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Solubility Formula – Definition, Formula and Solved Examples

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  • Last Updated : 05 Aug, 2022
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Solubility is defined as the maximum amount of solute that can be dissolved in a known amount of solvent at a particular temperature. A solution can be defined as a homogeneous mixture of one or more solutes in a solvent. Adding sugar cubes to a cup of tea or coffee is a very common example of a solution. And the property that helps sugar molecules dissolve is called solubility. Therefore, the term solubility can be defined as the property of a substance or solute that is soluble in a particular solvent. A solute is any component that can be liquefied in a solvent, either solid, liquid, or gas.

Solubility

Solubility is the maximum amount of solute that can be dissolved in a known amount of solvent at a particular temperature. Suppose you have salt and you are trying to dissolve it in a glass of water. Put sugar in water and stir to dissolve. If you repeat this process many times, the sugar will not dissolve, and the excess sugar will settle at the bottom of the jar. The amount of sugar that cannot be dissolved after it is added is its solubility. Solubility depends on many factors. Several factors include solute and solvent types, temperature, and pressure.

S = √Ksp 

The solubility product equation has the same general form as other equilibrium constant equations, and the Ksp is called the solubility product constant.

Factors affecting solubility 

  • Temperature effects: Le Chatelier’s principle shows that if the dissolution is endothermic, the solubility should increase with increasing temperature, and vice versa if the dissolution is exothermic. Solids are so incompressible that pressure does not have a significant effect on solid solutions in liquids. The solubility of the solute can be increased by changing the temperature. Generally, water dissolves solutes at 20°C or 100°C. The poorly soluble solids can be completely liquefied by raising the temperature. However, for gaseous substances, the temperature has the opposite effect on solubility. As the temperature rises, the gas expands and escapes from the solvent.
  • Power and Bonds: It is already known that intermolecular forces and bond properties vary from molecule to molecule. The possibility of dissolution between two different elements is more difficult compared to similar substances. For example, water is a protic solvent that easily dissolves polar solutes such as ethanol.
  • Pressure: Gaseous substances are much more sensitive to pressure than solid and liquid substances. As the partial pressure of a gas increases, so does its solubility. Soda bottles are one example of CO2 bottling under high pressure.

Solubility Product 

The solubilities of ionic compounds that dissociate in water to form cations and anions vary greatly. Some compounds are very soluble and can even absorb moisture from the atmosphere, while others are very insoluble. The solubility product is a kind of equilibrium constant whose value depends on temperature. Ksp usually increases in solubility and therefore increases with increasing temperature.

Solubility Product formula 

The solubility product constant is used to represent a saturated solution of an ionic compound with relatively low solubility. Saturated solutions are in dynamic equilibrium between ionic compounds and undissolved solids.

The Ksp expression is given in the form of the following expression:

MxAy (s) ⇢  xMy+(aq)+yAx−(aq)

A typical equilibrium constant is described as follows,

Kc = [My+]x[Ax−]y

Significance of solubility product 

  • The solubility of a substance is determined by the size of its solubility product.  
  • A specific form of non-uniform equilibrium constant or equilibrium constant is the solubility product. This is related to saturated solutions in which the ionic components are not completely dissolved.   
  • The solubility product changes with temperature, so the temperature at which they are measured should always be stated.

Solved Examples on Solubility

Example 1: The Ksp of copper bromide, CuBr, is  8× 10–10. Calculate the molar solubility of copper bromide.

Solution: 

CuBr (s) ⇢ Cu+(aq) + Br(aq)

Ksp =  Cu+ + Br

Ksp = S × S

8× 10–10 = S2

S = √8 × 10–10

S = 2.8 × 10-5mol/L.

Cu+ = S = 2.8 × 10-5mol/L.

Br = S = 2.8 × 10-5mol/L.

Example 2: The molar solubility of tin iodide SnI2 is 1.00 x 10-2 mol/L. Calculate the Ksp of this compound.

Solution:

The solubility equilibrium of SnI2 is,

SnI2(s) ⇢ Sn2+(aq) + 2I(aq)

The Ksp expression is written as,

Ksp = [Sn2+][I]2

1 mol of SnI2 Produces 1.0 mol of Sn2+, but 2.0 mol of I.

[Sn2+] = 1.00 × 10-2M

[I] = (2) × 1.00 × 10-2M

[I] = 2.00 × 10-2M

Putting these values in Ksp expression

Ksp = (1.00 × 10-2)(2.00× 10-2)2

Ksp = 4 × 10-6 M2.

Example 3: Calculate the solubility of Silver chloride, Ksp of AgCl = 4 × 10-8

Solution: 

The chemical formula of calcium chloride is CaCl2. When dissolved in a polar solvent, calcium chloride molecules dissociate into calcium cations and two chloride anions. This equilibrium reaction can be expressed as:

AgCl ⇢ Ag++ Cl

Therefore, the solubility product constant can be expressed as:

Ksp = [Ag+] [Cl] .

Ksp = S× S

Ksp = S2

4 × 10-8 = S2

S = √4 × 10-8

S = 2 × 10-4 mol/L.

[Ag+] = [Cl] =  2 × 10-4 mol/L.

Example 4: If solid CaCl2 equilibrates with pure water, what are [Ca2+] and [Cl] the solution at equilibrium? [Ksp(PbCl2) = 2.1 × 10-5].

Solution: 

CaCl2 ⇢ Ca2+ + 2Cl

Ksp = [Ca2+] + [Cl]2

Ksp = S × (2S)2

Ksp = 4S3

2.1 × 10-5 = 4S3

2.1 × 10-5 / 4 = S3

5.25 × 10-6 = S3

S = 1.73 × 10-2 mol/L

[Ca2+] = S = 1.73 × 10-2 mol/L

[Cl] = 2S = 2 × 1.73 × 10-2 

[Cl] = 3.46 × 10-2 mol/L.

FAQs on Solubility Formula 

Question 1: What does the term solubility mean? 

Answer:

The solubility of a substance is the maximum amount that can be dissolved in a particular amount of solvent at a particular temperature. Adding sugar cubes to a cup of tea or coffee is a very common example of a solution. And the property that helps sugar molecules dissolve is called solubility. A solute is any component that can be liquefied in a solvent, either solid, liquid, or gas.

Question 2: What is the difference between solubility and solubility product constants?

Answer:

The solubility of a particular solute in a particular solvent is given by the total amount of solute that can be dissolved in the solvent in equilibrium. The solubility product constant, on the other hand, is an equation, the equilibrium constant, that provides information about the equilibrium between the solute and the constituent ions that dissociate it in the solution. Determining the concentration of solute in a solvent after the substance has dissolved is important in determining how much it can dissolve in the solvent.

Question 3: How does temperature affect the solubility of a liquid in a liquid?

Answer:

Le Chatelier’s principle shows that if the dissolution is endothermic, the solubility should increase with increasing temperature, and vice versa if the dissolution is exothermic. The solid is so incompressible that the pressure does not significantly affect the solid solution in the liquid. The solubility of the solute can be increased by changing the temperature. Generally, water dissolves solutes at 20° C or 100° C. The poorly soluble solids can be completely liquefied by raising the temperature. However, for gaseous substances, the temperature has the opposite effect on solubility. As the temperature rises, the gas expands and escapes from the solvent.

Question 4: What are the factors that affect the value of Ksp?

Answer: 

Some important factors that affect the solubility product constant are:

  • Common-ion effect (the presence of common ions reduces the value of Ksp).
  • Diverse ionic effects (Ksp values ​​are high if solute ions are abnormal).
  • Existence of ion pair.

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