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# The solubility of AgCl will be minimum in:(a) \[0.01M\,AgN{O_3}\](b) Pure water(c) \[\,0.01M\,CaC{l_2}\](d) \[\,0.01M\,NaCl\]

Last updated date: 24th Jun 2024
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Hint: Before solving this question, we must recall the concept of common ion effect and how it affects the solubility of a particular solute in a solution.

Complete step by step solution:
The common-ion effect is the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate.
The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium. In simpler words, we can explain this as if several salts are present in a system; they all ionize in the solution. But if the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion.
Thus, when when AgCl is dissolved into a solution already containing \[CaC{l_2}\]the \[C{l^ - }\]ions come from the ionization of both AgCl and \[CaC{l_2}\]. The concentration of common ion \[C{l^ - }\] is maximum at 0.01M \[CaC{l_2}\]. Hence, it will suppress the ionization of AgCl and the solubility will be least in this case. \[NaCl\] too has \[C{l^ - }\] ions but we can see that its concentration is half of that in \[CaC{l_2}\].
Thus, we can conclude that Option (C) 0.01M \[CaC{l_2}\] is the correct answer.