Question

Selenium in a 10.0g soil sample is distilled as the tetrabromide, which is collected in aqueous solution where it is hydrolyzed to SeO₃^2-. The SeO₃^2-​ is determined iodometrically requiring 4.5ml of standard thiosulfate solution for the titration. If the theosulfate titer is 0.049mg K₂Cr₂O₇/ml, what is the concentration of selenium in the soil in ppm?

Thanks!

Answer 1

The reactions taking place are:

Se + 2 Br₂ ------> SeBr₄ …..(1)

2 SeBr₄ + 3 H₂O -------> Se₂O₃^2- + 6 H⁺ + 8 Br^- …..(2)

Se₂O₃^2- + 4 I^- + 6 H⁺ ------> 2 Se + I₂ + 3 H₂O ……(3)

I₂ + 2 S₂O₃^2- -------> 2 I^- + S₄O₆^2- …..(4)

As per the balanced stoichiometric reaction (4),

1 mole I₂ = 2 moles thiosulfate

Again, the standardization of thiosulfate with dichromate can be shown as:

6 I^- + Cr₂O₇^2- + 14 H⁺ -------> 3 I₂ + 2 Cr³⁺ + 7 H₂O ……(5)

I₂ + 2 S₂O₃^2- -------> 2 I^- + S₄O₆^2- ……(6)

As per equations (5) and (6),

1 mole Cr₂O₇^2- = 3 moles I₂

1 mole I₂ = 2 moles thiosulfate.

Therefore, 1 mole Cr₂O₇^2- = 6 moles thiosulfate.

Therefore, 1 mole thiosulfate = (1/6)*(moles of Cr₂O₇^2-)

The concentration of K₂Cr₂O₇ = 0.049 mg/mL.

Molar mass of K₂Cr₂O₇ = 294.185 g/mole.

Therefore, moles of K₂Cr₂O₇ = (0.049 mg)*(1 g/1000 mg)*(1 mole/294.185 g) = 1.6656*10^-7 mole.

Moles of thiosulfate neutralized = (1 mole K₂Cr₂O₇)*(6 moles thiosulfate/1 mole K₂Cr₂O₇) = 9.9936*10^-7 mole.

Concentration of thiosulfate = (9.9936*10^-7 mole)/[(1 mL)*(1 L/1000 mL)] = 9.9936*10^-4 mol/L = 9.9936*10^-4 M.

Volume of thiosulfate required for Se titration = 4.5 mL; therefore, moles of thiosulfate required for Se titration = (4.5 mL)*(9.9936*10^-4 mol/L) = 4.49712*10^-3 mmole.

As per stoichiometric equation (4),

Millimoles of I₂ generated in Se titration = (4.49712*10^-3 mmole)*(1 mole I₂/2 moles thiosulfate) = 2.24856*10^-3 mmole.

As per stoichiometric equation (3),

Millimoles of Se₂O₃^2- neutralized = millimoles of I₂ generated = 2.24856*10^-3 mmole.

As per stoichiometric equation (2),

2 mole SeBr₄ = 1 mole Se₂O₃^2-

Therefore, millimoles of SeBr₄ formed = (2.24856*10^-3 mmole)*(2 mole SeBr₄/1 mole Se₂O₃^2-) = 4.49715*10^-3 mmole.

As per stoichiometric equation (1),

1 mole Se = 1 mole SeBr₄.

Therefore, millimoles of Se in the sample = 4.49712*10^-3 mmole.

Atomic mass of Se = 78.96 g/mole.

Therefore, mass of Se in the soil = (4.49712*10^-3 mmole)*(1 mole/1000 mmole)*(78.96 g/1 mole)*(1000 mg/1 g) = 0.35509 mg.

Mass of soil taken = 10.0 g = (10.0 g)*(1 kg/1000 g) = 0.001 kg.

Therefore, concentration of Se in ppm = (0.35509 mg Se)/(0.001 kg soil) = 355.09 mg/kg ≈ 355.1 ppm (ans).