Question

A saturated solution of $A{{g}_{2}}S{{O}_{4}}$ ​ is $2.5\times {{10}^{-2}}$ M. The value of its solubility product is
A. $6.25\times {{10}^{-5}}M$
B. $6.25\times {{10}^{-2}}M$
C. $15.625\times {{10}^{-6}}M$
D. $3.125\times {{10}^{-6}}M$

Answer 1

Hint: Solubility product depends on the number of ions formed in a reaction at equilibrium. $A{{g}_{2}}S{{O}_{4}}$ dissociate into two silver cations and one sulphate anion.

Complete Step by Step Answer:
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.

$A{{g}_{2}}S{{O}_{4}}$ dissociate at equilibrium as follows:
$A{{g}_{2}}S{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+S{{O}_{4}}^{2-}$
Thus $A{{g}_{2}}S{{O}_{4}}$ dissociate into two silver cations and one sulphate anion at equilibrium.

Thus solubility product is given as follows:
$K_{sp}={{(2S)}^{2}}\times S$; where $Ksp$ is defined as the solubility product and $S$ is defined as the solubility for the cation and anion. Solubility of silver cation has a coefficient of $2$ .

The value of solubility is $2.5\times {{10}^{-2}}$. Putting the value of solubility in the above equation we get:
$K_{sp}=4\times {{(2.5\times {{10}^{-2}})}^{3}}$
$K_{sp}=6.25\times {{10}^{-5}}M$
Thus the value of the solubility product of silver sulphate is $6.25\times {{10}^{-5}}M$.
Thus the correct option is A.

Note: The chemical name of $A{{g}_{2}}S{{O}_{4}}$ is silver sulphate. It is an ionic compound. It dissociates into the corresponding ions in solution state. It is a white colored solid and it is insoluble in water.