Question

What is the ionic concentration of $A{{g}^{+}}$ and $Cr{{O}_{4}}^{2-}$ in a solution of $A{{g}_{2}}Cr{{O}_{4}}$ at $25{}^\circ C$ ? ${{K}_{sp}}$ of $A{{g}_{2}}Cr{{O}_{4}}$ is $1.9\times {{10}^{-12}}mo{{l}^{3}}d{{m}^{-9}}$ .

Answer 1

Hint: The question is based on the concepts of equilibrium and solubility equilibria. As, the name suggests equilibrium refers to the balance within the system of reactions or a reaction.

Complete answer:
Let us solve this illustration,
According to the given statement in the question we can say that the reaction taking place is;
\[A{{g}_{2}}Cr{{O}_{4\left( s \right)}}\rightleftarrows 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\]
Also, we have,
${{K}_{sp}}={{\left[ A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{2-} \right]=1.9\times {{10}^{-12}}mo{{l}^{3}}d{{m}^{-9}}$
If we consider S as the solubility of silver chromate under given condition we can say that;
${{K}_{sp}}={{\left( 2S \right)}^{2}}\times S=4{{S}^{3}}$
Thus,
\[S=\sqrt[3]{\dfrac{1.9\times {{10}^{-12}}mo{{l}^{3}}d{{m}^{-9}}}{4}}=7.802\times {{10}^{-5}}mold{{m}^{-3}}\]
From this we can tell the solubilities of ionic concentrations of given ions in the solution as;
${{S}_{Cr{{O}_{4}}^{2-}}}=S=7.802\times {{10}^{-5}}mol{{L}^{-1}}$ and
${{S}_{A{{g}^{+}}}}=2S=2\left( 7.802\times {{10}^{-5}} \right)=1.560\times {{10}^{-4}}mol{{L}^{-1}}$ .

Note:
Here, ${{K}_{sp}}$ is the solubility product constant which is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level of dissolution of solute in the solution.
Do note to properly use and analyse the units in such questions. Here, when cube root is done the units to get normalised and thus, are changed to mol/L.