Given that , the solubility product `K_(sp)` , of AgCl is `1.8 xx 10^(-10)` , the concentration of `Cl^(-)` ions that must just be exceeded before AgCl will precipitate from a solution containing `4 xx 10^(-3)` M `Ag^(+)` ions is

Options

A. `4.5 xx 10^(-8) M`
B. `4 xx 10^(-8) M`
C. `1.8 xx 10^(-8) M`
D. `1 xx 10^(-8) M`

Answer

A

Explanation

`[Ag^(+)][Cl^(-)]=1.8 xx 10^(-10)`
`[Cl^(-)]=(1. 8 xx 10^(-10))/(4 xx 10^(-3))=4.5 xx 10^(-8)`
Thus, if conc. Of `Cl^(-)` ions exceeds `4.5 xx 10^(-8)` M precipitation will occur because ionic product will exceed `K_(sp)`.