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

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3 years ago

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the solution?

1 answer:

lukranit [14] · Answer 1 · 2022-09-04T08:34:43+03:00

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.