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# Redox titrations - stuck :( watch

1. Hi, I'm trying to teach myself the redox titrations section from the OCR chemistry A2 spec. I'm stuck on the spread question on page 221 in the textbook, please can someone help me step by step.

The question is:
A sample of metal of mass 2.50g was thought to contain iron. The metal was reacted with sulphuric acid and the resulting solution made up to 250cm^3. 25.0cm^3 of this solution was titrated against 0.0180 mol dm^-3 potassium manganate (VII) solution and 24.80cm^3 was required. Calculate the percentage by mass of iron in the metal sample.

This is what I've attempted of it:

Amount of MnO4- = c x v/1000 = 0.0180 x 24.80/1000 = 4.64 x 10^-4

Moles of Fe^2+ = 4.64 x 10^-4 x 5 (molar ratio) = 2.32 x 10^-3 mol in 25cm^3
therefore in 250cm^3 there are 0.0232mol

Mass of Fe = 0.0232 x 55.8 = 1.29546g

% mass of iron in the tablet = mass of iron/mass of tablet x 100 = 1.29456/2.5 x 100
=51.7824%

However, the back of the book has different working out and thus a different final answer, it says:

Amount of MnO4- = 0.000446 mol
amount of Fe^2+ = 0.00223 mol
Amount of Fe2+ in 250cm^3 = 0.0223 mol
mass of Fe = 1.245g
%Fe = 49.8%

I don't understand how it gets 0.0223 instead of 0.0232. Please can someone tell me where I've gone wrong?
2. (Original post by kej817)
Hi, I'm trying to teach myself the redox titrations section from the OCR chemistry A2 spec. I'm stuck on the spread question on page 221 in the textbook, please can someone help me step by step.

The question is:
A sample of metal of mass 2.50g was thought to contain iron. The metal was reacted with sulphuric acid and the resulting solution made up to 250cm^3. 25.0cm^3 of this solution was titrated against 0.0180 mol dm^-3 potassium manganate (VII) solution and 24.80cm^3 was required. Calculate the percentage by mass of iron in the metal sample.

This is what I've attempted of it:

Amount of MnO4- = c x v/1000 = 0.0180 x 24.80/1000 = 4.64 x 10^-4

4.464 x 10-4

(Original post by kej817)

Moles of Fe^2+ = 4.64 x 10^-4 x 5 (molar ratio) = 2.32 x 10^-3 mol in 25cm^3
therefore in 250cm^3 there are 0.0232mol
4.464 x 10-4 x 5 = 2.232 x 10-3

therefore in 250cm^3 there are 2.232 x 10-2 mol

(Original post by kej817)

Mass of Fe = 0.0232 x 55.8 = 1.29546g
Mass of Fe = 2.232 x 10-2 x 55.8 = 1.245g

(Original post by kej817)

% mass of iron in the tablet = mass of iron/mass of tablet x 100 = 1.29456/2.5 x 100
=51.7824%
100 x 1.245/2.5 = 49.8%

(Original post by kej817)

However, the back of the book has different working out and thus a different final answer, it says:

Amount of MnO4- = 0.000446 mol
amount of Fe^2+ = 0.00223 mol
Amount of Fe2+ in 250cm^3 = 0.0223 mol
mass of Fe = 1.245g
%Fe = 49.8%

I don't understand how it gets 0.0223 instead of 0.0232. Please can someone tell me where I've gone wrong?
3. (Original post by charco)

4.464 x 10-4

4.464 x 10-4 x 5 = 2.232 x 10-3

therefore in 250cm^3 there are 2.232 x 10-2 mol

Mass of Fe = 2.232 x 10-2 x 55.8 = 1.245g

100 x 1.245/2.5 = 49.8%

Ah thank you I kept checking other parts of the calculation but not that part for some reason.

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Updated: March 19, 2011
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