Which form predominates at a given pH?
You will be able to: Compare pH with pKa to identify protonated and deprotonated forms.
Which form predominates at a given pH?
A weak acid can be mostly HA in one solution and mostly A⁻ in another. Compare the solution’s pH with that acid’s pKa to determine which form is more abundant.
A useful starting point: How does ammonia change when it accepts a proton? →
Words and symbols before equations
- Predominant
- Present at the larger concentration, not the only species present.
- Protonated form
- The conjugate acid, such as HA or BH⁺.
- Deprotonated form
- The conjugate base, such as A⁻ or B.
- Distribution fraction
- One form’s amount divided by the total of both forms.
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. An imposed pH and a single ideal conjugate pair determine the fractions. This is not a claim that a trace solute sets the solution pH.
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.
- pH=6.00, pKa=5.00: A⁻/HA=10. HA 9.1%; A⁻ 90.9%. Both forms coexist; predominance is not exclusivity.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
For HA ⇌ H⁺+A⁻, Ka≈[H⁺][A⁻]/[HA]. Rearranging this equilibrium ratio gives [A⁻]/[HA]≈10^(pH−pKa). This is a species comparison at an imposed solution pH, not an assumption that adding a trace acid sets that pH.
When pH<pKa, the ratio is below one and HA predominates. When pH>pKa, A⁻ predominates. At equality, both concentrations are equal in this ideal two-form model.
The same reasoning applies to BH⁺/B: compare pH to pKa of BH⁺, not pKb of B. Higher pH favors the form with one fewer proton.
Predominant is not exclusive. At pH one unit above pKa the ratio is 10:1, so the base fraction is 10/11≈91%, not 100%.
A worked example, step by step
For a weak acid with pKa=5.00 in a solution held at pH 6.00, find the ratio A⁻/HA and the predominant form.
- Calculate pH−pKa=1.00.
- [A⁻]/[HA]≈10¹=10.
- A⁻ is ten times as concentrated as HA.
- Its fraction is 10/(10+1)=0.909, about 91%; some HA remains.
Use the conjugate acid’s pKa even when the base is neutral, and do not turn predominance into an all-or-nothing claim.
For BH⁺ with pKa=9 at pH 7, which form predominates?
Compare with an explanation
BH⁺, the protonated acid form, because pH is below its pKa.
Predict. Change one thing. Explain.
Keep pKa=5 and move pH from 3 to 7. Read both fractions at pH 4, 5 and 6; explain why neither form disappears abruptly at pKa.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
pH=6.00, pKa=5.00: A⁻/HA=10. HA 9.1%; A⁻ 90.9%. Both forms coexist; predominance is not exclusivity.
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. An imposed pH and a single ideal conjugate pair determine the fractions. This is not a claim that a trace solute sets the solution pH.
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 BH⁺/B pair has pKa=8.00. At imposed pH 9.00, identify the predominant form, its ratio to the other form and its approximate fraction.
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Compare with the answer and four-point rubric
- 1 point: The pH is one unit above the conjugate-acid pKa.
- 1 point: B/BH⁺≈10.
- 1 point: The neutral base B predominates.
- 1 point: Its fraction is 10/11≈91%; BH⁺ is still present.
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 happens at pH=pKa?
Equal conjugate-form concentrations in the ideal model.
RECALL 2Which form is favored at higher pH?
The form with one fewer proton.
RECALL 3Does 10:1 mean 100%?
No; it means about 91% of the total.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Which form predominates at a given pH?
- pH<pKa: acid form predominates; pH>pKa: base form predominates.
- Base/acid≈10^(pH−pKa).
Remember: Use the conjugate acid’s pKa even when the base is neutral, and do not turn predominance into an all-or-nothing claim.
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. An imposed pH and a single ideal conjugate pair determine the fractions. This is not a claim that a trace solute sets the solution pH.
Refresh Kid · AP Chemistry Unit 8 · Objectives 8.7.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.7, objective 8.7.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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