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» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/solubility-1217018" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUBILITY">SOLUBILITY</a> of `PbCl_(2) = (11.84)/(207+2xx35.5) "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>" L^(-1) = 4.259 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-2) "mol " L^(-1)` <br/> `PbCl_(2) rarr Pb^(2+)+2 <a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>^(-), K_(sp)=[Pb^(2+)][Cl^(-)]^(2) = (4.259xx10^(-2))(2xx4.259xx10^(-2))^(2) = 3.09xx10^(-4)`</body></html>