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Premium experiment • saturation and common-ion effect

CHEMISTRY • EQUILIBRIUM

Solubility Product (Ksp) & Common Ion Effect

Compare intrinsic and common-ion-limited solubility, then test whether a chosen ion mixture is undersaturated or precipitation-favored.

● SOLUBILITY EQUILIBRIUMKsp and common ion
Molar solubility
Common-ion factor
Q/Ksp
Mixture state

Reasoning inspector

live model state
Ksp—
Ion quotient Q—
Q / Ksp—
log₁₀(Q/Ksp)—
Solubility with common ion—
No-common-ion solubility—
Solubility retained—
Equilibrium [M]—
Equilibrium [A]—
Saturation state—

Dissolution and precipitation

Comparing the ion product Q with Ksp shows whether a mixture is below, near or above its saturation threshold.

LEARN WITH THE SIMULATION

Understand Solubility Product (Ksp) & Common Ion Effect by experimenting.

START HERE

The solubility product Ksp is an equilibrium constant for dissolution, while molar solubility is the amount of solid that dissolves under a particular composition. This laboratory separates those ideas by solving equilibrium solubility for 1:1 and 1:2 salts, adding a common anion, and independently evaluating an arbitrary test mixture through its ion product Q. The premium inspector shows Ksp, Q/Ksp, common-ion suppression and equilibrium ion concentrations together.

BUILD THE IDEA

What you should understand first

  • Write Ksp expressions from dissolution stoichiometry for MX and MX₂ salts.
  • Solve how a common ion changes equilibrium molar solubility while Ksp itself remains fixed.
  • Compare Q with Ksp to classify undersaturation, near-saturation and precipitation-favored conditions.
  • Explain why Q/Ksp and equilibrium solubility answer different questions about the same chemical system.

TRY IT YOURSELF

How can adding a common ion reduce equilibrium solubility even though Ksp does not change?

  1. Choose an MX salt with fixed logKsp and zero common ion, then record molar solubility.
  2. Increase common anion concentration while keeping Ksp fixed.
  3. Compare new solubility with the no-common-ion reference and record the suppression factor.
  4. Separately adjust the test ion mixture until Q/Ksp crosses one and compare that precipitation criterion with the equilibrium-solubility calculation.
Read the visual:

Ksp is fixed for the modeled salt and temperature, but the equilibrium concentrations satisfying that product depend on ions already present. Adding a common ion forces the dissolved salt contribution lower. Q, by contrast, tests a specified mixture against the same equilibrium threshold.

WHY IT WORKS 01

Stoichiometry controls the algebra between Ksp and solubility

For MX dissolving as M + A, a pure-water equilibrium gives Ksp = s². For MX₂ dissolving as M + 2A, the ideal concentrations are [M]=s and [A]=2s, so Ksp = s(2s)² = 4s³. The same Ksp value therefore corresponds to different molar solubilities depending on dissolution stoichiometry.

WHY IT WORKS 02

The common-ion effect follows directly from the equilibrium product

If an anion is already present, the anion concentration in the Ksp expression becomes the common-ion concentration plus the amount contributed by new dissolution. Because the product must remain equal to Ksp at equilibrium, increasing the pre-existing anion concentration forces the allowed dissolved-solid contribution downward.

WHY IT WORKS 03

Q asks whether a current mixture is below or above the equilibrium boundary

The ion product Q uses the same concentration expression as Ksp but evaluates the mixture as it exists now. Q<Ksp indicates undersaturation, Q≈Ksp indicates saturation and Q>Ksp indicates precipitation is thermodynamically favored. This comparison is separate from solving the equilibrium solubility of a solid in contact with solution.

DO NOT MIX THESE UP

Common misconception

Ksp is not generally equal to molar solubility, especially when stoichiometric exponents are involved or a common ion is present. Q > Ksp predicts precipitation is thermodynamically favored in the ideal concentration model, but it does not specify precipitation rate. Activities, complexation and acid-base chemistry can also alter real solubility.

PRACTICE CHALLENGE

Can you tune a test mixture to a target log₁₀(Q/Ksp) and then explain why changing common ion alters solubility without changing that test Q?

  • Choose a target log₁₀(Q/Ksp), especially zero for the saturation boundary.
  • Adjust the test cation and anion concentrations until the target is reached.
  • Change the common-ion control while keeping test-mixture controls fixed.
  • Use the inspector to explain why equilibrium solubility changes while Ksp stays constant and the independent test Q can remain unchanged.
Reveal the reasoning

The test Q is calculated from the specified test concentrations, whereas the common-ion control belongs to the separate equilibrium-solubility calculation. Ksp is the same equilibrium constant in both views. Changing common ion suppresses the amount of additional salt that can dissolve, while Q/Ksp classifies whatever test mixture was entered.

MODEL BOUNDARIES

What this simulation assumes

Every simulation simplifies reality or a mathematical system. These are the assumptions and limits to keep in mind while interpreting the results.

Model assumptions

  • Dissolution is modeled for ideal MX or MX2 salts using concentration-based Ksp expressions.
  • The common ion is the anion and is added without changing volume.
  • Precipitation tendency is classified by comparing the ionic reaction quotient Q with Ksp.

Model limitations

  • Activity coefficients, ion pairing, complex formation, acid-base coupling and kinetic barriers are omitted.
  • The model does not perform mass-balance precipitation calculations after Q exceeds Ksp; it only classifies the tendency.

CONNECT IT

Where this matters

  • Selective precipitation
  • Water chemistry
  • Qualitative inorganic analysis
  • Designing separation conditions

WORDS THAT MATTER

Key vocabulary

Ksp
Equilibrium constant for dissolution of a sparingly soluble ionic solid.
Molar solubility
Moles of solid that dissolve per liter under specified equilibrium conditions.
Common ion effect
Suppression of dissolution when one dissolution-product ion is already present.
Reaction quotient Q
Current ion-product expression evaluated before requiring equilibrium.
Saturation
Condition where Q is approximately equal to Ksp in the ideal model.