Hong Kong Chemistry Olympiad for Secondary Schools (2013-2014) TEST THE TEST SHAU KEI WAN GOVERNMENT SECONDARY SCHOOL 筲箕灣官立中學 Group Members: LAM YAT LONG, ALEX 林日朗 LEUNG TSUM TONG, THOMAS 梁雋堂 WONG TING HEI, SHERMAN 黃庭熙 YIK KAI HEI, ANDREW 易啟希 Teacher Adviser: Mr Lau Chun-On p.1 Presentation material: photoalbum.ppt in CD
Abstract: SHAU KEI WAN GOVERNMENT SECONDARY SCHOOL LAM YAT LONG, ALEX LEUNG TSUM TONG, THOMAS WONG TING HEI, SHERMAN YIK KAI HEI, ANDREW To determine the concentration of sodium hypochlorite, iodometric method is used currently. However, is it the only method? The goal of this study is to determine whether two new-designed tests can be possible substitutes of the traditional test. In the traditional method, chlorine is collected through iodide solution. The resultant solution is then titrated against sodium thiosulphate. The concentration of sodium hypochlorite can be found quite accurately. In our studies, this tradition test will be used as a reference to compare the feasibilities of the proposed tests. In the traditional methods, multiple steps involve and it leads to errors. So, our first proposed test is a gravimetric test which only based on one reaction, the redox reaction between hydrogen peroxide and sodium hypochlorite. Our 1 st test can achieve a test result close to that from the traditional method. Our second proposed test is a redox back titration which requires the addition of iron(ii) sulphate and followed by the titration with potassium dichromate. This proposed test has similar numbers of steps to the original one with the use of different reagents. According to the experimental result, this test has a larger % difference than our 1 st proposed test. p.2
Contents Introduction P. 4 Principles and theories P. 5-8 Experiments P. 9-22 Data Analysis P. 23-33 Discussion P. 34-37 Further study P. 38-39 Conclusion P. 40 Reference P. 41 Acknowledgement P. 42 p.3
Introduction: Normally, to determine the sodium hypochlorite (NaClO) content in bleach, chemists test for the amount of chlorine in the hypochlorite compound with acid, potassium iodide (KI) and sodium thiosulphate(na 2 S 2 O 3 ) However, the test is actually quite complicated. So, is there any simpler yet accurate quantitative test for sodium hypochlorite? Our experiment is designed to investigate whether a gravimetric analytical test(with hydrogen peroxide),and a redox test (with iron(ii) sulphate, potassium dichromate) can test the concentration of sodium hypochlorite accurately and replace the traditional test. In the experiment, the traditional test will be carried out, which the result will act as a benchmark for comparison of the accuracy of the two new designed tests. Objectives: 1. Determine the sodium hypochlorite concentration in bleach using the traditional test. 2. Test for the feasibilities of using the two new-designed tests, i.e. the gravimetric analytical test with hydrogen peroxide and the redox test (with iron(ii) sulphate, potassium dichromate), to determine the sodium hypochlorite concentration in bleach. p.4
Principles and theories: (1) Principles and theories of the traditional test Determine the sodium hypochlorite concentration with sulphuric acid, potassium iodide, sodium thiosulphate and starch The amount of sodium hypochlorite(naclo) ingredient available can be determined by allowing a certain quantity of it to react with excess acidified potassium iodide(ki) solution, the liberated iodine(the amount is determined by the concentration of sodium hypochlorite), is titrated against the standardized sodium thiosulphate(na 2 S 2 O 3 ) solution. When hypochlorite ions mix with excess acidified potassium iodide solution, aqueous iodine is produced: OCl - + 2I - + 2H + I 2 + H 2 O + Cl -. Aqueous iodine reacts with thiosulphate ions to form iodide ions: I 2 + 2 S 2 O 3 2- S 4 O 6 2- + 2 I -, which the amount of iodide ions can help finding the amount of hypochlorite ions. Starch is used as indicator for the titration and is added when the solution in the conical flask turns pale yellow. When iodine is present, the starch-iodine complex turns the solution dark blue. But the addition of excess sodium thiosulphate solution will decolorize the solution, as the thiosulphate ions reduce all the iodine back into iodide ions. So, the first drop of sodium thiosulphate solution causing decolorization marks the end point of the titration. Since sodium thiosulphate cannot be a primary standard, standardization is needed to be performing simultaneously with this traditional test. It can be standardized with known amount of iodine(i 2 ), which is produced by the redox reaction of known amount of potassium iodate(kio 3 ) and excess acidified potassium iodide solution. When potassium iodate is mixed with acidified potassium iodide, iodine and water is produced: 6H + + IO 3 - + 5 I - 3I 2 + 3H 2 O. Known amount of potassium iodate will be added as the limited reagent, while the KI solution and dilute sulphuric acid added will be in excess amount. So known amount of iodine is produced, and is titrated against sodium thiosulphate solution, with starch solution as the indicator. Every molecule of aqueous iodine reacts with two thiosulphate ions to form iodide ions: I 2 + 2 S 2 O 3 2- S 4 O 6 2- + 2 I -. When iodine is present, the starch-iodine complex turns the solution dark blue. But p.5
the addition of sodium thiosulphate solution will decolourize the solution, as the thiosulphate ions reduce the iodine back into iodide ions. The decolorization of the solution marks the end point of the titration. Therefore, knowing the amount of iodine presence and the volume of sodium thiosulphate solution added, the molarity of the sodium thiosulphate solution can be calculated. (2) Principles and theories of our proposed tests TEST 1 Gravimetric analytical method Determine the sodium hypochlorite concentration with hydrogen peroxide Sodium hypochlorite is a common oxidizing agent which tends to accept two electrons in alkaline medium to produce sodium chloride. The half equation goes like this: NaClO + H 2 O + 2e - 2OH - + NaCl (E =+0.90). When hydrogen peroxide(h 2 O 2 ) is mixed with sodium hypochlorite, redox reaction occurs. Hydrogen peroxide acts as the reducing agent, and is oxidized to release oxygen gas. The half equation goes like this: H 2 O 2 O 2 + 2H + + 2e - (E =-0.70). The overall equation is: NaClO(aq) + H 2 O 2 (aq) H 2 O(l) + NaCl(aq) + O 2 (g) As a result, when excess hydrogen peroxide is added to bleach sample (sodium hypochlorite solution), oxygen gas is released, that is, the mass of the resultant solution decreases. Therefore, by calculating the weight difference of the solution, the amount of oxygen gas released can be found, so does the amount of sodium hypochlorite in the solution. TEST 2 Redox back titration Determine the sodium hypochlorite concentration with iron(ii) sulphate, acidified potassium dichromate and diphenylamine Sodium hypochlorite is a common oxidizing agent which tends to accept two electrons in alkaline medium to produce sodium chloride. The half equation goes like this: NaClO + H 2 O + 2e - 2OH - + NaCl (E =+0.90). While iron(ii) ion (Fe 2+ ), in this case, acts as a reducing agent, which lose electrons to form iron(iii) ions(fe 3+ ). The half equation goes like this: p.6
Fe 2+ Fe 3+ + e - (E =-0.77). When sodium hypochlorite is mixed with iron(ii) sulphate(feso 4) at an alkaline medium, redox reaction occurs with a standard voltage of +0.13V, which iron(ii) ions will be oxidized to iron(iii) ions, and sodium hypochlorite will be reduced to sodium chloride. The overall chemical equation: NaClO(aq) + H 2 O(l) + 2FeSO 4 (aq) NaCl(aq) + 2Fe 3+ (aq/s) # + 2OH - (aq/s)* + 2SO 4 2- (aq) # The state is either aqueous or solid due to formation of both iron(iii) chloride and iron(iii) hydroxide. * The state is either aqueous or solid due to formation of both sodium hydroxide and iron(iii) hydroxide. As a result, when excess iron(ii) sulphate of known concentration is added to bleach sample (sodium hypochlorite solution), iron(iii) ions are produced. If the amount of iron(iii) ions is found, the amount of sodium hypochlorite in the solution can be known. Since excess known concentration of iron(ii) sulphate is added, finding out the iron(ii) ions remained can let us calculate the amount of iron(iii) ions. Therefore, the next step is to find out the amount of iron(ii) ions by back titration with acidified potassium dichromate. Potassium dichromate(k 2 Cr 2 O 7 ) is a common oxidizing agent in the school laboratory, which tends to accept six electrons in acidic medium to produce chromium(iii) ions(cr 3+). The half equation goes like this: 14H + + 6e - + Cr 2 O 7 2-2Cr 3+ + 7H 2 O (E =+1.33). While iron(ii) ion is a reducing agent, which lose electron to form iron(iii) ion. The half equation goes like this: Fe 2+ Fe 3+ + e - (E =-0.77). It will be oxidized to iron(iii) ions when it is titrated against acidified potassium dichromate solution, with a standard voltage of +0.56V. Potassium dichromate is orange in color, but when it is added to the solution containing iron(ii) ions, it changes into green due to the presence of chromium(iii) ion, produced in the reduction of dichromate ions. The chemical equation is: 14H + (aq) + Cr 2 O 7 2- (aq) + 6Fe 2+ (aq) 2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O(l). p.7
However, the colour change from orange to green of dichromate is not sharp for end point detection. Diphenylamine is added as a redox indicator, with the chemical formula: (C 6 H 5 ) 2 NH, which turns purple when being oxidized. The equation is 2(C 6 H 5 ) 2 24 H 20 N 2 + 2H + + 2e - : 2 +2H + + 2e - It is used as redox indicator in this back titration. It will become oxidized (turns purple) immediately by acidified potassium dichromate when acidified potassium dichromate solution is added during the titration. But soon become colorless again due the reduction by iron(ii) ions. When the first drop of excess acidified potassium dichromate is added, intense color is produced due to oxidation of diphenylamine, and this signals the end point of the titration. Therefore, by calculating the amount of acidified potassium dichromate added, the amount of iron(ii) ions can be known and so as the iron (III) ions produced by sodium hypochlorite. So the amount of sodium hypochlorite can be found. However, iron(ii) sulphate cannot be a primary standard due to it reaction with oxygen in air. So standardization is needed to be performing simultaneously with this test. It can be standardized by titration with known amount of acidified potassium dichromate solution, with diphenylamine as redox indicator. The titration principle is actually same as the redox titration above. p.8
Experiments: (1) Traditional Test Test for the amount of ClO - ions with Potassium iodide solution KI(aq), Sodium thiosulphate solution Na 2 S 2 O 3 (s) by the following equations: NaClO(aq) + 2HCl(aq) NaCl(s) + Cl 2 (g) + H 2 O(l) Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (aq) I 2 (aq) +2Na 2 S 2 O 3 (aq) 2NaI(aq) +Na 2 S 4 O 6 (aq) Part 1 Standardization of sodium thiosulphate solution Na 2 S 2 O 3 (aq) by the following equation: IO 3 - + 6H + + 5I - 3I 2 + 3 H 2 O I 2 + 2 S 2 O 3 2- S 4 O 6 2- + 2 I - Reagents: ~2M H 2 SO 4 solution KI solid NaIO 3 solid ~0.01M Na 2 S 2 O 3 solution Starch solution Apparatus: 100.0 cm 3 and 250.0 cm 3 volumetric flasks Burette Magnetic stirrer Conical flasks Beakers Electronic balance 25.0 cm 3 pipettes Stand and clamps Procedures: Preparation of KI solution 1. Using an electronic balance, weigh about 14 g of KI(s) 2. Dissolve the KI(s) in a beaker containing about 150 cm 3 of distilled water 3. Pour all the KI solution into a 250.0 cm 3 volumetric flask. p.9
4. Add distilled water until the meniscus reaches the graduation mark. 5. Stopper the flask and turn it upside down for several times. Preparation of NaIO 3 solution 1. Using an electronic balance, weigh about 2 g of NaIO 3 (s), record the mass weighted. 2. Dissolve all the NaIO 3 (s) in a beaker containing about 150 cm 3 of distilled water 3. Pour all the NaIO 3 solution into a 250.0 cm 3 volumetric flask. 4. Wash the inner of the beaker and pour all the washing into the volumetric flask for several times. 5. Add distilled water until the meniscus reaches the graduation mark. 6. Stopper the flask and turn it upside down for several times. 7. Using a 25.0 cm 3 pipette, collect 25.0 cm 3 of the NaIO 3 solution from the 250.0 cm 3 volumetric flask to a 100.0 cm 3 volumetric flask for dilution. 8. Add distilled water until the meniscus reaches the graduation mark. 9. Stopper the flask and turn it upside down for several times. Titration 1. Using a measuring cylinder, collect about 15 cm 3 of KI solution to a conical flask. 2. Using a 25.0 cm 3 pipette, collect 25.0 cm 3 of the NaIO 3 solution to the conical flask 3. Using a measuring cylinder, collect about 5cm 3 of the H 2 SO 4 solution to the conical flask. 4. The color of the solution in the conical flask should have turned from colorless to deep brown. 5. Place the conical flask on a magnetic stirrer, turn on the stirrer. 6. Titrate the solution in the conical flask against the Na 2 S 2 O 3 solution, record the initial reading of the burette. 7. When the color of the solution in the conical flask turns to pale yellow, add starch as indicator, the color of the solution should turn into deep blue 8. Titrate until the color of the solution in the conical flask turns from deep blue to colorless. 9. Record the final reading 10. Repeat the steps and perform the titration for four times. p.10
Part 2 Test for the amount of NaClO(aq) with Potassium iodide solution KI(aq), Sodium thiosulphate solution Na 2 S 2 O 3 (aq) by the following equations: NaClO(aq) + 2HCl(aq) NaCl(s) + Cl 2 (g) + H 2 O(l) Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (aq) I 2 (aq) + 2Na 2 S 2 O 3 (aq) 2NaI(aq) + Na 2 S 4 O 6 (aq) Experimental Set-up: Fig.1 Explanation of the set-up: Acidified KI solution will be transferred to the filtering flask that contains diluted bleach solution by the syringe so that all Cl 2 gas produced can immediately react with the surrounding KI solution, which prevent the leakage of Cl 2 gas which may make the result of the test under-estimated. The closed system is to ensure that although there may be some Cl 2 gas that cannot react with the KI solution fast enough, it can still be transferred to the boiling tube containing KI solution and react with the KI solution, which can minimize the error of the test. Reagents: Bleach (NaClO solution) ~2M HCl solution KI solid ~0.07M standardized Na 2 S 2 O 3 solution Starch solution p.11
Apparatus: 100 cm 3 syringe Delivery tubes Stopper with a hole Stopper with two holes Burette Magnetic stirrer Conical flasks Boiling tubes 25.0 cm 3 pipettes Stand and clamps 250.0 cm 3 volumetric flasks Procedures: 1. Using an electronic balance, weight about 45 g of KI(s). 2. Dissolve the KI(s) in a beaker containing about 150 cm 3 of distilled water. 3. Pour all the KI solution into a 250.0 cm 3 volumetric flask. 4. Add distilled water until the meniscus reaches the graduation mark. 5. Stopper the flask and turn it upside down for several times. 6. Use a pipette to transfer 25.0 cm 3 of the bleach (NaClO solution) to another 250.0 cm 3 volumetric flask. 7. Add water into the volumetric flask until the meniscus reaches the graduation mark. 8. Prepare the set up (see Fig. 1). 9. Use a pipette to transfer 25.0 cm 3 of the diluted NaClO solution from the volumetric flask to the filtering flask of the set up. p.12
10. Add about 30 cm 3 1M KI solution into the boiling tubes of the set up. 11. Stopper the filtering flask and the boiling tube of the set up. 12. Mix about 50cm 3 1M KI solution with ~50cm 3 2M HCl in a beaker. 13. Using the syringe, collect the KI solution prepared in step 12, connect the rubber tubing and add the solution to the filtering flask containing bleach of the set up slowly. 14. The color of the solution in the filtering flask should become reddish brown in color. 15. Collect all the solution in a 250.0 cm 3 volumetric flask, use distilled water to wash the container and pour all the washing into the 250.0 cm 3 volumetric flask. 16. Add water into the volumetric flask until the meniscus reaches the graduation mark. 17. Use a 25.0 cm 3 pipette twice to collect 50.0 cm 3 of the solution from the volumetric flask to a small conical flask. 18. Place the conical flask on a magnetic stirrer, turn on the stirrer. 19. Titrate the solution in the conical flask against the standardized Na 2 S 2 O 3 solution, record the initial reading of the burette. p.13
20. When the color of the solution in the conical flask turns to pale yellow, add starch as indicator, the color of the solution should turn into deep blue. 21. Titrate until the color of the solution in the conical flask turns from deep blue to colorless. 22. Record the final reading. 23. Repeat the steps and perform the titration for four times. p.14
(2) Proposed test TEST 1 Gravimetric analytical method Determine the sodium hypochlorite concentration with hydrogen peroxide Gravimetric Analysis of ClO ion with H₂O₂ solution by the following equation: NaClO(aq) + H 2 O 2 (aq) NaCl(aq) + O 2 (g) + H 2 O(l) Experimental Set-up: Fig. 2 Apparatus: two electrical balances beakers 25.0 cm 3 pipette Reagents: Bleach (NaClO solution) ~ 0.5M Hydrogen Peroxide solution Procedures: 1. Set up the apparatus (refer to Fig.2). 2. Weigh an empty clean beaker with an electrical balance. Then, transfer 25.0 cm 3 bleach solution to the beaker that put on the electrical balance, record the weight. 3. Pour 100 cm 3 hydrogen peroxide into another beaker. p.15
4. Put the hydrogen peroxide-containing beaker on the electrical balance, record the reading. 5. Add the hydrogen peroxide solution to the bleach-containing beaker on the electrical balance, with a glass rod guiding the liquid. 6. Weigh the emptied hydrogen peroxide-containing beaker with an electrical balance after pouring out the hydrogen peroxide. 7. Record the mass of the mixture every 30 seconds in the first 20 minutes. 8. After 20 minutes, record the weight of the mixture-containing beaker every five minutes until a steady reading is received. p.16
TEST 2 Redox back titration Determine the sodium hypochlorite concentration with iron(ii) sulphate, acidified potassium dichromate and diphenylamine Redox Back-Titration of NaClO with FeSO 4, acidified K 2 Cr 2 O 7 and (C 6 N 5 ) 2 NH by the following reactions: 1. Reduction of NaClO with FeSO 4 NaClO(aq) + 2FeSO 4 (aq) + H 2 O(l) NaCl(aq) + 2Fe 3+ (aq) + 2SO 4 2+ (aq) + 2OH - (aq) 2. Oxidation of Fe 2+ ion with K 2 Cr 2 O 7 14H + (aq) + K 2 Cr 2 O 7 (aq) + 6FeSO 4 (aq) 2Cr 3+ (aq) + 6Fe 3+ (aq) + 6SO 2-4 (aq) + 2K + (aq) + 7H 2 O(l) Experimental Set-up: Fig. 3 p.17
Apparatus Required: 250 cm 3 conical flasks a 50 cm 3 burette a stand and burette clamp 10.0 cm 3 and 25.0 cm 3 pipettes a 250.0 cm 3 volumetric flask Beakers Reagent Required: Household bleach( same bottle as for other tests) FeSO 4 solution, i.e. dissolving about 20g FeSO 4 7H 2 O into approximately 500 cm 3 water ~0.014M K 2 Cr 2 O solution(acidified), i.e. dissolving about 1g anhydrous K 2 Cr 2 O 7 into a 250.0 cm 3 volumetric flask with p.18
~100 cm 3 deionized water and ~150 cm 3 2M H 2 SO 4 solution Diphenylamine (as a redox indicator, melted in hot concentrated sulphuric acid) Procedure: Part 1: Standardization of FeSO 4 1. Setup the apparatus (refer to Part 1 of Fig.3). 2. Transfer 18.0 cm 3 FeSO 4 into the conical flask. 3. Add a few drops of redox indicator into the solution. p.19
4. Record the initial reading of the acidified K 2 Cr 2 O 7 solution. 5. Titrate the mixture against the acidified K 2 Cr 2 O 7 until the solution just change from green to dark purple. 6. Record the amount of acidified K 2 Cr 2 O 7 solution added. 7. Repeat the procedure at least three times to obtain an accurate data of titrant required. Repeat the procedures 1-7 with 10.0 cm 3 FeSO 4 used in step 2. Part 2: Reduction of NaClO with FeSO 4 1. Wash a 25.0 cm 3 pipette first with distilled water and then with FeSO 4. Wash a 25.0 cm 3 pipette first with distilled water and then with bleach. 2. Transfer 25.0 cm 3 bleach solution into a 250.0 cm 3 volumetric flask. 3. Add distilled water until the meniscus reaches the graduation mark. 4. Stopper the flask and turn it upside down for several times. 5. Transfer 3.4 cm 3 diluted bleach into a conical flask using a 10.0 cm 3 pipette that was washed with the diluted bleach. 6. Transfer 25 cm 3 FeSO 4 Solution into the same conical flask. p.20
7. Wait 5 minutes for reaction to occur until red precipitate formed (i.e. Fe (OH) 3 ). Part 3: Oxidation of Fe 2+ ion with K 2 Cr 2 O 7 solution: 1. Wash the burette first with deionized water, subsequently with acidified K 2 Cr 2 O 7 solution. 2. Set up apparatus (refer to the part 3 of Fig.3) 3. Place the conical flask of resultant mixture under the burette and add a small amount of redox indicator into the mixture p.21
4. Record the initial reading of the acidified K 2 Cr 2 O 7 solution 5. Titrate the mixture against the acidified K 2 Cr 2 O 7 until the mixture just change from green to dark purple. 6. Record the amount of acidified K 2 Cr 2 O 7 solution added 7. Repeat the procedure at least three times to obtain an accurate data of titrant required p.22
Data and data analysis: (1) Traditional test Determine the sodium hypochlorite concentration with sulphuric acid, potassium iodide, sodium thiosulphate and starch Equation involves: NaClO(aq) + 2HCl(aq) NaCl(s) + Cl 2 (g) + H 2 O(l) Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (aq) I 2 (aq) + 2Na 2 S 2 O 3 (aq) 2NaI(aq) + Na 2 S 4 O 6 (aq) Part 1 Standardization of sodium thiosulphate solution Na 2 S 2 O 3 (aq) by the following equation: IO 3 - + 6H + + 5I - 3I 2 + 3 H 2 O I 2 + 2 S 2 O 3 2- S 4 O 6 2- + 2 I - Mass of NaIO 3(s) used /g 2.078 Number of mole of NaIO 3 = Mass of NaIO 3 (s) used / Molar mass of NaIO 3 (s) = (2.078) / (197.8924) =0.010500655mol All NaIO 3 (s) is dissolved in 250.0 cm 3 solution. Molarity of the NaIO 3 solution = Number of mole of NaIO 3 / Volume = (0.010500655) / (250/1000) = 0.042002623 mol dm -3 25.0 cm 3 of the NaIO 3 solution is diluted to 100.0 cm 3 Number of mole of NaIO 3 in the diluted solution=molarity of the initial NaIO 3 solution Volume used =(0.042002623) (25/1000) =1.050065591 10-3 mol Molarity of the diluted NaIO 3 solution= Number of mole of NaIO 3 used for dilution / p.23 Volume = (1.050065591 10-3 ) / (100/1000) =0.010500655 mol dm -3
25.0 cm 3 of the diluted solution is used for titration Number of mole of NaIO 3 used = Molarity of the diluted NaIO 3 solution Volume used = (0.010500655) (25/1000) = 2.625163978 10-4 mol Titration data Volume of Na 2 S 2 O 3 used Trial 1 st 2 nd 3 rd Initial reading /cm 3 1.60 23.50 3.70 12.40 Final reading /cm 3 23.50 45.40 25.60 34.30 Difference /cm 3 21.90 21.90 21.90 21.90 IO 3 - + 6H + + 5I - 3I 2 + 3 H 2 O By Number of mole of I 2 formed = Number of mole of NaIO 3 3 = 2.625163978 10-4 3 =7.875491934 10-4 mol By I 2 + 2 S 2 O 3 2- S 4 O 6 2- + 2 I - Number of mole of Na 2 S 2 O 3 reacted = Number of mole of I 2 2 = 7.875491934 10-4 2 =1.575098387 10-3 mol Molarity of Na 2 S 2 O 3 solution = Number of mole of Na 2 S 2 O 3 / volume used in titration = (1.575098387 10-3 ) / (21.90/1000) = 0.0719223 mol dm -3 Molarity of Na 2 S 2 O 3 solution: 0.0719223 mol dm -3 p.24
Part 2 Test for the amount of NaClO(aq) with Potassium iodide solution KI(aq), Sodium thiosulphate solution Na 2 S 2 O 3 (aq) by the following equations: NaClO(aq) + 2HCl(aq) NaCl(s) + Cl 2 (g) + H 2 O(l) Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (aq) I 2 (aq) + 2Na 2 S 2 O 3 (aq) 2NaI(aq) + Na 2 S 4 O 6 (aq) Titration data Volume of Na 2 S 2 O 3 used Trial 1 st 2 nd 3 rd Initial reading /cm 3 1.50 10.10 18.70 27.30 Final reading /cm 3 10.10 18.70 27.30 35.90 Difference /cm 3 8.60 8.60 8.60 8.60 Reagent data Molarity of Na 2 S 2 O 3 solution: 0.0719223 mol dm -3 Number of mole of Na 2 S 2 O 3 used in titration = Volume used in titration Molarity of Na 2 S 2 O 3 solution = (8.60 / 1000) (0.0719223) = 6.185317866 10-4 mol Let M NaClO be the molarity of NaClO in bleach. 25 cm 3 of bleach solution is collected and diluted to 250 cm 3 Number of mole of NaClO in the diluted bleach solution = molarity of NaClO in bleach Volume used p.25 = M NaClO (25/1000) = 0.025 M NaClO Molarity of NaClO in the diluted bleach solution = Number of mole of NaClO / Volume 25 cm3 of the diluted solution is used for reaction = 0.025 M NaClO / (250/1000) = 0.1 M NaClO Number of mole of NaClO reacted = Molarity of NaClO in the diluted bleach solution Volume used = 0.1 M NaClO (25/1000) = 0.0025 M NaClO
By NaClO (aq) + 2HCl (aq) NaCl(s) + Cl 2 (g) + H 2 O (l) NaClO is reacted to Cl 2 Number of mole of Cl 2 formed = Number of mole of NaClO reacted = 0.0025 M NaClO By Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (aq) Number of mole of I 2 formed = Number of mole of Cl 2 = 0.0025 M NaClO The solution of I 2 formed was diluted to 250 cm 3 for titration, 50 cm 3 of the diluted solution was used for titration each time. Molarity of the diluted solution of I 2 = Number of mole of I 2 formed / Volume = 0.0025 M NaClO / (250 / 1000) = 0.01 M NaClO Number of mole of I 2 used in each titration = Molarity of the diluted solution of I 2 Volume used = 0.01 M NaClO (50 / 1000) = 0.0005 M NaClO By I 2 (aq) + 2Na 2 S 2 O 3 (aq) 2NaI(aq) + Na 2 S 4 O 6 (aq) Number of mole of Na 2 S 2 O 3 used in titration = Number of mole of I 2 2 = (0.0005 M NaClO ) 2 = 0.001 M NaClO As calculated before, number of mole of Na 2 S 2 O 3 used in titration = 6.185317866 10-4 mol 0.001 M NaClO = 6.185317866 10-4 M NaClO = 0.618531786 mol dm -3 Molarity of NaClO in bleach = 0.618531786 mol dm -3 Molarity of NaClO in bleach : 0.618531786 mol dm -3 p.26
(2) Our proposed tests TEST 1 Gravimetric analytical method Determine the sodium hypochlorite concentration with hydrogen peroxide H 2 O 2 (aq) + NaClO(aq) H 2 O(l) + NaCl(aq) + O 2 (g) Experimental Data Mass of the beaker with H 2 O 2 /g 202.682 Mass of the empty beaker after pouring H 2 O 2 /g 100.859 Mass of 100 cm 3 H 2 O 2 solution/g 101.823 Mass of the empty beaker/g 108.629 Mass of the beaker with bleach/g 133.763 Mass of 25 cm 3 bleach /g 25.134g Sum of the mass of the above reactants with the 235.586 beaker/g p.27
Time / s Mass of the reacting mixture/g Time / s Mass of the reacting mixture/g 0 235.586 1680 235.469 30 235.574 1800 235.460 60 235.57 1920 235.453 90 235.567 2040 235.446 120 235.565 2160 235.44 150 235.562 2280 235.434 180 235.56 2400 235.429 210 235.558 2520 235.423 240 235.556 2640 235.419 270 235.554 2760 235.413 300 235.552 2880 235.407 330 235.55 3000 235.401 360 235.548 3300 235.39 390 235.547 3600 235.375 420 235.546 3900 235.362 450 235.544 4200 235.351 480 235.543 4500 235.292 510 235.541 5100 235.265 540 235.54 5700 235.247 570 235.538 6300 235.22 600 235.532 7500 235.19 720 235.526 7800 235.174 840 235.517 8100 235.16 960 235.51 8400 235.151 1080 235.503 9000 235.136 1200 235.496 9900 235.109 1320 235.488 10500 235.088 1440 235.481 10800 235.077 1560 235.474 p.28
By NaClO + H 2 O 2 NaCl + H 2 O + O 2 Change in mass of the solution = Mass of O 2 produced Mass of O 2 produced = 235.586 235.077 = 0.509 g Number of mole of O 2 produced = Mass of O 2 produced / Molar mass of O 2 = (0.509) / (32.0) = 0.01590625 mol Number of mole of NaClO = Number of mole of O 2 produced = 0.01590625 mol Molarity of NaClO = Number of mole of NaClO / Volume of the NaClO solution used = (0.01590625) / (25/1000) = 0.63625 mol dm -3 Molarity of NaClO in bleach : 0.63625 mol dm -3 p.29
TEST 2 Redox back titration Determine the sodium hypochlorite concentration with iron(ii) sulphate, acidified potassium dichromate and diphenylamine Equation involves: 1. Reduction of NaClO with FeSO 4 NaClO(aq) + 2FeSO 4 (aq) + H 2 O(l) NaCl(aq) + 2Fe 3+ (aq) + 2SO 4 2+ (aq) + 2OH - (aq) 2. Oxidation of Fe 2+ ion with K 2 Cr 2 O 7 14H + (aq) + K 2 Cr 2 O 7 (aq) + 6FeSO 4 (aq) 2Cr 3+ (aq) + 6Fe 3+ (aq) + 6SO 2-4 (aq) + 2K + (aq) + 7H 2 O(l) 14H + (aq) + Cr 2 O 2-7 (aq) + 6Fe 2+ (aq) 2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O(l) Part 1 Standardization of iron(ii) sulphate solution 19.993g of iron(ii) sulphate (heptahydrate) was dissolved into 500 cm 3 of water, and 1.017g of anhydrous potassium dichromate was dissolved into 250 cm 3 of acidic solution. Molarity of acidified potassium dichromate solution = (1.017/ 294.185)/0.25 W = 0.013828033 mol dm -3 First set of data collection 18 cm 3 of iron(ii) sulphate solution was titrated against 0.013828033 M acidified potassium dichromate solution. Volume of K 2 Cr 2 O 7 solution used Trial 1 st 2 nd 3 rd 4 th Initial burette reading /cm 3 1.00 6.00 7.10 9.20 8.60 Final burette reading /cm 3 31.20 35.50 36.10 38.50 37.80 Difference /cm 3 30.20 29.50 29.00 29.30 29.20 Average volume of acidified potassium dichromate solution added = (29.5+29+29.3+29.2)/4 = 29.25 cm 3 = 0.02925 dm 3 p.30
Number of moles of potassium dichromate used = 0.013828033 0.02925 = 0.0004044699653 mol Number of moles of iron(ii) sulphate in 18 cm 3 of iron(ii) sulphate solution = 0.0004044699653 6 = 0.002426819792 mol Therefore, Molarity of iron(ii) sulphate solution = 0.002426819792 / 0.018 = 0.134823321 mol dm -3 Second set of data collection 10 cm 3 of iron(ii) sulphate solution was titrated against 0.013828033 M acidified potassium dichromate solution. K 2 Cr 2 O 7 solution used Trial 1 st 2 nd 3 th Initial burette reading /cm 3 2.50 2.80 19.30 3.10 Final burette reading /cm 3 19.40 19.30 35.10 19.70 Difference /cm 3 16.90 16.50 16.80 16.60 Average volume of acidified potassium dichromate solution added = (16.5+16.8+16.6) / 3 = 16.633333 cm 3 = 0.016633333 dm 3 Number of moles of potassium dichromate used = 0.013828033 0.016633333 = 0.0002300062822 mol Number of moles of iron(ii) sulphate in 18 cm 3 of iron(ii) sulphate solution = 0.0002300062822 6 = 0.001380037693 mol Therefore, Molarity of iron(ii) sulphate solution = 0.001380037693 / 0.010 = 0.138003769 mol dm -3 p.31
Finally, take the average of the two results Molarity of iron(ii) sulphate solution = (0.134823321+ 0.138003769) / 2 = 0.136413545 mol dm -3 Part 3 Back titration of NaClO 3.4 cm 3 of bleach (10 times diluted) was added to 25 cm 3 iron(ii) sulphate solution. Then, the solution was titrated with 0.013828033 M acidified potassium dichromate solution. K 2 Cr 2 O 7 solution used Trial 1 st 2 nd 3 th 4 th 5 th Initial burette reading /cm 3 3.40 2.10 2.30 1.50 0.70 0.60 Final burette reading /cm 3 40.20 38.90 38.90 37.90 37.30 37.20 Difference /cm 3 36.80 36.80 36.60 36.40 36.60 36.60 Average volume of acidified potassium dichromate solution added = (36.8+36.6+36.4+36.6+36.6) / 5 = 36.6 cm 3 = 0.0366 dm 3 Number of moles of potassium dichromate used = 0.013828033 0.0366 = 0.0005061060078 mol Number of moles of iron(ii) sulphate remained after mixing with NaClO = 0.0005061060078 6 = 0.003036636047 mol Number of moles of iron(ii) sulphate in 25 cm 3 iron(ii) sulphate solution originally = molarity volume = 0.136413545 0.025 = 0.003410338625 mol Number of moles of iron(ii) sulphate being oxidized = 0.003410338625 0.003036636047 = 0.000373702578 mol p.32
Since 1 hypochlorite ion was reacted with 2 iron(ii) ions, Number of moles of sodium hypochlorite in 3.4 cm 3 10-times-diluted bleach = 0.000373702578 / 2 = 0.000186851289 mol And so, Molarity of sodium hypochlorite in 10-times-diluted bleach = 0.000186851289 / 0.0034 = 0.054956261 mol dm -3 By M 1 V 1 =M 2 V 2 Molarity of NaClO in the original bleach : 0.54956261 mol dm -3 As reference to the traditional test using by chemists, we can calculate the percentage differences of the two tests that we proposed by comparing with the traditional test. Traditional test result: Molarity of NaClO in bleach : 0.618531786 mol dm -3 Test 1 result: Molarity of NaClO in bleach : 0.63625 mol dm -3 Test 2 result Molarity of NaClO in bleach : 0.54956261 mol dm -3 Percentage difference of test 1 = ( Test 1 result - Traditional test result ) / Traditional test result 100% = ( 0.63625-0.618531786 ) / (0.618531786) 100% 2.864 % Percentage difference of test 2 = ( Test 2 result - Traditional test result ) / Traditional test result 100% = ( 0.54956261-0.618531786 ) / (0.618531786) 100% -11.150 % p.33
Discussions: (1)Discussion of TEST 1 Gravimetric Analytical method Determine the sodium hypochlorite concentration with hydrogen peroxide. The result of the gravimetric test of the NaClO(~0.636M) is similar to the one in the traditional test (~0.619M). Throughout the observation and data recording, there is first a higher rate of mass fall till 1000s.And the fall become less significant at the end of the measurement as the rate of gas evolvement is as reflected from the mass lost and the instantaneous slope of the graph. Such pattern implies that oxygen can be readily evolved when H 2 O 2 react with NaClO and the rate of reaction slow down as time passes due to the decrease in concentration of reactants. The lost mass can quite accurately reflect the molarity of NaClO ion, when compared to the traditional test. Furthermore, H 2 O 2 react rigorous once it mixed with NaClO, as stated in the first 10 min on the graph. Therefore, we are required to measure the mass of solution used separately before the mixing of reactants, in order to reduce the error in mass. By that, two electronic balances are utilized from that consideration. We can therefore accurately obtained the initial mass of reaction mixture. p.34
Possible Error in the test Concerning the gravimetric method, the error or differences could be resulted from the prolonged observation interval of the experiment. Water in the mixture can easily evaporate for the time being while oxygen evolved can re-dissolve in the water. This may explain the sudden change in evolvement for the interval at about time t=4000s. (2)Discussion of TEST 2 Redox Back titration Determine the sodium hypochlorite concentration with iron(ii) sulphate, acidified potassium dichromate and diphenylamine Comparing to the traditional test for ClO -, the test result over the concentration of NaClO was significantly differs and have a result of lower concentration (~0.619M>~0.550M) In terms of chemical principles applied, redox, same principle applied as in traditional test. This can be resulted from the release of Cl 2 during the reaction between the diluted bleach and FeSO 4. More iron(ii) ion would be left and concentration of bleach would be underestimated although we already allowed the reaction to be performed under alkaline medium in order to reduce this error. Choice of reagent: Originally, MnO - 4 ion (E o OA = +1.51)(in acidic medium) will be used as the oxidizing agent to give a clear end point when compared to Cr 2 O - - 7 ion. However, MnO 4 will react with the product Cl - ion from the reduction of Cl 2 (E o RA = +1.36), as of the high oxidation potential between the Cl - - ion and MnO ( 4 E o OX = E o OA -E o RA =+0.15). This will overestimate the use of permanganate as it reacts with both the excess FeSO 4 and the Cl - ion. Another drawback is that we have to standardize the permanganate solution as it easily decompose into manganese(iv) oxide while sodium salt of dichromate is so stable that it can be used as a primary standard. As a result, we decided to utilize the Cr 2 O - 7 ion (E o OA = +1.33) to act as the titrant in the redox titration of Fe 2+ ion (E o RA = -0.44) in FeSO 4, as it will not react spontaneously with the Cl - ion as of the small standard electrode potential ( E o OX = E o OA -E o RA =-0.03). Such figures ensure the oxidation of FeSO 4 by K 2 Cr 2 O 7 Preparation of redox indicator: In addition to the pre-test preparation of FeSO 4, Diphenylamine, as known as (C 6 N 5 ) 2 NH), the redox indicator, have to be dissolved in concentrated sulphuric acid instead of water due reagent s low solubility in polar solvent. Concentrated sulphuric acid can increase the solubility of diphenylamine so as to enhance the color intensity p.35
for the end point given by the reduction of Cr 2 O 7 - ion. Sulphuric acid can also further ensure the acidic environment for K 2 Cr 2 O 7 to involve in oxidation. Possible Error in the test The major error that led to the underestimation of NaClO could be the formation of Cl 2 during mixing NaClO with FeSO 4. During to the formation of iron(ii)/(iii) hydroxide, the decrease in free OH - would decrease the alkalinity of the solution. Upon mixing with NaClO, the ClO - ion may form HClO and may be reduced by the Fe 2+ to form Cl 2 according to the following side reaction pathway: 2H + (aq) + 2HOCl(aq) + FeSO 4 (aq) Fe 3+ (aq/s) + SO 4 2- (aq) + Cl 2 (g) + 2H 2 O(l) (E o = E RA o -E OA o = +1.67-(+0.77) = +0.9) If the Cl 2 can be conserved completely in the container and react completely with the iron(ii) ion, 2Fe 2+ + Cl 2 2Cl - + 2Fe 3+ the no. of moles of iron(ii) ions left over the reaction would be the same as the proposed pathway, i.e. NaClO(aq) + 2FeSO 4 (aq) + H 2 O(l) NaCl(aq) + 2Fe 3+ (aq) + 2SO 4 2+ (aq) + 2OH - (aq) But in fact, the reaction of gaseous chlorine under a not so alkaline environment may not be that efficient and it causes errors. Some Cl 2 was formed and lost in the process, allowing less FeSO 4 to involve in the intended reaction There is also an error, however, could lead to overestimation. The reduction of Fe 2+ by oxygen in air could result in experimental error as expected before experimental work. Before the test over NaClO, the FeSO 4 utilized in the experiment requires standardization. As FeSO 4 in the solution will be oxidized by the oxygen in air readily to form Fe 2 O 3 (insoluble) and Fe 2 (SO 4 ) 3 (soluble), it will reduce the FeSO 4 available for subsequent reduction of NaClO and the state of excessing reactant for NaClO. By that, the number of excess of FeSO 4 is reduced and thus less K 2 Cr 2 O 7 will be used to titrate excess FeSO 4. It will then led to overestimation of FeSO 4 involved this concern was stressed from the change in color from pale green to yellow for the FeSO 4 solution preparation. As a result, for every FeSO 4 solution preparation prior the test, approximately 15-20ml of FeSO 4 was titrated against K 2 Cr 2 O 7 before the experiment twice in order to p.36
obtain an accurate concentration for FeSO 4 solution. However, from the result and observation during solution preparation, the further standardization of FeSO 4 can t remedy the fact of oxidization of FeSO4 under the presence of air. This led to a lower amount of FeSO 4 in excess and thus lowers the amount of K 2 Cr 2 O 7 used. As a result, it could result in overestimation. Quantitative test which involves redox titration should utilize chemically more stable standard solution, for instance, low reactivity with oxygen. The choice of iron(ii) may lead to some limitations for our 2 nd proposed test. (3)Discussion on all test methods feasibility of application of gravimetric methods in every daily life To summarize, gravimetric method for NaClO tends to have a greater accuracy and ease on qualitative analysis, when compared to the redox method in our proposed test and traditional test respectively. Major advantage of the gravimetric test proposed is the minimized experimental procedure and utilization of reagents. In fact, gravimetric analysis involves less experimental preparation and procedure when compared to the redox tests tested previously, say standardization is not required for the reagent involved, nor it required multiple redox titrations. Fewer reagents and procedure reduce the amount of error resulted from experimental estimation of apparatus. Multiple measurements and reading taking can increase the experimental error significantly. Another advantage of gravimetric test would be the ease of application. The reagent utilized, H 2 O 2 can be easily obtained when compared to other reagent in other test, say Diphenylamine. This can allow easier and greater application of the test method within daily life. Household electrical balance might be utilized as average experimental electronic balance for gravimetric analysis. p.37
Further study: (1)Further improvements of our tests In our experiments, the two new-designed tests could not get the exact value of sodium hypochlorite concentration that found by the traditional test. In the hydrogen peroxide test, evaporation of water and solvation of oxygen are inevitable in our set-up. Although they seems are minor problems that will not hinders the experimental operation, they will lower the accuracy of the test. To solve the problem, it is suggested to cover the beaker with cotton wool that soaked in sodium hydroxide solution, and place the experimental set-up in environment of dehydrated air. Cotton wool that soaked in sodium hydroxide solution can absorb water that evaporated from the mixture solution of sodium hypochlorite and hydrogen peroxide, so that evaporated water will not escape from the system; placing the experimental set-up in environment of dehydrated air can ensure the water absorbed is not from the water vapor in air. These can help to eliminate the experimental error of mass loss due to evaporation of water. Yet, it is quite difficult to conduct in ordinary laboratory. Another solution is that we can increase the volume of reagent used. The percentage of decrease due to mass of water evaporated would be much smaller than the decrease of mass due to oxygen escaped. In the potassium dichromate test, the result, i.e. sodium hypochlorite concentration, is smaller than the sodium hypochlorite concentration that found by the traditional test. A reasonable explanation is that there was loss of chlorine during the reaction of sodium hypochlorite solution and iron(ii) sulphate solution in a neutral or even slightly acidic medium. Iron(II) ions will actually provide acidic environment by absorbing hydroxide ion (forming Fe(H 2 O) 5 (OH) 3+ complex). When excess amount of iron(ii) sulphate solution was mixed with slightly alkaline sodium hypochlorite solution, a neutral or slightly acidic medium might be created. Hypochlorite ions might had been turned into chlorine. Although chlorine can also reaction the same way, some of them might bubble out from the solution. These will affect the experimental result quite badly. To solve the problem, it is suggested to put a lid or a stopper that perfectly fit the conical flask*, onto the conical flask that contain bleach and iron(ii) sulphate solution, so that the chlorine gas cannot escape from the conical flask. After waiting for the reaction, put the liquid-tight conical flask upside down for a few times and each for one minute. So the remaining chlorine gas can react and the chlorine in hypochlorite can react completely. *make sure the lid is really tight so that it won't bump out by the air pressure from the flask inside. Also note that the flask should be strong enough to withstand the p.38
increasing air pressure. (2)Another new-designed test that have not been proposed the polarimetric analysis Beside the two tests that we have conducted, there is still a test that is possible to test for the sodium hypochlorite concentration. However, it cannot be conducted by us since there is no pure (+) butan-2-ol or pure (-) butan-2-ol in our school laboratory. This test involves the study of polarimetry. A chiral compound, butan-2-ol can be used in this test. Structures of two enantiomers of butan-2-ol are shown below. H 3 C OH CH 3 OH H 3 C CH 3 When polarized light passes through an optical isomer, also known as the enantisomer, rotation of plane of polarized light occurs. When passing though the (+) form of an enantisomer, the rotation is in clockwise direction. While passing though the (-) form of an enantisomer, there will be anticlockwise rotation. When butan-2-ol is mixed with sodium hypochlorite, it will be oxidized to butanone. Since butanone have no enantisomer (the central carbon atom is only attaching three different groups, which lack chirality due to the presence of plane of symmetry), it will not cause rotation of plane of polarized light. Structure of butanone: O H 3 C CH 3 The experiment goes like this: prepare solutions of pure (+) butan-2-ol or pure (-) butan-2-ol (can only choose one of them) of different concentrations, and then find their 'angle of rotation of plane of polarization of light' with a polarimeter. Plot a calibration curve of angles of rotation against concentrations of (+) / (-) butan-2-ol times a constant, i.e. α = k c ( α is angle, k is constant, c is concentration). Then let a known amount of butan-2-ol (in excess) to react with sample of sodium hypochlorite solution (bleach). Then check for the angle of rotation of the butan-2-ol remains. By referring to the graph plotted, the amount of butan-2-ol remained after reaction can be found, and so as the butan-2-ol reacted. Since C 4 H 9 OH + ClO - C 4 H 8 O + Cl - + H 2 O, the amount of sodium hypochlorite can be calculated. The idea is originally test the sodium hypochlorite concentration using glucose as p.39
D-glucose is easy to get. D-glucose is an enantisomer of glucose which is favorable for the polarimetric test. Glucose will react with sodium hypochlorite to form gluconic acid. Yet, gluconic acid will also cause rotation of plane of polarization of light. To solve this problem, two graph have to be plotted, one for different concentrations of glucose and one for that of gluconic acid. But, another problem appears, there is no pure gluconic acid in our school laboration for setting the calibration curve. So we have to synthesis a solution of gluconic acid with known concentration, by reacting known amount of glucose with some oxidizing agents. However, the final problem appears, there is a chance that gluconic acid may be further oxidized to dioic acid and not all oxidations stay at the gluconic level. This will affect the result seriously, so the use of glucose was rejected at last. As replacement of glucose-- pure (+) / (-) butan-2-ol cannot be found in our lab, this test was finally abandoned. Conclusion: As calculated before, percentage difference of test 1 is around 2.864 % while percentage difference of test 2 is around -11.150 % By comparing the percentage difference of the two tests to the traditional, we can conclude that the result of test one is nearer to the result of the traditional test. Test 1 may be a better substitute for determining the concentration of ClO - ions in solution. However, test 2, comparing to test1, has a much higher percentage difference. Therefore, it can t be used to find the concentration of ClO - ions in solution unless some improvements has been made. Test 1 is gravimetric analysis, which has lesser procedures comparing to the traditional test. Besides, the Na 2 S 2 O 3 solution used in the traditional test needed to be standardized. On the other hand, the H 2 O 2 solution use in test 1 does not need to be standardized, adding excess of it is praised instead. Not only can this enhance the convenience of the test, but it can also avoid doing titration, which increases the possible error. Therefore, we suggest that test 1 can replace the traditional test of finding the amount of ClO - ions in solution. p.40
References: 1. Walther F. Goebel, ON THE OXIDATION OF GLUCOSE IN ALKALINE SOLUTIONS OF IODINE,J. Biol. Chem. 1927, 72:801-807, http://www.jbc.org/content/72/2/801.full.pdf 2. C. Parameshwara Murthy, University Chemistry, Volume 1 (2002), New Age International, P.629-642 3. http://staff.buffalostate.edu/nazareay/che112/chromate.htm 4. http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt31.html 5. www.hach.com/asset-get.download.jsa?id=7639984029 6. http://en.wikipedia.org/wiki/hypochlorite p.41
Acknowledgement: We would like to express our heartfelt gratitude to Mr Lau Chun-On,Chemistry Teacher of SGSS and the teacher-advisor for our project, who advice and instruct us at times of difficulties and during experimental set-back. We would also like to thank Mr. Fung Wai-Ching, the Laboratory Technician of SGSS, for preparing the apparatus and reagents for our experiment. At last, we were grateful to our school, SGSS that provide us a superb site for scientific work and chance to participate in this competition on behalf of the school p.42