Analysis of a Commercial Bleach Solution Lab Report PDF

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Analysis of a Commercial Bleach Solution Lab Report: Date: 24th May 2017

Name: Caitlin Bettenay

There is not a formal lab report for this lab. Complete the below pages and submit them to your TA before leaving lab. Introduction: Include a brief statement of purpose for this experiment, relevant conceptual background, and include the relevant reactions and information as to how the concentration of sodium hypochlorite is determined. Indicate which species is oxidized and which one is reduced. The purpose of this experiment is to determine the moles of hypochlorite through reactions 1,2 and 3 and then to calculate the hypochlorite concentration. In order to complete this, background information is required. Commercial bleaches are made by bubbling chlorine gas into a sodium hydroxide solution. −¿ Some of the chlorine is oxidized to the hypochlorite ion present in a solution, Cl O¿ , and −¿ some is reduced to the chloride ion, Cl O¿ . Therefore, the solution remains strongly basic. The ionic balanced chemical equation for the process is: −¿(aq)+ H 2 O (l ) ¿ −¿(aq )+C l ¿ −¿(aq)→ Cl O ¿ C l2 (g)+2O H Through conducting a redox titration, the amount of hypochlorite ion present in a solution of bleach will be determined. The following experiment will utilize iodide and thiosulfate ions in −¿ the titrations. Iodide ion, I ¿ can be oxidized by almost any oxidizing agent. The reactions occur that analyze the titration: −¿( aq)+ I 2 (aq)+ H 2 O (l ) ¿ −¿( aq)→ C l ¿ −¿(aq)+2 I ¿ +¿ (aq)+C lO ¿ 2H

(1)

In the above reaction, an acidified iodine solution is added to hypochlorite ion solution and the iodine is oxidized to iodine. −¿(aq) −¿(aq)→ I 3¿ ¿ I 2 (aq)+ I

(2)

Iodine is slightly soluble in water, however it dissolves well in aq solution of iodine ion where it forms triiodide ion which is a complex ion and is a combination of a neutral I 2 molecule and −¿ an I ¿ ion. It is yellow in dilute solution and dark red-brown when it is concentrated. The reaction can be seen in (2) above.

2−¿(aq ) −¿(aq)+ S 4 O 6¿ ¿ 2−¿ (aq)→3 I (3) ¿ aq −¿ ( )+ 2 S2 O3 ¿ I3 In the above reaction, the triiodine is titrated with a standard solution of thiosulfate of thiosulfate ions, that reduce the iodine back to iodine ions. The third reaction has the triiodide ions (originally red-brown) fade to yellow and eventually clear (iodide ion). Addition of starch turns the solution into a deep blue colour which can be reversed. The disappearance of the blue color provides an effective method of determining the endpoint of the titration. This experiment will be conducted through using a redox titration process for three trials and by recording each of the volumes that the solution results in a colour change. Data: In tabular form, include the recorded concentration of the thiosulfate standard used for the titration, the labeled value of the concentration of commercial bleach and any data necessary to calculate the concentration of hypochlorite in commercial bleach. Any table must have a descriptive caption.

Mass of Kl (g): Concentratio

Titration 1: 1.002

Titration 2: 1.007

Titration 3: 1.002

Average: 1.00367

28.57 X (moles/L

28.57 X (moles/L

28.57 X (moles/L

28.57 X (moles/L

−¿ ¿ C lO Undiluted: Thiosulfate Concentratio n (M): Volume (mL): n of

diluted)

diluted)

diluted)

diluted)

0.10

0.10

0.10

0.10

8.89

8.81

8.01

8.57

Results: Report the concentration of hypochlorite in commercial bleach. Show annotated calculations. To find the concentration of hypochlorite: Hypochlorite is converted to chloride in the presence of excess iodide, producing brown iodine molecules. The complete the Redox titration of iodine molecules using aqueous sodium thiosulfate solution. So, the hypochlorite in the bleach solution if first reacted with an iodide ion under acidic conditions as can be seen below: −¿ ( aq )+ H 2 O (l) ¿ −¿ →Cl ¿ −¿ (aq)+2 e Reduction of Chlorine :ClO ¿ −¿ (aq ) → I2 ( aq ) ¿ Oxidation of Iodine :2 I

−¿ (aq )+ H 2 O ( l )+ I 2 ( aq ) ¿ −¿ ( aq )→ Cl ¿ −¿( aq) +2 I ¿ Redox Reaction :ClO Here it can be seen that the reaction goes to completion therefore the I 2 moles produced are equal to the moles of −¿ ¿ hypochlorite ClO that are present in the commercial bleach ¿ solution. In order to determine the number of I 2 moles, use titration with standardized sodium thiosulfate Na2 S2 O 3( aq ) : −¿ ( aq ) ¿ Reductionof Iodine : I 2 (aq ) → 2 I 2−¿( aq) 2−¿ ( aq )→ S4 O 6¿ ¿ Oxidation of Sulfur : S 2 O 3 −¿ ( aq ) ¿ 2−¿ ( aq )+2 I ¿ 2−¿ (aq )→ S 4 O6 ¿ Redox Reaction: I 2 ( aq ) +2 S 2 O3 Moles of Solute=Litres × Molarity Moles of Solute =0.0857 × 0.1 M

Moles of Solute=8.57 ×10−3

( n ) moles of solute ( V ) Volume of solution ( L ) −3 ( c ) Concentration of solution= 8.57 ×10 ( c ) Concentration of solution=0.1 0.0857 ( c ) Concentration of solution=

Report the percent error in concentration. Show annotated calculations. Actual commercial bleach was diluted with 6% hypochlorite.

( 0.06−0.1 ) % error=6.7 % ( 0.06) Further questions: 1. How would each of the following laboratory mistakes affect the calculated value of the percent hypochlorite in commercial bleach (too high, too low, no change)? Explain a. In step 3 the pipet had residual water inside it when used to measure the diluted bleach solution

% error=

If this were the case then the calculated percentage will be less than the expected value as the distilled water in the pipet will dilute the commercial bleach solution to an extent. Thus, the number of moles will also decrease and when divided with the mass of the commercial bleach, smaller percent of NaClO in the bleach is calculated. b. During the second step, 2 grams of KI were used instead of 1 gram If 2g of Kl were added instead of 1g, adding this extra gram will now show change in the calculated percentage of NaClO as Kl is already added in excess in the method, so by adding more it will not effect the percentage calculations. c. After adding starch too much of the thiosulfate solution was added The hypochlorite would increase 2. In the Introduction section, we showed you three equations that are important for the analysis of bleach. Combine the equations to give the net overall reaction between thiosulfate and the active ingredient in bleach.

2−¿ (aq) ¿ 2−¿( aq ) +2 O H ¿ −¿ (aq )+ S 4 O 6 2−¿ ( aq ) + H 2 O ( l )→ Cl ¿ ¿ −¿ ( aq ) +2 S 2 O 3 ¿ ClO

3. The reactions in the titrations are oxidation-reduction reactions. Examine the equations involved. a. What is the oxidation number of sulfur in S2O32-? (Hint: O has the oxidation number of 2.) The oxidation number of Sulfur in S2O32- is 2, as when added with the oxidation of Oxygen (-2), the overall charge has to equal -2 due to the -2 on the S2O32-. Therefore, 3X-2 = -6, +2 X +2 (as there are 2 Sulfurs) = -2. What is the oxidation number of sulfur in the S4O62-? (Hint, it may not be a whole number) Oxidation number of Sulfur in S4O62- is 2.5 as when added with the oxidation of Oxygen (-2), the overall charge has to equal -2 due to the -2 on the S4O62-. Therefore, 6X-2 = -12, +2.5 X 4 (as there are 4 Sulfurs) = -2. b. Is the sulfur in S2O32- gaining or losing electrons? Gaining electrons c. How many electrons are transferred per mole of S2O32-? 2 because it’s a 2:1 ratio. d. Is the S2O32- ion undergoing oxidation or reduction? The Sulfur is undergoing oxidation in this reaction.

There is not a formal lab report for this lab. Complete the above pages using the Microsoft version of this file that is available for download on the lab D2L page. Once the worksheet is complete, submit the worksheet on time and to your TA in the dropbox on D2L....