# The units of solubility product of silver chromate (${\text{A}}{{\text{g}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$) will be ______________.$(A) {\text{A}}{\text{. mo}}{{\text{l}}^{\text{2}}}{{\text{L}}^{{\text{ - 2}}}} \\(B) {\text{B}}{\text{. mo}}{{\text{l}}^{\text{3}}}{{\text{L}}^{{\text{ - 3}}}} \\(C) {\text{C}}{\text{. mol}}{{\text{L}}^{{\text{ - 1}}}} \\(D) {\text{D}}{\text{. mol}}{{\text{L}}^{{\text{ - 2}}}} \\$

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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. ${\left( {{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}} \right)^{\text{n}}}{\text{.}}$