Common-Ion Effect
Contrast buffer proton-transfer effects with an explicit common-ion source.
Before this connection: Chemical equilibrium: dissolution equations, Q versus K, and molar solubility.
AP Chemistry Topic 7.12 · CED printed p. 140 / PDF p. 147. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Ksp and molar solubility
- Le Châtelier response
- Approximation checks
A common ion is already in solution before the salt dissolves. For MX ⇌ M+ + X−, dissolving MX into NaX solution starts with X− already present. The larger product concentration suppresses further net dissolution. Ksp stays the same at the same temperature; the salt’s molar solubility decreases. If X begins at c, the concentrations are [M+] = s and [X−] = c + s, not two copies of s. Solve Ksp = s(c + s). You may replace c + s by c only after checking s/c is small. For MX2, keep the coefficient: the X concentration is c + 2s and the term is squared. A controlled comparison must hold temperature and solid identity fixed and avoid extra complexing or protonating species. Strong ionic-strength differences can make a concentration-only model imperfect; this AP calculation assumes activities are adequately represented by concentrations.
Let's work through it
- MX2 has Ksp = 1.00 × 10−12 and dissolves into a solution initially 0.0100 M in X−.
- Let its molar solubility be s. Write Ksp = s(0.0100 + 2s)2; the X term contains both sources.
- Assume 2s is small: s ≈ 1.00 × 10−12/(0.0100)2 = 1.00 × 10−8 M.
- Check 2s/0.0100 = 2.00 × 10−6. Neglect is justified. The ion product remains Ksp even though s is much lower than in pure water.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
What changes on adding a common ion at fixed temperature?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
What changes on adding a common ion at fixed temperature?
Worked answer and reason
The equilibrium concentrations and salt solubility change; Ksp does not.
The new composition must satisfy the same ion-product constant.
Retrieval 2
Why use c + s rather than s for the common ion?
Worked answer and reason
It includes the original dissolved ion plus the ion released by the salt.
Concentrations track all sources of that species.
Catch the tempting mistake
- More X must mean more MX dissolves.
- More X raises the product-side quotient and reduces net dissolution of MX.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 7 problem 14 · ap synthesis · multiple
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For the teacher
- In a synthetic investigation, ask students to choose a non-common-ion control and to explain why temperature must be controlled.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 7.12.A
- Identify the solubility of a salt, and/or the value of K_sp for the salt, based on the concentration of a common ion already present in solution.
- 7.12.A.1
- The solubility of a salt is reduced when it is dissolved into a solution that already contains one of the ions present in the salt. The impact of this “common-ion effect” on solubility can be understood qualitatively using Le Châtelier’s principle or calculated from the K_sp for the dissolution process.
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §15.1, Precipitation and Dissolution — Stoichiometric relation between molar solubility and Ksp; common-ion and precipitation reasoning.
Glossary and deeper reading · 3 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Solubility product
A solubility product is an equilibrium constant for a specified dissolution reaction. Its expression includes dissolved-species activities and omits the pure solid. For example, dissolving one formula unit that releases two anions gives an anion activity squared in the expression.
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Solubility
Solubility describes how much of a substance can dissolve in a specified solvent under stated conditions. It depends on the solute, solvent, temperature, and other relevant conditions and is distinct from how fast dissolution occurs. For example, two salts can have different equilibrium solubilities in water at the same temperature.
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Ion
An ion is an atom or group of atoms with a net electric charge. Ordinary ion formation changes electron count without changing the element's proton count. For example, a sodium atom losing one electron becomes a sodium ion with charge +1.