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. `1.0xx10^(-10)`
B. `1.0xx10^(-8)`
C. `1.0xx10^(-16)`
D. `1.0xx10^(-12)`

Answer

D

Explanation

`AgI to Ag^(+)+I^(-)`
For binary electrolyte
`K_(sp)=S^(2)`
Where S= solubility in mol/L
`1.0x10^(-16)=S^(2) or S=1x10^(-18)` mol/L
Solubility of Ag in Kl solution
`=1x10^(-8)x10^(-4)=1x10^(-2)` mol/L