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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

`5.0xx10^(-8)` g

B

`1.2xx10^(-10)` g

C

`1.2xx10^(-9)` g

D

`6.2xx10^(-5)` g

Text Solution

Verified by Experts

The correct Answer is:
C
`Br^(-)=(K_(sp)[AgBr])/(C_(Ag^(+)))`
`Br^(-)=(5xx10^(-13))/(0.05)=10^(-11)`
Conc. = `[KBr]=10^(-11)`
Moles of `KBr=10^(-11)`
Weight of `KBr=10^(-11)xx120=1.2xx10^(-9)` g .
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