Part A) Approximately how many mL of 5% BaCl2 solution would be required to prec

ID: 1047608 • Letter: P

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

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 SO42-

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 SO42-

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 SO42-

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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

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 %

Explanation / Answer

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).