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

Answer

B

Explanation

`[AgBr]=[Ag^(+)]=0.05 M`
`K_(sp)[AgBr]=[Ag^(+)][Br^(-)]`
`implies [Br^(-)]=(K_(sp)(AgBr))/([Ag^(+)])`
`=(5.0xx10^(-13))/(0.05)=10^(-11) M [mol L^(-1)]`
Moles of `KBr` needed to precipitate `AgBr`
`=[Br^(-)]xxV=10^(-11) mol L^(-1)xx1L=10^(-11) mol`
Therefore, amount of `KBr` needed to precipitate `AgBr`
`10^(-11) molxx120 g mol^(-1)=1.2xx10^(-9) g`