[Chemistry Class Notes] on Solubility Product Pdf for Exam

The solubility product constant is a simplified equilibrium constant denoted as K_sp which is defined for equilibrium between a solid and its respective ions in a given solution. Its value shows the degree to which a compound can dissociate in water. The greater the solubility product constant, the more soluble is the compound. The K_sp expression for a given salt is the product of the concentrations of the ions. 

 

Each concentration is raised to a power that is equal to the coefficient of that ion in a balanced equation to get the solubility equilibrium. The solubility product constants are used for describing the saturated solutions of ionic compounds of relatively low solubility. A saturated solution is in a dynamic equilibrium state between the dissolved and dissociated ionic compound and the undissolved solid. In this article, we will learn about what is the solubility product, what is K_sp, the solubility product definition, and some solubility product numerical.

 

Solubility Product Definition

The definition of the solubility product is given as follows:

Solubility is defined as the property of a substance known as solute to get dissolved in a solvent for the formation of a solution. The solubility of ionic compounds that dissociate and form cations and anions in water varies to a great extent. Some compounds are highly soluble and can even absorb moisture from the atmosphere whereas, on the other hand, the others are highly insoluble. 

 

Solubility Product Constant

Now, let us understand in detail what is the solubility product constant.

 

The solubility product is a type of equilibrium constant whose value depends on temperature. K_sp usually increases with the increase in temperature because of the increased solubility.

 

Most ionic compounds which are insoluble would still dissolve in water to a small extent. These insoluble compounds are considered to be strong electrolytes since whatever portion of the compound is dissolved also dissociates. For example, silver chloride dissociates to a small extent in the silver ions and chloride ions when added to water.

 

[AgCl (s)  leftrightarrow Ag^{+}(aq) + Cl^{-}(aq)]

 

This process is written as an equilibrium since the dissociation occurs only to a smaller extent. Hence,  an equilibrium expression is written for the process. You must keep in mind that the solid silver chloride does not have a variable concentration and is therefore not included in the expression. This is the solubility product principle.

 

[K_{sp} =[Ag^{+}][cl^{-}] ]

 

Solubility Product Formula

The solubility product constant is used to describe the saturated solutions of ionic compounds having relatively low solubility.  A saturated solution is said to be in a state of dynamic equilibrium between the ionic compound and the undissolved solid.

 

The K_sp formula is given in the form of the following equation:

[M_{x}A_{y} (s)  rightarrow  xM^{y+}(aq) + yA^{x-}(aq)]

 

The general equilibrium constant is written as follows:

[Kc = [M^{y+}]^{x}  [A^{x-}]^{y} ]

 

Significance of the Solubility Product

The solubility depends on several parameters which include the lattice enthalpy of salt and solvation enthalpy of ions in the solution. These two factors are the most important. Let us learn about the significance of the solubility product in detail.

1. Whenever a salt is dissolved in a solvent, the strong forces of attraction of solute that are the lattice enthalpy of its ions need to be overcome by the interactions between the ions and the solvent.

2. The solvation enthalpy of ions is always negative and this means that energy is released during the process.

3. The nature of the solvent finds the amount of energy that is released during the solvation which is solvation enthalpy.

4. Non-polar solvents have a small value of solvation enthalpy which means that this energy is not sufficient for overcoming the lattice enthalpy.

5. Hence the salts are not dissolved in the non-polar solvents. Therefore, for a salt to be dissolved in a solvent, its solvation enthalpy must be greater than its lattice enthalpy.

6. The solubility depends on the temperature and it is varied for every salt.

 

Salts are Classified According to their Solubility which is Given in the Table Below:

 

Solubility Product Numericals

Let us now take a look at some of the K_sp chemistry examples.

 

Example 1:

Determine the solubility product constant for [MgF_{2}].

 

Solution:

The relevant equilibrium is given by:

[MgF_{2} (s)  leftrightarrow Mg^{2+}(aq) + 2F^{-}(aq)]

Hence, the associated equilibrium constant is given by:

[K_{sp} =[Mg^{2+}][F^{2-}] ]

 

Example 2:

Give the solubility product constant for [Ag_{2}CrO_{4}].

 

Solution:

The relevant equilibrium is given by:

[Ag_{2}CrO_{4} (s)  leftrightarrow 2Ag^{+}(aq) + CrO_{4}^{2-}(aq)]

The associated equilibrium is therefore given by:

[K_{sp} =[Ag^{2+}]^{2}[CrO_{4}^{2-}] ]

 

The K_sp-Solubility Relationship

Because it is basically the product of the solubilities of the ions in moles per liter, K_sp is called the solubility product. A salt’s solubility product, or vice versa, can therefore be computed from its solubility.

The sensitivity of AgBr to light is used in photographic films. When light strikes an AgBr crystal, a small percentage of the Ag⁺ ions are reduced to silver metal. When the film is developed, the remaining Ag⁺ ions in these crystals are reduced to silver metal. To “fix” the image, AgBr crystals that do not absorb light are removed from the film.

Example: 3

To test if AgBr can be eradicated by merely washing the film, let’s calculate the solubility of AgBr in water in grams per liter.

Solution:

We start with the equilibrium equation of t
his salt in an aqueous solution, which is a balanced equation.

AgBr(s) [Longleftrightarrow] [Ag^{+}(aq) + Br^{-}(aq) ]

The solubility product equation for this reaction is then written.

[K_{sp} =[Ag^{+}][Br_{-}] = 5.0 times 10^{-13}]

 

For two unknowns, the Ag⁺ and Br^– ion concentrations, one equation cannot be solved. However, by noting that one Ag⁺ ion is emitted for every Br- ion, we can derive a second equation. Because there is no alternative source of either ion in this solution, their equilibrium concentrations must be equal. 

[ [Ag^{+}]=[Br^{-}] ]

The following result is obtained by substituting this equation into the K_sp expression.

[[Ag^{+}]^{2} = 5.0 times 10^{-13} ]

The equilibrium concentrations of Ag⁺ and Br^– ions can be determined by taking the square root of both sides of this equation.

[ [Ag^{+}]=[Br^{-}]] = [7.1 times 10^{-7} M ]

We can calculate the solubility in grams per liter after we know how many moles of AgBr dissolve in a liter of water.

[ frac{7.1 times 10^{-7} text{mol AgBr}}{1 L} times frac{187.8 g AgBr}{1 text{mol AgBr}} = 1.3 times 10^{-4} g text{AgBr}/L]

AgBr dissolves in water at a rate of 0.00013 grams per liter. Attempting to wash the unexposed AgBr off the photographic film with water is therefore impractical.

Calculating the solubility product for 1:1 salts like AgBr is relatively simple. We need to understand the relationship between a salt’s solubility and the concentrations of its ions at equilibrium in order to expand such calculations to compounds with more complex formulas. The amount of salt that dissolves in water will be denoted by the symbol Cs.

Get to know more about the concepts of solubility product at . The insights of the top subject matter experts will aid you to understand this topic better and use its concepts to solve problems in the exercises and exams.