A saturated solution of \[C{a_3}{\left( {P{O_4}} \right)_2}\] contains $2.0 \times {10^{ - 8}}{\text{M}}$ $C{a^{2 + }}$ and $1.6 \times {10^{ - 5}}{\text{M}}$ $PO_4^{3 - }$ at a certain temperature. The solubility product ${K_{sp}}$ of \[C{a_3}{\left( {P{O_4}} \right)_2}\]at that temperature is-
A. $2.048 \times {10^{ - 34}}$
B. $2.04 \times {10^{ - 33}}$
C. $3.20 \times {10^{ - 34}}$
D. $8 \times {10^{ - 34}}$
Answer
252.3k+ views
Hint: The solubility product is the product of concentration of ions in water. To find the solubility product, dissociate the compound\[C{a_3}{\left( {P{O_4}} \right)_2}\] into ions and then multiply the concentration of the ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation.
Formula used: ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$
Step-by-step Solution:
We are given that A saturated solution of \[C{a_3}{\left( {P{O_4}} \right)_2}\] contains $2.0 \times {10^{ - 8}}{\text{M}}$ $C{a^{2 + }}$ and $1.6 \times {10^{ - 5}}{\text{M}}$ $PO_4^{3 - }$ at a certain temperature.
We have to find its solubility product at that temperature. Solubility Product is the product of concentration of ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium. It is denoted by ${{\text{K}}_{{\text{sp}}}}$ . Now the given salt will dissociate into following ions-
\[ \Rightarrow C{a_3}{\left( {P{O_4}} \right)_2} \to 3C{a^{2 + }} + 2PO_4^{3 - }\]
Then using formula ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {C{a^{2 + }}} \right]^3} \times {\left[ {PO_4^{3 - }} \right]^2}$
We already know the concentration of the given ions so putting the values of their concentration in the formula we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {2.0 \times {{10}^{ - 8}}} \right]^3} \times {\left[ {1.6 \times {{10}^{ - 5}}{\text{M}}} \right]^2}$
On solving we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 8 \times 2.56 \times {10^{ - 10 - 24}}$ {as we know that ${a^m}.{a^n} = {a^{m + n}}$ }
On simplifying we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 20.4 \times {10^{ - 34}}$
We can also write it as-
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 2.04 \times {10^{ - 33}}$
Answer- Hence the correct answer is B.
Note: Difference between solubility and solubility product is-
The solubility of a substance is the total amount of the solute that can be dissolved in the solvent at equilibrium.
Solubility product constant (${{\text{K}}_{{\text{sp}}}}$ ) is an equilibrium constant which represents the level at which the solute dissolves in a solution or we can say it provides insight into the equilibrium between the solute and its constituent ions that are dissociated across the solution.
Formula used: ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$
Step-by-step Solution:
We are given that A saturated solution of \[C{a_3}{\left( {P{O_4}} \right)_2}\] contains $2.0 \times {10^{ - 8}}{\text{M}}$ $C{a^{2 + }}$ and $1.6 \times {10^{ - 5}}{\text{M}}$ $PO_4^{3 - }$ at a certain temperature.
We have to find its solubility product at that temperature. Solubility Product is the product of concentration of ions, with each concentration raised to the power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium. It is denoted by ${{\text{K}}_{{\text{sp}}}}$ . Now the given salt will dissociate into following ions-
\[ \Rightarrow C{a_3}{\left( {P{O_4}} \right)_2} \to 3C{a^{2 + }} + 2PO_4^{3 - }\]
Then using formula ${{\text{K}}_{{\text{sp}}}} = {\text{product of concentration of ions}}$
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {C{a^{2 + }}} \right]^3} \times {\left[ {PO_4^{3 - }} \right]^2}$
We already know the concentration of the given ions so putting the values of their concentration in the formula we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = {\left[ {2.0 \times {{10}^{ - 8}}} \right]^3} \times {\left[ {1.6 \times {{10}^{ - 5}}{\text{M}}} \right]^2}$
On solving we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 8 \times 2.56 \times {10^{ - 10 - 24}}$ {as we know that ${a^m}.{a^n} = {a^{m + n}}$ }
On simplifying we get,
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 20.4 \times {10^{ - 34}}$
We can also write it as-
$ \Rightarrow {{\text{K}}_{{\text{sp}}}} = 2.04 \times {10^{ - 33}}$
Answer- Hence the correct answer is B.
Note: Difference between solubility and solubility product is-
The solubility of a substance is the total amount of the solute that can be dissolved in the solvent at equilibrium.
Solubility product constant (${{\text{K}}_{{\text{sp}}}}$ ) is an equilibrium constant which represents the level at which the solute dissolves in a solution or we can say it provides insight into the equilibrium between the solute and its constituent ions that are dissociated across the solution.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Types of Solutions in Chemistry: Explained Simply

Difference Between Crystalline and Amorphous Solid: Table & Examples

Know The Difference Between Fluid And Liquid

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

[Awaiting the three content sources: Ask AI Response, Competitor 1 Content, and Competitor 2 Content. Please provide those to continue with the analysis and optimization.]

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Chemistry Chapter 9 - Hydrocarbons - 2025-26

CBSE Notes Class 11 Chemistry Chapter 5 - Thermodynamics - 2025-26

CBSE Notes Class 11 Chemistry Chapter 8 - Organic Chemistry Some Basic Principles And Techniques - 2025-26

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

