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The analysis of lawnsand. (REP as a reward, I don't mind multiple suggestions)
Quite long , please take time to read it. I know most people are familiar with this...just drop any hint or suggestions. I will always reward.
The analysis of lawnsand: Find the percentage of iron present.
Introduction
Potassium manganate (VII), KMnO4, in acidic solution is a strong oxidising agent. It accepts electrons easily and is reduced to Mn2+ ions which appears colourless in solution. The half equation is
MnO4- (aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O (l)
Purple colourless
The electrons provided by reducing agents such as iron (II) salts:
Fe2+ (aq) Fe2+ (aq) + e –
Lawnsand brought at a garden centre contains sand and iron(II) sulphate (ferroussulphate), a cheap and soluble iron (II) salt. The Fe2+(aq) ions are oxidised to Fe2+(aq) ions and makes the soil slightly acidic. The pH of the soil therefore falls and some plants such as clover cannot tolerate these conditions and die. The grass is unaffected.
Adding water to lawnsand dissolves the iron (II) sulphate and the salt will be slowly oxidised by the air to iron (III) unless dilute sulphuric acid is present. Heating to too high temperature will also encourage the oxidisation reaction. There are other soluble substances present in lawnsand but they do not interfere with any redox reaction of iron (II).
The problem
Given a packet of lawnsand, devise a plan to find the percentage of iron in it using 0.0200M potassium manganate solution.
Assume that all the iron in the lawnsand is the iron (II) form, and the amount of iron (II) sulphate in the lawnsand is no greater than 8%
Your answer should include relevant calculations and reasoning for your choice.
The analysis of lawnsand: Find the percentage of iron present.
Introduction
Potassium manganate (VII), KMnO4, in acidic solution is a strong oxidising agent. It accepts electrons easily and is reduced to Mn2+ ions which appears colourless in solution. The half equation is
MnO4- (aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O (l)
Purple colourless
The electrons provided by reducing agents such as iron (II) salts:
Fe2+ (aq) Fe2+ (aq) + e –
Lawnsand brought at a garden centre contains sand and iron(II) sulphate (ferroussulphate), a cheap and soluble iron (II) salt. The Fe2+(aq) ions are oxidised to Fe2+(aq) ions and makes the soil slightly acidic. The pH of the soil therefore falls and some plants such as clover cannot tolerate these conditions and die. The grass is unaffected.
Adding water to lawnsand dissolves the iron (II) sulphate and the salt will be slowly oxidised by the air to iron (III) unless dilute sulphuric acid is present. Heating to too high temperature will also encourage the oxidisation reaction. There are other soluble substances present in lawnsand but they do not interfere with any redox reaction of iron (II).
The problem
Given a packet of lawnsand, devise a plan to find the percentage of iron in it using 0.0200M potassium manganate solution.
Assume that all the iron in the lawnsand is the iron (II) form, and the amount of iron (II) sulphate in the lawnsand is no greater than 8%
Your answer should include relevant calculations and reasoning for your choice.
Lawnsand brought at a garden centre contains sand and iron(II) sulphate (ferroussulphate), a cheap and soluble iron (II) salt. The Fe2+(aq) ions are oxidised to Fe2+(aq) ions and makes the soil slightly acidic
this should read "oxidised to Fe3+(aq) ions"
its a standard problem:
you are told that it is about 8% in FeSO4 so calculate the mass needed to make up a solution of about 0.1M in FeSO4.
Making up 1dm3 you would need 0.1 mole of FeSO4 - Mr= 152 = 15.2g
but its only 8% in FeSO4 so weigh out about 100/8 x 15.2g
Now heat up your mixture with some water (not too much - you don't want to oxidise the Fe2+ ions) and transfer the filtrate into a 1dm3 volumetric. Wash the sand thoroughly and add this to the flask as well. Take care not to lose any FeSO4 solution at this stage.
Now titrate 25cm3 aliquots of the irons solution against the KMnO4 solution.
GOD DAMN ANOTHER TITRATION!
I remeber having to do this experiment with Irn Bru (chavs you see). There were to ways to do that experiment:
1. Stand for ages beating the gas out the Irn Bru with a glass stirer and titrating aliquots against permanganate.
2. Go outside to the can machine, put in 50p and look at the side of the ****ing can.
PS. In serious the person above has it right.
I remeber having to do this experiment with Irn Bru (chavs you see). There were to ways to do that experiment:
1. Stand for ages beating the gas out the Irn Bru with a glass stirer and titrating aliquots against permanganate.
2. Go outside to the can machine, put in 50p and look at the side of the ****ing can.
PS. In serious the person above has it right.
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