Why can one mole of a base supply two moles of hydroxide?
You will be able to: Use dissociation coefficients and pOH to calculate strong-base pH.
Why can one mole of a base supply two moles of hydroxide?
One dissolved formula unit of Ca(OH)₂ supplies two hydroxide ions. Counting formula units without the coefficient would miss half of the hydroxide.
A useful starting point: How do you calculate the pH of a strong acid? →
Words and symbols before equations
- Dissociation
- Separation of a dissolved ionic substance into ions.
- Coefficient
- Balanced multiplier giving the mole relationship.
- Hydroxide, OH⁻
- The ion whose concentration sets pOH.
- Dissolved concentration
- Concentration actually in solution, not undissolved solid added.
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. The coefficient applies only to the base. Water autoionization is included so extreme dilution approaches neutrality from the correct side.
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.
- Dissolved formula concentration 0.01 M; 2 OH⁻ per formula unit. pH=12.301, pOH=1.699. Water is included near neutrality.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
NaOH(aq) supplies one OH⁻ per formula unit; Ca(OH)₂(aq) supplies two. Use the dissolved concentration and balanced dissociation before taking a logarithm.
First find [OH⁻], then pOH=−log[OH⁻]. At 25 °C convert with pH=14.00−pOH.
A strong base reacts extensively when dissolved, but this does not mean any amount of its solid will dissolve. Some group 2 hydroxides have limited solubility.
The numerical model includes water near neutrality. At ordinary concentrations the coefficient method gives the same rounded result; dilution of a base approaches neutral pH from above.
A worked example, step by step
A solution contains 0.0020 M dissolved Ca(OH)₂ at 25 °C. Find [OH⁻], pOH and pH.
- Write Ca(OH)₂ → Ca²⁺ + 2OH⁻.
- [OH⁻]≈2×0.0020=0.0040 M.
- pOH=−log(0.0040)=2.40.
- pH=14.00−2.40=11.60; the solution is basic.
The logarithm of hydroxide gives pOH, not pH. Use the dissolved amount, not the amount of solid originally added.
What does 0.0050 M NaOH contribute?
Compare with an explanation
Approximately 0.0050 M OH⁻ because the dissociation coefficient is one.
Predict. Change one thing. Explain.
Select strong hydroxide and compare coefficients 1 and 2 at the same concentration. Explain why doubling hydroxide raises pH by about 0.30, not by 2.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Dissolved formula concentration 0.01 M; 2 OH⁻ per formula unit. pH=12.301, pOH=1.699. Water is included near neutrality.
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. The coefficient applies only to the base. Water autoionization is included so extreme dilution approaches neutrality from the correct side.
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 questionFind the approximate pH of 0.0010 M dissolved Ba(OH)₂ at 25 °C. State the concentration assumption.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Assume the stated 0.0010 M is dissolved and dissociates completely.
- 1 point: [OH⁻]≈0.0020 M.
- 1 point: pOH≈2.70.
- 1 point: pH≈11.30, using pKw=14.00.
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 1Which concentration goes into pOH?
Hydroxide concentration.
RECALL 2What sets the OH⁻ multiplier?
The balanced dissociation equation.
RECALL 3Does strong imply unlimited solubility?
No.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why can one mole of a base supply two moles of hydroxide?
- [OH⁻]≈νC for dissolved strong hydroxides; ν is the OH⁻ coefficient.
- At 25 °C: pH=14.00−pOH.
Remember: The logarithm of hydroxide gives pOH, not pH. Use the dissolved amount, not the amount of solid originally added.
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. The coefficient applies only to the base. Water autoionization is included so extreme dilution approaches neutrality from the correct side.
Refresh Kid · AP Chemistry Unit 8 · Objectives 8.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.2, objective 8.2.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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