If 55.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 742 mg sample of na2so4, what is the molarity of the solution?

Question

Mollie Frye

Chemistry

23 January, 07:21

If 55.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 742 mg sample of na2so4, what is the molarity of the solution?

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Yareli Walter · Answer 1

Answer is: the molarity of the solution is 0.0936 M.

Balanced chemical reaction: BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaCl (aq).

Net ionic reaction: Ba²⁺ + SO₄²⁻ → BaSO₄ (s).

V (BaCl₂) = 55.8 mL; volume of barium chloride solution.

V (BaCl₂) = 55.8 mL : 1000 mL/L = 0.0558 L.

m (Na₂SO₄) = 742 mg : 1000 mg/g.

m (Na₂SO₄) = 0.742 g; mass of sodium sulfate.

n (Na₂SO₄) = m (Na₂SO₄) : M (Na₂SO₄).

n (Na₂SO₄) = 0.742 g : 142.04 g/mol.

n (Na₂SO₄) = 0.00522 mol; amount of substance.

From chemical reaction: n (Na₂SO₄) : n (BaCl₂) = 1 : 1.

n (BaCl₂) = 0.00522 mol.

c (BaCl₂) = n (BaCl₂) : V (BaCl₂).

c (BaCl₂) = 0.00522 mol : 0.0558 L.

c (BaCl₂) = 0.0936 M.