`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

`K_(sp)=[Pb^(+)]^(2)[IO_(3)^(-)]^(2)=[s(2s)^(2)]=4s^(3)`
`s=3 sqrt((K_(sp))/(4))=3sqrt((3.2 xx 10^(-14))/(4))`
`3sqrt(8 xx 10^(-15))=2xx10^(-5)mol L^(-1)`