Which ions remain after strong acid and base are mixed?
You will be able to: Use moles, limiting reactant and total volume before calculating mixture pH.
Which ions remain after strong acid and base are mixed?
Mixing equal volumes does not guarantee neutrality. If the acid is twice as concentrated as the base, equal volumes leave excess acid.
A useful starting point: Why can a salt solution be acidic or basic? →
Words and symbols before equations
- Neutralization
- H₃O⁺ + OH⁻ → 2H₂O in this strong-acid/base case.
- Millimole, mmol
- One thousandth of a mole; M×mL gives mmol.
- Excess
- Amount remaining after the limiting reactant is consumed.
- Total volume
- Combined solution volume, assumed additive here.
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 sample: 25.0 mL of 0.100 M HCl. Volumes are additive; no other reactive solutes. At equivalence water supplies finite ion concentrations.
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.
- Excess strong base. Added titrant 3.00 mmol; original analyte 2.50 mmol. Total volume=55.0 mL; pH=11.959. Reactive excess 0.50 mmol.
- 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 strongly favored neutralization removes equal amounts of hydronium and hydroxide. Determine their initial moles using concentration times volume, including any hydroxide coefficient.
Subtract moles, not concentrations or pH values. The larger initial amount determines the excess ion.
Divide the excess amount by the combined volume, then calculate pH or pOH. At exact equivalence, use water equilibrium instead of taking log(0).
This method assumes strong monoprotic acid, a strong hydroxide, additive volumes and no other acid/base-active solutes. At very small excess, water’s contribution matters; the explorer handles that boundary.
A worked example, step by step
Mix 20.0 mL of 0.100 M HCl with 30.0 mL of 0.100 M NaOH at 25 °C. Find pH.
- H₃O⁺ amount=2.00 mmol; OH⁻ amount=3.00 mmol.
- Neutralization leaves 1.00 mmol OH⁻.
- Total volume=50.0 mL; [OH⁻]≈1.00/50.0=0.0200 M.
- pOH=1.699 and pH=12.301; the larger base amount controls the result.
Do not average the two starting pH values. React amounts first, then account for dilution.
Is a mixture neutral whenever acid and base volumes match?
Compare with an explanation
Only if the reactive mole amounts also match, with the correct stoichiometry.
Predict. Change one thing. Explain.
Keep 25 mL of 0.100 M acid fixed and vary the volume of 0.100 M base through 25 mL. Predict the excess species on either side and inspect exact equivalence.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Excess strong base. Added titrant 3.00 mmol; original analyte 2.50 mmol. Total volume=55.0 mL; pH=11.959. Reactive excess 0.50 mmol.
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 sample: 25.0 mL of 0.100 M HCl. Volumes are additive; no other reactive solutes. At equivalence water supplies finite ion concentrations.
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 questionMix 10.0 mL of 0.100 M HCl and 20.0 mL of 0.0200 M NaOH at 25 °C. Calculate the resulting pH.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Acid=1.00 mmol and base=0.400 mmol.
- 1 point: Excess hydronium=0.600 mmol.
- 1 point: [H₃O⁺]≈0.600/30.0=0.0200 M.
- 1 point: pH≈1.699; total volume must include both solutions.
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 do you subtract?
Reactive mole amounts.
RECALL 2Which volume is used after mixing?
The total solution volume.
RECALL 3Why is equivalence not log(0)?
Water still supplies hydronium and hydroxide.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Which ions remain after strong acid and base are mixed?
- n=CV; use consistent units.
- After reaction: excess concentration=excess amount/total volume.
Remember: Do not average the two starting pH values. React amounts first, then account for dilution.
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 sample: 25.0 mL of 0.100 M HCl. Volumes are additive; no other reactive solutes. At equivalence water supplies finite ion concentrations.
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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