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

Answer

`5xx10^(-11) "mol" L^(-1)`

Explanation

Suppose solubility of AgBr in 0.01 M NaBr = s mol `L^(-1)`. Then as `AgBr rarr Ag^(+) + Br^(-)`,
`[Ag^(+)]=s "mol " L^(-1) and "Tota" [Br^(-)]=0.01 + s ~~ 0.01 M`
`K_(sp) = [ Ag^(+)][Br^(-)], i.e., 5xx10^(-13) = s xx0.01 or s = 5 xx 10^(-11) "mol" L^(-1)`