
The solubility of $$$S{{b}_{2}}{{S}_{3}}$in water is $1.0\times {{10}^{-5}}mol/litre$at 298K. What will be its solubility product?
A. $108\times {{10}^{-25}}$
B. $1.0\times {{10}^{-25}}$
C. $144\times {{10}^{-25}}$
D. $126\times {{10}^{-25}}$
Answer
164.7k+ views
Hint: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It usually increases with the increase in temperature because of the increased solubility.
Complete Step by Step Answer:
Consider a saturated solution of silver chloride $(AgCl)$that is in contact with solid silver chloride.
$AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}$
It is shown by the equilibrium. As the solid silver chloride does not have a variable concentration and is therefore not included in the above expression. This is the solubility product principle.
${{K}_{sp}}$= $\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
Let x be the solubility of $\left[ A{{g}^{+}} \right]\,\,and\,\,\left[ C{{l}^{-}} \right]$
We can written the above equation as
${{K}_{sp}}=[x][x]$
By multiplying the above equation, we get
Then ${{K}_{sp}}={{x}^{2}}$
Similarly the solubility equilibrium of $S{{b}_{2}}{{S}_{3}}$can be represented as follow:
$S{{b}_{2}}{{S}_{3}}\rightleftharpoons 2S{{b}^{3+}}+3{{S}^{-}}$
Let x be the solubility of $[S{{b}^{3+}}]\,\,and\,[{{S}^{-}}]$
Then $$${{K}_{sp}}={{[2x]}^{2}}{{[3x]}^{3}}$
then the value of $$${{K}_{sp}}=108\times {{(1\times {{10}^{-5}})}^{-5}}$
then ${{K}_{sp}}=108\times {{10}^{-25}}$
Thus the solubility product is $108\times {{10}^{-25}}$
Thus, Option (A) is correct.
Note: Remember that solubility and the solubility product are different from each other. Solubility of a substance in a solvent is the highest amount of solute that can be dissolved in a solvent whereas the solubility product is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution. Both the quantities are dependable upon the temperature. If we increase the temperature, both the solubility and the solubility product increases.
Complete Step by Step Answer:
Consider a saturated solution of silver chloride $(AgCl)$that is in contact with solid silver chloride.
$AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}$
It is shown by the equilibrium. As the solid silver chloride does not have a variable concentration and is therefore not included in the above expression. This is the solubility product principle.
${{K}_{sp}}$= $\left[ A{{g}^{+}} \right]\left[ C{{l}^{-}} \right]$
Let x be the solubility of $\left[ A{{g}^{+}} \right]\,\,and\,\,\left[ C{{l}^{-}} \right]$
We can written the above equation as
${{K}_{sp}}=[x][x]$
By multiplying the above equation, we get
Then ${{K}_{sp}}={{x}^{2}}$
Similarly the solubility equilibrium of $S{{b}_{2}}{{S}_{3}}$can be represented as follow:
$S{{b}_{2}}{{S}_{3}}\rightleftharpoons 2S{{b}^{3+}}+3{{S}^{-}}$
Let x be the solubility of $[S{{b}^{3+}}]\,\,and\,[{{S}^{-}}]$
Then $$${{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.
Recently Updated Pages
Sandmeyer Reaction Mechanism: Steps, Diagram & Exam Notes

Environmental Chemistry Chapter for JEE Main Chemistry

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Get P Block Elements for JEE Main 2025 with clear Explanations

Sets, Relations and Functions Chapter For JEE Main Maths

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
