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The estimation of iron(II) and iron(III) in a mixture containing both

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Introduction

The estimation of iron(II) and iron(III) in a mixture containing both I have been given a solution that contains between 1.1g and 1.3g of iron ions; they are in a mixture of both Fe2+ and Fe3+. My aim is to work out the percentage of each of them in the solution. With my research I have found out that the best method would be to find out how many grams of Fe2+ ions there are first. I must then convert all the Fe3+ ions in to Fe2+ ions by reducing them with granulated zinc and H2SO4. This would enable me to then work out the total mass of all the iron ions in the solution. Total Mass - Mass of Fe2+ = Mass of Fe3+ After I know the mass of each different iron ion in the solution I can then convert these into percentages. Equipment 200cm3 of Fe2+/Fe3+ solution H2SO4 (1 mol dm-3) KMnO4 (0.0100 mol dm-3) Clamp stand Beaker Eye protection Burette White tile Graduated pipette Accuracy I am using 20cm3 of my solution each time as opposed to using 10cm3 as this will give me a lower percentage error. ...read more.

Middle

Procedure to work out the mass of Fe2+ ions 1. Set up the burette, clamp stand, 200cm3 beaker and white tile as in the picture on the right. 2. Make sure the tap is close on the burette and then fill it to the top amount line with KMnO4. 3. I will then add 20cm3 of my iron solution to the beaker. 4. The solution needs to be added to an acid, this is due to the fact that it will stop any atmospheric oxidation of iron (II) to iron (III). Therefore I will add 10cm3 of H2SO4 to the beaker as well. 5. I will then turn the tap on the burette to let out the KMnO4. I will do this slowly so I can record the amount on the burette being let out. I will also stir the beaker at the same time. 6. In my first titre I am looking for the rough amount of KMnO4 needed to make the colour change happen. 7. After I have found the rough amount needed I will do a lot more accurate ones, letting out the KMnO4 drop by drop when close the amount. ...read more.

Conclusion

I can do this as I know the volume that I will have titrated and the concentration of it. The equation I will use is: Moles = Concentration * (Volume/1000) I will then look at the balanced equation and see how many moles Fe2+ are made with the moles of KMnO4. The equation is: KMnO4- + 5Fe2+ >>> KMn2+ + 5Fe3+ I can then see that 1 mole of KMnO4 makes 5 of Fe2+ so would times the moles of KMnO4 by 5. I will then times it by 10 as I have a 200cm3 solution but have only used 1/10th of that. I will then use the equation: Mass = Moles * Mr I will now have the mass of Fe2+ I will then reduce the Fe2+ to Fe3+ and use the exact same calculations as above to work out the total mass, but use the titre recordings from the second lot of titrations. To get the final percentage I will do the equation: (Mass of Fe2+/Total Mass)*100 This will give the the percentage of Fe2+ in the solution. ...read more.

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