Question

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
A. `6.2xx10^(-5) g`
B. `5.0xx10^(-8) g`
C. `1.2xx10^(-10) g`
D. `1.2xx10^(-9) g`

Answer 1

Correct Answer - D
`[Ag^(+)]=0.05 M `
`K_(sp)(AgBr)=[Ag^(+)][Br^(-)]`
`:. [Br^(-)]=(K_(sp))/([Ag^(+)])=(5.0xx10^(-13))/(0.05)=10^(-11)M`
i.e., amount of KBr to be added `=10^(-11)` mole
`=10^(-11)xx120 g = 1.2 xx 10^(-9) g`.