Question

To $ A{{g}_{2}}Cr{{O}_{4}} $ ​ solution over its precipitate, $ CrO_{4}^{2-} $ ions are added. This results in:
A. Increase in $ A{{g}^{+}} $ concentration
B. Decrease in $ A{{g}^{+}} $ concentration
C. Increase in solubility product
D. Shifting of $ A{{g}^{+}} $ ions from the precipitate into the solution

Answer 1

Hint : We know that some ionic compounds dissolve in water, which arises because of the attraction between positive and negative charges. For example, the salt's positive ions (e.g. $ A{{g}^{+}} $ ) attract the partially negative oxygen in the water. Likewise, the salt's negative ions (e.g. $ \text{ }C{{l}^{-}} $ ) attract the partially positive hydrogen in water. Oxygen is partially negative because it is more electronegative than hydrogen, and vice versa
 $ AgCl\left( s \right)\rightleftharpoons A{{g}^{+}}\left( aq \right)\text{ }+\text{ }C{{l}^{-}}\left( aq \right) $

Complete Step By Step Answer:
There is a limit to how much salt can be dissolved in a given volume of water. This amount is given by the solubility product, $ {{K}_{sp}} $ This value depends on the type of salt (AgCl vs. NaCl, for example), temperature, and the common ion effect.
At equilibrium, the saturated solution is dissolved as solids and its constituent ions. This can be represented as follows: $ A{{g}_{2}}Cr{{O}_{4}}\rightleftharpoons 2A{{g}^{+}}+CrO_{4}^{2-} $
From this reaction, we can write the solubility product $ {{K}_{sp}}. $
 $ \Rightarrow {{K}_{sp}}={{[A{{g}^{+}}]}^{2}}[CrO_{4}^{2-}] $
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.
As $ {{K}_{sp}} $ ​ remains constant. So on the addition of $ CrO_{4}^{2-} $ ​ concentration $ CrO_{4}^{2-} $ ​ will increase. So this will result in a decrease in $ A{{g}^{+}} $ concentration.
Therefore, Correct option is B i.e. To $ A{{g}_{2}}Cr{{O}_{4}} $ ​ solution over its precipitate, $ CrO_{4}^{2-} $ ions are added. This results in decrease in $ A{{g}^{+}} $ concentration

Note :
Note that solubility of a substance is useful when separating mixtures. For example, a mixture of salt (sodium chloride) and silica may be separated by dissolving the salt in water and filtering off the undissolved silica. The synthesis of chemical compounds, by the milligram in a laboratory, or by the ton in industry, both make use of the relative solubility of the desired product, as well as unreacted starting materials, byproducts, and side products to achieve separation.