The solubility product of `AgI` at `25^(@)C` is `1.0xx10^(-16) mol^(2) L^(-2)`. The solubility of `AgI` in `10^(-4) N` solution of `KI` at `25^(@)C` is approximately ( in `mol L^(-1)`)

Options

A. `1.0 xx 10^(-8)`
B. `1.0 xx 10^(-16)`
C. `1.0 xx 10^(-12)`
D. `1.0 xx 10^(-10)`

Answer

C

Explanation

Let solubility of AgI in `10^(-4)` N (or `10^(-4)M)`
KI at `25^(@)C` is x ( mol `L^(-1))`
Here `[I^(-)] =10^(-4)M` ( from `10^(-4)M KI)`
`[Ag^(+)]=xM`
`K_(sp)=[Ag^(+)][I^(-)]`
`10^(-16)=x xx 10^(-4)`
`x=10^(-12) mol L^(-1)`