7.12 Common Ion Effect

The common-ion effect: when you dissolve a slightly soluble salt into a solution that already contains one of its ions, the salt’s solubility drops. Use Le Châtelier: adding a common ion shifts the dissolution equilibrium (MxAy(s) ⇌ xM^n+ + yA^m−) left, so less solid dissolves. Practically, you can predict or calculate this with Ksp: set up the Ksp expression, include the preexisting ion concentration, solve for the new molar solubility. Example: AgCl(s) in plain water gives [Ag+] = s, [Cl−] = s with Ksp = [Ag+][Cl−]; but in 0.10 M NaCl, [Cl−] ≈ 0.10 so [Ag+] = Ksp/0.10 (much smaller than in pure water)—precipitation is also predicted by comparing Q to Ksp. This idea is used for selective precipitation and understanding spectator ions. For AP context see CED LO 7.12.A (Topic 7.12) and our Topic study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc). For extra practice problems, check (https://library.fiveable.me/practice/ap-chemistry).

When you add a common ion, you’re increasing the concentration of one of the ions already produced when the salt dissolves, so Le Châtelier’s principle pushes the dissolution equilibrium back toward the solid. Example: for AgCl(s) ⇌ Ag+(aq) + Cl−(aq) with Ksp = [Ag+][Cl−], adding extra Cl− raises [Cl−]; to keep Ksp constant the equilibrium lowers [Ag+] by shifting toward AgCl(s), so less solid dissolves (solubility decreases). Practically that means Q (the ion product) can exceed Ksp and precipitation occurs until the solution is saturated again. On the AP exam you should be able to state this qualitatively with Le Châtelier and/or set up the Ksp expression to calculate the new molar solubility (learning objective 7.12.A). For worked examples and practice problems, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and more practice (https://library.fiveable.me/practice/ap-chemistry).

Le Châtelier’s principle says a system at equilibrium shifts to oppose a change. For a salt that dissociates (e.g., AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−]), adding more of a “common” ion (Cl−) increases its concentration. The equilibrium responds by shifting left—more Ag+ combines with the extra Cl− to form solid AgCl—so less Ag+ stays dissolved. In other words, the molar solubility (amount that dissolves) decreases even though Ksp stays the same. Qualitatively: add common ion → Q > Ksp → precipitation or reduced solubility; remove common ion → more dissolves. This is exactly what AP CED Topic 7.12 expects: explain reduced solubility using Le Châtelier and Ksp (7.12.A). For worked examples and practice problems on common-ion effect, see the Topic 7 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and the AP Chem practice bank (https://library.fiveable.me/practice/ap-chemistry).

Regular solubility = how much of a salt dissolves in pure water (molar solubility s found from Ksp). Example: AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−] = s·s = s^2. With a common ion already present (common-ion effect) the solubility goes down. If the solution already contains [Cl−] = c, dissolution becomes Ksp = [Ag+][Cl−] = s·(s + c). If c ≫ s you can approximate s ≈ Ksp / c, so molar solubility is much smaller than in pure water. Qualitatively, Le Châtelier’s principle explains it: added Cl− shifts the equilibrium left, reducing dissolution. Also check Q (ion product)—if Q > Ksp, precipitation occurs. On the AP exam you’ll be asked to identify or calculate reduced solubility or Ksp given a common ion (LO 7.12.A). For worked examples and practice problems, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and AP practice sets (https://library.fiveable.me/practice/ap-chemistry).

Short answer: write the dissolution equation, express ion concentrations including the common ion, plug into Ksp, and solve for the unknown solubility (x). Steps: 1. Write dissociation and Ksp. Example: MX(s) ⇌ M+ + X−, Ksp = [M+][X−]. 2. If the solution already has [M+]0 (common ion), let s = molar solubility of MX. At equilibrium [M+] = [M+]0 + s and [X−] = s. So Ksp = ([M+]0 + s)(s). 3. If [M+]0 ≫ s, you can approximate Ksp ≈ [M+]0·s => s ≈ Ksp / [M+]0 (common-ion approximation). If not, solve the quadratic Ksp = s([M+]0 + s). 4. Check units and that the physically meaningful root (positive, small) is chosen. Use Le Châtelier qualitatively: added common ion reduces s. This is exactly the AP skill in 7.12.A—practice these algebra steps and approximations (see the Topic 7.12 study guide) (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

You add a salt that already supplies one of the ions to intentionally shift a solubility equilibrium by the common-ion effect. Adding a common ion increases that ion’s concentration, so by Le Châtelier the dissolution equilibrium (e.g., AgCl(s) ⇌ Ag+ + Cl–) shifts left and the salt becomes less soluble—molar solubility drops even though Ksp stays the same. Practically this is used to force selective precipitation (separate ions), to suppress ion concentrations, or to control equilibria in titrations. You can calculate the new solubility from Ksp by plugging the added ion concentration into the ICE setup and solving for the unknown [ion]. This is an AP Topic 7.12 idea (Ksp, molar solubility, precipitation)—review the study guide for worked examples (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and find extra practice problems at (https://library.fiveable.me/practice/ap-chemistry).

If a solution already contains one of a salt’s ions, adding that salt shifts the dissolution equilibrium left (less dissolves)—that’s the common-ion effect. Use Le Châtelier: for AgCl(s) ⇌ Ag+(aq) + Cl−(aq), extra Cl− pushes equilibrium toward solid AgCl, lowering Ag+ (and molar solubility). Quantitatively, Ksp = [Ag+][Cl−] stays constant, so increasing [Cl−] forces [Ag+] down; if the ion product Q > Ksp, precipitation occurs. This is exactly what AP CED 7.12.A expects you to identify and (if needed) calculate from the Ksp expression. Common examples: AgCl, CaCO3, PbI2. For practice on problems and worked examples, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and lots of practice questions (https://library.fiveable.me/practice/ap-chemistry).

AgCl(s) ⇌ Ag+ (aq) + Cl− (aq). Because Ksp = [Ag+][Cl−] is fixed, adding NaCl raises [Cl−], so Le Châtelier’s principle shifts the equilibrium left. That reduces Ag+ in solution—i.e., the molar solubility of AgCl decreases—and extra solid may precipitate until Q ≤ Ksp. If the ion product Q = [Ag+][Cl−] becomes greater than Ksp right after adding NaCl, precipitation occurs until equilibrium is reestablished. This is the common-ion effect (CED 7.12.A). For worked examples and practice problems, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and more practice at (https://library.fiveable.me/practice/ap-chemistry).

Adding sodium acetate adds the common ion acetate (Ac–). For calcium acetate the dissolution is: Ca(C2H3O2)2(s) ⇌ Ca2+ + 2 Ac– Ksp = [Ca2+][Ac–]^2. If you add NaC2H3O2, [Ac–] rises, and Le Châtelier says the equilibrium shifts left so less Ca2+ is in solution—solubility of Ca(C2H3O2)2 decreases. Mathematically, if the salt’s molar solubility is s and the added acetate concentration is c (c ≫ s), Ksp ≈ s·c^2 so s ≈ Ksp / c^2—bigger c → much smaller s. If the ion product Q = [Ca2+][Ac–]^2 exceeds Ksp after adding acetate, precipitation will occur until Q = Ksp. This is exactly Topic 7.12 in the CED (use Le Châtelier or Ksp calculations on the exam). For a worked review, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and try practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Start by writing the dissolution equation and Ksp. Then build an ICE table that includes the common ion as an initial concentration (not zero). Steps: 1. Write dissociation: AgCl(s) ⇌ Ag+ + Cl− and Ksp = [Ag+][Cl−]. 2. ICE table columns: Initial, Change, Equilibrium. Put the known common-ion [Cl−] (e.g., 0.100 M from added NaCl) under Initial; [Ag+] initial = 0 (if none added). 3. Change: if s = molar solubility, Ag+ increases by +s and Cl− increases by +s. So Change row: Ag+ +s, Cl− +s. 4. Equilibrium: [Ag+] = 0 + s = s; [Cl−] = 0.100 + s ≈ 0.100 (common-ion approximation if s ≪ 0.100). 5. Plug into Ksp: Ksp = (s)(0.100); solve for s. Check approximation: if s < 5% of 0.100, it’s fine; otherwise solve quadratic. Why it matters: this uses the common-ion effect from the CED (7.12.A) and Le Châtelier. For more examples and practice, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and Unit 7 resources (https://library.fiveable.me/ap-chemistry/unit-7). For lots of practice problems, try the Fiveable practice page (https://library.fiveable.me/practice/ap-chemistry).

Ksp and the common-ion effect are directly connected. Ksp is the equilibrium expression for a salt’s dissolution (e.g., AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−]). If the solution already contains a “common ion” (like extra Cl−), Le Châtelier’s principle shifts the equilibrium left, so less solid dissolves and the molar solubility drops. Quantitatively, you plug the added ion into the Ksp expression: with [Cl−]initial = 0.100 M, [Ag+] at saturation ≈ Ksp / 0.100. If the ion product Q = [Ag+][Cl−] exceeds Ksp, precipitation occurs. This is exactly what AP CED objective 7.12.A expects you to identify and calculate. For a clear worked example and practice, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and practice problems (https://library.fiveable.me/practice/ap-chemistry).

Think of the dissolution as an equilibrium: AgCl(s) ⇌ Ag+ + Cl− with Ksp = [Ag+][Cl−]. If you add more of a common ion (say extra Cl−), you raise [Cl−]. Ksp must stay constant, so the only way to satisfy Ksp is for [Ag+] to drop. That means less AgCl dissolves—i.e., solubility goes down. You can also use Le Châtelier’s principle: adding a product (Cl−) shifts the equilibrium left (toward the solid), reducing the amount of salt that stays dissolved. Or check Q = [Ag+][Cl−]: if Q > Ksp after adding Cl−, precipitation occurs until Q returns to Ksp. This is exactly the Common-Ion Effect in the AP CED (7.12.A). If you want worked examples and practice problems to calculate molar solubility with a common ion, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7). For lots of practice problems, check (https://library.fiveable.me/practice/ap-chemistry).

Start with the dissolution equation and Ksp, then account for the ion already present. Steps: 1. Write dissociation and Ksp. Example: AgCl(s) ⇌ Ag+ + Cl−, Ksp = [Ag+][Cl−]. 2. Let s = molar solubility of the salt in the solution with a preexisting common ion concentration c (the ion that appears in the salt). On dissolving, the salt adds s of the ions, but the common-ion concentration becomes (c + s). 3. Substitute into Ksp and solve for s: Ksp = (s)(c + s). If c >> s you can approximate c + s ≈ c, so s ≈ Ksp / c. Always check that s << c after solving; if not, solve the quadratic exactly: s^2 + c s − Ksp = 0. 4. Qualitative check: adding a common ion reduces solubility (Le Châtelier), so s in the common-ion solution < s in pure water. Quick numeric example: Ksp(AgCl) ≈ 1.8×10^−10, [Cl−]initial = 0.010 M → s ≈ (1.8×10^−10)/(0.010) = 1.8×10^−8 M. This is exactly the AP Topic 7.12 approach—see the Topic 7.12 study guide for more examples (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc). For extra practice, try problems at (https://library.fiveable.me/practice/ap-chemistry).

The common-ion effect matters because it changes how much of a salt dissolves when one of its ions is already in solution—useful for controlling real processes. For example: - Water treatment & chemistry: adding a common ion (like Cl–) helps predict or force precipitation (AgCl, PbI2) to remove toxic ions or recover metals (selective precipitation). - Scale control: CaCO3 solubility drops when carbonate or Ca2+ is present, so predicting scale formation depends on Ksp and the common-ion effect. - Pharmaceuticals & formulation: solubility of ionic drugs can be reduced or increased (with complex ion formation) to adjust bioavailability or stability. - Lab analysis: qualitative separation uses common ions to precipitate one ion but leave others in solution (identification schemes on the exam often test this idea). You’ll apply Le Châtelier’s principle and Ksp/molar-solubility math on the AP (Topic 7.12, LO 7.12.A). For a focused review and practice problems, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc), the Unit 7 overview (https://library.fiveable.me/ap-chemistry/unit-7), and hundreds of practice questions (https://library.fiveable.me/practice/ap-chemistry).

Quick step-by-step you can use on the AP exam: 1. Write the dissociation equation for the salt (e.g., AgCl(s) ⇌ Ag+ + Cl–) and the Ksp expression (Ksp = [Ag+][Cl–]). 2. Identify any common-ion initial concentrations already in solution (given or from mixing). Put those as starting values in an ICE setup. 3. Let s = molar solubility change for the salt and add/subtract s in the ICE table for the ions produced/consumed. 4. Substitute equilibrium concentrations into Ksp and solve for s. If an ion has a large initial concentration relative to s, use the common-ion approximation (drop s from the sum) to simplify; check the % error (if s is <5% of the initial common-ion, approximation is fine). 5. If approximation isn’t valid, solve the quadratic algebraically. 6. For qualitative questions, use Le Châtelier: adding a common ion reduces solubility; compare Q to Ksp to predict precipitation. 7. When asked about selective precipitation or complex formation, include any ligand/complex equilibria (adjust expressions accordingly). Use the AP-style ICE/Math approach and show algebra clearly for full credit. For a focused review, see the Topic 7.12 study guide (https://library.fiveable.me/ap-chemistry/unit-7/common-ion-effect/study-guide/z1tr3Mg2Afz6kj9yQGtc) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).