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2 years ago

9

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 soluti

on with an excess of Ag+ ions . The Ksp of AgBr is 5.0×10−13 .

2 answers:

amid [387]2 years ago

6 0

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) = $\frac{2.96 x 10^{-3} }{45 x 10^{-3} } = 0.0658 mol/l$

bazaltina [42]2 years ago

4 0

<span>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.</span>

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