[whitnos]

Aim: To determine the amount of Fe2+ in a dissolved wire solution.

Method: 25cm^3 of a sol. Of iron wire in dilute sulphuric acid(some in the form of Fe3+ and some in the form of Fe2+) was pipetted and placed into a conical flask to which 10cm^3 of dil. H2SO4 was added. The contents were titrated using Potassium Manganate (3.16 gdm^-3 KmnO4) until a pale pink colour was observed.

(1M H2SO4 was used)

Calculations:

Title value =22.73

Calculate the concentration in mol/dm^3 Potassium Manganate, KMnO4

Mass ok KmnO4 = 3.16 gdm^-3

Molecular mass of KMnO4 = 158g

Concentration = Mass/Mol. Mass

= 3.16/158

= 0.02 mol dm^-3

What is the concentration in mol/dm^3 of MnO4- ions in the sol.

Mass of MnO4/mass of KMnO4 x conc. Of KMnO4

= 119/158 x 0.02

= 0.015 mol dm^-3

Calculate the no. of moles of MnO4- ions run from the burette into the titration flask

If in 1000cm^3 of KMnO4 sol, there is 0.015 mol dm^-3 of MnO4-

Then in 22.7 cm^3 (title value), there will be: 1000/22.73 x 0.015 mol dm^-3

= 0.0003 mol

I have to write half equations for the reaction:

MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l)

and Fe2+(aq) -----> Fe3+(aq) + e-

To combine the no. of electrons used must be equal to the no. formed so the second equation needs to be multiplied by five:

MnO4-(aq) + 8H+(aq) + 5e- ------> Mn2+(aq) + 4H2O(l)

5Fe2+(aq) -----> 5Fe3+(aq) + 5e-

Then I add them together to give:

MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) ------> Mn2+(aq) + 4H2O(l) + 5Fe3+(aq)

Find the no. of moles Fe2+ ions present in the titration flask:

Ratio MnO4- : 5Fe2+

So no. of moles Fe2+ present = 0.0003 x 5

= 0.0015 mol

Calculate the mass of Fe2+ ions in 1 dm^3:

If in 25 cm^3, there was 0.0015 mol

Then in 1000cm^3, there will be: 1000/25 x 0.0015

= 0.06 mol dm^-3

PLEASE CORRECT ME!

[Hunter2]

The only thing what is wrong is your moles of MnO₄^-

0.02 mol/l KMnO₄ = 0.02 mol/l MnO₄^- Different are the masses but the moles have to be equal. So 0,015 mol/l is wrong.

Think KMnO₄ => K⁺ + MnO₄^-

[Borek]

THe way you are rounding and reporting results is wrong.

= 3.16/158

= 0.02 mol dm^-3

You have three significant figures in both numbers, so you should report concentration with three significant figures. But for further calculations you should use additional 2 or 3 so called guard digits (or just use whatever your calculator has in its register, with full accuracy). Otherwise you will get hit by rounding errors (actually your final result is incorrect for that reason).

Rule of thumb is: use full accuracy for calculations, report results with a correct number of significant digits.

[whitnos]

The only thing what is wrong is your moles of MnO₄^-

0.02 mol/l KMnO₄ = 0.02 mol/l MnO₄^- Different are the masses but the moles have to be equal. So 0,015 mol/l is wrong.

Think KMnO₄ => K⁺ + MnO₄^-

My impatience got the best of me so I did some research online and was able to ascertain my error: an issue with the cation/anion ratio (your correction essentially) Thanks for replying anyway.

[whitnos]

THe way you are rounding and reporting results is wrong.

= 3.16/158

= 0.02 mol dm^-3

You have three significant figures in both numbers, so you should report concentration with three significant figures. But for further calculations you should use additional 2 or 3 so called guard digits (or just use whatever your calculator has in its register, with full accuracy). Otherwise you will get hit by rounding errors (actually your final result is incorrect for that reason).

Rule of thumb is: use full accuracy for calculations, report results with a correct number of significant digits.

Actually my final answer is inaccurate due to my approach to the second question. I disregarded anion/cation ratio and somehow came up with a method and an answer that I deemed suitable. From further research however, I was able to note my mistake and subsequently correct it. While my approximating may lead to inaccuracies, I can assure you that they weren't the reason for my incorrect final answer. Anyway, I thank you for taking the time to reply and I'll certainly take your advice.

[Borek]

While my approximating may lead to inaccuracies, I can assure you that they weren't the reason for my incorrect final answer.

Perhaps I wasn't clear enough. Your final result was wrong for two reasons - one was the incorrect calculation of permanganate concentration, no doubt about it.

But if we ignore this step and assume concentration of permanganate was 0.015M, calculations without premature rounding should yield 0.00170 moles of iron, not 0.0015 moles. That's 13% off due to rounding errors alone.