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. `6.2xx10^(-5) g`
B. `5.0xx10^(-8) g`
C. `1.2xx10^(-10) g`
D. `1.2xx10^(-9) g`

Answer

D

Explanation

`[Ag^(+)]=0.05 M `
`K_(sp)(AgBr)=[Ag^(+)][Br^(-)]`
`:. [Br^(-)]=(K_(sp))/([Ag^(+)])=(5.0xx10^(-13))/(0.05)=10^(-11)M`
i.e., amount of KBr to be added `=10^(-11)` mole
`=10^(-11)xx120 g = 1.2 xx 10^(-9) g`.