K_{sp} is defined as the product of the equilibrium concentration of ions, each raised to a power that is its coefficient in the balanced chemical equation. In other words, K_{sp} is the product of the molar solubilities of ions at equilibrium. It is dependent on temperature. Additionally, it is normally a small quantity at room temperature. In addition to it, the solubility product (K_{sp}) value can be determined from the solubility of a compound, and vice versa.

K_{sp} refers to the solubility product constant. It is an application of the law of mass action to find the solubility of a sparingly soluble salt. For example, when a slightly soluble ionic compound is added to water, some of it dissolves and the remaining settles at the bottom. Afterward, dynamic equilibrium is established between this undissolved solid and its ions in the saturated solution.

The solubility of an ionic compound generally increases by increasing the temperature. In fact, this is the case for ionic compounds whose enthalpy of solution is endothermic.

## General Expression and Example of Solubility product

Generally, K_{sp} expression of a sparingly soluble ionic compound A_{a}B_{b} can be written as

K_{sp} = [A^{+b}]^{a} [B^{-a}]^{b}

For the general equation,

A_{a}B_{b(s) }⇌ aA^{+b}_{(aq)} + bB^{-a}_{(aq)}

This shows that the solubility product constant, K_{sp}, is equal to the product of the concentration of ions each having an exponent equal to the number of its ions in the formula unit of the compound.

**Example**

An example would be to add CaF_{2} to water. The following equilibrium will be established:

CaF_{2(s) }⇌ Ca^{2+}_{(aq)} + 2F^{–}_{(aq)}

The kc expression for this equilibrium will be written as

K_{c} = [Ca^{+2}] [F^{–}]^{2 }/ [CaF_{2}]

As CaF_{2} is sparingly soluble, its concentration almost remains constant.

K_{c} [CaF_{2}] = [Ca^{2+}] [F^{–}]^{2}

So we get the following solubility product (K_{sp}) expression as given below:

K_{sp} = [Ca^{2+}] [F^{–}]^{2}

## Application of Solubility product (K_{sp})

A reaction that takes place when two or more solutions, when mixed, yield an insoluble solid precipitate, is known as a precipitation reaction. Ion product (Q’) is obtained by using given concentrations of ions rather than the equilibrium concentrations in the K_{sp} expression. For CaF_{2}, it is written as

Q’ = [Given concentration of Ca^{+2}] [Given concentration of F^{–}]^{2}

Comparing K_{sp} value with Q’ can help us to predict whether precipitation will occur or not:

If K_{sp} > Q’, precipitation will not occur, and the solution is unsaturated.

If K_{sp} < Q’, the solution is momentarily supersaturated. Therefore, precipitation will occur until the concentration of ions reaches of equilibrium.

If K_{sp} = Q’, the solution is saturated with the ions but there will be no precipitation.

**Related Resources**

- Strong vs. Weak Electrolytes: How to Categorize the Electrolytes?
- Le-Chatelier Principle: Examples and Applications
- Common Ion Effect: Principle, Importance and Examples

## Concepts Berg

**What is the Solubility Product, K _{sp}?**

The solubility product constant, K_{sp}, is the product of the concentration of ions at equilibrium, where each concentration of ion has an exponent equal to that of their coefficient in the balanced chemical equation.

**What are K _{sp} and Q?**

The solubility product, K_{sp }is the product of the concentration of ions in a saturated solution (at equilibrium). On the other hand, the ion product quotient, Q, is the product of the concentration of ions in a supersaturated, saturated, or unsaturated solution.

If Q > K_{sp}, supersaturated solution, and precipitation occur until Q = K_{sp}.

If Q < K_{sp}, unsaturated solution and more solid can be dissolved.

If Q = K_{sp}, saturated solution but no precipitation (system at equilibrium).

**What is meant by solubility product?**

It is the product of the concentration of ions at equilibrium with its solid ionic compound in a saturated solution.

**What does low K _{sp} mean?**

Low K_{sp} for an ionic compound indicates that the compound is less soluble. On the contrary, the higher the value of the solubility product constant, K_{sp}, more soluble the compound will be.

**How do you write K _{sp} equations?**

For the general equation,

A_{m}B_{n(s) }⇌ mA^{+n}_{(aq)} + nB^{-m}_{(aq)}

K_{sp} expression can be written as,

K_{sp} = [A^{+n}]^{m} [B^{-m}]^{n}

As an example, in the balanced equation, MgF_{2(s) }⇌ Mg^{2+}_{(aq)} + 2F^{–}_{(aq)}, the coefficient of Mg^{2+} is 1 whereas the coefficient of F^{–} is 2. These values will be put as their respective exponents. So the K_{sp} expression will be written as

K_{sp} = [Mg^{2+}] [F^{–}]^{2}

**Why is understanding K _{sp} important?**

K_{sp} gives a general idea of how soluble an ionic compound is. A higher K_{sp} value corresponds to higher solubility. In addition, it can be compared with the ion product quotient to predict whether precipitation will occur or not.

**How do you calculate K _{sp} from the saturated solution?**

From a saturated solution, we can find out the molar solubilities of each ionic specie. In order to calculate the K_{sp}, we take the product of the molar solubilities, each raised to an exponent that corresponds to its coefficient in the balanced chemical equation.

**How does K _{sp} relate to solubility?**

K_{sp} is directly proportional to solubility. Hence, a higher K_{sp} value corresponds to higher solubility.

**Is K _{sp} the solubility product constant for a saturated solution?**

Yes, K_{sp} is the equilibrium constant. However, an equilibrium can only be established when the solution is saturated.

**How do you find solubility given K _{sp} and molarity?**

If the K_{sp} of an ionic compound is provided, the solubility of its component ions can be found by taking the relevant root of the K_{sp} value.

**Reference**

- Solubility Product Constant, Ksp (chem.libretexts.org)