1.

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

Answer»

`1.2xx10^(-10)g`<BR>`1.2xx10^(-9)g`
`6.2xx10^(-5)g`
`5.0xx10^(-8)g`

Solution :`AgBr HARR Ag^(**) ** Br^(**)`
`K_(sp) =([Ag^(+)][Br^(-)]`
For precipitation to occur
Ionic product `gt` Solubility product
`[Br^(**)]** (K_(sp))/([Ag^(**)])**(5** 10^(-13))/(0.05) 10^(-11)`
i.e. precipitation just STARTS when `10^(-11)` moles of KBR is added to `1//AgNO_(3)` solution
`therefore` Number of moles of `Br^(-)` needed from `KBr=10^(-11)`
`therefore` Mass of `KBr=10^(-11)xx120=1.2xx10^(-9) g`


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