What is the `K_(sp)` of `AgCl` if it is given that solubility of `AgCl` at `20^(@)C` is `1.435xx10^(-5)g//L`?

Options

A. `1xx10^(-14)`
B. `1xx10^(-10)`
C. `1.435xx10^(-12)`
D. `1.08xx10^(-3)`

Answer

A

Explanation

Solubility of `AgCl`
`=(1.435xx10^(-5))/(143.5)` mol/L
`:.=0.01xx10^(-5)`
`K_(sp)` of `AgCl=[Ag^(+)][Cl^(-1)]`
`=[0.01xx10^(-5)][0.01xx10^(-5)]`
`=1.0xx10^(-14)`