Gdetermine 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.

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.

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