Oxidation

Topics: Chlorine, Bleach, Redox Pages: 3 (480 words) Published: November 17, 2011
Experiment 39
Oxidation – Reduction Titrations II : Analysis of Bleach

A.Standardization of 0.05 M Na2S2O3 Solution
KIO3 concentration 0.01 M
Volume of KIO3 Trial #1 Trial #2

Final burette reading 15.01 mL 30.00 mL
Initial burette reading 0.00 mL 15.01 mL
Milliliters of KIO3 used 15.01 mL 14.99 mL

Volume of Na2S2O3 Trial #1 Trial #2

Final burette reading 13.40 mL 25.78 mL
Initial burette reading 0.03 mL 13.37 mL
Milliliters of Na2S2O3 used 13.37 mL 12.41 mL

Molarity of Na2S2O3
Trial #1
0.01501L x 0.01M = 1.5 ee -4 KIO3 ….. 6 x 1.5 ee -4 = 9ee -4 ….. 9 x 10^-3 mol / 0.01337 L = 0.0674 M Na2S2O3
Trial #2
0.01499 L x 0.01M = 1.499 ee -4 ……. 1.499 ee -4 x 6 = 8.994 ee -4 8.994 ee -4 mol / 0.01241 L = 0.0725 M Na2S2O3

Average Molarity
(0.0674 + 0.0725) / 2 = 0.0700 M

Standard Deviation
L1 = .0674 .0725
n = 2 1 Var
standard deviation = 0.0036

B. Determination of the Oxidizing Capacity of an Unknown Liquid Bleach Mass of beaker, test tube, and bleach 56.225 g 54.548 g
Mass of beaker and test tube 55.767 g 54.012 g
Mass of unknown bleach 0.458 g 0.536 g

Volume of Na2S2O3
Final burette reading 32.35 mL 38.93 mL
Initial burette reading 25.78 mL 32.35 mL
mL of Na2S2O3 6.57 mL 6.58 mL

Mass NaOCL
Trial #1
½ ( 0.05M )( 0.00657 L) = 1.6425 ee -4 mol
1.6425 ee -4 mol x 56.225 g = 0.004 g NaOCL

Trial #2
½ (0.05M)(0.00658 L) = 1.645 ee -4 mol
1.645 ee -4 mol x 54.548 g = .009 g NaOCL

Percent of NaOCL

Trial #1
(0.004 g / 0.458 g ) x 100 = .87 %

Trial #2
(0.009 g / 0.536 g) x 100 = 1.68 %

Questions and Answers

1) In the standardization of the Na2S2O3 solution, if you did not titrate the liberated iodine immediately, how would this likely have affected the value of molarity you calculated?

If we did not titrate the liberated iodine immediately the liberated I2 would go out of the system,...
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