at equivalence, not a ratio with zero denominator.
Predict. Change one thing. Explain.
Select weak acid + strong base. Inspect 0, 12.5, 25 and 35 mL added base. Identify weak-acid, buffer, conjugate-base and excess-base regions.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Before equivalence: conjugate pair forms. Added titrant 1.25 mmol; original analyte 2.50 mmol. Total volume=37.5 mL; pH=5.000. 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 questionStart with 12.0 mmol HA, pKa=4.00. Add 4.00 mmol NaOH. Estimate the buffer pH and explain why the reaction must precede the logarithm.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: The hydroxide consumes 4.00 mmol HA.
- 1 point: Remaining HA=8.00 mmol and formed A⁻=4.00 mmol.
- 1 point: pH≈4.00+log(4/8)=3.70.
- 1 point: The logarithm needs the conjugate-pair amounts after neutralization, not the original reactant amounts.
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 creates A⁻ here?
Neutralization of HA by hydroxide.
RECALL 2What defines half-neutralization?
Half the initial HA has converted to A⁻.
RECALL 3What governs exact equivalence?
Hydrolysis of the conjugate base.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How can adding base to a weak acid create a buffer?
- Before equivalence: nHA=nHA,initial−nOH; nA=nOH.
- When both forms are substantial: pH≈pKa+log(nA/nHA).
Remember: Use the amounts after the strong reaction, not the initial acid and the initially added hydroxide as though they were conjugate partners.
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 8.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.4, objective 8.4.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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