Question

The solubility product of AgCl is \[4.0\text{ }\times \text{ }{{10}^{-10}}~\]at 298 K. The solubility of AgCl in\[0.04\text{ }M~CaC{{l}_{2}}~\]​ will be
(A) \[2.0\text{ }\times \text{ }{{10}^{-5}}M\]
(B) \[1.0\text{ }\times \text{ }{{10}^{-4}}M\]
(C) \[5.0\text{ }\times \text{ }{{10}^{-9}}M\]
(D) \[2.2\times \text{ }{{10}^{-4}}M\]

Answer 1

Hint: Solubility product is the product of the concentration of ions (each raised to the power of their stoichiometry coefficient) produced in equilibrium or saturated solution and it is represented as \[{{K}_{sp}}\]. In the given question solubility product of AgCl is given, \[4.0\text{ }\times \text{ }{{10}^{-10}}\]. When AgCl is dissolved, it will dissociate into \[A{{g}^{+}}\]ions and \[C{{l}^{-}}\]ions in a saturated solution. The concentration of \[C{{l}^{-}}\] can be found by the given mol of \[CaC{{l}_{2}}\], \[0.04M\], and from this, we can easily find solubility (s) of AgCl.

Complete Step by Step Solution:
Given equation is given as:
\[AgCl(s)\rightleftarrows AgCl(aq)\]

In this reaction, AgCl is dissolved in a solvent and as it is a slightly soluble salt thus, most of its part will get precipitate (in solid form) and a very less amount of AgCl will get dissolved in solvent (aq).

Now, AgCl is a very good electrolyte so, it will easily and quickly dissociate to give ions such as
\[AgCl(s)\rightleftarrows A{{g}^{+}}(aq)+C{{l}^{-}}(aq)\]
\[AgCl(aq)\rightleftarrows A{{g}^{+}}(aq)+C{{l}^{-}}(aq)\]

Now equilibrium set between precipitated AgCl (s) and ionised AgCl at a saturated point such as
\[AgCl(s)\rightleftarrows A{{g}^{+}}(aq)+C{{l}^{-}}(aq)\]
As per hint, the solubility product \[({{K}_{sp}})\]of AgCl is given \[\left( 4.0\text{ }\times \text{ }{{10}^{-10}} \right)\]which is equal to the product of the concentration of both ions produced in saturated solution such as
\[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\]

The \[0.04M\]of \[CaC{{l}_{2}}\] tends to give \[0.08M\]of chloride ion and the solubility of \[A{{g}^{+}}\] is represented as S then
\[\left( 4.0\text{ }\times \text{ }{{10}^{-10}} \right)=S\times 0.08\]
\[S=(4.0\times {{10}^{-10}})/0.08\]



Now solving it we will get the solubility of AgCl which is \[5.0\text{ }\times \text{ }{{10}^{-9}}M\]
Thus, the correct option is C.

Note: It is important to note that calcium chloride \[CaC{{l}_{2}}\] will break into\[C{{a}^{2+}}\]and \[2C{{l}^{-}}\]. As the mol of calcium chloride is \[0.04M\]so, the mol of chloride ion which is multiplied by 2 will be \[\text{2 }\times \text{ }0.04\text{ }=\text{ }0.08\]. Thus, the number of mol of chloride ions is \[0.08M\].