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

Answer

C

Explanation

`K_(sp) =[Ag^(+)][Br^(-)] =5.0 xx 10^(-13) M^(2)`
`[Ag^(+)] =0.05 M`
`[Br^(-)] =((5.0xx 10^(-13)M^(2)))/((0.5 M)) =10^(-11)M`
Moles of KBr =`10^(-11)M`
Moles of KBr = `(10^(-11) "mol") xx (120.0 g "mol"^(-1))`
`=1.20 xx 10^(-9) g`