Precipitation, Complexation, Redox Titration

Question

A 20.0 mL sample of commercial bleach was diluted with water in a 250.0 mL volumetric flask. A 50.0 mL aliquot of the diluted sample was treated with excess KI and starch. The liberated iodine was titrated with 0.399 M Na2S2O3, requiring 13.9 mL to reach the endpoint.

(a) Calculate the % w/v NaOCl in the sample.

Answer 1

1) The moles of iodine formed with the reaction of hypochlorite is 1:1

2) The moles of iodine required to react with the thiosulfate is 1:2

(The redox equation is I₂ + 2S₂O₃^2- ⇒ 2I^- + S₄O₆^2-

.0139 liters of sodium thiosulfate x (.399 moles/liter sodium thiosulfate)(1mole thiosulfate/mole of sodium thiosulfate)(1 mole iodine/2 moles of sodium thiosulfate)(1mole hypochlorite/mole iodine) = 2.773 x 10^-3 moles of hypochlorite.

The moles per volume of the aliquot and the 250 ml dilution concentration is 2.773 x 10^-3 moles / 50 ml which we can turn into weight per 100 ml by multiplying by 2/2 and by molar mass of NaOCl (74.4 g/mole):

2(2.773 x 10^-3)(74.4)/(2x50 ml) = .413 g/100 ml

Finally this concentration was based on a 250/20 dilution, so the concentration in the bleach was (250/20) more concentrated or 5.16 W/V %

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