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. `2.5 xx 10^(-8) molL^(-1)`
B. `2.8 xx 10^(-12) molL^(-1)`
C. `5.6 xx 10^(-8) molL^(-1)`
D. `2.8 xx 10^(-4) molL^(-1)`

Answer

A

Explanation

`K_(sp) = [Ag^(+)][Cl^(-)]`
`2.8 xx 10^(-10) = [Ag^(+)]xx 0.01`
`[Ag^(+)] = (2.8 xx 10^(-10))/(10^(-2)) = 2.8 xx 10^(-8)"mol" L^(-1)`