The minimum mass of NaBr which should be added in 200 ml of `0.0004M-AgNO_(3)` solution just to start precipitation of `AgBr.K_(sp)` of `AgBr=4xx10^(-13)(Br=80)`

Options

A. `1.0xx10^(-9)g`
B. `2xx10^(-10)g`
C. `2.06xx10^(-8)g`
D. `1.03xx10^(-7)g`

Answer

C

Explanation

`[Br^(-)]=(4xx10^(-13))/(4xx10^(-4))=10^(-9)M`
mass of NaBr added of 200 ml`=103xx(10^(-9))/(1000)xx200`
`=2.06xx10^(-8)g`