Solubility product of a salt AB is `1xx10^(-8)M^(2)` in a solution in which the concentration of `A^(+)` ions is `10^(-3)M`. The salt will precipitate when the concentration of `B^(-)` ions is kept

Options

A. between `10^(-8)M` to `10^(-7)M`
B. between `10^(-7)M` to `10^(-8)M`
C. `gt 10^(-5)M`
D. `lt 10^(-8)M`

Answer

C

Explanation

`K_(sp)=1xx10^(-8)M^(2)`
`[A^(+)]=10^(-3)M`
`therefore [B^(-)]=(1xx10^(-8))/(10^(-3))M=10^(-5)M`.
AB will be precipitated if ionic product will be greater than solubility product. That means when the concentration of `B^(-)` ions will be more than `10^(5)M`, then AB will be precipitated.