The Common Ion Effect in Analysis

Introduction

The common-ion effect is a practical tool in qualitative analysis. Adding an ion already present in a dissolution equilibrium generally lowers how much of the sparingly soluble salt can remain dissolved. An analyst can exploit that shift to precipitate one group while keeping another ion in solution. The result must be reasoned with free-ion concentrations and Ksp, because complex formation or protonation can change the apparent outcome.

Core explanation

For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻] in a dilute concentration model. If soluble chloride is added, free [Cl⁻] rises. At saturation with AgCl solid, free [Ag⁺] must fall approximately as Ksp/[Cl⁻]. More silver enters the solid phase. This is not a decrease in Ksp; the same equilibrium constant is satisfied with a different pair of free-ion concentrations.

The effect can suppress molar solubility. In pure water, let AgCl solubility be s, so Ksp ≈ s². In a solution containing c M chloride from another source, the exact simple expression is Ksp = s(c+s), assuming silver comes only from AgCl. If c ≫ s, s ≈ Ksp/c. The approximation should be checked by comparing the calculated s with c. If the ratio is not small, solve the quadratic instead.

Analytical group separations use more than a memorized cation list. Chloride concentration can be adjusted so very sparingly soluble chlorides precipitate while highly soluble chlorides remain aqueous. Similarly, sulfide ion concentration is controlled by pH; in acidic solution free S²⁻ is very low, so only extremely insoluble metal sulfides form. In a more alkaline solution, higher available sulfide allows additional metals to precipitate. This is common-ion and acid-base reasoning together.

Adding a common ion can also control an unwanted precipitate. For a metal hydroxide M(OH)₂, a sufficiently high metal concentration lowers its compatible free OH⁻ level through Ksp = [M²⁺][OH⁻]². Conversely, adding OH⁻ can precipitate it. But if M forms a stable hydroxo complex in excess base, the simple common-ion prediction may reverse at high pH. Always list the important dissolved species.

An analyst must consider reagent contamination and downstream tests. A reagent that introduces chloride may make a later silver-nitrate halide test ambiguous. A deliberate common-ion addition can improve one separation while complicating another. Systematic schemes either use separate sample portions or remove interferences before later steps.

Temperature and ionic strength matter for quantitative use. A fixed thermodynamic Ksp is defined with activities. At substantial salt concentration, concentration products do not equal activity products because activity coefficients depart from one. A course calculation may intentionally use a dilute approximation; a precise laboratory interpretation needs the medium's effective constants.

The common-ion effect is a shift of a reversible dissolution equilibrium, not a guarantee that every target ion is removed completely. The residual free-ion concentration remains finite and can be estimated from Ksp. If the goal is a separation, calculate how much of the first ion remains when the second begins precipitating; an early onset alone does not ensure a useful purity.

Step-by-step reasoning

1. Write the solid dissolution and Ksp expression. 2. Identify which added species is a free common ion. 3. Express each free-ion level after addition or mixing. 4. Solve for the residual dissolved target concentration. 5. Check assumptions about complexation, acid-base speciation and dilution.

Visual explanation

Draw AgCl(s) on one side and Ag⁺ plus Cl⁻ on the other. Add a large Cl⁻ arrow into solution and show Ag⁺ moving into the solid to restore the Ksp product.

Real-world analogy

A seesaw must keep a fixed product of two loads. If one side's effective load rises, the other must fall to preserve the balance; common-ion addition raises one free-ion factor.

Real-world example

Chloride can help isolate a classical group of sparingly soluble metal chlorides while sodium and many other cations remain in the filtrate. The actual group boundary depends on reagent concentration and temperature.

Why?

Why does free silver decrease when chloride is raised in contact with AgCl solid? The equilibrium product Ksp constrains [Ag⁺][Cl⁻]; more chloride requires less free silver.

Common misconception

“A common ion always lowers total dissolved metal.” If the added reagent also creates a strong soluble complex, total metal can rise even while free metal falls; both equilibria must be included.

Worked example

Take a hypothetical 1:1 salt MX with Ksp = 1.0×10⁻⁹. In pure water, s ≈ √Ksp = 3.2×10⁻⁵ M. In a 0.010 M X⁻ background, s ≈ Ksp/0.010 = 1.0×10⁻⁷ M. Since s/c = 10⁻⁵, the approximation is justified. The molar solubility is much lower, but not zero.

Quick check

1. Does adding chloride to saturated AgCl change Ksp or free [Ag⁺] at fixed temperature? Answer: It lowers free [Ag⁺] and usually AgCl molar solubility while thermodynamic Ksp remains unchanged.

Exam focus

Write the equilibrium product and check small-s assumptions. Keep free-ion suppression distinct from total dissolved species when complexes exist.

Advanced insight

The common-ion effect described by OpenStax at https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution is one application of mass action. In analytical schemes, pH often controls the effective concentration of a common anion rather than the reagent label alone.

Summary

A common ion shifts dissolution toward the solid by forcing the other free-ion concentration down at fixed Ksp. This can make a group precipitation selective, but residual ions remain and competing complexation or protonation may alter the result. Quantitative work requires post-mixing free-ion levels.

Practice questions

1. What is approximate Ag⁺ at AgCl saturation when free chloride is c and c dominates dissolution? Answer: [Ag⁺] ≈ Ksp/c in a dilute concentration model. 2. Does common-ion suppression imply absolute insolubility? Answer: No. Finite dissolved ions remain at equilibrium. 3. Why can pH change sulfide-group precipitation? Answer: Protonation equilibria alter free S²⁻ concentration, changing each metal-sulfide ion product. 4. What check justifies replacing c+s with c? Answer: The calculated dissolution contribution s must be small compared with background c.