Solubility product of silver bromide is `5.0 xx10^(-13)`. The quantity of potassium bromide (molar mass taken as 120 g of `"mol"^(-1)`) to be added to 1 litre of 0.05 M solution of silver nitrate of start the precipitation of AgBr is

Options

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

Answer

B

Explanation

`[AgBr]=[Ag^(+)]=0.05M`
`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) molL^(-1) xx1L=10^(-11)mol`
Therefore, amount of KBr needed to precipitate `AgBr=10^(-11) molxx120gmol^(-)=1.2xx10^(-9)g`