`K_(sp)` for lead iodate `[Pb(IO_(3))_(2) is 3.2 xx 10^(-14)` at a given temperature. The solubility in `molL^(-1)` will be

Options

A. `2.0 xx 10^(-5)`
B. `(3.2 xx 10^(-7))^(1//2)`
C. `(3.8 xx 10^(-7))`
D. `4.0 xx 10^(-6)`

Answer

A

Explanation

`{:([Pb(IO_(3))_(2)]hArr,Pb^(2+)+,2IO_(3)^(Theta),),(,2S,S,):}`
`K_(sp) = 4S^(3)`.
`S =root3((K_(sp))/(4))root3((3.2xx10^(-14))/(4))=2xx10^(-5)M`.