Question

At ${{20}^{o}}C$, the $A{{g}^{+}}$ion concentration in a saturated solution of $A{{g}_{2}}Cr{{O}_{4}}$is$1.5\times {{10}^{-4}}$moles per litre. At ${{20}^{o}}C$, the solubility product of $A{{g}_{2}}Cr{{O}_{4}}$would be:
(A) $3.3750\times {{10}^{-14}}$
(B) $1.6875\times {{10}^{-10}}$
(C) $1.6875\times {{10}^{-12}}$
(D) $1.6875\times {{10}^{-11}}$

Answer 1

Hint: The solubility product of a sparingly soluble salt forming a saturated solution in water is calculated as the product of the concentrations of the ions raised to a power equal to the number of the ions occurring in the equation representing the dissociation of the electrolyte. The solubility product is denoted by ${{K}_{sp}}$.

Formula Used: The formula for the solubility product depends on the number of ions formed after dissociation of the compound. The solubility product is given by ${{K}_{sp}}={{\left[ {{A}^{+}} \right]}^{a}}{{\left[ {{B}^{-}} \right]}^{b}}$ where ${{A}^{+}}$ are the cations, ${{B}^{-}}$are the anions, and a and b are the number of cations and anions formed in the reaction.

Complete Step by Step Solution:
 $A{{g}_{2}}Cr{{O}_{4}}$is dissociated as

This means \[A{{g}_{2}}Cr{{O}_{4}}\]on dissociation gives 2$A{{g}^{+}}$ ions and 1 $Cr{{O}_{4}}^{2-}$ ion.
Given the $A{{g}^{+}}$ion concentration in a saturated solution of $A{{g}_{2}}Cr{{O}_{4}}$ is $1.5\times {{10}^{-4}}$ moles per litre.
As \[A{{g}_{2}}Cr{{O}_{4}}\] gives 2$A{{g}^{+}}$ ions so the concentration of each of the $A{{g}^{+}}$ ions is $1.50\times {{10}^{-4}}$
And the concentration of $Cr{{O}_{4}}^{2-}$is $\frac{1.50\times {{10}^{-4}}}{2}$ $=0.75\times {{10}^{-4}}$

Now, the solubility product of \[A{{g}_{2}}Cr{{O}_{4}}\]is given by ${{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{2-} \right]$
${{K}_{sp}}={{\left( 1.50\times {{10}^{-4}} \right)}^{2}}\times \left( 0.75\times {{10}^{-4}} \right)$
${{K}_{sp}}=\left( 2.25\times {{10}^{-8}} \right)\times \left( 0.75\times {{10}^{-4}} \right)$
${{K}_{sp}}=1.6875\times {{10}^{-12}}mo{{l}^{3}}{{L}^{-3}}$
Correct Option: (C) $1.6875\times {{10}^{-12}}$.

Additional Information: 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 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. The greater the value of a solubility product, the greater the solubility of the compound and vice-versa.