October 22, 2020, 08:49:59 AM
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Topic: Back Titration Question  (Read 156 times)

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Offline S_100

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Back Titration Question
« on: September 25, 2020, 10:48:38 AM »
50 cm3 of 0.2 mol dm-3  Ce4+ added to solution of 25cm3 of Fe2+,
excess Ce4+ reduced to Ce3+ by adding 30 cm3 of 0.1mol dm-3  of C2O42-.

Calculate concentration of Fe2+ in its initial solution (25 cm3).  My Q is up to finding mol of Fe2+

Back titration Equation:
-------------------------------------
Reaction 1:
Fe2+ + Ce4+ -----> Fe3+ + Ce3+
Reaction 2:
C2O42- + 2Ce4+ ------->2 CO2 + 2Ce3+
------------------------------------
Calculation

Fe2+ react with excess Ce4+,  and this leftover Ce4+ from this first reaction,  titrated against C2O42-

mol of titre C2O42-  = 0.003
thus mol of  leftover Ce4+ that reacts with this oxidising agent = 0.006 (2 x 0.003)

total mol of Ce4+ added to solution of Fe2+  is given as 0.01

thus initial mol of Ce4+ that react with Fe2+  is 0.01 - 0.006 = 0.004

 mol of Fe2+ = ???


What I am confused about is calculating mol of Fe2+. A markscheme indicates that the mol of Fe2+ = 1/3 x mol of  Ce4+  which reacts with it.

I assumed they got this from the overall equation of the back titration

Fe2+ + 3Ce4+ +C2O42----> Fe3+ + 3Ce3+ + 2 CO2



But I don't understand why this is necessary when we know from reaction 1 mol of Fe2+  is oxidised by 1 mol Ce4+, thus mol of Ce4+ which reacts in reaction 1 is also  = mol of Fe2+ from 25cm3


Why do we need the overall reaction equation to work out how many mols of Fe2+ initially added?

Offline chenbeier

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Re: Back Titration Question
« Reply #1 on: September 25, 2020, 01:07:08 PM »
Your calculation is correct. The Statement that 1/3 Ce4+ = Fe2+ is also correct according the total equation,  but cannot be used, because we have a subtraction of the cerium solution.
You dont know  how much  Cerium will be used for each in advance.
You have 0,01 mol cerium total, but can not say its one third of iron.

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