Common-Ion Suppression of Solubility

Introduction

Adding an ion already produced by a sparingly soluble solid usually reduces how much of that solid dissolves. This common-ion effect follows from the solubility-product expression and mass action. The calculation is simple only if background ion concentration remains nearly constant as the solid dissolves. Advanced problems require checking that approximation and keeping the correct dissolution stoichiometry.

Core explanation

For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻] in the dilute concentration model. In pure water, if molar solubility is s, both ion concentrations are approximately s and Ksp = s². In a solution initially containing c mol L⁻¹ chloride from a soluble salt, dissolution contributes s more chloride, so [Cl⁻] = c+s while [Ag⁺] = s. The exact simple expression is Ksp = s(c+s). If c ≫ s, then s ≈ Ksp/c, much smaller than √Ksp in pure water.

The common ion can be either cation or anion. Adding Ag⁺ from another soluble source similarly suppresses AgCl dissolution. The statement is about free ion concentrations or activities, not merely the label of the added salt. If an added reagent also binds silver strongly in a complex, its effect may oppose the simple common-ion prediction. In an advanced calculation, identify all coupled equilibria before assuming one-ion mass action is the entire story.

Stoichiometry matters for salts such as Mg(OH)₂. Its dissolution is Mg(OH)₂(s) ⇌ Mg²⁺ + 2 OH⁻, giving Ksp = [Mg²⁺][OH⁻]². In pure water, the crude solubility relation is Ksp ≈ s(2s)² = 4s³ if water-derived hydroxide is negligible. In a strong-base solution with background [OH⁻] ≈ c and c ≫ 2s, Ksp ≈ s c², so s ≈ Ksp/c². Doubling c approximately quarters s in that regime. Using Ksp/c rather than Ksp/c² would miss the stoichiometric exponent.

The initial background concentration itself can change on mixing. If 50 mL of a chloride solution is combined with 50 mL of water, chloride concentration halves before considering dissolution. Use concentrations after mixing for the ion product or equilibrium expression. If the system already contains a precipitate, Ksp describes free dissolved ions at equilibrium with that solid; it does not state how much total solid exists.

Check the small-s assumption numerically. After estimating s ≈ Ksp/c, calculate s/c. If it is not small enough for the required accuracy, solve s(c+s) = Ksp as a quadratic. A common classroom threshold is a few percent, but the acceptable error depends on the question. Also check whether other ion sources, acid-base protonation or complex formation could change c.

At non-negligible ionic strength, the thermodynamic Ksp is defined with activities. Adding a background electrolyte can change activity coefficients even without a common ion, so concentration-based trends can be nuanced. The usual common-ion suppression remains a strong qualitative guide when the added species simply raises one free ion and no competing chemistry dominates.

Step-by-step reasoning

1. Write the balanced dissolution equation and Ksp expression. 2. Compute any common-ion concentration after mixing. 3. Let s be molar solubility and express each free ion as background plus stoichiometric contribution. 4. Approximate only after comparing background with expected contribution. 5. Substitute the result back into Ksp to check error and chemistry assumptions.

Visual explanation

Draw a balance between solid AgCl and dissolved Ag⁺/Cl⁻. Add a large chloride arrow from a second source; show the equilibrium moving toward solid while Ksp remains fixed at a given temperature.

Real-world analogy

If a room already contains many people of one role, fewer additional matched pairs can fit under a fixed product limit. A common ion fills part of the allowed ion-product capacity before the solid dissolves.

Real-world example

Silver chloride is less soluble in a chloride-containing solution than in pure water under the simple model. This principle helps predict precipitation and wash-solution choices in qualitative chemistry.

Why?

Why does adding chloride reduce AgCl solubility? Ksp fixes the equilibrium product of free Ag⁺ and Cl⁻ activities; higher free chloride requires lower free silver, so less AgCl can dissolve.

Common misconception

“A common ion changes the value of Ksp.” At fixed temperature the thermodynamic Ksp is unchanged; the equilibrium concentrations shift. Activity effects can alter concentration-based apparent values, but not the underlying constant.

Worked example

Suppose Ksp(AgCl) = 1.8×10⁻¹⁰ and chloride background c = 0.010 M, with no complex formation. Estimate s = Ksp/c = 1.8×10⁻⁸ M. Check s/c = 1.8×10⁻⁶, so neglecting s in c+s is excellent. In pure water the simple solubility would be √Ksp ≈ 1.3×10⁻⁵ M, far larger than in the chloride background.

Quick check

1. For Mg(OH)₂ in a strong-base background c = [OH⁻], what is the approximate dependence of solubility s on c? Answer: s ≈ Ksp/c² when c dominates the hydroxide released by dissolution, because the Ksp expression contains [OH⁻] squared.

Exam focus

Use the correct ion stoichiometric exponents and post-mixing background concentration. Check every neglected s term after estimating.

Advanced insight

OpenStax treats common-ion suppression at https://openstax.org/books/chemistry-atoms-first-2e/pages/15-1-precipitation-and-dissolution. Coupled complexation or acid-base chemistry can change free-ion concentrations enough to reverse a naive total-concentration prediction.

Summary

A common free ion suppresses dissolution through the fixed Ksp product. Express each ion as background plus dissolution contribution, then justify any c ≫ s simplification. Stoichiometric exponents, mixing dilution and competing equilibria control the quantitative result.

Practice questions

1. What is Ksp for CaF₂(s) ⇌ Ca²⁺ + 2 F⁻? Answer: Ksp = [Ca²⁺][F⁻]² in a dilute concentration model. 2. Does adding fluoride generally increase or decrease CaF₂ solubility in the simple model? Answer: Decrease it, because fluoride is a common ion. 3. Why must c+s be used before approximating [Cl⁻] ≈ c for AgCl? Answer: Dissolution itself adds chloride; the approximation is valid only if that addition is small relative to c. 4. Can a ligand that complexes a metal complicate the common-ion prediction? Answer: Yes. Complexation changes the free-metal concentration that enters Ksp.