Question

The solubility of $PbC{{l}_{2}}$ is
(A) $\sqrt{{{K}_{sp}}}$
(B) ${{\left( {{K}_{sp}} \right)}^{\frac{1}{3}}}$
(C) ${{\left( \frac{{{K}_{sp}}}{4} \right)}^{\frac{1}{3}}}$
(D) $\sqrt{8{{K}_{sp}}}$

Answer 1

Hint: Solubility means the ability of the solute to get dissolved in a solvent to form a solution. When a solute is dissolved in water, it can be completely soluble, sparingly soluble, or completely insoluble. The solubility product is calculated as the product of the concentration of each of the ions, each raised to power to its coefficient in a balanced equation. More is the solubility product; more is the solubility of the compound and vice-versa. It is denoted by ${{K}_{sp}}$.

Complete Step by Step Solution:
$PbC{{l}_{2}}$dissociates into ions as

On dissociation, $PbC{{l}_{2}}$gives 1$P{{b}^{2+}}$ ion and 2 $C{{l}^{-}}$ions. So, the solubility product of $PbC{{l}_{2}}$is
${{K}_{sp}}=\left[ P{{b}^{2+}} \right]{{\left[ C{{l}^{-}} \right]}^{2}}$

Let the solubility be s. So,
${{K}_{sp}}=\left( s \right)\times {{\left( 2s \right)}^{2}}$
${{K}_{sp}}=4{{s}^{3}}$
${{s}^{3}}=\frac{{{K}_{sp}}}{4}$
$s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{\frac{1}{3}}}$
Hence, the solubility of $PbC{{l}_{2}}$ is $s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{\frac{1}{3}}}$
Correct Option: (C) ${{\left( \frac{{{K}_{sp}}}{4} \right)}^{\frac{1}{3}}}$.

Additional Information: When a salt dissolves in a solvent, the interactions between the ions and the solvent must overcome the strong forces of attraction of the solute, which are the lattice enthalpies of its ions. For a salt to dissolve in a solvent, the solvation energy of ions must be greater than the lattice enthalpy. A non-polar solvent does not dissolve a salt as its solvation energy is very low and it is not sufficient to overcome the lattice enthalpy.

Note: The solubility product depends upon temperature, common-ion effect, diverse-ion effect, and the presence of ion pairs. The units of solubility are moles per litre. The units of the solubility product depend upon the number of ions formed after dissociation of the electrolyte. It is generally given as ${{(mol{{L}^{-1}})}^{n}}$ where n is the number of ions formed in the reaction.