Question

Solubility product of a salt AB is ${{10}^{-8}}$ in a solution in which the concentration of ${{A}^{+}}$ ions is ${{10}^{-3}}M$ . The salt will precipitate when the concentration of ${{B}^{-}}$ ions is kept:
A. between ${{10}^{-8}}to{{10}^{-7}}M$
B. between ${{10}^{-7}}to{{10}^{-8}}M$
C. \[>{{10}^{-5}}M\]
D. \[<{{10}^{-8}}M\]

Answer 1

Hint: A salt forms precipitate only when the product of the concentration of its cation and anion is greater than its solubility product. Solubility product of a salt depends on the number of ions formed in a reaction or the concentration of the ions at equilibrium.

Complete Step by Step Answer:
A salt with chemical formula AB forms one cation and one anion. Thus the solubility product depends on concentration of the cations and anions.

Solubility product of a species is defined as the product of the concentration of the cation and anion formed from a species at equilibrium. At this stage the coefficients before the ions exist as a power to the concentration of the corresponding cation and anion.

AB dissociate at equilibrium as follows:
$AB\rightleftharpoons {{A}^{+}}+{{B}^{-}}$
Thus solubility product is given as follows-
$Ksp=[{{A}^{+}}][{{B}^{-}}]$; where $K_{sp}$ is defined as the solubility product.

The solubility product of the salt $AB$ is $1\times {{10}^{-8}}$ the concentration of ${{A}^{+}}$ ions is ${{10}^{-3}}M$. To precipitate the product of concentration should be greater than the solubility product.
$K_{sp}<[{{A}^{+}}][{{B}^{-}}]$
${{10}^{-8}}<{{10}^{-3}}\times [{{B}^{-}}]$
$[{{B}^{-}}]>{{10}^{-5}}M$
Thus to precipitate the concentration of the anion should be greater than ${{10}^{-5}}$ M.
Thus the correct option is C.

Note: Precipitate is an insoluble solid substance that is formed from a solution. The process of forming precipitate is called precipitation. It is only formed when the concentration of the species in solution is higher in amount.