Titration question doesn't make sense????

A solution of iron(II) sulfate was prepared by dissolving 10.00 g ofFeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution. The solutionwas left to stand, exposed to air, and some of the iron(II) ions became oxidised toiron(III) ions. A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in thepresence of an excess of dilute sulfuric acid.Calculate the percentage of iron(II) ions that had been oxidised by the air.

I've got up to the last stage where I have to calculate the number of moles of iron(ii) that ha been oxidised. I don't understand why we perform a subtraction.

I have the total moles of iron(ii), I have the number of moles of iron(iii)

Surely the amount of moles of Fe(ii) oxidised is the SAME as the amount of moles of Fe(iii)... isn't that logical???????

Reply 1

WhisperingTide

Do you have any iron (iv) or iron (v) etc?

Do the number of moles of iron(iii) + (iron(ii) when you're finished add up to the original number of iron(ii) moles? If not, then you probably have iron with oxidation numbers above (iii) or something else like hydrated iron(iii)oxide etc.

Reply 2

ps1265A

Bump!!!!!!!!!!!!!!!!!

Reply 3

charco

Study Forum Helper

Original post by ps1265A

A solution of iron(II) sulfate was prepared by dissolving 10.00 g ofFeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution. The solutionwas left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions. A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in thepresence of an excess of dilute sulfuric acid.Calculate the percentage of iron(II) ions that had been oxidised by the air.

I've got up to the last stage where I have to calculate the number of moles of iron(ii) that ha been oxidised. I don't understand why we perform a subtraction.

I have the total moles of iron(ii), I have the number of moles of iron(iii)

Surely the amount of moles of Fe(ii) oxidised is the SAME as the amount of moles of Fe(iii)... isn't that logical???????

read the question ...

Reply 4

ps1265A

Original post by charco

read the question ...

Doesn't the subtraction give the amount of moles that hasn't been oxidised?

I have got to the stage where I worked out the moles of iron (ii) OXIDISED by dichromate in 250cm3

The question asks to work out the percentage of iron (ii) OXIDISED

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(edited 8 years ago)

Reply 5

charco

Study Forum Helper

Original post by ps1265A

A solution of iron(II) sulfate was prepared by dissolving 10.00 g of FeSO4.7H2O (Mr = 277.9) in water and making up to 250 cm3 of solution.

moles of Fe²⁺ = 10/277.9 = 0.036 mol

Original post by ps1265A

The solution was left to stand, exposed to air, and some of the iron(II) ions became oxidised to iron(III) ions.

moles of Fe²⁺ = 0.036 - x mol
moles of Fe³⁺ = x mol

Original post by ps1265A

A 25.0 cm3 sample of the partially oxidised solution required 23.70 cm3 of 0.0100 mol dm–3 potassium dichromate(VI) solution for complete reaction in the presence of an excess of dilute sulfuric acid.

This oxidises the iron(II) which has NOT been oxidised previously.
mol dichromate(VI) ions = 0.0237 x 0.01 = 0.000237 mol

These react in a ratio of 1:6 with iron(II) ions

Hence mol iron(II) = 0.000237 x 6 = 0.001422 mol

This is from a 25ml sample so the 250 ml sample contains 0.001422 x 10 mol = 0.01422 mol

Original post by ps1265A

Calculate the percentage of iron(II) ions that had been oxidised by the air.

Initially the sample contained 0.036 mol
and after oxidation contains 0.01422 mol

hence 0.036 - 0.01422 mol have been oxidised = 0.0218 mol

percentage oxidation = (0.0218/0.036) x 100 = 60.5 %

Original post by ps1265A

I've got up to the last stage where I have to calculate the number of moles of iron(ii) that ha been oxidised. I don't understand why we perform a subtraction.

I have the total moles of iron(ii), I have the number of moles of iron(iii)

Surely the amount of moles of Fe(ii) oxidised is the SAME as the amount of moles of Fe(iii)... isn't that logical???????

Do you see now?

Reply 6

ps1265A

Original post by charco

moles of Fe²⁺ = 10/277.9 = 0.036 mol

moles of Fe²⁺ = 0.036 - x mol
moles of Fe³⁺ = x mol

This oxidises the iron(II) which has NOT been oxidised previously.
mol dichromate(VI) ions = 0.0237 x 0.01 = 0.000237 mol

These react in a ratio of 1:6 with iron(II) ions

Hence mol iron(II) = 0.000237 x 6 = 0.001422 mol

This is from a 25ml sample so the 250 ml sample contains 0.001422 x 10 mol = 0.01422 mol

Initially the sample contained 0.036 mol
and after oxidation contains 0.01422 mol

hence 0.036 - 0.01422 mol have been oxidised = 0.0218 mol

percentage oxidation = (0.0218/0.036) x 100 = 60.5 %

Do you see now?

Fantastic explanation! Thanks :smile:

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Reply 7

thevish_23

These react in a ratio of 1:6 with iron(II) ions

How do you get this ratio?

Reply 8

charco

Study Forum Helper

Original post by thevish_23

These react in a ratio of 1:6 with iron(II) ions

How do you get this ratio?

Chromium(VI) gets reduced to chromium (III) and there are two chromium(VI) atoms in the dichromate ion.
Cr₂O₇^2- + 14H⁺ + 6e ===> 2Cr³⁺ + 7H₂O

and iron(II) only loses one electron to become iron(III)
Fe²⁺ ===> Fe³⁺ + 1e

So the two species react in a ratio of 1:6