Question

At 298 K the solubility of \[PbC{{l}_{2}}\]​ is \[2\times {{10}^{-2}}mol/litre\], then \[{{K}_{sp}}=\]?
(A) \[1\text{ }\times \text{ }{{10}^{-7}}\]
(B) \[3.2\text{ }\times \text{ }{{10}^{-7}}\]
(C) \[1\text{ }\times \text{ }{{10}^{-5}}\]
(D) \[3.2\text{ }\times \text{ }{{10}^{-5}}\]

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 (represented as S) of lead chloride is \[2\times {{10}^{-2}}mol/litre\]. Solubility of solute is to dissolve and dissociate in solvent at a given temperature at equilibrium or saturated point. Solubility of ions is represented as S.

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).

Now \[PbC{{l}_{2}}\]is a weak electrolyte so, it will dissociate into ions with difficulty such as\[PbC{{l}_{2}}\left( aq \right)~\rightleftarrows P{{b}^{2+}}+\text{ }2C{{l}^{-}}\]

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

Now at equilibrium, solubility product which is represented as \[{{K}_{sp}}\] is equal to the product of the concentration of both ions, \[P{{b}^{2+}}\]and \[2C{{l}^{-}}\]. And solubility of both ions is S and 2S. The relationship between solubility product and solubility is given as
\[{{K}_{sp}}=\text{ }\left[ P{{b}^{2+}} \right]\left[ C{{l}^{-}} \right]{}^\text{2}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }\left( 2S \right){}^\text{2}\]
\[{{K}_{sp}}=\text{ }S\text{ }\times \text{ }4{{S}^{2}}\]
\[{{K}_{sp}}=\text{ }4{{S}^{3}}\]
Putting the value of S which is given in question we get the required value of solubility product such as
\[{{K}_{sp}}=4\times (2\times {{10}^{-5}})\]
\[{{K}_{sp}}=3.2\times {{10}^{-5}}\] \[{{K}_{sp}}=3.2\times {{10}^{-5}}\]
Thus, the correct option is D.

Note: It is important to note that \[{{K}_{sp}}\], solubility product for the salts of type MX2 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 \[{{K}_{sp}}\rightleftarrows {{S}^{2}}\].