Answer
Verified
44.1k+ 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
To get a maximum current in an external resistance class 1 physics JEE_Main
silver wire has diameter 04mm and resistivity 16 times class 12 physics JEE_Main
A parallel plate capacitor has a capacitance C When class 12 physics JEE_Main
A series combination of n1 capacitors each of value class 12 physics JEE_Main
When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main
Which of the following is not true in the case of motion class 12 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main