MCAT topics → Water and Solutions → Solubility, Ksp, and the Common Ion Effect

Solubility, Ksp, and the Common Ion Effect

MCAT trap: Incorrectly includes the solid's concentration in the Ksp equilibrium expression. The concentration of a pure solid is constant and omitted from the Ksp expression; Ksp includes only the dissolved ion concentrations.

Ksp (the solubility product constant) is an equilibrium concept the MCAT tests in a few distinct ways. It describes the equilibrium between a slightly soluble ionic solid and its dissolved ions — a specific application of Keq that follows naturally if you have equilibrium locked in.: straightforward Ksp-to-solubility calculations, passage-based scenarios where you predict whether a precipitate forms, and mechanism questions about why adding a common ion changes solubility. The math is rarely brutal, but the conceptual traps are everywhere.

The trickiest part for most students is keeping three related but distinct ideas separate: the Ksp expression itself, molar solubility (how many moles of salt dissolve per liter), and the reaction quotient Q (used to predict precipitation). These concepts connect, but students who blur them together make systematic errors. The MCAT loves to present a passage where two solutions are mixed and asks whether a precipitate forms — if you don't know the Q vs. Ksp comparison cold, you'll guess.

Two misconceptions dominate: students incorrectly include the solid's concentration in the Ksp expression (it's omitted, just like water in Ka), and students predict that adding a common ion increases solubility because 'more ions means more dissolved.' Both errors come from not fully internalizing what the equilibrium expression represents. Fix these two mental models and most Ksp questions become straightforward.

Common misconceptions

Common mistake
Wrong: The Ksp expression includes the concentration of the undissolved solid.
Right: The concentration of a pure solid is constant and omitted from the Ksp expression; Ksp includes only the dissolved ion concentrations.
Pure solids have a defined, constant 'concentration' that gets absorbed into the equilibrium constant — it doesn't appear as a variable in the expression, exactly like water is omitted from Ka in dilute solution. So for CaF2 dissolving into Ca²⁺ and 2F⁻, Ksp = [Ca²⁺][F⁻]² — no [CaF2] term anywhere. Including the solid would make Ksp appear to change with the amount of solid present, which contradicts the definition of an equilibrium constant.
Common mistake
Wrong: Adding a common ion to a saturated solution increases solubility by providing more ions.
Right: Adding a common ion shifts the dissolution equilibrium left (Le Chatelier), decreasing the solubility of the salt.
Adding a common ion increases the concentration of a product ion that's already at equilibrium. By Le Chatelier's principle, the system responds by shifting left — back toward the undissolved solid — to reduce the stress. This means less of the salt dissolves, not more. Think of it this way: you're making it harder for the solid to dissolve by pre-loading the solution with one of its own products.
Common mistake
Wrong: Precipitation occurs whenever ions of a sparingly soluble salt are mixed together.
Right: Precipitation occurs only when the ion product Q exceeds Ksp; if Q < Ksp the solution is unsaturated and no precipitate forms.
Mixing ions of a sparingly soluble salt does not automatically produce a precipitate — it depends on whether the ion product Q exceeds Ksp. If Q < Ksp, the solution is unsaturated and can still dissolve more salt; no solid forms. If Q = Ksp, the solution is exactly saturated. Only when Q > Ksp does the system drive back toward the solid and precipitation occurs. Always calculate Q from the actual concentrations after mixing before predicting what happens.
Free Deck audit

See if your Anki deck covers this topic.

Upload your deck →
Guided session

Stuck on this? An AI tutor that probes your understanding.

Start a session →

What the exam tests

  1. Know the definition of Ksp as the equilibrium constant for the dissolution of a slightly soluble salt — written in terms of dissolved ion concentrations only, with the solid omitted from the expression.
  2. Be able to calculate molar solubility from a given Ksp (and vice versa), correctly accounting for the stoichiometric coefficients in the dissolution equation (e.g., for AB2 salts, solubility appears as 's' and '2s' in the expression).
  3. Explain mechanistically why adding a common ion to a saturated solution decreases the solubility of the salt, using Le Chatelier's principle to show that the equilibrium shifts left toward the solid.
  4. Predict whether precipitation will occur when two solutions are mixed by calculating the ion product Q and comparing it to Ksp — precipitate forms only if Q > Ksp, not simply because relevant ions are present.

Can you avoid these mistakes?

Write the Ksp expression for Ag2CrO4 (which dissolves to give 2 Ag⁺ and CrO4²⁻). If the molar solubility is s, express Ksp in terms of s and solve for s given Ksp = 1.1 × 10⁻¹².
A saturated solution of PbSO4 (Ksp = 1.6 × 10⁻⁸) is placed in contact with 0.10 M Na2SO4. Without calculating, predict whether the solubility of PbSO4 increases or decreases, and explain the mechanism using Le Chatelier's principle.
25 mL of 0.002 M BaCl2 is mixed with 25 mL of 0.004 M Na2SO4. Given Ksp for BaSO4 = 1.1 × 10⁻¹⁰, calculate Q for the mixed solution and determine whether a precipitate forms. (Remember to account for dilution when mixing.)
A student writes Ksp for Ca3(PO4)2 as [Ca²⁺]³[PO4³⁻]²/[Ca3(PO4)2]. Identify the error, explain why it's wrong, and write the correct expression.

Related topics

Chemical Equilibrium and Keq Water — Polarity, Hydrogen Bonding, Anomalous Density Brønsted-Lowry and Lewis Acids and Bases pH, pOH, and the Ion Product of Water Ka, Kb, pKa, pKb and Acid Strength

See how your Anki deck covers this topic.

Upload your deck for a free audit →