Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A student followed the procedure of this experiment to determine the percent NaO

ID: 879270 • 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 mass of NaOCl present in the diluted bleaching solution titrated

Explanation / Answer

first Calculate the number of moles of S2O32- 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

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at drjack9650@gmail.com
Chat Now And Get Quote