redox Titration help! Watch

Deyn_08
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hey

rep for anyone who does a worked solution for this
many thanks, i'm tearing my hair out over this!!

Ammonium iron (II) sulphate crystals have the following formula:
(NH4)2SO4.FeSO4.nH2O. In an experiment to determine n, 8.492g of the salt were dissolved and made up to 250 cm3 of solution with distilled water and dilute sulphuric acid. A 25 cm3 portion of the solution was further acidified and titrated against potassium manganate (VII) solution of concentration 0.0150 moldm-3. A volume of 22.5 cm3 was required. Determine n.
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qwerty11
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Hi

I've been trying to get my head around these questions today too but I think I get it so i'll explain it as best as I can

Umm well to start off with you need to write two redox half equations (for the iron 2+ ions and the manganate - ions) and then combine then to give an overall equation:
Fe2+ --> Fe3+ + e- (oxidation)
MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O (reduction of Mn)
Balance the electrons to give an overall equation:
MnO4- + 5Fe2+ + 8H+ --> 5Fe3+ + Mn2+ + 4H2O

MOLES IN
You now work out the number of moles of potassium manganate (VII) which has reacted using the formula n= c x v / 1000:
n= (0.0150)(22.5)/1000
n= 3.375 x10^-4

RATIO
You then look at your overall redox equation to look at how many moles of iron react for every mole of potassium manganate. Since there's 5Fe2+ ions for every MnO4- ion, five times as many iron ions have reacted.

MOLES OUT
The moles of iron that have reacted= 5(3.375 x 10^-4)
= 1.6875 x 10^-3 moles that have reacted.
However, since that was only in 25cm3 of the original 250cm3, there must have been ten times as many moles of iron in the original solution, so the number of moles
= (1.6875 x 10^-3) x 10
= 1.6875 x 10^-2

(Phew we're nearly there...)

Now you use the formula n=m/Mr and rearrange it to give Mr=m/n.
We got told the mass in the question and we've just worked out the number of moles of iron that must have been in the original solution.

Mr = 8.492/(1.6875 x 10^-2)
Mr= 503.2296296
Mr= 503

This is the Mr of the ammonium sulphate crystals (including the water molecules). Now you can work out the Mr of the crystals (excluding the water) i.e. (NH4)2SO4.FeSO4 by just adding the relative atomic masses of the elements from the periodic table.

Mr= 2(14 + 4) + 32 + 4(16) + 56 + 32 + 4(16)
Mr= 284

So the Mr of the crystals with water= 503
And the Mr of the crystals without water= 284
So the water (nH2O) = 503-284
= 219

Now all you need to do is divide by the Mr of water which is 18 to give n.
219/18 = 12.1666667
= 12

Wow this is a really long post but I hope it helps (and makes some sort of sense to anyone but myself )
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