Ial chemistry unit 5 edexcel

Hey guys!
Iron(II) ions in solution are oxidized by the air to iron(III) ions.
The amounts of iron(II) and iron(III) ions in solution can be found by using titrations
with acidified potassium manganate(VII) which only reacts with the iron(II) ions.
MnO4
−
+ 5Fe2+ + 8H+ ĺ Mn2+ + 5Fe3+ + 4H2O
• 25.0cm3
portions of a solution A, containing a mixture of iron(II) and iron(III)
ions, were acidified and titrated with potassium manganate(VII) solution of
concentration 0.0195mol dm−3.
• The mean titre was 16.80cm3
.
• About 150 cm3
of solution A was reacted with excess zinc, which reduced the
iron(III) ions to iron(II) ions. The excess zinc was filtered off.
• 25.0cm3
portions of this reduced solution, which contained iron(II) ions but no
iron(III) ions were acidified and titrated with potassium manganate(VII) solution
of concentration 0.0195 mol dm−3.
• The mean titre was 18.20cm3
.
Calculate the mass of iron(II) ions and the mass of iron(III) ions in 500cm3
of the
original solution A.

In this question,I got 4/5 marks.
But at the very end,my mass of Fe3+ comes 1.98g,but in the mark scheme it comes 0.152g.
I am not understanding why.
Please somone explain it!

Reply 1

UtterlyUseless69

I’ve highlighted in bold some important things I think you’ve missed.

The solution contains both Fe^2+ and Fe^3+ - the first titration tells you how much Fe^2+ there is and the second tells you how much Fe^2+ AND Fe^3+ there is.

The titre in the first titration (i.e volume of KMnO4 dropped from the burette) is 16.80 cm^3 and the concentration of KMnO4 is 0.0195 mol dm^-3.

The moles of KMnO4 = 16.80/1000 dm^3 x 0.0195 mol dm^-3 = 0.0003276 mol

But from the equation, we can see that for every KMnO4 used, 5Fe^2+ ions are consumed.

So moles of Fe^2+ = 5 x 0.0003276 mol = 0.001638 mol

But this is in just a 25 cm^3 portion - the total amount of solution is 500 cm^3, so there are actually (500 cm^3)/(25 cm^3) = 20 times the calculated amount of Fe^2+ in the entire solution:

Total mol Fe^2+ = 20 x 0.001638 mol = 0.03276 mol

We can now repeat the above procedure using a titre of 18.20 cm^3 instead of 16.80 cm^3 to find the total amount of both Fe^2+ and Fe^3+. Doing this shows that the total amount of iron in the solution is 0.03549 mol.

The amount of Fe^3+ is the total amount minus the amount of Fe^2+, so 0.03549 mol - 0.03276 mol = 0.00273 mol.

We can look up the relative mass of iron from the periodic table (55.8 g/mol) and we find that the mass of Fe^3+ is

0.00273 mol x 55.8 g/mol = 0.152 g (3 sf)

My suspicion is you just calculated the total moles of iron and then multiplied by 55.8 g/mol. This is the total mass of iron in the solution - i.e the mass of both Fe^2+ and Fe^3+.