If 56.0 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 758 mg sample of na2so4, what is the molarity of the solution?

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Chemistry

Herring

19 November, 00:54

If 56.0 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 758 mg sample of na2so4, what is the molarity of the solution?

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Dylan Hester · Answer 1

Answer is: molarity of solution is 0,0951 M.

Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl.

m (Na₂SO₄) = 758 mg : 1000 mg/g = 0,758 g.

n (Na₂SO₄) = m (Na₂SO₄) : M (Na₂SO₄).

n (Na₂SO₄) = 0,758 g : 142 g/mol.

n (Na₂SO₄) = 0,00533 mol.

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

n (BaCl₂) = 0,00533 mol.

V (BaCl₂) = 56,0 mL = 0,056 L.

c (BaCl₂) = n (BaCl₂) : V (BaCl₂).

c (BaCl₂) = 0,00533 mol : 0,056 L.

c (BaCl₂) = 0,0951 mol/L.