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) ⇄ +
.
45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains ×1000 = 14.411 gm-mole. Thus the solubility product
of AgBr = [
]×
.
Or, 5.0× =
(the given value of solubility product of AgBr is 5.0×
and the charge of the both ions are same).
Thus S = (5.00×)
= 7.071×
g/mL.
Thus the concentration of Br or HBr is 7.071×
g/mL.
