Why does an existing ion reduce how much salt dissolves?
You will be able to: Explain common-ion suppression without changing Ksp.
Why does an existing ion reduce how much salt dissolves?
A solution already containing X⁻ needs less additional X⁻ from an MX crystal to reach saturation. Adding an ion that the solid also releases changes the starting composition.
A useful starting point: Will a precipitate form when two solutions are mixed? →
Words and symbols before equations
- Common ion
- An ion already present that is also produced by the salt’s dissolution.
- Background concentration C
- Common-ion concentration supplied before additional salt dissolves.
- Suppressed solubility
- Less solid dissolves under the stated solution conditions.
- Fixed Ksp
- The same dissolution constant at the same temperature in the ideal model.
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. MX Ksp=10⁻⁸, initially no M⁺; enough solid for saturation. Exact positive solution of s(C+s)=Ksp. No complexing or ion-consuming processes.
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.
- Exact s=9.999×10⁻⁷ M; s(C+s)=1×10⁻⁸=Ksp. Pure-water s=10⁻⁴ M.
- 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 MX(s)⇌M⁺+X⁻, a background X⁻ concentration raises the ion product for a given [M⁺]. Equilibrium is reached with less M⁺ and hence less newly dissolved MX.
Ksp does not decrease because a common ion is present. The concentrations readjust to satisfy the same product under ideal dilute conditions.
If a saturated solution with solid present receives extra common ion at negligible volume change, Qsp rises above Ksp and net precipitation can follow.
The effect assumes no strong complex formation, acid–base consumption or important activity changes. Such coupled processes can alter the simple prediction and require a broader model.
A worked example, step by step
A one-to-one salt has Ksp=1.0×10⁻⁸. Compare pure-water solubility with approximate solubility in 0.010 M X⁻; check the background approximation.
- In pure water s=√Ksp=1.0×10⁻⁴ M.
- With background X⁻, Ksp=s(0.010+s).
- If s is small, s≈Ksp/0.010=1.0×10⁻⁶ M.
- s/0.010=0.0001=0.01%, so the approximation is well justified; solubility is about 100 times smaller while Ksp stays fixed.
A common ion changes solubility, not the fixed-temperature Ksp in this stated ideal model.
Why can [X⁻] be high while [M⁺] is low?
Compare with an explanation
Their product, rather than equality of the individual concentrations, must match Ksp.
Predict. Change one thing. Explain.
Increase background X⁻ from zero. Watch the exact molar solubility fall while the calculated equilibrium ion product remains 10⁻⁸.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Exact s=9.999×10⁻⁷ M; s(C+s)=1×10⁻⁸=Ksp. Pure-water s=10⁻⁴ M.
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. MX Ksp=10⁻⁸, initially no M⁺; enough solid for saturation. Exact positive solution of s(C+s)=Ksp. No complexing or ion-consuming processes.
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 questionA saturated MX solution with solid present receives additional X⁻ at fixed T and negligible volume change. Explain immediate Qsp, net response, final Ksp relation and a limitation of the simple model.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Raising [X⁻] immediately raises Qsp.
- 1 point: Qsp>Ksp promotes net precipitation.
- 1 point: Remaining dissolved ions return to a product equal to the unchanged Ksp if solid remains.
- 1 point: The simple model assumes no complex formation, ion-consuming reaction or substantial nonideal effects.
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 is a common ion?
An ion supplied both by the background solution and the dissolving salt.
RECALL 2Does lower solubility imply lower Ksp?
No, not at fixed T in this ideal model.
RECALL 3Why state no side reactions?
A reaction consuming or complexing an ion can change the simple prediction.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does an existing ion reduce how much salt dissolves?
- MX with background X⁻: Ksp=s(C+s).
- For s≪C, s≈Ksp/C; check the assumption.
Remember: A common ion changes solubility, not the fixed-temperature Ksp in this stated ideal model.
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. MX Ksp=10⁻⁸, initially no M⁺; enough solid for saturation. Exact positive solution of s(C+s)=Ksp. No complexing or ion-consuming processes.
Refresh Kid · AP Chemistry Unit 7 · Objectives 7.12.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.12, objective 7.12.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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