Question

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solution with an excess of ag ions. the ksp of agbr is 5.0 × 10–13.

Answer 1

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) ⇄$Ag^{+}$ + $Br^{-1}$.

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$of AgBr = [$Ag^{+}$]×$Br^{-}$.

Or, 5.0×$10^{-13}$ = $S^{2}$ (the given value of solubility product of AgBr is 5.0×$10^{-13}$ and the charge of the both ions are same).

Thus S = (5.00×$10^{-13}$)$^{1/2}$ = 7.071×$10^{-7}$ g/mL.

Thus the concentration of Br$^{-1}$ or HBr is 7.071×$10^{-7}$ g/mL.