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

`AgI hArr underset((s))(Ag^(+)) + underset((s))(I^(-)) , K_(sp) = S^(2) = 10^(-4) xx S`
`S = (1.0 xx 10^(-16))/(10^(-4)) = 1 xx 10^(-12) (mol)/(1)`.