When solid lead iodide is added to water, the equilibrium concentration of `I^(-)` becomes `2.6 xx 10^(-3)M` . What is the `K_(sp)` for `PbI_(2)` ?

Options

A. `2.2xx10^(-9)`
B. `8.8xx10^(-9)`
C. `1.8xx10^(-8)`
D. `3.5xx10^(-8)`

Answer

B

Explanation

`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)`.