Question

The solubility of $$$S{{b}_{2}}{{S}_{3}}$in water is $1.0\times {{10}^{-5}}mol/litre$at 298K. What will be its solubility product?
A. $108\times {{10}^{-25}}$
B. $1.0\times {{10}^{-25}}$
C. $144\times {{10}^{-25}}$
D. $126\times {{10}^{-25}}$

Answer 1

Hint: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It usually increases with the increase in temperature because of the increased solubility.

Complete Step by Step Answer:
Consider a saturated solution of silver chloride $(AgCl)$that is in contact with solid silver chloride.
$AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}$
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.
${{K}_{sp}}$= $\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
Let x be the solubility of $\left[ A{{g}^{+}} \right]\,\,and\,\,\left[ C{{l}^{-}} \right]$
We can written the above equation as
${{K}_{sp}}=[x][x]$
By multiplying the above equation, we get
Then ${{K}_{sp}}={{x}^{2}}$

Similarly the solubility equilibrium of $S{{b}_{2}}{{S}_{3}}$can be represented as follow:
$S{{b}_{2}}{{S}_{3}}\rightleftharpoons 2S{{b}^{3+}}+3{{S}^{-}}$
Let x be the solubility of $[S{{b}^{3+}}]\,\,and\,[{{S}^{-}}]$
Then $$${{K}_{sp}}={{[2x]}^{2}}{{[3x]}^{3}}$
then the value of $$${{K}_{sp}}=108\times {{(1\times {{10}^{-5}})}^{-5}}$
then ${{K}_{sp}}=108\times {{10}^{-25}}$
Thus the solubility product is $108\times {{10}^{-25}}$
Thus, Option (A) is correct.

Note: Remember that solubility and the solubility product are different from each other. Solubility of a substance in a solvent is the highest amount of solute that can be dissolved in a solvent whereas the solubility product is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution. Both the quantities are dependable upon the temperature. If we increase the temperature, both the solubility and the solubility product increases.