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

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

1) Calculate the number of moles of S2O32- ion required for the titration

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

3) Calculate the number of moles of OCl-- ion required for the titration

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

5) Determine the volume of commercial bleaching solution present in the diluted bleaching solution titrated

Explanation / Answer

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

1) Calculate the number of moles of S2O32- ion required for the titration

moles of S2O3-2 = moles of NaS2O3 = MNa2SO3 * VNa2SO3 = (35.46/1000)*0.1052 = 0.00373 moles of Na2SO3

1 mol of S2O3-2 = 1 mol of Na2SO3

0.00373 mol of S2O3-2 must be present

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

The reaciton is given as: 2 Na2S2O3 + I2 = Na2S4O6 + 2 NaI

therefore,

0.00373 mol of Na2S2O3 --> 1/2*0.00373 = 0.001865 mol of I2 will be reacted

3) Calculate the number of moles of OCl-- ion required for the titration

The reaction:

OCl-(aq) + 2 H+ (aq) + 2 I- (aq) <--> I2(aq) + Cl-(aq) + H2O(l)

ratio is 1 mol of OCl- = 1 mol of I2

0.001865 mol of I2 --> 0.001865 mol of OCl-

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

mol of NaOCl --> mol of OCl-

0.001865 mol of OCl- = 0.001865 mol of NaOCl

mass of NaOCl = mol*Mw

mass = (0.001865 )(74.44)

mass of NaOCl = 0.1388306 g

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