A student followed the procedure of this experiment to determine the percent NaO
ID: 61373 • Letter: A
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 number of moles of OCl-- ion required for the titration
Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated
Explanation / Answer
Answer:
Calculate the number of moles of OCl-- ion required for the titration:
Cl2 + 2 OH------------> ClO- + Cl- + H2O
Oxidation-reduction titration is done in to know the hypochlorite ion present in a solution.General method used is iodine-thiosulfate titration. hypochlorite ions oxidize iodide ions to form iodine, I2, this iodine is then titrated with standard called sodium thiosulfate.
to caliculate number of moles of OCl-- ion, moles of S2O3-2 and I2 are required
Moles of S2O3-2 = 3.546x10-2 in litres x 0.1052
= 0.373 moles
I3- + 2 S2O3-2----------------------------> 3 I- + S4O6-2
Moles of I = 3.730x10-3 ( S2O3-2 ) (1 moles of I2/ 2 moles of S2O3-2)
= 1.865x10-3 moles of Iodine
Now moles of OCl-- =
=1.865x10-3 moles of I2
= 1 mol OCl-1/ 1 mol I2
= 1.865x10-3 moles
Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated
1.865x10-3 moles of NaOCl = mass/ 74.45 (GMW)
= 1.865x10-3 x 74.45
= 0.1388 grams of commercial bleaching
(74.45g/mol) = 0.1388g NaOCl
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