Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as 120 g `mol^(-1)`) to be added to 1 litre 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

For `AgBr, K_(sp)=5.0xx10^(-13)`
Precipitation of AgBr will occur when ionic product `[Ag^(+)][Br^(-)]` becomes larger than `K_(sp)`
`AgNO_(3)hArr Ag^(+)+NO_(3)^(-)`
`K_(sp)=[Ag^(+)][Br^(-)]`
`[Ag^(+)]=0.05 M`
The concentration of `Br^(-)` required to start precipitation.
`[Br^(-)]=(K_(sp))/([Ag^(+)])=(5.0xx10^(-13))/(0.05)=1.0xx10^(-11)`
Now `[Br^(-)]=[KBr]`
Molar mass of KBr = 120
`therefore` Amount of KBr required
`=1.0xx10^(-11)xx120=1.20xx10^(-9)g`