Question

4) 350.3 mg of BaCl2 (208.23 g/mol) is dissolved in 100 mL of water. It is then mixed with a solution made by dissolving 400 mg of ammonium sulfate (132.14 g/mol) in 200 mL of water. Assuming at this juncture that the BaSO4 formed is totally insoluble. What is the final concentrations (in ppm and mM) of Ba (137.327 g/mol) and sulfate (96.06 g/mol) in the solution.

Answer 1

4)

moles of BaCl2 = 350.3 x 10^-3 / 208.23 = 1.68 x 10^-3

moles of (NH4)2SO4 = 400 x 10^-3 / 132.14 = 3.03 x 10^-3

BaCl2    +   (NH4)2SO4   -----------------> BaSO4 + 2 NH4Cl

   1                   1                                         1

1.68 x 10^-3    3.03 x 10^-3

here limiting reagent is BaCl2

moles of (NH4)2SO4 remians = 3.03 x 10^-3 - 1.68 x 10^-3

                                               = 1.35 x 10^-3 mol

moles of SO42- = 1.35 x 10^-3 mol

concentration of SO42- = 1.35 x 10^-3 / (0.100 +0. 200)

                                      = 4.49 x 10^-3 M

concentration of SO42- = 4.49 mM

                                       = 431.3 ppm

BaSO4 Ksp = 1.08 x 10^-10

BaSO4   ---------------> Ba+2   +   SO42-

Ksp = [Ba+2][SO42-]

1.08 x 10^-10 = [Ba+2] [4.49 x 10^-3]

[Ba+2] = 2.41 x 10^-8 M

[Ba+2] = 2.41 x 10^-5 mM

            = 3.30 x 10^-3 ppm