Answer
64.8k+ views
Hint: The solubility product is the equilibrium constant for saturated solutions of Ionic compounds. The units of solubility product depend upon the stoichiometric coefficients of concentration terms. It is generally expressed as\[{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}\].
Complete step by step answer: At equilibrium, the saturated solution is dissolved as solids and its constituent ions.
This can be represented as follows:
\[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}} \rightleftarrows {\text{2A}}{{\text{g}}^{\text{ + }}}{\text{ + Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}\]
From this reaction, we can write the solubility product \[{K_{sp}}\].
${{\text{K}}_{{\text{sp}}}}{\text{ = }}\dfrac{{{{{\text{[2Ag]}}}^{\text{ + }}}{{{\text{[CrO4]}}}^{{\text{2 - }}}}}}{{{\text{[A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{]}}}}$
Since \[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}\] is a solid the concentration is taken to be unity.
Thus, we have, \[{{\text{K}}_{{\text{sp}}}}{\text{ = [2Ag}}{{\text{]}}^{\text{ + }}}{{\text{[CrO4]}}^{{\text{2 - }}}}\]
Solubility products have units of concentration raised to the power of stoichiometric coefficients of the ions in the equilibrium.
For Ag2CrO4 the unit of solubility product is \[{\left( {{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}} \right)^{\text{3}}}\]or \[{\text{mo}}{{\text{l}}^{\text{3}}}{{\text{L}}^{{\text{ - 3}}}}\].
Since litre L is also equal to \[{\text{d}}{{\text{m}}^{{\text{ - 3}}}}\] the unit of solubility product of \[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}\] can also be expressed as \[{\text{mol d}}{{\text{m}}^{{\text{ - 6}}}}\].
Correct option: B
Additional Information: Solubility product is used to measure the solubility of the ion in the solution. High \[{K_{sp}}\]value indicates high solubility.
By knowing the solubility product, \[{K_{sp}}\]we can also predict whether a precipitate will be obtained or not for the given solutions.
The solubility, s, can be calculated by knowing the \[{K_{sp}}\] value.
Note: The easiest way to get the unit of solubility product is to find the stoichiometric coefficients of the ions involved in the equilibrium and substitute it in the value of n in the general unit expression i.e. \[{\left( {{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}} \right)^{\text{n}}}{\text{.}}\]
Complete step by step answer: At equilibrium, the saturated solution is dissolved as solids and its constituent ions.
This can be represented as follows:
\[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}} \rightleftarrows {\text{2A}}{{\text{g}}^{\text{ + }}}{\text{ + Cr}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}\]
From this reaction, we can write the solubility product \[{K_{sp}}\].
${{\text{K}}_{{\text{sp}}}}{\text{ = }}\dfrac{{{{{\text{[2Ag]}}}^{\text{ + }}}{{{\text{[CrO4]}}}^{{\text{2 - }}}}}}{{{\text{[A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}{\text{]}}}}$
Since \[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}\] is a solid the concentration is taken to be unity.
Thus, we have, \[{{\text{K}}_{{\text{sp}}}}{\text{ = [2Ag}}{{\text{]}}^{\text{ + }}}{{\text{[CrO4]}}^{{\text{2 - }}}}\]
Solubility products have units of concentration raised to the power of stoichiometric coefficients of the ions in the equilibrium.
For Ag2CrO4 the unit of solubility product is \[{\left( {{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}} \right)^{\text{3}}}\]or \[{\text{mo}}{{\text{l}}^{\text{3}}}{{\text{L}}^{{\text{ - 3}}}}\].
Since litre L is also equal to \[{\text{d}}{{\text{m}}^{{\text{ - 3}}}}\] the unit of solubility product of \[{\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}\] can also be expressed as \[{\text{mol d}}{{\text{m}}^{{\text{ - 6}}}}\].
Correct option: B
Additional Information: Solubility product is used to measure the solubility of the ion in the solution. High \[{K_{sp}}\]value indicates high solubility.
By knowing the solubility product, \[{K_{sp}}\]we can also predict whether a precipitate will be obtained or not for the given solutions.
The solubility, s, can be calculated by knowing the \[{K_{sp}}\] value.
Note: The easiest way to get the unit of solubility product is to find the stoichiometric coefficients of the ions involved in the equilibrium and substitute it in the value of n in the general unit expression i.e. \[{\left( {{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}} \right)^{\text{n}}}{\text{.}}\]
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)