Solubility Product - Definition, Formula, Significance

Solubility is defined as the amount of solute required to form a saturated solution in a given amount of solvent at a specific temperature. When a solute is added to a solvent it dissolves up to a certain limit after that, no more solute dissolves this point is called saturation.

The solubility product (Ksp) is a constant value that represents this equilibrium. The product of the concentrations of the ions present in a saturated solution, each raised to the power of their coefficients.

Example:

AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)

Ksp = [Ag⁺] [Cl⁻]

At equilibrium:

• Solid AgCl is present
• Ions (Ag⁺ and Cl⁻) are present in solution

Formula for Solubility Product

The solubility product (Ksp) is written using the balanced chemical equation of the dissociation of a sparingly soluble salt. Ksp is product of concentrations of ions, each raised to the power of its coefficient.

For a salt:

A_xB_y (s) ⇌ xA ^y+ (aq) + yB ^x- (aq)

The Ksp expression is:

Ksp = [A^y+] ^x [B^x-] ^y

Significance of Solubility Product

The solubility product (Ksp) is very useful in understanding the behavior of sparingly soluble salts in solutions. It has several important applications in chemistry.

• Predicting Solubility: It helps us know whether a salt is highly soluble or sparingly soluble. Low Ksp refers to low solubility and High Ksp means higher solubility.
• Predicting Precipitation: By comparing ionic product (Q) with Ksp: Q < Ksp → no precipitation (unsaturated), Q = Ksp → saturated solution , Q > Ksp → precipitation occurs.
• Calculation of Solubility: It is used to calculate the solubility of sparingly soluble salts.
• Common Ion Effect: This shows how the presence of a common ion reduces solubility.
• Selective Precipitation: It helps to separate ions from a mixture. Different salts have different Ksp values, so they precipitate at different conditions.

Solubility Product Constant

Ksp is defined as the product of the molar concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients, at a given temperature.

• At a fixed temperature, the value of Ksp remains constant
• It does not depend on the amount of solid present
• It changes only when temperature changes

Example:

AgCl (s) ⇌ Ag⁺ (aq) + Cl⁻ (aq)

Ksp = [Ag⁺] [Cl⁻]

Steps to Calculate Solubility Product Constant

To calculate Ksp, follow these steps:

Step 1: Write the Dissociation Equation

Write the balanced chemical equation of the salt dissolving in water.

Example:
CaF₂ (s) ⇌ Ca²⁺ (aq) + 2F^-(aq)

Step 2: Assume Solubility (s)

Let the solubility of the salt = s mol/L. It means Amount of salt dissolved = s

Step 3: Write Ion Concentrations

Use the equation to express ion concentrations in terms of s.

Example:

For CaF₂:

• [Ca ²⁺] = s
• [F ^-] = 2s

Step 4: Write Ksp Expression

Ksp = product of ion concentrations (with powers)

For CaF₂:

Ksp = [Ca²⁺] [F ^-] ²

Step 5: Substitute Values

Put the values in terms of s:

Ksp = s × (2s) ²
Ksp = 4s ³

Step 6: Calculate Ksp

• If s is given, substitute and calculate Ksp
• If Ksp is given, solve to find solubility (s)

Relation between Solubility and Solubility Product

The solubility (s) of a salt is directly related to its solubility product (Ksp). This relation depends on how the salt dissociates. The relation depends on the number of ions formed. Higher powers of s appear when more ions are produced

Using this relation, we can:

• Find solubility from Ksp
• Find Ksp from solubility

Examples:

For 1 : 1 Electrolyte

AgCl ⇌ Ag⁺ + Cl⁻

Ksp = s²

Relation:
s = √Ksp

Solved Examples

Example 1: Calculate the solubility product constant for silver chloride (AgCl) given that the concentration of Ag⁺ ions is 1.5×10 ⁻⁵M and Cl⁻ ions is 2.0×10⁻⁵M.

Solution:

K_sp = [Ag⁺]×[Cl^-]

K_sp = (1.5×10⁻⁵)×(2.0×10⁻⁵)

K_sp = 3.0×10 ⁻¹⁰

Example 2: Calculate the Molar Solubility for calcium hydroxide (Ca(OH)₂) given that the solubility product constant for calcium hydroxide is 6.7 X 10^-6

Solution:

Ca(OH)² = Ca ²⁺ + 2OH ^-

K_sp = [Ca ²⁺]+[OH ^-] ²

Let x be molar solubility and Ca²⁺ = x and OH ^-= 2x

6.7 × 10 ^-6= 4x ³

Now substitute x

x³= 6.7 × 10 ^-6 /4

x = 0.012 M (approx)

Example 3: Calculate the solubility product constant for silver chloride (AgCl) given that the concentration of Ag⁺ ions is 0.5×10 ⁻⁶M and Cl⁻ ions is 3.0×10⁻⁹M.

Solution:

K_sp =[Ag⁺]×[Cl^-]

K_sp =( 0.5×10 ⁻⁶M)×(3.0×10⁻⁹M)

Ksp = 1.5 x 10 ^-15

Example 4:How to Calculate Solubility Product Constant? Demystify the calculation of Ksp with an example. Consider silver chloride (AgCl) with [Ag+]=2.4×10⁻⁵M and [Cl-]=1.0×10⁻⁵M.

Solution:

K_sp = Ag ⁺ × Cl ^-

K_sp = (2.4×10⁻⁵)×(1.0×10⁻⁵)

K_sp = 2.4×10⁻¹⁰

Related Articles:

• Solubility Equilibria 
• Factors affecting Solubility
• Factors Affecting Chemical Equilibrium