Calamine, is a mixture of zinc and iron oxides. A 1.0220g sample of dried calami

Question

Calamine, is a mixture of zinc and iron oxides. A 1.0220g sample of dried calamine was dissolved in acid and diluted to 250 mL. Potassium fluoride was added to a 10mL aliquot of the dissolved solution to mask the iron: after suitable adjustment of the pH, Zn2+ consumed 38.71 mL of 0.01294 M EDTA. A second 50mL aliquot was suitably buffered and titrated with 2.40mL of 0.002727 M ZnY2- solution. Calculate the percentages of ZnO and Fe2O3 in the sample. Calamine, is a mixture of zinc and iron oxides. A 1.0220g sample of dried calamine was dissolved in acid and diluted to 250 mL. Potassium fluoride was added to a 10mL aliquot of the dissolved solution to mask the iron: after suitable adjustment of the pH, Zn2+ consumed 38.71 mL of 0.01294 M EDTA. A second 50mL aliquot was suitably buffered and titrated with 2.40mL of 0.002727 M ZnY2- solution. Calculate the percentages of ZnO and Fe2O3 in the sample.

Explanation / Answer

Calamine sample = dried 1.022 g (mixture of ZnO and Fe2O3)

We are required to find the % of ZnO and Fe2O3 in the dried calamine sample

In 10mL aliquot of the dissolved solution after suitable adjustment of the pH, Zn2+ consumed 38.71 mL of 0.01294 M EDTA

so, (Molarity x volume)ZnO = (Molarity x volume)EDTA

      MZnO x 10 mL = 0.01294 X 38.71 mL

      MZnO = 0.05 M

Now, weight of ZnO = (MZnO X Volume after diluting in mL X molecular weight of ZnO)/1000

                             = (0.05 x 250 x 81.4)/1000

                             = 1.0175 g

% ZnO = (1.0175/1.022) X 100 = 99.56 %

% ZnO = 99.56 %

Then, 50mL aliquot was suitably buffered and titrated with 2.40mL of 0.002727 M ZnY2- solution

so, (Molarity x volume)Fe = (Molarity x volume)ZnY2-

      MFe x 50 mL = 0.002727 X 2.4 mL

      MFe = 1.3 X 10-4 M

Now, since, each Fe2O3 molecule has 2 Fe molecules, so, number of Moles of Fe2O3 equal ½ that of Fe

MFe2O3 = MFe/ 2 = 6.5 x 10-5

Weight of Fe2O3 = (MFe2O3 x volume after dilution in mL x molecular weight of Fe2O3) / 1000

                         = (6.5 x 10-5 x 250 x 159.69) /1000

                         = 2.59 x 10-3 g

% of Fe2O3 = (2.59 x 10-3 / 1.022) x 1000 = 0.253 %

% of Fe2O3 = 0.253 %

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