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Author Topic: Back titration water with k2mn2o7  (Read 373 times)

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Amanyy

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Back titration water with k2mn2o7
« on: December 29, 2018, 01:13:30 AM »

I have 100ml of water mixed with 25.00ml of 0.02517M k2cr2o7 (m.w=294.2g/mol)to reduce Fe+3 to Fe+2 ..the exess from k2cr2o7 back titration with 5.30ml to 0.0949 Fe+2 ...need to found concentration Fe+2 (55.8g/mol) in ppm ..I don't know how to solve it and how can use 100 water
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chenbeier

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Re: Back titration water with k2mn2o7
« Reply #1 on: December 29, 2018, 02:56:26 AM »

Some errors. In the headline it is written K2Mn2O7. I think you mean K2Cr2O7. Second it is an oxidation of Iron-II to iron-III, not a reduction.


You need the redox reaction. How much mole  iron correspond to the given molarity of the dichromate. With the difference you can find out the original concentration.
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Amanyy

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Re: Back titration water with k2mn2o7
« Reply #2 on: December 29, 2018, 11:23:07 AM »

Ya I know sorry about that mistakes in rewriting..there' 6 mole of Fe+3 and 6mole of fe+2 ..
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Borek

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Re: Back titration water with k2mn2o7
« Reply #3 on: December 30, 2018, 07:54:29 AM »

All back titrations are identical: follow the stoichiometry of the titration to calculate amount of excess substance, subtract the excess from the initial amount and you know how much of the substance reacted.

Once you calculate amount of Fe2+ in the initial solution, its volume (100 mL) will be needed to find the concentration.
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