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
`(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.