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Solubility product of silver bromide is ...

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

A

`1.2xx10^(-10)g`

B

`1.2xx10^(-9)g`

C

`6.2xx10^(-9)g`

D

`5.0xx10^(-8)g`

Text Solution

Verified by Experts

The correct Answer is:
B
`[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`
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