A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constantat room temperature. (At. wt . : Pb= 207, Cl = 35.5 )

A saturated solution of sparingly soluble lead chloride on analysis wa

1.	A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constantat room temperature. (At. wt . : Pb= 207, Cl = 35.5 )
Answer» Solution :SOLUBILITY of `PbCl_(2) = (11.84)/(207+2xx35.5) "MOL" L^(-1) = 4.259 XX 10^(-2) "mol " L^(-1)` `PbCl_(2) rarr Pb^(2+)+2 CL^(-), K_(sp)=[Pb^(2+)][Cl^(-)]^(2) = (4.259xx10^(-2))(2xx4.259xx10^(-2))^(2) = 3.09xx10^(-4)`