Can you rank solubility by comparing Ksp numbers alone?
You will be able to: Compare molar solubilities using each salt’s dissolution stoichiometry.
Can you rank solubility by comparing Ksp numbers alone?
Two bottles list different solubility-product constants. The larger printed number does not always belong to the solid that dissolves more formula units per liter.
A useful starting point: How do ion ratios change the solubility calculation? →
Words and symbols before equations
- Molar-solubility comparison
- Comparison of formula-unit mol/L under matching conditions.
- Stoichiometry
- Balanced ratio connecting solid to dissolved ions.
- Like-for-like conditions
- Same temperature and stated solvent/common-ion assumptions.
- Ksp magnitude
- Value of a reaction-specific product of ion concentrations.
- Base-10 logarithm, log₁₀
- The exponent on 10: log₁₀(0.001)=−3 because 0.001=10⁻³. For a concentration slider, first divide by 1 M so the logarithm uses a numerical ratio.
What this picture assumes
Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. Compare MX, fixed Ksp=10⁻⁸, with NX₂ at its supplied Ksp. Pure water, same stated T, no complex formation or acid–base reactions. Comparison is mol/L, not g/L.
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.
- MX: Ksp=10⁻⁸, s=1×10⁻⁴ M. NX₂: Ksp=1×10⁻¹², s=(Ksp/4)^(1/3)=6.2996×10⁻⁵ M. Rank computed s, not raw Ksp values across different dissolution patterns.
- 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 salts with the same dissolution pattern in pure water and no side reactions, larger Ksp corresponds to larger s. Their formulas relating Ksp to s have the same form.
Across different ion ratios, Ksp involves different powers. Calculate s separately instead of ranking the raw constants.
For MX use s=√Ksp, whereas MX₂ uses s=(Ksp/4)^(1/3). A smaller Ksp can therefore correspond to a larger molar solubility.
The comparison is about mol/L. Ranking grams per liter additionally depends on molar masses, and common ions can change the ranking under different solution conditions.
A worked example, step by step
Salt P is one-to-one with Ksp=1.0×10⁻⁸. Salt R is one-to-two with Ksp=4.0×10⁻¹². Compare their molar solubilities in pure water.
- For P: sP=√(1.0×10⁻⁸)=1.0×10⁻⁴ M.
- For R: sR=(4.0×10⁻¹²/4)^(1/3)=1.0×10⁻⁴ M.
- They have equal molar solubility despite unequal Ksp values.
- Their different powers and ion ratios explain why the raw constants cannot be directly ranked here.
Different dissolution equations require different Ksp-to-s relationships.
Can equal s values produce very different Ksp values?
Compare with an explanation
Yes, when the dissolution stoichiometries differ.
Predict. Change one thing. Explain.
Keep the one-to-one salt fixed; change the supplied Ksp of the one-to-two salt. Compare computed s values rather than raw Ksp.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
MX: Ksp=10⁻⁸, s=1×10⁻⁴ M. NX₂: Ksp=1×10⁻¹², s=(Ksp/4)^(1/3)=6.2996×10⁻⁵ M. Rank computed s, not raw Ksp values across different dissolution patterns.
Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. Compare MX, fixed Ksp=10⁻⁸, with NX₂ at its supplied Ksp. Pure water, same stated T, no complex formation or acid–base reactions. Comparison is mol/L, not g/L.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using relative rates, particle conservation, the Q/K comparison or the stated dissolution 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 questionCompare a one-to-one salt with Ksp=9×10⁻¹⁰ and a one-to-two salt with Ksp=1.08×10⁻¹³. Calculate each s and explain the comparison.
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Compare with the answer and four-point rubric
- 1 point: One-to-one s=√(9×10⁻¹⁰)=3×10⁻⁵ M.
- 1 point: One-to-two s=(1.08×10⁻¹³/4)^(1/3).
- 1 point: The latter is also 3×10⁻⁵ M.
- 1 point: Equal s arises despite different Ksp values because the concentration products use different powers.
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 1When is direct Ksp ranking useful?
For matched dissolution patterns and conditions.
RECALL 2Does mol/L ranking automatically rank g/L?
No; molar masses also matter.
RECALL 3What should be calculated across different patterns?
Each molar solubility separately.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Can you rank solubility by comparing Ksp numbers alone?
- Compare s under the same conditions.
- For equal ion patterns only, a larger Ksp directly implies larger s in this model.
Remember: Different dissolution equations require different Ksp-to-s relationships.
Conditions: Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. Compare MX, fixed Ksp=10⁻⁸, with NX₂ at its supplied Ksp. Pure water, same stated T, no complex formation or acid–base reactions. Comparison is mol/L, not g/L.
Refresh Kid · AP Chemistry Unit 7 · Objectives 7.11.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.11, objective 7.11.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 7: Equilibrium, Topics 7.1–7.12. 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. Converting between Kc and Kp and calculations for a dissolved species in equilibrium with its gas phase are excluded from assessed Unit 7 scope. Concentrations use mol/L and gas partial pressures use the stated pressure convention. Supplied constants are teaching data at fixed temperature unless otherwise specified. Ideal dilute-solution and ideal-gas approximations are stated. 3D views show inventories, not molecular trajectories, measured structures or proof of equilibrium from a single snapshot. Approximation checks are explicit; a small K alone does not justify neglecting every change.
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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