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

Answer

B

Explanation

Equilibrium expression is
`AgI(s)hArr Ag^(+)(aq.)+I^(-)(aq.)`
Let the solubility of AgI in `10^(-4)` N or `10^(-4)` M solution of KI be x mol `L^(-1)`. The total concentration of `I^(-)(aq.)`. Will be
`C_(I^(-))=x+10^(-4)`
Since the solubility product of AgI is very small and due to the presence of KI, equilibrium shifts backward (common ion effect), we have
`C_(I^(-))~~10^(-4)mol L^(-1)`
Now,
`K_(sp)=C_(Ag^(+)C_(I^(-))`
`1.0xx10^(-16)=(x)(10^(-4))`
`x=1.0xx10^(-12)`