Chem lab help. This analysis should be performed on two different brands of blea

Question

Chem lab help.

This analysis should be performed on two different brands of bleach; preferably one name brand and one generic. Prepare one sample of each bleach solution by determining the mass of 10.00 mL of the bleach placed into a 100 mL volumetric flask. The masses will be used to determine the density of bleach. Dilute the sample to the 100 mL mark with deionized water and mix well. Rinse the pipet out with a little of the diluted bleach solution twice and pipet a 10.00 mL aliquot into a 250 mL Erlenmeyer flask. Add 1.0 g of potassium iodide and swirl the resulting mixture. Add 5.0 mL of 6 M HCI to the mixture. Titrate with the standardized sodium thiosulfate solution until the amber iodine color begins to fade to a pale yellow. At this point add 2 mL of 0.4% starch solution and resume titrating until the dark color of the starch-iodine complex just disappears. Be sure to use ½ drop quantities as you near the endpoint by swiftly turning the stopcock 180° or by splitting a drop. An abrupt color change from dark blue to colorless marks the endpoint. Repeat the analysis with a second brand of bleach.

Reults

Clorox

A-1

Mass of 10 mL

9.697 g

10.72 g

Density

.9697 g/mL

1.072 g/mL

Avg density

1.021 g/mL

1.021 g/mL

Vol of diluted bleach

10 mL

10 mL

Vol of Na2S2O3

12.7 mL

12mL

Moles of Na2S2O3

.0014 mol

.00124 mol

Calculate the molarity of the diluted aliquot of bleach, the concentrated bleach solution and mass% of the bleach solutions.

Calculate average, average deviation, and the percent error.

Clorox

A-1

Mass of 10 mL

9.697 g

10.72 g

Density

.9697 g/mL

1.072 g/mL

Avg density

1.021 g/mL

1.021 g/mL

Vol of diluted bleach

10 mL

10 mL

Vol of Na2S2O3

12.7 mL

12mL

Moles of Na2S2O3

.0014 mol

.00124 mol

Explanation / Answer

Bleaching action is due to sodium hypochlorite(NaOCl). The balanced reactions occuring during titration are

NaOCl(aq) -------- > Na+(aq) + OCl-(aq)

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

2S2O32-(aq) + I2 ------- > S4O62-(aq) + 2I-(aq)

(a): CloroX:

From the above balanced chemical reaction it is clear that 1 mole of OCl-(aq) reacts with 2 mol S2O32-(aq).

Hence moles of OCl-(aq) in cloroX

= 0.0014 mol S2O32-(aq) x [1 mol I2 / 2 mol S2O32-(aq)] x [1 mol OCl-(aq) / 1 mol I2]

= 0. 0007 mol  OCl-(aq)

Hence moles of NaOCl in the diluted aliquot of cloroX = 0. 0007 mol

Hence molarity of diluted aliquot of cloroX = 0. 0007 mol / 0.010 L = 0.07 M (answer)

Since the10 mL of concentrated bleach of chloroX is diluted 10 times to 100 mL, concentration of diluted bleach of cloroX = 0.07 x 10 = 0.7 M (answer)

moles of NaOCl in cloroX = MxV(L) = 0.7 mol/L x 0.010 L = 0.007 mol

Hence mass of NaOCl in 10 mL of concentrated bleach of chloroX = 0.007 mol x 74.44 g/mol = 0.5211 g

Hence mass percentage of cloroX = 0.5211 g / 10 mL = 0.05211 g/mL (answer)

(a): A- 1:

moles of OCl-(aq) in A-1

= 0.00124 mol S2O32-(aq) x [1 mol I2 / 2 mol S2O32-(aq)] x [1 mol OCl-(aq) / 1 mol I2]

= 0. 00062 mol  OCl-(aq)

Hence moles of NaOCl in the diluted aliquot of A-1 = 0. 00062 mol

Hence molarity of diluted aliquot of A-1 = 0. 00062 mol / 0.010 L = 0.062 M (answer)

Since the10 mL of concentrated bleach of A-1 is diluted 10 times to 100 mL, concentration of diluted bleach of A-1 = 0.062 x 10 = 0.62 M (answer)

moles of NaOCl in A-1 = MxV(L) = 0.62 mol/L x 0.010 L = 0.0062 mol

Hence mass of NaOCl in 10 mL of concentrated bleach of A-1 = 0.0062 mol x 74.44 g/mol = 0.4615 g

Hence mass percentage of A-1 = 0.4615 g / 10 mL = 0.04615 g/mL (answer)

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