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. `2.2 xx 10^(-9)`
B. `8.8 xx 10^(-9)`
C. `1.8 xx 10^(-8)`
D. `3.5 xx 10^(-8)`

Answer

B

Explanation

`PbI_(2) hArr Pb^(2+)+2I^(-)`
On the basis of this equation, the concentration `Pb^(2+)` ions will be half of the concentration of `I^(-)` ions . Thus,
`[I^(-)]=2.6 xx 10^(-3)M` and `[Pb^(2+0]=1.3 xx 10^(-3)M`
`:. K_(sp)=[Pb^(2+)][I^(-)]^(2)`
`=(1.3 xx 10^(-3))xx (2.6 xx 10^(-3))^(2) =8.8 xx 10^(-9)`