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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added to the mixture which will emphasize the changes. This chemical also returns iodine to iodine ions. 2S2O32-aq + I2aq -----------> S4062-aq + 2I-aq As the reactants of this reactions are colourless it is difficult to tell when the colour change will happen, and when it does it is quite sudden. This happens because the thiosulphate is being used up in the reactions, and when it is all used up the starch-iodine complex is formed resulting in the colour change. Equipment and apparatus needed: "¢ Test tubes "¢ Boiling tubes "¢ Thermometer 0-110°C "¢ Stopwatch "¢ Burettes "¢ 15cm3 Potassium iodide solution,KI, 1.00 mol dm-3 "¢ 10cm3 Potassium peroxodisulphate solution,K2S2O8, 0.0400 mol dm-3 "¢ 10cm3 Sodium thiosulphate solution, Na2S2O3, 0.0100 mol dm-3 "¢ 5cm3 fresh Starch solution "¢ Test tube rack Method "¢ Read and understand carefully the table below illustrating the different mixtures of concentrations for individual experiments. mixture Volume of KLaq/ cm3 Volume of water / cm3 Volume of Na2S2O3aq / cm3 Volume of starch solution/cm3 Volume of K2S2O8aq / cm3 1 5 0 2 1 2 2 4 1 2 1 2 3 3 2 2 1 2 4 2 3 2 1 2 5 1 4 2 1 2 "¢ When making the mixtures, make sure that all chemicals are added to a boiling tube, except the potassium peroxodisulphate which must be added separately as this is what will trigger the reaction. "¢ Make up mixture 1 and place a thermometer in the boiling tube. Add the potassium peroxodisulphate at begin the timing with the clock "¢ Stop timing when the solution turns blue-black "¢ Constantly stir the solution. "¢ Take a temperature reading "¢ Repeat procedure for each mixture of different concentrations. Results A Table mixture Concentration of I-aq / mol dm-3 Clock time / seconds Rate / mol dm-3 s-1 Temperature / °C 1 0.5 28.8 3.45 X10-7 23 2 0.4 40.9 2.44 X10-7 23 3 0.3 58.0 1.72 X10-7 26 4 0.2 136.7 7.32 X10-8 26 5 0.1 455.5 2.20 X10-8 24 Questions B Rate = concentration of I2 Time = [ I2 ] = mol dm-3 s-1 S [ I2 ] = 1.0 X10-5 Mixture: 1. = 1.0 X10-5 = 3.45 X10-7 28.8 2. = 1.0 X10-5 = 2.44 X10-7 40.9 3. = 1.0 X10-5 = 1.72 X10-7 58.0 4. = 1.0 X10-5 = 7.32 X10-8 136.7 5. = 1.0 X10-5 = 2.20 X10-8 455.5 C The Iodine ions are in excess in the reaction mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced. D Moles = mol dm-3 x dm3 = 0.04 x 2/1000 = 8x10-5 mols of S2O8 2- E i 2cm3 of Na2S2O3 0.0100 mol dm-3 Moles of Na2S2O3 = 0.01 x 2/1000 = 2x10-5 moles of S2O3 ii Moles of = 1.0 X10-5 I2 iii Percentage % of reaction = 1.0 X10-5 x100 8.0 X10-5 = 12.5 %
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Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added...
mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced.

D Moles = mol dm-3 x dm3

= 0.04 x 2/1000

= 8x10-5 mols of S2O8 2-

E i 2cm3 of Na2S2O3 0.0100 mol dm-3

Moles of Na2S2O3 = 0.01 x 2/1000

= 2x10-5 moles of S2O3

ii Moles of = 1.0 X10-5

I2

iii Percentage % of reaction = 1.0 X10-5 x100

8.0 X10-5

= 12.5 %

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Determination of the relative atomic mass...Determination of the relative atomic mass of Lithium Implementing For all my results I have decided to use three significant figures. This is because the apparatus I was told to use for the experiment gave me results to three significant figures. There was, therefor, no point in calculate to any further accuracy as this would just define a figure that has already been calculated to a specific depth of accuracy. Results Method 1 Mass of Lithium used 1.09 g Volume of gas produced 175 cm3 Method 2 1 2 3 Final Burette Reading 42.1 42.3 42.3 Initial Reading 0 0 0 Volume added cm3 42.1 42.3 42.3 Average 42.2 cm3 Analysing Method 1 Calculate the number of moles of hydrogen. 1 175 = 0.00729 Number of moles = Amount of gas produced 2400 2400 This was calculated to find out the number of moles in the Hydrogen gas that had produced. Deduce from this the number of moles of Lithium. 2 The number of moles of Lithium = 0.00729 x 2= 0.0146 If you look at the chemical equation¡V 2Lis + 2H2Ol ------"žÂ³ 2LiOHaq + H2g You can see that the molar ratio of number of moles of hydrogen to number of moles of lithium is 1:2. As I had already calculated the number of moles for Hydrogen, I can multiply it by two and this will give me the number of moles of Lithium. Using your values above and the original mass of Lithium, calculate the relative atomic mass of lithium. 3 1.09 = 7.13 Relative atomic mass= Mass used in grams 0.146 Number of moles From the amount of lithium used and the number of moles of lithium. I used the above equation to calculate the relative atomic mass of lithium. Method 2 Calculate the number of moles of HCl used in the titration. Average volume 1 42.7 x 0.1= 0.00422 Number of moles of HCl = concentration x volume 1000 1000 The formula I used is the one shown above. This enabled me to calculate the number of moles of HCl used in the titration experiment, as the HCl was a solution. Deduce the number of moles of LiOH used in the titration. 2 The number of moles of LiOH used in the titration = 0.00422 LiOHaq + HClaq -----"žÂ³ LiCl + H2g This calculation was much the same as calculation 2 of the last experiment but this time the molar ratio was 1:1 between LiOH and HCl, as shown in the equation above¡V Calculate the number of moles of LiOH present in 100cm3 of the solution from method 1. 3 The number of moles of LiOH present in 100 cm3 is = 0.00422 x 4 = 0.0169 The first equation shows that in 25 cm3 of LiOH there are 0.00422 moles. 100 cm3 is 4 times as much as 25cm3, therefor if I multiply 0.00422 by 4 then this would give me the number of moles on 100 cm3 Use the result and the original mass of lithium to calculate the relative atomic mass of Lithium. 4 0.104 = 6.15 Relative atomic mass = Mass used in g 0.0169 Number of moles I used the formula above to calculate the relative atomic mass. My average atomic mass for Lithium = 6.64 The actual relative atomic mass for Lithium = 6.90 Difference = 0.26 % difference = 3.77 Risk assessment The first safety hazard I encountered in my experiment was the Li that I placed into the waster in the first method. I was told it reacted violently with water to form lithium hydroxide. The Li was contained at almost all times throughout the experiment but as there was a chance of me being in proximity of the reaction then I wore goggles. Also I was shown that I was not to use a cube that has sides of more than 3mm. So I used a cube as near this number as possible but I ensured that I did not exceed this 3mm constraint at all times. The second safety hazard I encountered was the hydrogen gas. This is an extremely flammable gas and so I made sure that there were no naked flames around the experiment at all times. Thirdly I encountered lithium hydroxide this is only a problem when it exceeds 6.5 molar and as I used no where near this molarity all that this substance was, was a mild irritant. Therefor all I needed was eye protection and in the case of it getting on my skin I was to wash it off with water. HCl was the last harmful substance that I used in my experiment and the procedure for this was exactly the same as lithium hydroxide. Evaluation There are two ways that there could have been inaccuracies in my experiment, the first of which could have been my experimental procedure and the second could have been problems with the apparatus I used to implement my experiment. The first inaccuracy that I encountered really cancels itself out in some respects. There is oil used to stop the Li reacting with the air and this is helpful in some respects but not in others. This creates a multiple number of inaccuracies, as you have to remove the oil from the Li so that you weight the Li by itself to stop inaccuracies in later calculations. This though will then let the Li oxidise with the oxygen in the atmosphere, therefor I will, if the experiment is to be repeated use an inert atmosphere to stop any oxidisation from taking place. Also the only effective way to remove the oil from the Li is to cut the outside layer off the Li and this will ensure there is no oil left on the Li when it is weighed. Also this will stop the oil from being in the Li hydroxide and creating any inaccuracies in the second method. Hydrogen is another source for an inaccuracy. This is because in can be effected and lost. It can be lost in two ways, either when the bung is replaced onto the conical flask when the Li starts to react or when the hydrogen goes through the rubber tubing it can be lost. There are no ways to improve the experimental procedure for this and so they are inaccuracies that I cannot change. The third inaccuracy is the fact the gas expanded when it came from the Li and water as it was hot so the final gas reading would be greater than it should be and so it effected my results. I would of used a percentage error type of evaluation but as I don¡¦t have specific measurements that could of shown me how inaccurate the data I collected was I cannot use percentage errors for some data¡¦s and not for others.   

Determination of the relative atomic mass of Lithium Implementing For all my results I have decided to use three significant figures. This is because the apparatus I was told to use for the experiment gave me results to three significant figures. There was, therefor, no point in calculate...

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