Question

The solubility of \[PbC{{l}_{2}}~\]at \[25{}^\circ C~\]is\[6.3\times {{10}^{-3}}mole/litre\]. Its solubility product at that temperature is_______
(A) \[\left( 6.3\times {{10}^{-3}} \right)\times \left( 6.3\times {{10}^{-3}} \right)\]
(B) \[\left( 6.3\times {{10}^{-3}} \right)\times \left( 12.6\times {{10}^{-3}} \right)\]
(C) \[\left( 6.3\times {{10}^{-3}} \right)\times \left( 12.6\text{ }\times {{10}^{-3}} \right){}^\text{2}\]
(D) \[\left( 12.6\times {{10}^{-3}} \right)\times \left( 12.6\text{ }\times {{10}^{-3}} \right)\]

Answer 1

Hint: In the given question, solubility (represented as S) of \[PbC{{l}_{2}}~\] is \[6.3\times {{10}^{-3}}mole/litre\]. Solubility is defined as the ability of solute to dissolve and dissociate in solvent at a given temperature when equilibrium or saturated point reaches.

Complete Step by Step Solution:
The equation is given as
\[PbC{{l}_{2}}\left( s \right)~\rightleftarrows PbC{{l}_{2}}\left( aq \right)\]

In this reaction, \[PbC{{l}_{2}}~\] 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 \[PbC{{l}_{2}}~\]will get dissolved in solvent (aq).
And PbCl2 is a weak electrolyte so, it dissociate to give ions with difficulty such as
\[PbC{{l}_{2}}\left( aq \right)~\rightleftarrows P{{b}^{2+}}+\text{ }2C{{l}^{-}}\]

Now, indirectly equilibrium set between precipitated \[PbC{{l}_{2}}~\]and ionized \[PbC{{l}_{2}}~\]at a saturated point such as
\[PbC{{l}_{2}}\left( s \right)~\rightleftarrows P{{b}^{2+}}+\text{ }2C{{l}^{-}}\]

Now at equilibrium, solubility product of \[PbC{{l}_{2}}~\]which is represented as \[{{K}_{sp}}\]is equal to the product of concentration of both ions, \[P{{b}^{2+}}\]and \[2C{{l}^{-}}\]produced in saturated solution. And solubility of both ions is S and 2S. The relationship between solubility product and solubility is given as
\[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }\left( 2S \right){}^\text{2}\]

Putting the value of S which is given in question (\[6.3\times {{10}^{-3}}\]) we get the required value of solubility product such as
\[{{K}_{sp}}=\text{ 6}\text{.3}\times \text{1}{{\text{0}}^{-3}}\text{ }\times \text{ }\left( 2\times \text{6}\text{.3}\times \text{1}{{\text{0}}^{-3}} \right){}^\text{2}\]
\[{{K}_{sp}}=\text{ 6}\text{.3}\times \text{1}{{\text{0}}^{-3}}\text{ }\times \text{ }\left( \text{12}\text{.6}\times \text{1}{{\text{0}}^{-3}} \right){}^\text{2}\]
Thus, the correct option is C.

Note: It is important to note that \[{{K}_{sp}}\], solubility product for the salts of type \[M{{X}_{2}}\]\[(PbC{{l}_{2}})\]is always same and that is \[{{K}_{sp}}=\text{ }4{{S}^{3}}\]and if a salt is formed with one anion and one cation such as MX, then the solubility product is equal to square of solubility S such as \[{{K}_{sp}}\rightleftarrows {{S}^{2}}\].