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

Answer

D

Explanation

`|Ag^(+)|=0.05 M ,K_(sp)(AgBr)=[Ag^(+)][Br^(-)]`
`:. Br^(-) =(K_(sp))/(|Ag^(+)|)=(5.0 xx 10^(-13))/(0.05)=10^(-11)M`
`i.e., ` amount of KBr to be added to `=10^(-11)` mole
`=10^(-11)xx120=1.2 xx 10^(-9) ` g