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

Answer

C

Explanation

`K_(sp)` for `AgI=1xx10^(-16)mol^(2)L^(-2)`
In solution of `KI,I^(-)` would be due to the both AgI and KI, `10^(-4)` solution KI would provide `=10^(-4)I^(-)`
AgI would provide, say `=xI^(-)` (x is solubility of AgI)
Total `I^(-)=(10^(-4)+x),K_(sp)` of `AgI=(10^(-4)+x)x`
`impliesK_(sp)=10^(-4)x+x^(2)`
as x is very small `:.x^(2)` can be ignored
`:.10^(-4)x=10^(-16)`
or `underset(("solubility"))(x)=(10^(-16))/(10^(-4))=10^(-12)(molL^(-1))`