The `K_(b)` for `AgCl` is `2.8 xx 10^(-10)` at a given temperature. The solubility of `AgCl` in `0.01` molar `HCl` solution at this temperature will be :

Options

A. `2.8xx10^(-8)"mol"L^(-1)`
B. `2.8xx10^(-12)"mol"L^(-1)`
C. `5.6xx10^(-8)"mol"L^(-1)`
D. `2.8xx10^(-4)"mol"L^(-1)`

Answer

A

Explanation

`K_(SP)=[Ag^(+)][Cl^(-)]`
`2.8x10^(-10)=[Ag^(+)]x0.01`
`[Ag^(+)]=(2.8xx10^(-10))/(10^(-2))=2.8xx10^(-8)"mol" L^(-1)`