Solubility of `AgCl` at `20^(@)C` is `1.435xx10^(-3) g per litre`. The solubility product of `AgCl` is

Options

A. `1.0 xx 10^(-10)`
B. `2 xx 10^(-10)`
C. `1.035 xx 10^(-5)`
D. `1.8 xx 10^(-3)`

Answer

A

Explanation

Solubility of AgCl `1.435 xx 10^(-3)gL^(-1)`
`=(1.435)/(143.5)xx10^(-3)mol L^(-1)`
`=10^(-5)M`
`:. [Ag^(+)]=[Cl^(-)]=10^(-5)M`
`:. [Ag^(+)]=[Cl^(-)]=10^(-5)M`
`K_(sp)=[Ag^(+)][Cl^(-)]`
`=10^(-5)xx10^(-5)=10^(-10)`