At `18^(@)C`, the solubility product of AgCI is `1.8xx10^(-10)`. In the solution, the value of `Ag^(+)` is `4xx10^(-3)` mol `L^(-1)`. The value of `[CI^(-)]` to precipitate AgCI from this solution should be greater than

Options

A. `4.5xx10^(-8)mol L^(-1)`
B. `7.2xx10^(-12)mol L^(-1)`
C. `4.0xx10^(-3)mol L^(-1)`
D. `4.5xx10^(-7)mol L^(-1)`

Answer

A

Explanation

Equilibrium expression is
`AgCl(s)hArr Ag^(+)(aq.)+Cl^(-)(aq.)`
`K_(sp)=C_(Ag^(+))C_(Cl^(-))`
`1.8xx10^(-10)=(4xx10^(-3))C_(CI^(-)`
or `C_(CI^(-))= (1.8xx10^(-10))/(4.0xx10^(-3))`
`= 0.45xx10^(-7)mol L^(-1)`
`=4.5xx10^(-8)mol L^(-1)`
Since precipitation takes place whenever ionic product just exceeds solubility product, `C_(CI^(-))` should be greater than `4.5xx10^(-8)mol L^(-1)` to precipitate AgCI.