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 rarr Ag^(+)+I^(-)`
For binary electrolyte
`K_(sp)=S^(2)`
where, S = solubility in mol/L
`1.0xx10^(-16)=S^(2)`
or `S=1xx10^(-8)` mol/L
Normality of KI solution `=10^(-4)N`
Here change is one
`M = 10^(-4)M " " [n=1]`
or S for KI solution `=10^(-4)M`
Solubility of AgI in KI solution
`=1xx10^(-8)xx10^(-4)`
`=1xx10^(-12)` mol/L