Question

If $ {{\text{S}}_0},{{\text{S}}_1},{{\text{S}}_2}{\text{ and }}{{\text{S}}_3} $ are the solubility of $ {\text{AgCl}} $ in water. $ 0.01{\text{M CaC}}{{\text{l}}_2},0.01{\text{ M NaCl and }}0.5{\text{M AgN}}{{\text{O}}_3} $ Solutions, respectively, then which of the following is true?
(A) $ {{\text{S}}_0} > {{\text{S}}_2} > {{\text{S}}_1} > {\text{ }}{{\text{S}}_3} $
(B) $ {{\text{S}}_0} = {{\text{S}}_2} = {{\text{S}}_1} > {\text{ }}{{\text{S}}_3} $
(C) $ {{\text{S}}_3} > {{\text{S}}_1} > {{\text{S}}_2} > {\text{ }}{{\text{S}}_0} $
(D) None of these

Answer 1

Hint: The present of strong electrolyte with the sparingly soluble salt having the common ion decreases the solubility of sparingly soluble salt. More is the concentration of strong electrolyte with common ions added, lower will be the solubility.

Complete step by step solution:
Silver chloride is a sparingly soluble salt, that is it does not completely dissociate into its ion but exists in equilibrium with the ions.
The dissociation of silver chloride that is $ {\text{AgCl}} $ occurs as:
 $ {\text{AgCl}} \rightleftharpoons {\text{A}}{{\text{g}}^ + } + {\text{C}}{{\text{l}}^ - } $
Water contains only hydrogen ions and hydroxide ions, it does not contain any common ions with silver chloride. Hence silver chloride will be maximum soluble in water.
Calcium chloride is a strong salt and hence dissociate complete as:
 $ {\text{CaC}}{{\text{l}}_2} \to {\text{C}}{{\text{a}}^{2 + }} + 2{\text{C}}{{\text{l}}^ - } $
As we can see that the common ion that is chloride is given by the dissociation of calcium chloride. Hence it increases the concentration of chloride ions. According to Le Chaterliar’s principle, if concentration of any species in equilibrium is increased then the reaction will oppose the change. So the chloride ion reacts back to form silver chloride and decreases the solubility of silver chloride. $ {\text{NaCl}} \to {\text{N}}{{\text{a}}^ + } + {\text{C}}{{\text{l}}^ - } $
Sodium chloride also is a strong electrolyte and $ 0.01{\text{ M}} $ of sodium chloride will give $ 0.01{\text{ M}} $ of sodium and chloride ion each. In the same way the presence of $ {\text{NaCl}} $ will decrease the solubility. $ {\text{CaC}}{{\text{l}}_2} $ gives twice the number of chloride ion as given by $ {\text{NaCl}} $ hence it will decrease the solubility more than $ {\text{NaCl}} $ .
The concentration of $ {\text{AgN}}{{\text{O}}_3} $ is $ 0.5\,{\text{ M}} $ . The dissociation of $ {\text{AgN}}{{\text{O}}_3} $ occurs as:
 $ {\text{AgN}}{{\text{O}}_3} \to {\text{A}}{{\text{g}}^ + } + {\text{N}}{{\text{O}}_3}^ + $
It gives $ 0.5\,{\text{ M}} $ of silver ions which is a common ion and highest among all. Hence it will decrease the concentration maximum. So the order will be:
 $ {{\text{S}}_0} > {{\text{S}}_2} > {{\text{S}}_1} > {\text{ }}{{\text{S}}_3} $
Thus, the correct option is A.

Note:
The solubility product is defined as the product of ions that are given by dissociation of sparingly soluble salt in water each concentration raised to the power of their stoichiometric coefficients. If the common ion is present then its concentration is also taken into account while calculating the solubility product.