Solubility Formula: Definition, Ksp, Henry’s Law and Solved Examples

The Solubility Formula is used to calculate the maximum amount of solute that dissolves in a given amount of solvent or solution at a fixed temperature and pressure.
For basic calculations, solubility can be written as Solubility = Mass of solute / Volume of solution or s = Moles of solute / Volume of solution.

The Solubility Formula helps students measure how much of a substance dissolves in a solvent. In simple Chemistry questions, solubility is calculated using mass per litre or moles per litre. In ionic equilibrium, solubility is linked with the solubility product constant, Ksp. For gases, solubility depends on pressure and is calculated using Henry’s law.

In Class 10, Class 11 and Class 12 Chemistry, solubility appears in solutions, concentration, equilibrium, ionic compounds, gases in liquids and Ksp numericals. CBSE, ICSE, state board, NEET and JEE foundation questions may ask students to calculate mass solubility, molar solubility, Ksp, gas solubility or conversions between g/L and mol/L.

Key Takeaways

• Solubility Formula: Solubility = Mass of solute / Volume of solution.
• Molar Solubility Formula: s = Moles of solute / Volume of solution.
• Ksp Formula: Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ for AₘBₙ.
• Henry’s Law Formula: C = kH × P.
• Main Unit: Solubility is commonly expressed in g/L or mol/L.
• Condition: Solubility depends on temperature, pressure and the nature of solute and solvent.

Solubility Formula Structure 2026

What is Solubility Formula?

The Solubility Formula calculates the amount of solute that dissolves in a given volume of solution at a particular temperature and pressure.

Mass solubility formula:

Solubility = Mass of solute / Volume of solution

Unit:

g/L

Molar solubility formula:

s = Moles of solute / Volume of solution

Unit:

mol/L

Where:

• s = molar solubility
• Solute = substance that dissolves
• Solvent = substance in which solute dissolves
• Solution = solute + solvent

Solubility tells how much solute can dissolve before the solution becomes saturated.

Solubility

Solubility is the maximum amount of solute that can dissolve in a specific amount of solvent at a given temperature and pressure. A solution that contains the maximum dissolved solute is called a saturated solution.

Example:

If 20 g of a salt dissolves completely in 1 L of water at a fixed temperature, its solubility is:

Solubility = 20 g / 1 L

Solubility = 20 g/L

This means 20 g of that salt dissolves in 1 L of solution under those conditions.

Solubility Calculation Formula

The solubility calculation formula depends on whether the question gives mass, moles, volume, Ksp or pressure.

For mass-based questions:

Solubility = Mass of solute / Volume of solution

For mole-based questions:

s = Moles of solute / Volume of solution

For gases:

C = kH × P

For sparingly soluble salts:

Ksp = Product of ion concentrations raised to their powers

Each formula is used in a different type of solubility problem.

Mass Solubility Formula

The mass solubility formula gives solubility in grams per litre.

Formula:

Solubility = Mass of solute (g) / Volume of solution (L)

Unit:

g/L

Example:

If 15 g of solute dissolves in 0.5 L of solution, then:

Solubility = 15 / 0.5

Solubility = 30 g/L

Answer:

The mass solubility is 30 g/L.

Molar Solubility Formula

The molar solubility formula gives solubility in moles per litre. It is represented by s.

Formula:

s = Moles of solute (mol) / Volume of solution (L)

Unit:

mol/L

Example:

If 0.2 mol of a solute dissolves in 1 L of solution, then:

s = 0.2 / 1

s = 0.2 mol/L

Answer:

The molar solubility is 0.2 mol/L.

Molar solubility is commonly used in Ksp calculations.

Concentration Formula and Solubility

Solubility is a type of concentration when the solution is saturated. The general concentration formula is:

Concentration = Amount of solute / Volume of solution

For mass concentration:

Concentration = Mass of solute / Volume of solution

For molar concentration:

Molarity = Moles of solute / Volume of solution

When the solution is saturated, this concentration becomes the solubility of the substance at that temperature.

Solute and Solvent in Solubility Formula

The solubility formula depends on the amount of solute and the volume of solution or solvent.

Example:

In sugar water:

• Sugar = solute
• Water = solvent
• Sugar water = solution

Solubility Formula Chemistry

In Chemistry, solubility is used in solutions, ionic equilibrium, precipitation reactions, gas solubility and concentration calculations.

Common solubility formula chemistry forms:

Solubility = Mass of solute / Volume of solution

s = Moles of solute / Volume of solution

Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ

C = kH × P

These formulas help students calculate how much solute dissolves and whether an ionic compound will dissolve, precipitate or stay in equilibrium.

Solubility Product Formula

The solubility product formula is used for sparingly soluble ionic compounds. It is represented by Ksp.

For a general salt:

AₘBₙ(s) ⇌ mAⁿ⁺(aq) + nBᵐ⁻(aq)

The Ksp formula is:

Ksp = [Aⁿ⁺]ᵐ[Bᵐ⁻]ⁿ

Where:

• Ksp = solubility product constant
• [Aⁿ⁺] = concentration of positive ion
• [Bᵐ⁻] = concentration of negative ion
• m and n = coefficients from the balanced dissociation equation

Only dissolved ions are included in the Ksp expression. The solid compound is left out because its concentration remains constant.

Ksp Formula

The Ksp formula changes according to the ionic compound and its dissociation.

Case 1: AB type salt

For:

AB(s) ⇌ A⁺(aq) + B⁻(aq)

If molar solubility = s, then:

[A⁺] = s

[B⁻] = s

Ksp = [A⁺][B⁻]

Ksp = s × s

Ksp = s²

So:

s = √Ksp

Case 2: A₂B type salt

For:

A₂B(s) ⇌ 2A⁺(aq) + B²⁻(aq)

If molar solubility = s, then:

[A⁺] = 2s

[B²⁻] = s

Ksp = [A⁺]²[B²⁻]

Ksp = (2s)²(s)

Ksp = 4s³

Case 3: AB₂ type salt

For:

AB₂(s) ⇌ A²⁺(aq) + 2B⁻(aq)

If molar solubility = s, then:

[A²⁺] = s

[B⁻] = 2s

Ksp = [A²⁺][B⁻]²

Ksp = (s)(2s)²

Ksp = 4s³

Molar Solubility from Ksp

Molar solubility can be calculated from Ksp by writing the dissociation equation and replacing ion concentrations with s.

Example:

For AgCl:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Let molar solubility = s.

Then:

[Ag⁺] = s

[Cl⁻] = s

Ksp = [Ag⁺][Cl⁻]

Ksp = s²

So:

s = √Ksp

If Ksp = 1.6 × 10⁻¹⁰, then:

s = √(1.6 × 10⁻¹⁰)

s = 1.26 × 10⁻⁵ mol/L

Answer:

The molar solubility of AgCl is 1.26 × 10⁻⁵ mol/L.

Henry’s Law Formula

Henry’s law is used for the solubility of gases in liquids. It states that gas solubility is directly proportional to the partial pressure of the gas above the liquid.

Formula:

C = kH × P

Where:

• C = solubility of the gas
• kH = Henry’s law constant
• P = partial pressure of the gas

If pressure increases, gas solubility increases.

Example:

If kH = 2 × 10⁻³ mol/L·atm and P = 3 atm, then:

C = kH × P

C = 2 × 10⁻³ × 3

C = 6 × 10⁻³ mol/L

Answer:

The gas solubility is 6 × 10⁻³ mol/L.

Solubility of Gases Formula

The solubility of gases formula is based on Henry’s law.

Formula:

C = kH × P

Where:

• C = concentration or solubility of gas in liquid
• kH = Henry’s law constant
• P = partial pressure of the gas

This formula is useful in questions about gases dissolved in water, soft drinks, oxygen in water and carbon dioxide under pressure.

Example:

Carbon dioxide remains dissolved in sealed soda because pressure is high. When the bottle is opened, pressure decreases and gas escapes.

Conversion from g/L to mol/L

To convert mass solubility from g/L to mol/L, divide by molar mass.

Formula:

mol/L = (g/L) / Molar mass

Example:

A solute has solubility 20 g/L and molar mass 40 g/mol.

mol/L = 20 / 40

mol/L = 0.5 mol/L

Answer:

The molar solubility is 0.5 mol/L.

Conversion from mol/L to g/L

To convert molar solubility from mol/L to g/L, multiply by molar mass.

Formula:

g/L = mol/L × Molar mass

Example:

A solute has molar solubility 0.25 mol/L and molar mass 80 g/mol.

g/L = 0.25 × 80

g/L = 20 g/L

Answer:

The mass solubility is 20 g/L.

Factors Affecting Solubility

Solubility depends on the nature of the solute, solvent, temperature and pressure.

Main factors:

• Nature of solute
• Nature of solvent
• Temperature
• Pressure for gases
• Common ion effect for ionic compounds

For many solids, solubility increases as temperature increases. For gases, solubility usually decreases when temperature increases and increases when pressure increases.

Saturated, Unsaturated and Supersaturated Solutions

Solubility is closely connected with the type of solution formed.

The solubility formula is usually applied to saturated solutions because solubility means the maximum amount that can dissolve.

Difference Between Solubility and Molar Solubility

Solubility and molar solubility are related but expressed in different units.

Example:

If 10 g of a substance dissolves in 1 L, its solubility is 10 g/L. If this equals 0.1 mol, its molar solubility is 0.1 mol/L.

How to Calculate Solubility

To calculate solubility, first identify what values are given in the question.

Case 1: When mass and volume are given

Use:

Solubility = Mass of solute / Volume of solution

Example:

Mass = 25 g

Volume = 0.5 L

Solubility = 25 / 0.5

Solubility = 50 g/L

Case 2: When moles and volume are given

Use:

s = Moles of solute / Volume of solution

Example:

Moles = 0.4 mol

Volume = 2 L

s = 0.4 / 2

s = 0.2 mol/L

Case 3: When Ksp is given

Use the ionic dissociation equation.

Example:

For AB type salt:

Ksp = s²

So:

s = √Ksp

Case 4: When gas pressure is given

Use Henry’s law:

C = kH × P

Example:

kH = 1.5 × 10⁻³ mol/L·atm

P = 2 atm

C = 1.5 × 10⁻³ × 2

C = 3.0 × 10⁻³ mol/L

Solved Examples on Solubility Formula

Solubility Formula questions usually test mass solubility, molar solubility, Ksp and Henry’s law.

Example 1: Find mass solubility

A 12 g solute dissolves in 0.3 L solution. Find its solubility.

Given:

Mass of solute = 12 g

Volume of solution = 0.3 L

Formula:

Solubility = Mass of solute / Volume of solution

Substitute:

Solubility = 12 / 0.3

Solubility = 40 g/L

Answer:

The solubility is 40 g/L.

Example 2: Find molar solubility

A 0.6 mol solute dissolves in 3 L solution. Find molar solubility.

Given:

Moles of solute = 0.6 mol

Volume of solution = 3 L

Formula:

s = Moles of solute / Volume of solution

Substitute:

s = 0.6 / 3

s = 0.2 mol/L

Answer:

The molar solubility is 0.2 mol/L.

Example 3: Convert g/L to mol/L

A compound has solubility 30 g/L and molar mass 60 g/mol. Find molar solubility.

Formula:

mol/L = (g/L) / Molar mass

Substitute:

mol/L = 30 / 60

mol/L = 0.5 mol/L

Answer:

The molar solubility is 0.5 mol/L.

Example 4: Find Ksp from molar solubility for AB salt

For AB salt, molar solubility is 2 × 10⁻⁴ mol/L. Find Ksp.

Dissociation:

AB(s) ⇌ A⁺(aq) + B⁻(aq)

If molar solubility = s:

[A⁺] = s

[B⁻] = s

Ksp = s²

Substitute:

Ksp = (2 × 10⁻⁴)²

Ksp = 4 × 10⁻⁸

Answer:

Ksp = 4 × 10⁻⁸.

Example 5: Find gas solubility using Henry’s law

Given:

kH = 4 × 10⁻³ mol/L·atm

P = 5 atm

Formula:

C = kH × P

Substitute:

C = 4 × 10⁻³ × 5

C = 20 × 10⁻³

C = 2.0 × 10⁻² mol/L

Answer:

The solubility of the gas is 2.0 × 10⁻² mol/L.

Common Mistakes in Solubility Formula

Many solubility formula mistakes happen when students mix g/L with mol/L or use the wrong Ksp expression.

Important checks:

• Use mass solubility formula for g/L.
• Use molar solubility formula for mol/L.
• Convert grams to moles before using Ksp.
• Raise ion concentrations to the correct powers in Ksp.
• Use coefficients from the balanced dissociation equation.
• Use Henry’s law only for gases dissolved in liquids.
• Mention temperature and pressure when solubility conditions are important.

For Ksp questions, always write the dissociation equation before substituting values.

Applications of Solubility Formula

The Solubility Formula is useful in Chemistry, Biology, medicine, environmental science and industry. It helps explain how substances dissolve and form solutions.

Main applications:

• It calculates how much solute dissolves in a solvent.
• It helps prepare solutions of known concentration.
• It is used in Ksp and precipitation reactions.
• It explains gas solubility in liquids.
• It helps study water quality and dissolved gases.
• It is used in medicines and drug formulation.
• It supports crystallisation and purification processes.
• It helps students understand saturated solutions.