Question

A sample of 500. mL of 0.0400 M barium ion is mixed with 1000. mL of 0.0600 M oxalate ions.

a) will barium oxalate precipitate?

b) if there is precipitation, determine the mass of barium oxalate precipitated.

Answer 1

Moles of barium ions = Volume * Molarity = 0.5*0.04 = 0.02 mol

Moles of oxalate ions = 0.06*1.0 = 0.06 mol

As moles of barium ions are less, the barium is limiting reactant here. Barium oxalate will precipitate according to the following reaction.

Ba²⁺ + C₂O₄^2- → BaC₂O₄

As one mole of barium ions react with 1 mole of oxalate ions to produce 1 mole of barium oxalate, 0.02 moles of barium ions will produce 0.02 moles of barium oxalate.

Mass of barium oxalate = 0.02*225.34 = 4.5068 g