) If 55.8 mL of a BaCl2 solution is needed to precipitate all the sulfate ion in a 521 mg sample of Na2SO4, what is the molarity of the BaCl2 solution?

Answer 1

BaCl₂(aq) + Na₂SO₄(aq) ==> BaSO₄(s) + 2 NaCl(aq)

Find moles Na₂SO₄ present:

521 mg Na₂SO₄ x 1 mmole/142.04 mg = 3.668 mmoles Na₂SO₄

mmoles BaCl₂ needed = 3.668 mmoles Na₂SO₄ x 1 mol BaCl₂/mmol Na₂SO₄ = 3.668 mmoles BaCl₂

molarity of BaCl₂ = 3.668 mmole/0.0558 L = 65.7 mmol/L = 65.7 mM = 0.0657 M