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Topic: Redox Titrations  (Read 4937 times)

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jasonmclean1321

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Redox Titrations
« on: May 27, 2006, 11:53:34 PM »
Hi, I am rather new at these forums so please excuse me if I do not use proper etiquette or I am not specific enough.

OK, I am having trouble with my Year 12 TEE redox titrations homework, I know how the basic idea works, but I am having trouble with the more complex exam type questions. Here is an example out of the Science Teachers' Association of Western Australia (Inc) book titled Chemistry Problems For Senior Secondary School:

SET 22
10) The percentage by mass of chromium as a mineral is determined by converting a sample of known mass into sodium dichromate, and titrating an acidified solution of the sodium dichromate with a standard solution of iron (II) sulfate. Using this method, 1.27g of a chromium-containing mineral was converted into an acidified solution of sodium dichromate, which required 37.5mL of 4.00x10-1molL-1 iron (II) sulfate to reach the end-point. Calculate the percentage by mass of chromium in the mineral.

OK, that is the question I am having trouble with. Firstly I don't know wheather I am supposed to use Na2Cr2O7 or Na2Cr2O7.2H2O, and I also don't know how to work out the equation. Once I do know these I have to work out the mole ratio between sodium dichromate and iron (II) sulfate. Then work out the concentration of the iron (II) sulfate by using c=n/V and therefore the number of moles of sodium dichromate by using the mole ratio. Then I must work out the mass of sodium dichromat using n=m/M and therefore of chromium. Then simply divide the mass of chromium by 1.27g, times by 100 and done!

I just need help with the first few steps. Could someone please help me?

Thank you,
Jason

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Re: Redox Titrations
« Reply #1 on: May 28, 2006, 08:55:08 AM »
You're on the right track, but you sound slighty muddled up.

To get the balanced ionic equation you have to know what the reaction of dichromate (VI) with Fe (II) produces.

I'll give you a hint, The Fe(II) will reduce the dichromate (VI) to chromium (III) in the presence of acid. So what do you reckon the ionic equation is?

It doesn't matter which formulation of sodium dichromate you use, as long as you use the same one throughout your calculations.

1. You titrate the dichromate with a solution of iron (II) sulfate of known concentration
2. From the volume of iron (II) sulfate solution required, you can work out moles of Fe(II) reacted
3. From that you can work out moles of dichromate (using your balanced equation)
4. from this you can work out mass of dichromate (I'd use this formulation of dichromate ion Cr2O72-)
5. From the mass of dichromate ion, you can work out the mass of chromium
6. Work out the percentage chromium by mass in the mineral using the result from 5, and the known mass of mineral used.

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