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. `35xx10^(-8)`

Answer

B

Explanation

`{:(PbI_2 hArr , Pb^(2+)+ , 2I^(-)),(,s,2s):}`
`K_(sp)=(s)(2s)^2 =4s^3`
`[I^-]=2.6xx10^(-3) M`=2s
`therefore s=(2.6xx10^(-3))/2 M=1.3xx10^(-3)`M
`K_(sp)=4s^3 =4xx(1.3xx10^(-3)M)^3`
`=8.8xx10^(-9)M^3`