---
title: "Solubility Product Constant (Ksp) — AP Chem Definition"
description: "Ksp is the equilibrium constant for a sparingly soluble salt dissolving into its ions. Learn how to convert between Ksp and molar solubility for AP Chem Unit 7."
canonical: "https://fiveable.me/ap-chem/key-terms/solubility-product-constant-ksp"
type: "key-term"
subject: "AP Chemistry"
unit: "Unit 7"
---

# Solubility Product Constant (Ksp) — AP Chem Definition

## Definition

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt into its ions in a saturated solution at a given temperature; it equals the product of the ion concentrations, each raised to its stoichiometric coefficient.

## What It Is

Ksp is what you get when you treat dissolving a [salt](/ap-chem/key-terms/salt "fv-autolink") as an [equilibrium](/ap-chem/unit-7/reaction-quotient-le-chateliers-principle/study-guide/JFx1InPfZCZ9SugPKDCE "fv-autolink") problem. When a sparingly soluble salt like AgCl or CaF₂ sits in water, a tiny amount dissolves and the rest stays solid. At saturation, the rate of dissolving equals the rate of precipitating, so the system is at equilibrium (EK 7.11.A.1). Ksp is just the equilibrium constant for that dissolution reaction. For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), Ksp = [Ca²⁺][F⁻]². The solid never appears in the expression because pure solids are left out of equilibrium constants.

The big idea is that Ksp puts a number on the solubility rules you memorized back in [Unit 4](/ap-chem/unit-4 "fv-autolink"). "Insoluble" salts have tiny Ksp values, while Ksp values greater than 1 correspond to soluble salts (EK 7.11.A.3). One warning the CED makes explicit (EK 7.11.A.2): you can't compare two salts' solubilities just by comparing their Ksp values unless they have the same dissolution stoichiometry. A 1:1 salt and a 1:2 salt with the same Ksp do not have the same molar solubility.

## Why It Matters

Ksp lives in Topic 7.11 (Introduction to Solubility Equilibria) in [Unit 7](/ap-chem/unit-7 "fv-autolink"). The learning objective is direct. [AP Chem](/ap-chem "fv-autolink") 7.11.A asks you to calculate the solubility of a salt from its Ksp value, and the reverse, calculating Ksp from measured molar solubility (EK 7.11.A.4). This is one of the most calculation-heavy topics in Unit 7, and it's where the abstract idea of an equilibrium constant turns into something you can measure with a balance and a volumetric flask. It also closes a loop from earlier in the course. The qualitative solubility rules from Topic 4.7 get a quantitative upgrade here, so a question can now ask exactly how much BaSO₄ dissolves, not just whether it's "insoluble."

## Connections

### [Molar Solubility (Unit 7)](/ap-chem/key-terms/molar-solubility)

[Molar solubility](/ap-chem/key-terms/molar-solubility "fv-autolink") (s) is the moles of salt that dissolve per liter; Ksp is the constant you build from it. For AgCl, Ksp = s². For CaF₂, Ksp = (s)(2s)² = 4s³. Going back and forth between s and Ksp is the core skill of Topic 7.11.

### [Stoichiometric Coefficients (Unit 4 & 7)](/ap-chem/key-terms/stoichiometric-coefficients)

The [coefficients](/ap-chem/key-terms/coefficients "fv-autolink") in the balanced dissolution equation become the exponents in the Ksp expression and the multipliers on s. Mess up the 2 in CaF₂ ⇌ Ca²⁺ + 2F⁻ and your answer is off by a factor of 4 or more. This is why EK 7.11.A.2 stresses that the Ksp-solubility relationship depends on stoichiometry.

### [Precipitation Reaction (Unit 4)](/ap-chem/key-terms/precipitation-reaction)

[Precipitation](/ap-chem/key-terms/precipitation "fv-autolink") reactions from Unit 4 are dissolution reactions running in reverse. Ksp tells you which direction is favored. A tiny Ksp means the precipitate-forming direction wins, which is the quantitative reason behind the solubility rules you used to predict precipitates.

### [Precipitation Prediction (Unit 7)](/ap-chem/key-terms/precipitation-prediction)

When you mix two solutions, you compare the ion product Q to Ksp. If Q > Ksp, a precipitate forms; if Q < Ksp, everything stays dissolved. Same Q vs. K logic you learned for gas-phase equilibria, just applied to ions in solution.

## On the AP Exam

Expect MCQs in two directions. One gives you experimental data (like 0.0152 g of CaF₂ dissolving in 100.0 mL of water) and asks you to convert to molar solubility, set up the ion concentrations, and compute Ksp. The other gives you Ksp and asks for the solubility. Either way, the stoichiometry trap is the main thing being tested, so write the balanced dissolution equation first, every time. Questions also probe whether you know what data is sufficient to find Ksp (the concentration of just one ion in a saturated solution is enough, since stoichiometry gives you the rest) and whether you can identify Ksp as the right constant for a dissolution equilibrium. Trickier items change the conditions, like dissolving CaF₂ in an acidic buffer instead of pure water; the acid removes F⁻ ions, shifting the equilibrium and increasing the measured solubility, even though Ksp itself doesn't change at constant temperature.

## Solubility product constant (Ksp) vs Molar solubility

Molar solubility is an amount (mol/L of salt that dissolves), while Ksp is an equilibrium constant built from ion concentrations raised to their coefficients. They're related by the stoichiometry of the dissolution equation, but they are not interchangeable. Two salts can have the same Ksp and different molar solubilities if their formulas differ (a 1:1 salt vs. a 1:2 salt). Comparing Ksp values directly to rank solubility only works for salts with matching stoichiometry.

## Key Takeaways

- Ksp is the equilibrium constant for a sparingly soluble salt dissolving into its ions in a saturated solution, and the solid itself never appears in the expression.
- Write the balanced dissolution equation first, because the coefficients become both the exponents in the Ksp expression and the multipliers on molar solubility s.
- For a 1:1 salt like AgCl, Ksp = s²; for a 1:2 salt like CaF₂, Ksp = 4s³, so identical Ksp values do not mean identical solubilities.
- You can calculate Ksp from the measured concentration of just one ion in a saturated solution, since stoichiometry fixes the ratio of the other ions.
- Ksp values greater than 1 correspond to soluble salts, which quantitatively backs up the solubility rules from Unit 4.
- Ksp only changes with temperature; adding acid or a common ion changes how much salt dissolves, but not the value of Ksp itself.

## FAQs

### What is the solubility product constant (Ksp) in AP Chem?

Ksp is the equilibrium constant for a sparingly soluble salt dissolving into its ions in a saturated solution at a given temperature. For CaF₂, Ksp = [Ca²⁺][F⁻]², with the solid left out of the expression.

### Does a bigger Ksp always mean a salt is more soluble?

Not necessarily. Comparing Ksp values only ranks solubility correctly when the salts have the same dissolution stoichiometry. A 1:2 salt like CaF₂ follows Ksp = 4s³ while a 1:1 salt follows Ksp = s², so the math differs. Always convert to molar solubility before comparing different salt types.

### What's the difference between Ksp and molar solubility?

Molar solubility (s) is how many moles of salt dissolve per liter, a measurable amount. Ksp is the equilibrium constant computed from the ion concentrations raised to their coefficients. For AgCl with s = 1.3 × 10⁻⁵ M, Ksp = s² = 1.7 × 10⁻¹⁰.

### Does adding acid or another ion change the Ksp of a salt?

No. Ksp depends only on temperature. Dissolving CaF₂ in a pH 4 buffer increases its measured solubility because H⁺ consumes F⁻ and shifts the equilibrium, but the Ksp value itself stays the same.

### How do I calculate Ksp from experimental data?

Convert the dissolved mass to moles, divide by liters to get molar solubility s, then use stoichiometry for ion concentrations. For 0.0152 g CaF₂ in 100.0 mL, s ≈ 1.95 × 10⁻³ M, so Ksp = (s)(2s)² = 4s³ ≈ 3.0 × 10⁻⁸.

## Related Study Guides

- [7.11 Introduction to Solubility Equilibria](/ap-chem/unit-7/intro-solubility-equilibria/study-guide/lkqnvEmZIxrvjX1sogng)

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