How many moles of `AgBr(K_(w) =5 xx 10^(-13))` will dissolve in a 0.01 M NaBr solution ? (NaBr is completely ionised in solution)

Answer

`5xx 10^(-11) "mol"//L`

Explanation

The ionisation of NaBr in solution may be represented as :
`NaBr (s) to Na^(+) (aq) + Br^(-) (aq) `
`:.` concentration of `Br^(-)` ions in solution `[Br^(-) (aq)] =0.01 M =10^(2) M`
The solubility product of AgBr i.e., `K_(sp) =[Ag^(+) (aq)][Br^(-)(aq)]`
`:." " [Ag^(+)(aq)] =(K_(sp))/[[Br^(-)(aq)]] =(5xx 10^(-13))/(1xx 10^(-2)) =5xx 10^(-11) M`
Thus , `5xx 10^(-11)` mole of AgBr will dissolve per litre of the solution .