Analysis of Commercial Bleach Lab

Topics: Titration, Chlorine, Iodine Pages: 6 (1694 words) Published: May 24, 2011
Analysis of Commercial Bleach Lab
I. Purpose
In this experiment, the amount of sodium hypochlorite in a commercial bleach will be determined by reacting it with sodium thiosulfate in the presence of iodide ions and starch. A solution of sodium thiosulfate of known concentration will be added to the bleach using a buret in a titration procedure. The disappearance of the dark blue color of the starch-iodine complex will signal the end point. II. Procedures

Pre-Lab Questions
1. What is meant by a “titration”?
Titration is a technique where a solution of known (acid or base) concentration is used to determine the concentration of an unknown (acid or base) solution.

2. A solution of household vinegar (a mixture of acetic acid and water) is to be analyzed. A pipet is used to measure out 10.0 mL of the vinegar, which is placed in a 250-mL volumetric flask. Distilled water is added until the total volume of solution is 250 mL. A 25.0-mL portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide.

The neutralization reaction is as follows:
HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l)
It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar. Calculate the molarity of the diluted vinegar.
Moles = Molarity x Volume
(.0500 M NaOH)(16.7 mL) = ( x M diluted vinegar)( 25.0 mL)
x M diluted vinegar = .0334 M diluted vinegar

3. Calculate the molarity of the household vinegar.
Molarity = Moles/Volume
Moles = Molarity x Volume
(.0334 M diluted vinegar)(250 mL of diluted vinegar) = ( x M household vinegar)( 10.0 mL of Household vinegar) x M household vinegar = (.0334 M diluted vinegar x 250 mL of diluted vinegar)/ 10.0 mL of Household Vinegar x M household vinegar = .835 M

4. The household vinegar has a density of 1.05g/mL. Calculate the percent by mass of acetic acid in the household vinegar. (.835 mol/1 L) x (1L/1000mL) x (1 mL/1.05g) x (59.05g C2H3O2/1 mol C2H3O2) x 100% = 4.70% Materials – Chemicals

* Commercial Bleach, NaClO
* Hydrochloric Acid, HCl, 3M, 6 mL
* Potassium Iodide, KI, 6 g
* Sodium Thiosulfate Solution, Na2S2O3, 0.100 M, 70 mL
* Starch Solution, 2%, 3 mL
Materials – Equipment
* Balance, 0.1-g precision
* Buret
* Buret Clamp
* Erlenmeyer Flask
* Pipet Bulb
* Ring Stand
* Transfer Pipets
* Volumetric Flask with Stopper
* Wash Bottle
* Water, distilled
* Weighing Dish
* Measure 5.00 mL of commercial bleach into a 100-mL volumetric flask. * Dilute the bleach with distilled or deionized water to the mark and mix well. * Weigh out 2 g of solid KI. Pipet 25.0 mL of the dilute bleach into an Erlenmeyer flask. * Add the solid KI along with 25 mL of distilled or deionized water. * Add 2 mL of 3 M HCl while stirring the solution.

* Fill the buret with 0.100 M sodium thiosulfate solution until the iodine color fades to light yellow. * Add one dropperful of starch solution.
* Continue the titration until one final drop of Na2S2O3 causes the color to disappear. * Record the final buret reading.
* Repeat the titration procedure two more times to receive more accurate results. III. Data Collection
Titration of Iodine Solution
| Trial 1| Trial 2| Trial 3|
Initial Buret Reading| 0| 19| 0|
Final Buret Reading| 19| 36| 19|
Volume Na2S2O3| 19| 17| 19|

IV. Data Analysis
1. Determine the number of moles of sodium thiosulfate that are equivalent to one mole of sodium hypochlorite. 1 mol ClO- x (1 mol I2/1 mol ClO-) x (1 mol I3-/1 mol I2) x (2 mol S2O32-/1 mol I3-) = 2 mol S2O32-. Therefore, 1 mol ClO- = 2 mol S2O32- 2. Calculate the average volume of Na2S2O3 needed for the titration of 25.0 mL of diluted bleach. 19 mL + 17 mL + 19 mL = 55/3 = 18.33

3. Use the average volume and molarity of Na2S2O3 to determine the molarity of...
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