Question

Determine the concentration of an HBr solution if a 45.00 mL aliquot of the solution yields 0.5555 g AgBr when added to a solution with an excess of Ag+ ions . The Ksp of AgBr is 5.0×10−13 .

Answer 1

Molecular weight of AgBr = 187.7
moles of Ag = $\frac{0.5555}{187.7} = 2.96 x 10^{-3}$
moles of Br = moles of Ag = 2.96 x 10⁻³ mol
concentration of HBr (Molarity) = conc. of Br (strong acid) = $\frac{2.96 x 10^{-3} }{45 x 10^{-3} } = 0.0658 mol/l$

Answer 2

Mole wt of AgBr = 187.7 

Moles of AG = 0.5555/ 187.7 = 0.002960.

 Moles of Br = 0.00296.

Molarity = 0.00296/ 45.0 = 0.00658M.