The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄[tex]Ag^{+}[/tex] + [tex]Br^{-1}[/tex].
45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains [tex]\frac{0.6485}{45}[/tex]×1000 = 14.411 gm-mole. Thus the solubility product [tex]K_{sp}[/tex]of AgBr = [[tex]Ag^{+}[/tex]]×[tex]Br^{-}[/tex].
Or, 5.0×[tex]10^{-13}[/tex] = [tex]S^{2}[/tex] (the given value of solubility product of AgBr is 5.0×[tex]10^{-13}[/tex] and the charge of the both ions are same).
Thus S = (5.00×[tex]10^{-13}[/tex])[tex]^{1/2}[/tex] = 7.071×[tex]10^{-7}[/tex] g/mL.
Thus the concentration of Br[tex]^{-1}[/tex] or HBr is 7.071×[tex]10^{-7}[/tex] g/mL.
