Question

What is $K_{sp}$?
(a) It is the solubility constant.
(b) It is the saturation constant.
(c) It is the concentration constant.
(d) It is the solubility of the substance

Answer 1

Hint: It is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the $K_{sp}$ value it has.

Complete step by step answer:

Let us first understand what solubility means.
- Solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble form the 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.
Now, let us try to understand about the solubility constant.
- The solubility product constant describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant defines the degree to which the compound can dissociate in water. The higher the $K_{sp}$, the more soluble the compound is.
$K_{sp}$ 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.
- Let us try to understand better with the help of an example!
So, for instance let us find the solubility constant of common salt.
As we know the chemical formula of common salt is NaCl.
$NaCl\rightleftharpoons { Na }^{ + }+{ Cl }^{ - }$
From the above equation we can see that, when dissolved in solvents one sodium chloride molecule dissociates into one sodium cation and one chloride anion.
Therefore, the solubility product constant can be expressed as,
$K_{sp}$ = ${ Na }^{ + }{ Cl }^{ - }$
Hence, $K_{sp}$ is the solubility constant. Therefore, option A is the required answer.

Note: With an ionic pair (a cation and an anion), the $K_{sp}$ value calculated is less than the experimental value due to ions involved in pairing. To reach the calculated $K_{sp}$ value, more solute must be added.