1.

Solubility product of silver bromide is 5.0 xx 10^(-13). The quantity of potassium bromide (molar mass taken as 120 g mol^(-1)) to be added to 1 litre of 0.5 M solution of silver nitrate to start the precipitation of AgBr is

Answer»

`5.0 xx 10^(-8) G`
`1.2 xx 10^(-10) g`
`1.2 xx 10^(-9) g`
`6.2 xx 10^(-5) g`

Solution :`Br^(-) = (K_(sp)(AGBR))/(C_(AG^(+)))`
`Br^(-) = (5 xx 10^(-13))/(0.05) = 10^(-11)`
Conc. `= [KBr] = 10^(-11)` ltbgt Moles of KBr `= 10^(-11)`
Weight of KBr `= 10^(-11) xx 120 xx 1.2 xx 10^(-9)g`.


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