Question

Part A) Approximately how many mL of 5% BaCl2 solution would be required to precipitate all the sulfate if we assume that your samples are pure Magnesium sulfate? Assume that the density of the barium chloride solution is 1.00 g/mL.

Part B) If the samples were pure potassium sulfate would you require a smaller or larger volume of barium chloride solution than the amount calculated in (Part A) above?

Crucible/sample Number	1	2	3
Mass of unknown sample (g)	0.3623 g	0.3461 g	0.3607 g
Mass of crucibles (g)	32.6072 g	33.0450 g	32.4736 g
Mass of crucible + BaSO4 (g)	32.8294 g	33.2525 g	32.6823 g
Mass of BaSO4 (g)	0.2222 g	0.2075 g	0.2087 g
percentage of SO42-	25.2 %	24.7 %	23.8 %

Average Percentage	24.6 %
St Dev	0.709

Mass of Sulfate,

Trial 1) 0.0914 grams

Trial 2) 0.0854 grams

Trial 3) 0.0859 grams

These were determined by stiochemetric calculations

0.2222g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moe BaSO4) x (96.0576 g Sulfate / 1 mole sulfate) = 0.0914 grams SO₄^2-

0.2075g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moe BaSO4) x (96.0576 g Sulfate / 1 mole sulfate) = 0.0854 grams SO₄^2-

0.2087g x (1 mole BaSO4 / 233.39 g) x (1 mole sulfate / 1 moe BaSO4) x (96.0576 g Sulfate / 1 mole sulfate)= 0.0859 grams SO₄^2-

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Part A) Approximately how many mL of 5% BaCl2 solution would be required to precipitate all the sulfate if we assume that your samples are pure Magnesium sulfate? Assume that the density of the barium chloride solution is 1.00 g/mL.

Part B) If the samples were pure potassium sulfate would you require a smaller or larger volume of barium chloride solution than the amount calculated in (Part A) above?

Answer 1

a) The balanced chemical equation for the reaction between pure MgSO₄ and BaCl₂ is

MgSO₄ + BaCl₂ -------> MgCl₂ + BaSO₄

The molar ration of BaCl₂ and BaSO₄ is 1:1. The molar mass of BaSO₄ is 233.43 g/mol; the molar mass of BaCl₂ is 208.23 g/mol

Now let use find out the moles of BaSO₄ produced and the moles of BaCl₂ required in the three trials.

	Trial 1	Trial 2	Trial 3
Mass of BaSO4 produced (g)	0.2222	0.2075	0.2087
Moles of BaSO4 produced	0.2222/233.43 = 9.519*10-4	0.2075/233.43 = 8.889*10-4	0.2087/233.43 = 8.940*10-4
Moles of BaCl2 required	9.519*10-4	8.889*10-4	8.940*10-4
Average moles of BaCl2 required	9.116*10-4
Mass of BaCl2 required (g)	9.116*10-4*208.23 = 0.1898

We have a 5% BaCl₂ solution, which means that 100 mL solution contains 5 g BaCl₂. The density of the solution is 1 g/mL; therefore, mass of 100 mL BaCl₂ solution = 100 g.

Therefore, concentration of BaCl₂ solution = 5 g/100 g.

0.1898 g BaCl₂ is present in (0.1898 g BaCl₂)/(5 g/100 g) = 3.796 g ≈ 3.8 g; since the density is 1 g/mL, the volume of BaCl₂ required = 3.8 mL (ans).