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# The basic aim of this experiment is to determine the percentage concentration of Iron (II) that is contained in lawn sand. By carrying out a series of titrations as outlined below, hopefully, the percentage of iron can be determined.

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Introduction

Skill 1: Planning; the Analysis of Lawnsand Basic Aim The basic aim of this experiment is to determine the percentage concentration of Iron (II) that is contained in lawn sand. By carrying out a series of titrations as outlined below, hopefully, the percentage of iron can be determined. Based upon the following equation: MnO4 (aq) + 8H+(aq) + 5Fe2+(aq) (((Mn2+ (aq) + Fe3+(aq) + 4H2O (L) We will titrate Potassium Manganate against a solution of the lawnsand. Apparatus needed: - Beaker (250 cm3) - Pipette (25 and 10cm3) - Pipette filler - Glass rod - Burette with the clamp and stand and boss - Funnel - Filter paper - Conical flask (250 cm3) - Graduated volumetric flask (250 and 1000cm3) - White tile - Heating wooden mat - Volumetric flask (250 cm3) and bung - Balance - Glass weighing bottle (if available, if not a weighing plastic cup) Agents and Reagents - 38.75 grams of lawnsand If we were to titrate the solution, we would want to obtain a result within a few cm3 difference from 20 cm3,1 therefore, according to the equation: MnO4 (aq) + 8H+(aq) + 5Fe2+(aq) ...read more.

Middle

Conclusion

1 This is the standard average titre results, hence the fact we would like to end up closer to this value 2 if it does become hydrolysed, iron (II) would be oxidised into iron (III) slowly by the air, it would jeopardise the calculations, hence the fact that the dilute sulphuric acid needs to be in excess 3 High concentration of hydrogen ions is required; a strong acid has a larger concentration of hydrogen ions, as it dissociates more. 4 If it was, it may react with the reductant and affect titration results 5 it may be oxidised by the manganate (VII) ions and therefore affect the titration results 6 to ensure that any colour change in the titration would be apparent 7 25 cm3 is the standard value used in titrations, that ensures that the volume would contain enough of the element that is titrated, to be accurate enough and it would not require too much of the solution it is titrated against 8 an amount of 10 cm3 ensures that the acid is in excess 9 that occurs when 2 or more results are within 0.1 cm3 of each other, or exactly the same ...read more.

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