DETERMINATION of the iron(ii) content of fertiliser by redox titration

[helloworld123]

Hi could someone please desperately help me with this i have the SAc tommorow and im not sure what the equations will be nor how to get them! someone plz tell me the equations and other information i might need to know about the SAC.

Method:

1. crush and weigh accurately approx. 5.50g of the fertiliser on a watchglass.
2. using a funnel and deionised water, pour this into a conical flask, swirling to dossolve. add water to approximately 60ml
3.using a pipette, add 30ml of 1M h2s04 to the flask
4. add 0.01M KMnO4 solution to the burette and titrate it against the fertiliser solution. when the solution turns a pinkish blue, the reaction is complete.
5. repeat steps 1 to 4 until 3 concordant titres are achieved.

also what errors could occur leaving to an over/underestimation

thanks so much!!!!!!

[Somye]

So you have a redox equation and your two half equations would be

MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O (Reduction)
Fe(2+) --> Fe(3+) + e(-) (Oxidation)

Therefore your full equation becomes

MnO4(-) + 8H(+) + 5Fe(2+) --> Mn(2+) + 4H2O + 5Fe(3+)

you would need to make calculations on percentage composition, which you would do through calculating n(KMnO4) through n=cv. You would then make a calculation on n(Fe2+) as it is 5*n(KMnO4). Then work out the mass of Fe2+ through n*M and then percentage comp.

[helloworld123]

thanks alot !

also what would be another underestimation? i already have a loss of fertiliser through transfer but cant think of another. also when i am finding the mass of fe2+ which equation do  i use to get the stoichoiometric ratios? m(fe)=nxm >> = x*(5*x) or will it just be fe on its own?

thanks

[helloworld123]

So you have a redox equation and your two half equations would be

MnO4(-) + 8H(+) + 5e(-) --> Mn(2+) + 4H2O (Reduction)
Fe(2+) --> Fe(3+) + e(-) (Oxidation)

Therefore your full equation becomes

MnO4(-) + 8H(+) + 5Fe(2+) --> Mn(2+) + 4H2O + 5Fe(3+)

you would need to make calculations on percentage composition, which you would do through calculating n(KMnO4) through n=cv. You would then make a calculation on n(Fe2+) as it is 5*n(KMnO4). Then work out the mass of Fe2+ through n*M and then percentage comp.

thanks alot !

also what would be another underestimation? i already have a loss of fertiliser through transfer but cant think of another. also when i am finding the mass of fe2+ which equation do  i use to get the stoichoiometric ratios? m(fe)=nxm >> = x*(5*x) or will it just be fe on its own?

thanks

[Shenz0r]

Try to come up with errors relating to the tools that you're going to use. Like, if your burette wasn't rinsed properly, then the KMnO4 solution in the burette will become diluted, and hence more titre will be required to neutralise the fertiliser solution. Since you used more volume, you have used a greater amount of mol, and hence the calculated mass of Fe will be greater than expected.

Or if your pipette wasn't rinsed properly, the H2SO4 solution in the pipette would have been diluted and hence less mol of FeSO4 are formed in the conical flask, meaning that less titre will be required of permanganate to neutralise Fe2+. This would lead to less mols of permanganate being used in the titration, and hence the calculated mass of Fe will be less than expected.

You use the molar ratios of the overall equation, as Somye has written. So n(Fe2+) = 5/1 x n(MnO4). Then you can work out the mass of iron.