Solubility product of silver bromide is 5.0xx10^(-13). The quantity of potassium bromide (molar mass taken as 120 g "mol"^(-1)) to beadded to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is

Solubility product of silver bromide is 5.0xx10^(-13). The quantity of

1.	Solubility product of silver bromide is 5.0xx10^(-13). The quantity of potassium bromide (molar mass taken as 120 g "mol"^(-1)) to beadded to 1 litre of 0.05M solution of silver nitrate to start the precipitation of AgBr is
Answer» `6.2xx10^(-5) g` `5.0xx10^(-8) g` `1.2xx10^(-10) g` `1.2xx10^(-9) g` Solution :`[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`.