How do you calculate solubility with a common ion already present?
You will be able to: Solve the complete common-ion expression and verify approximate results.
How do you calculate solubility with a common ion already present?
A starting solution already contains some of one dissolution product. An ICE table must preserve that starting amount instead of pretending both ions begin at zero.
A useful starting point: Why does an existing ion reduce how much salt dissolves? →
Words and symbols before equations
- Initial common-ion concentration C
- Concentration before additional solid dissolves.
- Additional solubility s
- Formula-unit molarity contributed by the solid being studied.
- Exact quadratic
- s²+Cs−Ksp=0 for the one-to-one case.
- Back-substitution
- Check that final concentrations reproduce the supplied Ksp.
- 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. MX Ksp=10⁻⁸, no initial M⁺. Compare exact s(C+s)=Ksp with s≈Ksp/C only for C>0. The approximation requires a small relative change; 5% is a chosen screening tolerance.
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=6.1803×10⁻⁵ M; s(C+s)=1×10⁻⁸=Ksp. Pure-water s=10⁻⁴ M. Shortcut s≈Ksp/C=1×10⁻⁴ M; its change/C=100%, fails the chosen 5% screen.
- 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 into a solution containing X⁻ only, the final concentrations are [M⁺]=s and [X⁻]=C+s. Thus s(C+s)=Ksp.
The positive quadratic root gives s. A numerically stable equivalent is s=2Ksp/(C+√(C²+4Ksp)), which also works at C=0.
When C is much larger than s, use s≈Ksp/C and check s/C. At small background concentrations this shortcut can overestimate solubility substantially.
For a different salt ratio, rebuild the table. For MX₂ in background X⁻, [M²⁺]=s, [X⁻]=C+2s and Ksp=s(C+2s)². If the background is M²⁺ instead, the expression becomes (C+s)(2s)².
A worked example, step by step
MX has Ksp=1.0×10⁻⁸ and initial [X⁻]=1.0×10⁻⁴ M. Find exact s and evaluate the small-s shortcut.
- Use s(1.0×10⁻⁴+s)=1.0×10⁻⁸.
- The positive root is s≈6.180×10⁻⁵ M.
- Final [X⁻]≈1.6180×10⁻⁴ M; their product is 1.0×10⁻⁸.
- The shortcut gives 1.0×10⁻⁴ M, which is not small compared with C and overestimates s; use the exact result.
Include the common ion’s initial concentration and the additional ion contribution from dissolution.
What happens when C=0?
Compare with an explanation
The exact expression reduces to s=√Ksp; the shortcut Ksp/C is invalid.
Predict. Change one thing. Explain.
Vary background C across zero, 10⁻⁴ and 10⁻² M. Compare exact s with the shortcut; explain why division by C is unavailable at zero.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Exact s=6.1803×10⁻⁵ M; s(C+s)=1×10⁻⁸=Ksp. Pure-water s=10⁻⁴ M. Shortcut s≈Ksp/C=1×10⁻⁴ M; its change/C=100%, fails the chosen 5% screen.
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, no initial M⁺. Compare exact s(C+s)=Ksp with s≈Ksp/C only for C>0. The approximation requires a small relative change; 5% is a chosen screening tolerance.
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 questionFor MX, C=3.0×10⁻⁴ M and Ksp=4.0×10⁻⁸. Test s=1.0×10⁻⁴ M, compute both final concentrations, and assess the approximation s≈Ksp/C.
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Compare with the answer and four-point rubric
- 1 point: At s=1.0×10⁻⁴ M, [M⁺]=1.0×10⁻⁴ M.
- 1 point: [X⁻]=C+s=4.0×10⁻⁴ M.
- 1 point: Their product is 4.0×10⁻⁸, so the supplied s is exact for this model.
- 1 point: Ksp/C≈1.33×10⁻⁴ M is not small relative to C; its 44% check fails a 5% tolerance.
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 does s measure with a background ion?
Additional dissolved formula units per liter.
RECALL 2Why use the positive root?
Negative solubility is not physical for this dissolution setup.
RECALL 3What changes with another salt ratio?
The ion increments and powers; rebuild the expression.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you calculate solubility with a common ion already present?
- MX: Ksp=s(C+s), positive root only.
- MX₂ with background X⁻: Ksp=s(C+2s)².
Remember: Include the common ion’s initial concentration and the additional ion contribution from dissolution.
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, no initial M⁺. Compare exact s(C+s)=Ksp with s≈Ksp/C only for C>0. The approximation requires a small relative change; 5% is a chosen screening tolerance.
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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