Question

A student followed the procedure of this experiment to determine the percent NaOCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-mL aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na2S2O3 solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g/mL

Calculate the mass of NaOCl present in the diluted bleaching solution titrated

Answer 1

first Calculate the number of moles of S₂O₃^2- ion required for the titration.

The titration required 35.46 mL of 0.1052M Na2S2O3 solution.

volume = 35.46 ml = 0.03546 L

molarity = 0.1052 M

molarity = moles / volume

Mole S₂O₃^-2 = 0.0354 x 0.1052 = 3.72 x10^-3 mole S₂O₃^-2

Calculate the number of moles of I₂ produced in the titration mixture.

1 mol of I₂    ---------------- 2 moles of S₂O₃^-2

   ? mol of I₂ ---------------- 3.730x10^-3mole S₂O₃^-2

moles of I₂ = 3.730x10^-3 / 2

                    = 1.865x10^-3 mol I₂

Calculate the number of moles of OCl^- ion present in the diluted bleaching solution titrated.

1 mol of I₂        ---------------- 1 mole of OCl^-1

1.865x10^-3 mol I₂ -------------- ?

moles of OCl^-1 = 1.865x10^-3 moles

Calculate the mass of NaOCl present in the diluted bleaching solution titrated (the 20.00mL sample).

moles of NaOCl = 1.865x10^-3

molecular weight = 74.45g/mol

moles = mass / molecular weight

mass of NaOCl = 1.865x10^-3 x 74.45

                         = 0.1388g

mass of NaOCl = 0.1388g