
Which is the correct representation for the solubility product constant \[A{{g}_{2}}Cr{{O}_{4}}\]?
A. \[{{\left[ A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{-2} \right]\]
B. \[\left[ A{{g}^{+}} \right]\left[ Cr{{O}_{4}}^{-2} \right]\]
C. \[\left[ 2A{{g}^{+}} \right]\left[ Cr{{O}_{4}}^{-2} \right]\]
D. \[{{\left[ 2A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{-2} \right]\]
Answer
164.4k+ views
Hint: Solubility (S) and solubility product (\[{{K}_{sp}}\]) are determined for sparingly soluble salts (sparingly soluble means saturated or equilibrium point attain earlier between precipitate(s) of salt and dissolved part of salt (aq) in solution) and which is a good electrolyte. Almost all the salts are good electrolytes, which means all the salts easily dissociate into ions in an aqueous solution. How much ions dissociate is measured by solubility and maximum number of ions which can form in the solution is measured by solubility product.
Complete Step by Step Answer:
Given compound is a salt which is slightly soluble in the solvent (say water), \[A{{g}_{2}}Cr{{O}_{4}}\]. If we add silver chromate to water, it starts to dissolve in water slightly but at any instant, silver chromate stops to get dissolve in water and settles down to the base in the form of precipitates (insoluble silver chromate) which is in solid form (generally salts are solid).
In other words, at any instant, the solution attains a saturated point and equilibrium set up between the insoluble part of silver chromate which is in solid form (precipitates), and the soluble part of silver chromate which is in an aqueous form such as:
\[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows A{{g}_{2}}Cr{{O}_{4}}(aq)\]
Now, being a strong electrolyte, silver chloride does not exist in the molecular form in solution but dissociates in ions. In short, we can say equilibrium exists between an insoluble part of silver chloride which is in solid form (ppt.) and ions of the soluble part of silver chloride such as:
\[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\]
Solubility product is the mathematical expression which is defined as the product of concentrations of both ions that present in saturated solution both raised to the power of their stoichiometric coefficient if any ( 2 moles of silver cation and one mole of chromate anion) such as:
\[~{{K}_{sp}}\left( solubility\text{ }product \right)\text{ }=\text{ }{{\left[ A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{2-} \right]\]
Thus, the correct option is A.
Note: Given equation is \[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\] \[\]. The equilibrium constant of a given reaction is given as
\[{{K}_{c}}={{[A{{g}^{+}}]}^{2}}[Cr{{O}_{4}}^{2-}]/[A{{g}_{2}}Cr{{O}_{4}}]\]. As in denomination, the concentration of solid is needed to calculate equilibrium constant value but the concentration of product which is in solid form is always 1 so equilibrium constant is the same as solubility product constant such as \[{{K}_{c}}={{[A{{g}^{+}}]}^{2}}[Cr{{O}_{4}}^{2-}]={{K}_{sp}}\].
Complete Step by Step Answer:
Given compound is a salt which is slightly soluble in the solvent (say water), \[A{{g}_{2}}Cr{{O}_{4}}\]. If we add silver chromate to water, it starts to dissolve in water slightly but at any instant, silver chromate stops to get dissolve in water and settles down to the base in the form of precipitates (insoluble silver chromate) which is in solid form (generally salts are solid).
In other words, at any instant, the solution attains a saturated point and equilibrium set up between the insoluble part of silver chromate which is in solid form (precipitates), and the soluble part of silver chromate which is in an aqueous form such as:
\[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows A{{g}_{2}}Cr{{O}_{4}}(aq)\]
Now, being a strong electrolyte, silver chloride does not exist in the molecular form in solution but dissociates in ions. In short, we can say equilibrium exists between an insoluble part of silver chloride which is in solid form (ppt.) and ions of the soluble part of silver chloride such as:
\[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\]
Solubility product is the mathematical expression which is defined as the product of concentrations of both ions that present in saturated solution both raised to the power of their stoichiometric coefficient if any ( 2 moles of silver cation and one mole of chromate anion) such as:
\[~{{K}_{sp}}\left( solubility\text{ }product \right)\text{ }=\text{ }{{\left[ A{{g}^{+}} \right]}^{2}}\left[ Cr{{O}_{4}}^{2-} \right]\]
Thus, the correct option is A.
Note: Given equation is \[A{{g}_{2}}Cr{{O}_{4}}\left( s \right)\rightleftarrows 2A{{g}^{+}}+Cr{{O}_{4}}^{2-}\] \[\]. The equilibrium constant of a given reaction is given as
\[{{K}_{c}}={{[A{{g}^{+}}]}^{2}}[Cr{{O}_{4}}^{2-}]/[A{{g}_{2}}Cr{{O}_{4}}]\]. As in denomination, the concentration of solid is needed to calculate equilibrium constant value but the concentration of product which is in solid form is always 1 so equilibrium constant is the same as solubility product constant such as \[{{K}_{c}}={{[A{{g}^{+}}]}^{2}}[Cr{{O}_{4}}^{2-}]={{K}_{sp}}\].
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