Redox Titrations

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Report Thread starter 5 years ago
I've been given a redox titration worksheet from chemsheets

(I don't know if that link will work)

I've done the first 2 questions, but am confused with regards to details on question 3 and 4.

I have the mark scheme available to me because I subscribed and I can't even figure out how to do it from their working out shown.

Specifically Question 3
Calculate x in the formula FeSO4.xH2O from the following data: 12.18 g of iron (II) sulphate crystals were made up to 500 cm3 acidified with sulphuric acid. 25.0 cm3 of this solution required 43.85 cm3 of 0.01 mol dm-3 KMnO4 for complete oxidation.

I get some of it but come undone after the initial calculation of moles... Would anyone mind explaining it step by step for me?
I just cannot get my head around it

Also, Question 4....
A tablet weighing 0.940 g was dissolved in dilute sulphuric acid made up to 250 cm3 with water. 25.0 cm3 of this solution was titrated with 0.00160 M K2Cr2O7 requiring 32.5 cm3 of the K2Cr2O7. Calculate the percentage by mass of Fe2+ in the tablet.

I've done the initial steps:
- Calculated moles of K2Cr2O7
- Used the molar ratio to find the number of Fe2+ moles which i thought should have been 0.0003072

And then you'd proceed to link that with Mass = moles x Mr and work out percentage from there (comes out at approx 1.82%)

However, the mark scheme multiplies the moles of Fe2+ by 10(?); causing the final answer to be 18.25%
I don't get why? Is it to do with the volumes? E.g 250cm to 25cm?
Even still, I don't understand why....

Sorry. I realise this post sounds as confused as I am :P
Any help would be massively appreciated!
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Report 5 years ago
The 12.18 g is the mass of hydrated crystals. The 43.85 cm3 allows you to work out the amount of MnO4-, which (if you have the overall redox equation) allows you to work out the amount of Fe2+ in the 25 cm3 sample from there you'd work out the amount in the 500 cm3 stock solution and hence the amount of FeSO4 in the crystals. There are a couple of ways of doing the next bit, but I'd work out the mass of FeSO4 in the crystals, then then it is a standard water of crystalisation calculation from then on.

The multiplies by 10 bit is caused by the stock solution being 250 cm3 and the titration volume being 25 cm3, i.e. the stock solution contains 10x the number of mol.

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