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Considering the relation between solubility and Ksp . K𝑠𝑝 is important when describing the solubility of slightly ionic compounds. However, this article discusses ionic compounds that are difficult to dissolve; they are considered "slightly soluble" or "almost insoluble." Solubility product constants (Ksq) are given to those solutes, and these constants can be used to find the molar solubility of the compounds that make the solute. This relationship also facilitates finding the Ksq of a slightly soluble solute from its solubility.
Introduction
Solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. Solubility product can be determined by slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:
The solubility product can be calculated by taking solid's solubility expressed in units of moles per liter (mol/L), and known as molar solubility .As the example of Calculation Ksp from Equilibrium Concentration .
Example 1:
Fluorite (CaF2) is slightly soluble solid that dissolves according to the equation :
CaF2 (s) −⇀ Ca2+ (aq)+2 F–(aq)
The concentration of Ca2+ in a saturated solution of CaF2 is 2.1 × 10–4 M; therefore, that of F– is 4.2 × 10–4 M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?
Solution :
First , write out the Ksp expression, then substitute in concentrations and solve for Ksp:
CaF2 (s) −⇀ Ca2+ (aq)+2 F–(aq)
A saturated solution is a solution at equilibrium with the solid. Thus:
Ksp =[Ca2+] [F–]2
Ksp =(2.1 x 10-4) (4.2x10–4)2
= 3.7 x10-11
As with other equilibrium constants, we do not include units with Ksp
Example 2 :
The Ksp of copper(I) bromide, CuBr, is 6.3 × 10–9. Calculate the molar solubility of copper bromide
Solution
The solubility product constant of copper(I) bromide is 6.3 × 109.The reaction is:
CuBr(s)⇌ Ca2+(aq)+ Br –(aq)
First, write out the solubility product equilibrium constant expression:
Ksp =[Cu+] [Br–]
Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr CuBr column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
Therefore, the molar solubility of CuBr is 7.9 x 10-5
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium. Equilibrium is the state at which the concentrations of products and reactant are constant after the reaction has taken place. The solubility product constant (Ksp) describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the Ksp, the more soluble the compound is. Ksq is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. Ksp is used to describe the saturated solution of ionic compounds. (A saturated solution is in a state of equilibrium between the dissolved, dissociated, undissolved solid, and the ionic compound.)
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