Question

The solubility product of $PbC{{l}_{2}}$ at ${{20}^{\circ }}C$is $1.5\times {{10}^{-4}}$. Calculate the solubility.
A. $3.75\times {{10}^{-4}}$
B. $3.34\times {{10}^{-2}}$
C. $3.34\times {{10}^{2}}$
D. None of these

Answer 1

Hint: Solubility product usually increases with the increase in temperature because of the increased solubility. By using the formula of solubility and putting the values, we are able to find the solubility of $PbC{{l}_{2}}$.

Formula Used:
Formula of solubility product is: $S=\sqrt[3]{\dfrac{{{K}_{sp}}}{4}}$; where S is the solubility.
And ${{K}_{sp}}$ is the solubility product.

Complete Step by Step Solution:
Given ${{K}_{sp}}$= $1.5\times {{10}^{-4}}$. We are asked to find solubility.

We know most ionic compounds which are insoluble in water still dissolve in water to some extent. For example silver chloride dissociates to a small extent in the silver ions and the chloride ions.
$AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}$

It is shown by the equilibrium. As the solid silver chloride does not have a variable concentration and is therefore not included in the above expression. This is the solubility product principle.
${{K}_{sp}}$= $\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$

The relevant equilibrium of $PbC{{l}_{2}}$ is given by:
$PbC{{l}_{2}}$$\rightleftharpoons P{{b}^{2+}}+2C{{l}^{-}}$

We know the formula of solubility is :
${{K}_{sp}}$= $S\times {{(2S)}^{2}}=4{{S}^{3}}$
Then $S=\sqrt[3]{\dfrac{{{K}_{sp}}}{4}}$

By putting the values in the above equation, we get
$S=\sqrt[3]{\dfrac{1.5\times {{10}^{-4}}}{4}}$

Further solving, we get
S = $3.34\times {{10}^{-2}}$
Thus the solubility of $PbC{{l}_{2}}$is $3.34\times {{10}^{-2}}$
Thus, Option (B) is correct.

Note: Solubility depends on the various parameters which include the lattice energy and the solvation enthalpy of ions in the solution. Whenever a salt is dissolved in the solvent, the strong forces of attraction of the solute that is the lattice enthalpy of its ions needs to be overcome by the interactions between the ions and the solvent.