When solid lead iodide is added to water, the equilibrium concentration of I^(-) becomes 2.6xx10^(-3)M. What is K_(sp) for PbI_(2) ?

When solid lead iodide is added to water, the equilibrium concentratio

1.	When solid lead iodide is added to water, the equilibrium concentration of I^(-) becomes 2.6xx10^(-3)M. What is K_(sp) for PbI_(2) ?
Answer» `2.2xx10^(-9)` `8.8xx10^(-9)` `1.8xx10^(-8)` `3.5xx10^(-8)` Solution :`PbI_(2)hArrPb^(2+)+2I^(-)` Thus, `PB^(2+)` ion concentration is half of the `I^(-)` ion concentration `[I^(-)]=2.6xx10^(-3)M:.[Pb^(2+)]=1.3xx10^(-3)M` `K_(SP)=[Pb^(2+)][I^(-)]^(2)` `=(1.3xx10^(-3)) (2.6xx10^(-3))^(2)=8.8xx10^(-9)`.