Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as `120g mol^(-1)`) to be added to `1 L` of `0.05 M` solution of silver nitrate to start the precipitation of `AgBr` is

Options

A. `1.2 xx 10^(-10) g`
B. `1.2 xx 10^(-9) g`
C. `6.2xx 10^(-5) g`
D. `5.0 xx 10^(-8) g`

Answer

B

Explanation

`k_(SP) = [Ag^(+)][Br^(-)]`