Solubility Product Tables

Introduction

Ksp tables describe equilibria between a specified solid and dissolved ions. A low Ksp often suggests low solubility, but Ksp alone is not a direct concentration because stoichiometry, common ions, pH and complexing ligands matter. Read the exact dissolution equation and medium before comparing salts or predicting precipitation.

Core explanation

For MX(s) ⇌ M⁺ + X⁻, the activity-based solubility product is Ksp = a(M⁺)a(X⁻); the pure solid has unit activity by convention. A dilute concentration approximation gives [M⁺][X⁻] with appropriate dimensionless standard-state factors understood. For M₂X(s) ⇌ 2M⁺ + X²⁻, Ksp involves a(M⁺)²a(X²⁻). Thus equal Ksp values do not imply equal molar solubility for salts with different stoichiometry. OpenStax's solubility-equilibria summary gives the stoichiometric exponents and coupled-equilibrium context.

In pure water with no side reaction, a 1:1 salt of molar solubility s gives free-ion concentrations near s and Ksp ≈ s². For a 2:1 salt under the same simplified assumptions, free ions are 2s and s, so Ksp ≈ (2s)²(s) = 4s³. These are algebraic relations, not universal formulas for real solutions. Ionic-strength activity corrections can matter even when the solid is sparingly soluble if supporting electrolyte is present.

A common ion suppresses dissolution by increasing one product-ion activity. A ligand may have the opposite effect on total solubility: by binding free M⁺ to form ML or ML₂, it reduces free M⁺ activity, allowing more solid to dissolve while Ksp for the free-ion equilibrium remains the same at fixed temperature. OpenStax's coupled-equilibria example demonstrates enhanced AgCl solubility with ammonia complexation. Acid-base reactions can similarly consume anions and increase dissolution.

For precipitation, compare the current ion activity product Qsp with Ksp under matching conventions. Qsp > Ksp suggests supersaturation and a thermodynamic tendency to precipitate; nucleation may delay visible solid. Qsp < Ksp indicates undersaturation for that solid. A clear solution is not conclusive proof that Qsp < Ksp, because precipitation kinetics and colloids complicate observations. Tables of Ksp are temperature-specific and sometimes use differing activity or concentration conventions.

Step-by-step reasoning

1. Write the exact solid dissolution equation and identify free-ion species. 2. Build Ksp with stoichiometric exponents and a consistent activity convention. 3. Combine it with mass and charge balances to find molar solubility. 4. Add common-ion, acid-base or complex-formation equilibria when present. 5. Compare Qsp and Ksp for precipitation while noting kinetic limitations.

Visual explanation

Draw a crystal releasing M⁺ and X⁻ into water. A common-ion arrow adds X⁻ and shifts the equilibrium toward solid. A ligand L wraps around M⁺ to make ML, lowering free M⁺ and drawing more ions from the solid. Label the unchanged free-ion Ksp beside the crystal and the increased total metal concentration beside the complex.

Real-world analogy

Imagine a queue leaving a theater through a door. If the exit area is crowded, fewer people leave; if a shuttle quickly carries people away, more can exit. Common ions crowd the dissolved products, while a ligand “carries away” free metal into a complex. The door rule, analogous to Ksp at a fixed temperature, has not changed.

Real-world example

An analyst predicts that a sparingly soluble silver salt will precipitate. The sample contains ammonia, which binds silver ions. A simple Ksp-only free-ion calculation may overpredict solid formation unless the complexation equilibrium and total silver balance are included. The same solid can show very different apparent solubility in pure water and ligand-rich solution.

Why?

Why can total dissolved metal rise while free metal-ion activity stays constrained by Ksp? Complex formation converts free metal into a bound species. The solid dissolves to replace free ions until the free-ion activity product again reaches Ksp. Total analytical metal includes both free and complexed forms.

Common misconception

“Ksp equals molar solubility” ignores stoichiometry. “A ligand changes the intrinsic Ksp of the same solid at fixed temperature” confuses equilibrium constant and coupled speciation. “Any Qsp > Ksp instantly produces visible precipitate” ignores nucleation. “A solid's formula alone determines its Ksp expression” is risky without the actual dissolution species.

Worked example

For hypothetical MX(s) ⇌ M⁺ + X⁻ with Ksp = 1.0 × 10⁻⁶ in a dilute approximation, pure-water solubility is s ≈ √Ksp = 1.0 × 10⁻³ M. If a solution initially contains [X⁻] = 0.10 M and added X⁻ dominates, the free [M⁺] at equilibrium is approximately Ksp/[X⁻] = 1.0 × 10⁻⁵ M. The common ion reduces free-metal concentration by about a factor of 100. If ligand binds M⁺ strongly, total dissolved M can exceed 10⁻⁵ M even while free [M⁺] stays near this value. The example ignores activity corrections and ligand balance to isolate the ideas.

Quick check

1. For M₂X(s) ⇌ 2M⁺ + X²⁻, what powers appear in Ksp? Answer: Ksp = a(M⁺)²a(X²⁻) under an activity-based convention.

Exam focus

Derive Ksp from a balanced dissolution equation and relate s to free-ion concentrations. Explain common-ion suppression and ligand-enhanced total solubility. Compare Qsp and Ksp while distinguishing thermodynamic tendency from precipitation speed.

Advanced insight

Several solid phases may compete, and the stable phase can change with pH, redox potential or temperature. A single Ksp entry cannot describe that whole phase diagram. In concentrated electrolyte, activity coefficients and complex ions can dominate measured concentrations, so thermodynamic modeling needs a consistent speciation framework.

Summary

Ksp constrains the activities of free dissolution ions for a specified solid and temperature. Molar solubility follows only after stoichiometry and coupled equilibria are considered. Common ions, ligands and kinetics explain why observed dissolved amounts or precipitation differ from a simple table lookup.

Practice questions

1. For a simple 1:1 salt in pure dilute water, what is s in terms of Ksp? Answer: Approximately s = √Ksp when no side reactions or activity corrections matter. 2. Why does adding a common ion often reduce dissolution? Answer: It raises a product-ion activity and shifts the dissolution equilibrium toward solid. 3. Does complexation necessarily alter intrinsic Ksp of the same solid at fixed temperature? Answer: No. It changes free-ion speciation and total solubility while the free-ion equilibrium constant remains defined. 4. Does Qsp > Ksp guarantee immediate visible precipitation? Answer: No. Nucleation and growth can delay observable solid formation.