Analysis of a Commercial Bleach
Purpose: The purpose of this lab is to determine the amount of sodium hypochlorite (NaClO) in commercial bleach. This can be done by forming triiodide ions. To make the measurement more accurate, starch was added to help determine the endpoint of the solution. The significance of this lab is that industry can use these techniques to determine the amount of NaClO in the bleach of the rival industry and improve it. Hypothesis: The hypothesis is that, an accurate determination of NaClO in commercial bleach can be done. By mixing the acidified iodide ion to the hypochlorite solution, the iodide is oxidized to iodine which forms complex triiodide ions that give the red-brown color to the solution.
Because the endpoint of the titration of triiodide is hard to determine, starch is added to give the solution a dark blue color. If starch was not added, the color would be turning from yellow to clear. It is quite hard to distinguish between the two colors and therefore, starch was added so the color would turn from yellow to dark blue. This makes it a lot easier to determine the endpoint. Materials:
Distilled water- 100 mL
5% bleach (NaClO)- 5 mL
3 M Hydrochloric acid (HCl)- 6 mL
0.100 M Sodium thiosulfate (Na2S2O3)- 100 mL
2% Starch solution- 3 mL
Potassium iodide (KI)- 6g
Burette
Burette clamp
Ring stand
Small funnel
6 pipettes
0.0001-g precision balance
3 wash glasses
Spatula
Rubber spatula
6 125-mL Erlenmeyer flasks
Size 2 rubber stopper
Stirring rod
10-mL graduated cylinder
25-mL beaker
Procedures:
1.) Measure out 5 mL of NaClO into a 125-mL Erlenmeyer flask 2.) Add 95 mL of distilled water to the flask
3.) Mass out 2 g of solid KI
4.) Transfer 25 mL of the diluted NaClO in the flask to another 125-mL Erlenmeyer flask 5.) Repeat step 3 and 4 for two more times
6.) Label the flask: Trial 1, Trial 2, Trial 3
7.) Add 2 g of solid KI to the Trial 1 flask
8.) Swirl the flask to dissolve the KI
9.) Working in the fume hood, stir the solution
10.)Add 2 mL of 3 M HCl to the flask while stirring
11.)The solution should become red-brown color
12.)Rinse a burette with distilled water
13.)Re-rinse the burette with 0.100 M Na2S2O3
14.)Attach the burette to the ring stand
15.)Fill the burette with 0.100 M Na2S2O3 until the volume goes above the 0-mL line 16.)Put a 25-mL beaker under the burette
17.)Turn the burette cork so that the solution flows into the 25-mL 18.)Turn the cork off when the volume of the Na2S2O3 reaches the 0-mL line 19.)Titrate the solution in the Trial 1 flask until the solution turns from red-brown to light yellow 20.)Add a dropperful of starch solution to the Trial 1 flask 21.)The solution should become blue
22.)Swirl the flask
23.)Titrate the solution in the Trial 1 flask until the blue color disappear 24.)Record the final burette reading
25.)Repeat step 7 to 24 for Trial 2 and Trial 3 flasks
Results: During the first titration, the solution turns from red-brown to orange-brown before becoming yellow. The starch solution cannot be stored
for a long period, and it feels somewhat sticky. The blue color after adding the starch solution is a dark shade of blue. The solution looks similar to ink. During the second titration, the blue color turns to orange then to yellow if more Na2S2O3 is added. The ratio of sodium hypochlorite (NaClO-) to sodium thiosulfate (Na2S2O3) is 1:2
Titration of Iodine Solution
Molarity of Na2S2O3= 0.100M
Molar mass of NaClO= 74.44g
Volume of original bleach= 5mL
Volume of diluted bleach= 25mL
Mass of KI=
Trial 1- 2.0021g
Trial 2- 2.0007g
Trial 3- 2.0000g
Initial burette reading=
Trial 1- 0mL
Trial 2- 14.5mL
Trial 3- 29mL
Final burette reading=
Trial 1- 14.5mL
Trial 2- 29mL
Trial 3- 41.5mL
Volume of Na2S2O3 added=
Trial 1- 14.5mL
Trial2- 14.5mL
Trial 3- 12.5mL
Average volume added= 13.8mL
Moles of Na2S2O3 added=
Trial 1- 1.45 × 10-3moles
Trial 2- 1.45 × 10-3moles
Trial 3- 1.25 × 10-3moles
Average moles added= 1.38 × 10-3
Moles of ClO- in diluted bleach= 6.92 × 10-4
Average molarity of ClO- in diluted bleach= 0.0277M
Molarity of the original bleach= 0.553M
Mass of NaClO= 0.2g
Mass of commercial bleach= 5.4g
Percent of NaClO in the commercial bleach= 3.81%
Standard deviation of each trial= 0.89mL
Percent error= 23.7%
Volume if Na2S2O3 added = (Final burette reading) – (Initial burette reading) Trial 1: 14.5 – 0 = 14.5 mL
Trial 2: 29 – 14.5 = 14.5 mL
Trial 3: 41.5 – 29 = 12.5 mL
Average volume added = ?(Volume of Na2S2O3 added) ÷ 3
(14.5 + 14.5 + 12.5) ÷ 3 = 13.83 mL
Moles of Na2S2O3 added = (Volume of Na2S2O3 added) × (Molarity of Na2S2O3) Trial 1: 0.0145 × 0.100 = 1.45 × 10-3 moles Na2S2O3
Trial 2: 0.0145 × 0.100 = 1.45 × 10-3 moles Na2S2O3
Trial 3: 0.0125 × 0.100 = 1.25 × 10-3 moles Na2S2O3
Average moles added = (Average volume added) × (Molarity of Na2S2O3) 0.01383 × 0.100 = 1.383 × 10-3. moles Na2S2O3
Moles of ClO- in diluted bleach = (Average moles added) ÷ 2 (1.383 × 10-3) ÷ 2 = 6.915 × 10-4 moles ClO-
Average molarity of ClO- in diluted bleach = (Moles of ClO- in diluted bleach) ÷ (Volume of diluted bleach) (6.915 × 10-4) ÷ 0.025 = 0.02766 M ClO-
Molarity of the original bleach = [(Average molarity of ClO- in diluted bleach) × (0.100 L)] ÷ (Volume of original bleach) (0.02766 × 0.100) ÷ 0.005 = 0.5532 M ClO-
Mass of NaClO = [(Molarity of the original bleach) × (Volume of original bleach)] × (Molar mass of NaClO) (0.5532 × 0.005) × 74.44 = 0.2059 g NaClO
Mass of commercial bleach = (Density of commercial bleach) × (Volume of original bleach) 1.08 × 5 = 5.4 g
Percent of NaClO in the commercial bleach = [(Mass of NaClO) ÷ (Mass of commercial bleach)] × 100% (0.2059 ÷ 5.4) × 100% = 3.813%
Standard deviation of each trial = ?|(Volume of Na2S2O3 added) – (Average volume added)| ÷ 3 (|14.5 – 13.83|+|14.5 – 13.83|+|12.5 – 13.83|) ÷ 3 = 0.89
mL Percent error = {[(Actual percent of NaClO in the commercial bleach) – (Percent of NaClO in the commercial bleach)] ÷ (Actual percent of NaClO in the commercial bleach)]} × 100% [(5 – 3.81) ÷ 5] × 100 = 23.74%
Analysis: The hypothesis could not be verified true. This is because the results acquired from the experiment have the percent error of 23%, which shows that it is not accurate. Oxidation refers to the losing of electron, resulting in an element having a more positive oxidation number. Reduction, on the other hand, refers to the gaining of electron, resulting in an element having a more negative oxidation number.
