Experiment Title : Oxidimetry and Its Application Date of Experiment : Friday November 28th, 2014 Purpose : 1. Making and determining (standardization) Na2S2O3 solution 2. Determining the level of Cl2 in bleaching Basic Theory A. Redox Titration Titration, also known as titrimetry, is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Since volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand to determine concentration. The volume of titrant reacted is called titre. Redox Titration Redox titrations are based on a reduction-oxidation reaction between an oxidi-zing agent and a reducing agent. A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, as when one of the constituents is the oxidizing agent potassium dichromate. The color change of the solution from orange to green is not definite, therefore an indicator such as sodium diphenylamine is used. Analysis of wines for sulfur dioxide requires iodine as an oxidizing agent. In this case, starch is used as an indicator; a blue starch-iodine complex is formed in the presence of excess iodine, signalling the endpoint. Some redox titrations do not require an indicator, due to the intense color of the constituents. For instance, in permanganometry a slight persisting pink color signals the endpoint of the titration because of the color of the excess oxidizing agent potassium permanganate. Even more distinct is the endpoint in iodometry, which goes from a deep brown to colourless. B. Iodometry Iodometry, also known as iodometric titration, is a method of volumetric chemical analysis, a redox titration where the appearance or disappearance of elemen-tary iodine indicates the end point. Note that iodometry involves indirect titration of iodine liberated by reaction with the analyte, whereas iodimetry involves direct titration using iodine as the titrant. Iodine can be used as an oxidizing agent in many oxidation-reduction titrations and iodide can be used as a reducing agent in other oxidation-reduction titrations: I2 + 2e- = 2 I- If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to quantitatively reduce a chemical species while simultaneously forming iodine, and if the iodine is subsequently titrated with thiosulfate, the technique is iodometry. Iodometry is an example of an indirect determination since a product of a preliminary reaction is titrated. The use of iodine as a titrant suffers from two major disadvantages. First, iodine is not particularly soluble in water, and second, iodine is somewhat volatile. Consequently, there is an escape of significant amounts of dissolved iodine from the solution. Both of these disadvantages are overcome by adding iodide (I-) to iodine (I2) solutions. In the presence of iodide, iodine reacts to form triiodide (I3-) which is highly soluble and not volatile. I2 + I- = I3- The major chemical species present in these solutions is triiodide. The reduction of triiodide to iodide is analogous to the reduction of iodine. I3- + 2e- = 3I- Dilute triiodide solutions are yellow, more concentrated solutions are brown, and even more concentrated solutions are violet. Iodide solutions are colorless. In iodimetric titrations, free iodine is used. Since it is difficult to prepare the solution of iodine (iodine sublimates and is less soluble in water) it is dissolved in KI solution. KI+I2 → KI3 This solution is first standardized before use. With the standard solution of I2 , substances such as sulfite, thiosulphate, arsenite are estimated. Iodometry is commonly used to analyse the concentration of oxidizing agent in water samples, such as oxygen saturation in ecological studies or active chlorine in swimming pool water analysis. To a known volume of sample, an excess but known amount of iodine is added, which the oxidizing agents oxidizes iodide to iodine. Iodine dissolves in the iodide-containing solution to give triiodide ions, which have a dark brown color. The triiodide ion solution is then titrated against standard thiosulphate solution to give iodide again using starch indicator: I3- + 2 e- ⇌ 3 I- (Eo = + 0.5355 V) Together with reduction potential of thiosulfate: S4O62- + 2 e- ⇌ 2 S2O32- (Eo = + 0.08 V) The overall reaction is thus: I3- + 2 S2O32- → S4O62- + 3 I- (Eo = + 0.4555 V) KIO3+KI20%+HCl⟶ brown solution Solution after titrated by Na2S2O3⟶ yellow solution Solution after addition of starch indicator⟶ blackish blue After titration ⟶ colourless solution For simplicity, the equations will usually be written in terms of aqueous molecular iodine rather than the triiodide ion, as the iodide ion did not participate in the reaction in terms of mole ratio analysis.The disappearance of deep blue color due to the decomposition of the iodine-starch clathrate marks the end point. At the end point the blue or violet color disappears when iodine is completely changed to iodide. There are two reasons why the indicator is not added at the beginning of the titration when the iodine concentration is high. First, a diffuse endpoint would result from the slow dissociation of the starch-iodine complex if a large amount of iodine were absorbed in the starch. Second, iodometric titrations are carried out in strongly acid media, a situation that promotes the reaction between oxidizing agents and iodide. Unfortunately starch has a tendency to hydrolyze (decompose) in acidic media, destroying its indicator qualities. The reducing agent used does not necessarily need to be thiosulfate,stannous chloride, sulphites, sulfides, arsenic (III) and antimony (III) are commonly used alternatives. At higher pH (> 8) At low pH would also react with the thiosulfate S2O32− + 2 H+ → SO2 + S + H2O C. Bleaching Bleach refers to a number of chemicals which remove colour, whiten or disin-fect, often by oxidation. The bleaching process has been known for millennia, but the chemicals current-ly used for bleaching resulted from the work of several 18th century scientists. Chlorine is the basis for the most commonly used bleaches, for example, the solution of sodium hypochlorite, which is so ubiquitous that most simply call it "bleach", and calcium hypochlorite, the major compound in "bleaching powder". Oxidizing bleaching agents that do not contain chlorine most often are based on peroxides, such as hydrogen perox-ide, sodium percarbonate and sodium perborate. While most bleaches are oxidizing agents, some are reducing agents such as sodium dithionite and sodium borohydride. Bleaches are used as household chemicals to whiten clothes and remove stains and as disinfectants, primarily in the bathroom and kitchen. Many bleaches have strong bactericidal properties, and are used for disinfecting and sterilizing and thus are used in swimming pool sanitation to control bacteria, viruses and algae and in any institution where sterile conditions are needed. They are also used in many industrial processes, notably in the bleaching of wood pulp. Bleach is also used for removing mildew, killing weeds and increasing the longevity of flowers The level of Sodium Hypochlorite (NaClO), the active ingredient in Household Bleach, is determined iodometrically by reacting it with an excess of Iodide and then titrating the Iodine produced with standard sodium thiosulfate. ClO-(aq) + 3 I-(aq) + 2 H+(aq) → Cl-(aq) + I3-(aq) + H2O I3-(aq) + 2S2O32−(aq) → 3 I-(aq) + S4O62-(aq) Note the Hypochlorite is reduced to Chloride, Cl+ to Cl-, in this process. Bleach, as the name implies, is a substance that will whiten.“Chlorine” bleaches contain Sodium Hypochlorite as the oxidizing agent that causes whitening. Commercial chlorine bleach will be at least 5.25% Sodium Hypochlorite. Other bleaches contain Calcium Hypochlorite (Bleaching Powder)or Peroxides. In chlorine bleaches, the hypochlorite is in equilibrium with Chlorine: HClO + Cl- + H+ → Cl2 + H2O An acidic environment favors the production of Chlorine. Tools and Materials A. Tools B. Materials Ipi vitamin bottle Coloured bottle Measuring flask 100 mL Burette Erlenmeyer 250 mL Pipettes Volumetric pipette Pikno mass Measuring glass Distilled water Procline bleaching Na2S2O3 KIO3 KI 20% HCl 4N Starch solution KI H2SO4 1:6 Ammonium Molybdate Flow Chart A. Making Sodium Thiosulphate ±0,1 N Solution ±25 g of Na2S2O3.5H2O Na2S2O3 ±0,1 N Solution Dilute with 1 liter of water which have been boiled and cooled Added 0,2 g of Na2CO3 as the preservative Keep in the coloured bottle B. Standardization Na2S2O3 ±0,1 N Solution with KIO3 Standard ± 0,3577 g of KIO3 Weighed Moved to volumetric flask 100 mL Dilute with distilled water until boundary line Shake well KIO3 ± 0,1 N Solution Na2S2O3 Solution KIO3 ± 0,1 N Solution Entered to Erlenmeyer 250 mL Added 2 ml of KI 20% solution and 2,5 ml of HCl 4N Entered to burette Titrated Colour changes to yellow Added 7 drops of starch solution Titrated Blue colour disappear Read and write down the volume of Na2S2O3 Repeated 3 times Average concentration of Na2S2O3 Repeated 3 times C. Determination the Level of Cl2 in Bleaching Na2S2O3 Solution 2 mL of Sample Write down the brand Entered to Erlenmeyer Added 75 ml of distilled water Added 3 g of KI Added 8 ml of H2SO4 1:6 Added 3 drops of ammonium molybdate 3% Entered to burette Titrated Colour changes to yellow Added 7 drops of starch solution Titrated Blue colour disappear Read and write down the volume of Na2S2O3 Repeated 3 times Average concentration of Na2S2O3 Repeated 3 times   Result Of The Experiment Experiment Procedure Result of The Experiment Assumption/Reaction Conclusion Before After A. Making Sodium Thiosulphate ±0,1 N Solution ±25 g of Na2S2O3.5H2O Na2S2O3 ±0,1 N Solution Dilute with 1 liter of water which have been boiled and cooled Added 0,2 g of Na2CO3 as the preservative Keep in the coloured bottle Na2S2O3.5H2O+ distilled water⟶ colorless solution +Na2CO3→ colorless solution From the experiment and calculation, the average concentration of Na2S2O3 is 0,12965 N B. Standardization Na2S2O3 ±0,1 N Solution with KIO3 Standard ± 0,3577 g of KIO3 Weighed Moved to volumetric flask 100 mL Dilute with distilled water until boundary line Shake well KIO3 ± 0,1 N Solution C. Determination the Level of Cl2 n Bleaching Na2S2O3 Solution KIO3 ± 0,1 N Solution Entered to Erlenmeyer 250 mL Added 2 ml of KI 20% solution and 2,5 ml of HCl 4N Entered to burette Titrated Colour changes to yellow Added 7 drops of starch solution Titrated Blue colour disappear Read and write down the volume of Na2S2O3 Repeated 3 times Average concentration of Na2S2O3 Repeated 3 times KIO3: white powder Distilled water: colorless solution Na2S2O3: colorless solution KIO3: colorless solution KI20%: colorless solution HCl 4N: colorless solution Starch solution: colorless solution KIO3+ Distilled water→ colorless solution Na2S2O3+ KIO3→ colorless solution + HCl 4N⟶ brown solution and precipitate After titration: yellow solution +7 drops of starch ⟶blackish blue After titration: colorless solution Volume of Na2S2O3 V1=19,4 ml V2=19,2 ml V3=19,3 ml 2I- + 2e x5 10e + 2IO3- +12H+ I2 + 6H2Ox1 10I- + 2IO3- + 12H+ 6I2 + 6H2O :2 5I- + IO3- + 6H+ 3I2 + 3H2O I3- + 2 e- ⇌ 2 I- 2 S2O32- ⇌ S4O62- + 2 e- I3- + 2 S2O32- → S4O62- + 2 I- Na2S2O3 Solution 2 mL of Sample Write down the brand Entered to Erlenmeyer Added 75 ml of distilled water Added 3 g of KI Added 8 ml of H2SO4 1:6 Added 3 drops of ammonium molybdate 3% Entered to burette Titrated Colour changes to yellow Added 7 drops of starch solution Titrated Blue colour disappear Read and write down the volume of Na2S2O3 Repeated 3 times Average concentration of Na2S2O3 Repeated 3 times Bleach brand : procline Bleach: colorless solution Distilled water: colorless solution KI: White crystal H2SO4: colorless solution Ammonium molybdate: colorless solution Starch solution: colorless solution Na2S2O3: colorless solution Density: 1,11033 Sample+ Distilled water⟶ colorless solution +3g of KI⟶ colorless solution +8 ml of H2SO4 ⟶Yellow solution +ammonium molybdate ⟶ Yellow solution +5 ml of starch solution ⟶ blackish blue solution After titration: colorless solution Volume of Na2S2O3 V1=0,5 ml V2=0,5 ml V3=0,5 ml ClO-(aq) + 3 I-(aq) + 2 H+(aq) → Cl-(aq) + I3-(aq) + H2O I3-(aq) + 2S2O32−(aq) → 3 I-(aq) + S4O62-(aq) From the experiment and calculation, the level of Cl2 in Procline bleaching is 0,31072%  Analysis and Explanation A. Standardization Na2S2O3 ±0,1 N Solution with KIO3 Standard In the analysis of standarization Na2S2O3 with KIO3 as standard we get the average concentration of Na2S2O3 is 0,12965N. Thus analysis start with making a KI standard solution from weighing 0,357gram of KIO3 then pour into volumetric flask 100ml, after that dilute it with distilated water until the boundary line. The reaction is : IO3- + 6H+ + 5I- → 3I2 + 3H2O Pippete 25ml of standard solution into erlenmeyer flask and added 2ml of KI 20% and 2,5ml HCl 4N. The function adding of KI20% in this experiment is because KI oxidized iodide quantitatively to iodine in acidic solution where I2 bond by KI. It is produce lower vapour pressure than pure water. The higher the level of KI the higher I2 solubility and the function of adding HCl 4N is to makes the solution in the acidic ambience and also as a catalyst. The color is brown and form black precipitate, it is appropriate with the theory and then titrated with Na2S2O3 solution and the color become yellow which appropriate with theory where after KIO3 is titrated with Na2S2O3 the colour changes to yellow, it’s indicate the existence of iod which have been released. The reaction is: I2 + 2 S2O32− → S4O62- + 2 I- (left) based on the theory; (right) our experiment After the solution is titrated by Na2S2O3 the solution changes colour to yellow (left) based on the theory; (right) our experiment The addition of KI 20% and HCl in KIO3 solution makes the solution turn to brown Added starch solution indicator and the color changing into blackist blue but in the theory when we aded starch solution it will changed to blue or violet. and then titrated it, the solution become colorless. At the end point the blue or violet color disappears when iodine is completely changed to iodide. There are two reasons why the indicator is not added at the beginning of the titration when the iodine concentration is high. First, a diffuse endpoint would result from the slow dissociation of the starch-iodine complex if a large amount of iodine were absorbed in the starch. Second, iodometric titrations are carried out in strongly acid media, a situation that promotes the reaction between oxidizing agents and iodide. Unfortunately starch has a tendency to hydrolyze (decompose) in acidic media, destroying its indicator qualities. (left) based on the theory; (right) our experiment The addition of starch indicator makes the solution turn to blackish blue (left) based on the theory; (right) our experiment After the solution is titrated again the solution changes colour to colourless From the experiment we get the volume of Na2S2O3 that use for titration are 19,4ml ; 19,2ml ; and 19,3ml and we calculate the concentration of Na2S2O3 using equality : Molek S2O32− = molek IO3- And from the calculation we get the concentration of Na2S2O3 are 0,12898N ; 0,13033N and 0,12965N. So, the average concentration of Na2S2O3 is 0,12965N B. Determination the Level of Cl2 in Bleaching Here we use procline bleaching as the sample. In the analysis of determining the level of Cl- in bleaching we get pikno mass is 1,11033 and the average level of Cl- in proclean is 0,31072%. Thus analysis start with pippete 2ml of proclean solution and pour into erlenmeyer 100ml, then added 75ml of distilated water, the addition of water is used to make the solution not too concentrate. After that, added 3gram of KI and the color is colorless and its doestn match with the theory that has yellow solution. and then added 8ml of H2SO4 1:6 the color is yellow it doesnt match with the theory which should be produce colorless solution. It might because of KI that we use had been oxidized because we let it in the opened air too long, the extra iodine will be formed by the former reaction. Nitrite should not exist, because based on R.A Day, A.L. Underwood. 2002 if we shouldn’t let the solution too long in opened air because it will be reduced by iodide ions to nitrogen (II) oxide which is subsequently oxidized back to nitrite by oxygen from the air: 2HNO2 + 2H + + 2I- → 2NO + I2 + 2H2O + O2 + 2H2O 4NO → 4HNO2 (left) based on the theory; (right) our experiment The addition of KI in the sample should turn the solution to brown but in the experiment the colour still yellow We make H2SO4 1:6 solution by 1ml H2SO4 2N added with 6ml of distilated water. The function of adding KI is oxidized the iodide quantitatively to iodine in acidic solution and the iodate reaction works fast enough and just need a bit of hydrogen ion excess to complete the reaction. And the function of adding H2SO4 2N is to give acidic ambience. Then, added ammonium molybdate 3% solution for a catalyst. The color is yellow and it’s not appropriate with the theory because the level of Cl2 in the sample is too small it makes the addition of starch indicator is done before titration. The color become blackish blue and its match with the theory, the color changing into blackist blue showing that in the solution there are I2 and then titrated again until the color become colorless, it shows that I2 disappear and changes with I-. (left) based on the theory; (right) our experiment The addition of KI and H2SO4 1:6 in the sample should turn the solution to brown but in our experiment the colour is yellow (left) based on the theory; (right) our experiment The addition of starch indicator makes the solution turn to blackish blue (left) based on the theory; (right) our experiment After the solution is titrated again the solution changes colour to colourless From the experiment we get the volume of Na2S2O3 that use for titration are the same that is 0,5ml and we calculate the concentration of Na2S2O3 using equation : Molek S2O32− = molek Cl2 And from the calculation we get gram of Cl2 is 0,0069gram. And from the calculation we get the level of Cl2 in proclean bleaching is 0,31072% Discussion The first experiment is based on the theory. But, in the second experiment is not suitable with the theory, it happened because we let KI in opened air too long and the level of Cl2 in the sample is too small which can be proven by the result of calculation of Cl2 level on the procline bleaching is 0,31072%. It makes the comparison of the titration result between first experiment and second experiment is too different. This makes the result is too risky. To deal with this situation we can add more sample as the analyte, example: 2 ml of sample changes to 10 ml of sample Conclusion From the experiment, we can conclude that : The concentration of Na2S2O3 standard solution is 0,12965 N The level of Cl2 in Procline bleaching is 0,31072% Question and Answer Write the reaction that occurs in the titration Permanganometry, if the reductor is ferrous ion! Each mole of ferrous ions is equal to how many equivalent? Answer: Fe2+ Fe3+ + e (reduktor) MnO4- + 8H+ + 5e Mn2+ + 4H2O (oksidator) 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O mol Fe2+ 1 ekivalen MnO4- Why Permanganometry titration does not need addition of another indicator? Answer: Because the system potential Fe2 + Fe3 + in the titration vessel can be measured at any point to make these vessels into half galvanic cell. Another cell is half the standard electrode. A slow platinum electrode used in the titration vessel as an indicator electrode. What is the difference iodometric titration and iodimetri? Answer:  Iodometric: - I- as a reductant - Reaction: substance + I2 + I- .... - I2 formed reacted with S2O32- - S2O32- standardized with K2Cr2O7 / KIO3 Iodimetri: - I2 as the oxidant - Reaction: substances I- + I2 + .... - Starch indicator.        Kanji + I2 purplish blue        Kanji + I- colorless - S2O32- standardized with HAsO2 How is the reaction between potassium iodidat + potassium iodide + hydrochloric acid? Every 1 mole of potassium iodate is equal to how many equivalent?      Answer: 2I- I2 + 2e x 5 10e + 2IO3- +12H+ I2 + 6H2O x 1 10I- + 2IO3- + 12H+ 6I2 + 6H2O setiap 1 mol KIO3 = 6 ekivalen APPLICATIONS Iodo-Iodometric Titration 1. Describe some of the shortcomings of starch is used as an indicator!      Answer: is non-soluble in cold water instability suspension in water with iodine to give a complex insoluble in water, so that the starch should not be added too early in the titration. (Hence, the iodine titration, starch solution should not be added until just before the end point, when the color starts to fade) sometimes there is a final point that 'float', which is striking when the aqueous solution. Why is the iodometric titration starch indicator is added at the approach of the equivalence point? Answer: Because if it is added at the beginning, iodine gives a complex insoluble in water Why is the addition of Na2S2O3 solution using distilled water boiling? Answer: To prevent excessive carbon dioxide content; This can cause slow decomposition accompanied by the formation of sulfur:                                  S2O3- + H + → HSO3- + S     Moreover, the decomposition can also be caused by the action of bacteria (eg, thiobacilus thioparus), especially if the solution was allowed to stand for some time. Referencess Day, R.A, Jr & Underwood, A.L. 2002. Analisis Kimia Kuantitatif: Edisi keenam.. Translated Iis Sopyan. Jakarta: Erlangga Noname.2013.Bleaching, (online),(http://www.en.wikipedia.org/wiki/bleaching, accessed on December 7th, 2014) Noname.2013.Iodometry, (online),(http://www.en.wikipedia.org/wiki/iodometry, accessed on December 7th, 2014) Noname.2013.Redox Titration, (online),(http://www.en.wikipedia.org/wiki/redox_titration, accessed on December 7th, 2014) Svehla, G.(1985). Vogel: Buku Teks Analisis Oraganik Kualitatif Makro dan Semimikro.(first edition). Translated Setiono, L and Handayana, P.A Jakarta: Kaliman Media Pusaka. Tim Kimia Dasar. 2014. Panduan Praktikum Kimia Analitik I Dasar-Dasar Kimia Analitik. Surabaya: Jurusan Kimia FMIPA Unesa ATTACHMENT I A. Standardization Na2S2O3 ±0,1 N Solution with KIO3 Standard ±0,357 g of KIO3 ±0,357 g of KIO3 entered to volumetric flask Shaking mixture of KIO3 + distilled water Colorless solution ±0,357 g of KIO3 + distilled water Colorless solution  Starch Solution Colorless Titration with Na2S2O3 KIO3+ 2 ml KI 20%+2,5 ml HCl 4 N Brown solution and precipitate Addition of 2 ml KI 20% Addition of 2,5 ml HCl 4 N  Third Try After titration process Yellow Solution After addition of starch Blackish blue Solution After titration V1=19,3 ml Colorless solution After titration V1=19,2 ml Colorless solution After addition of starch Blackish blue Solution After titration process Yellow Solution Second Try First Try After titration process Yellow Solution After addition of starch Blackish blue Solution After titration V1=19,4 ml Colorless solution  B. Determination the Level of Cl2 in Bleaching 2 ml Sample+ 75 ml distilled water+ 3g KI+8 ml H2SO4 1:6 ⇒ yellow solution Entering H2SO4 1:6 to Erlenmeyer Entering KI to Erlenmeyer 3g of KI White crystal ρ=1,11033 colorless solution 2 ml Sample+ 75 ml distilled water (colorless solution) Weighing the sample density using piknometer  After titrated with Na2S2O2; V1=V2=V3=0,5 mL Colorless solution After titrated with Na2S2O2 Colorless solution Addition of starch solution Blackish blue solution Addition of 3 drops of ammonium molybdate 3% Yellow solution  ATTACHMENT II 2I- I2 + 2e x 5 10I- 5I2 + 10e 10e + 2IO3- +12H+ I2 + 6H2O x 1 10e + 2IO3- +12H+ I2 + 6H2O 10I- + 2IO3- + 12H+ 6I2 + 6H2O I3- + 2 e- ⇌ 2 I- 2 S2O32- ⇌ S4O62- + 2 e- I3- + 2 S2O32- → S4O62- + 2 I- Mass of Na2S2O3 = 0,357 g Mr of Na2S2O3 = 158,99 Mr of KIO3 = 214 V Na2S2O3 in standardization V Na2S2O3 in application V1 = 19,4 ml V1 = 0,5 ml V2 = 19,2 ml V2 = 0,5 ml V3 = 19,3 ml V3 = 0,5 ml V=V1=V2=V3=0,5 ml Determination of concentration of Na2S2O3 First try Molek Na2S2O3 = Molek KIO3 N1V1 = N2 V2 N1 x 19,4 = N1 x 19,4 = N1 x 19,4 = 0,10009 x 25 N1 = N1 = N1 = 0,12898 N Second Try Molek Na2S2O3 = Molek KIO3 N1V1 = N2 V2 N1 x 19,2 = N1 x 19,2 = N1 x 19,2 = 0,10009 x 25 N1 = N1 = N1 = 0,13033 N Third Try Molek Na2S2O3 = Molek KIO3 N1V1 = N2 V2 N1 x 19,3 = N1 x 19,3 = N1 x 19,3 = 0,10009 x 25 N1 = N1 = N1 = 0,12965 N Average concentration of Na2S2O3 =0,12898 N+0,13033 N+0,12965 N 3 =0,12965 N Determination the level of Cl2 in bleaching V=V1=V2=V3=0,5 ml ρ=1,11033 Molek Na2S2O3 = Molek Cl2 N1V1 = 0,12965 x 0,5x10-3= g = 0,12965 x 0,5x10-3x71 2 g =0,0069 g Bleaching mass = ρ x v =1,11033 x 2 =2,22066 g The level of Cl2 = g Cl2 x 100% g bleaching = 0,0069 g x 100% 2,22066 g = 0.31072 % Group IV | Oxidimetry and Its Application 8