[doctorwho23]

hello,
basically i'm extremely confused about the sub-part c) of this question, i was wondering of i could please get any help on this?

3. Iron(II) ions are oxidised by manganate(VII) ions.
A 25cm3 aliquot of solution containing Fe2+ and Fe3+ ions was acidified and titrated against potassium manganate(VII) solution. 15.0cm3 of a 0.020 moldm-3 solution of potassium manganate(VII) was required.

a) Use the following half-equations to construct an ionic equation representing this reaction.
MnO4- + 8H+ + 5e-    Mn2+ + 4H2O
and Fe2+  Fe3+ + e-
I got this answer: Mn04- + 8H+ + 5Fe2+ ---> Mn2+ + 4H2O + 5Fe3+

A second 25cm3 aliquot was reduced with zinc and titrated against the same manganate(VII) solution. 19cm3 of the oxidant solution were required.
b) Write an ionic equation to represent this reaction between iron ions and zinc metal
My answer: 2Fe3+ + Zn ------> 2Fe2+ + Zn2+

c) Now calculate the concentration of the Fe3+ ions in the aliquot of solution.
I'm having a lot of trouble with part d), because I don't understand firstly how the Fe3+ ions come about and I also don't really know what to do with the information I've been given. Thank you!

[sjb]

hello,
basically i'm extremely confused about the sub-part c) of this question, i was wondering of i could please get any help on this?

3. Iron(II) ions are oxidised by manganate(VII) ions.
A 25cm3 aliquot of solution containing Fe2+ and Fe3+ ions was acidified and titrated against potassium manganate(VII) solution. 15.0cm3 of a 0.020 moldm-3 solution of potassium manganate(VII) was required.

a) Use the following half-equations to construct an ionic equation representing this reaction.
MnO4- + 8H+ + 5e-    Mn2+ + 4H2O
and Fe2+  Fe3+ + e-
I got this answer: Mn04- + 8H+ + 5Fe2+ ---> Mn2+ + 4H2O + 5Fe3+

A second 25cm3 aliquot was reduced with zinc and titrated against the same manganate(VII) solution. 19cm3 of the oxidant solution were required.
b) Write an ionic equation to represent this reaction between iron ions and zinc metal
My answer: 2Fe3+ + Zn ------> 2Fe2+ + Zn2+

c) Now calculate the concentration of the Fe3+ ions in the aliquot of solution.
I'm having a lot of trouble with part d), because I don't understand firstly how the Fe3+ ions come about and I also don't really know what to do with the information I've been given. Thank you!

Initially you have a mixture of iron(II) and iron(III). Only the iron(II) reacts with permanaganate. Once you treat the mixture with zinc, all the iron(III) is converted to iron(II).

[vmelkon]

2Fe3+ + Zn ------> 2Fe2+ + Zn2+

I would have written
2 Fe3+ + 3 Zn => 2 Fe + 3 Zn2+

[maivu]

hello,
basically i'm extremely confused about the sub-part c) of this question, i was wondering of i could please get any help on this?

3. Iron(II) ions are oxidised by manganate(VII) ions.
A 25cm3 aliquot of solution containing Fe2+ and Fe3+ ions was acidified and titrated against potassium manganate(VII) solution. 15.0cm3 of a 0.020 moldm-3 solution of potassium manganate(VII) was required.

a) Use the following half-equations to construct an ionic equation representing this reaction.
MnO4- + 8H+ + 5e-    Mn2+ + 4H2O
and Fe2+  Fe3+ + e-
I got this answer: Mn04- + 8H+ + 5Fe2+ ---> Mn2+ + 4H2O + 5Fe3+

A second 25cm3 aliquot was reduced with zinc and titrated against the same manganate(VII) solution. 19cm3 of the oxidant solution were required.
b) Write an ionic equation to represent this reaction between iron ions and zinc metal
My answer: 2Fe3+ + Zn ------> 2Fe2+ + Zn2+

c) Now calculate the concentration of the Fe3+ ions in the aliquot of solution.
I'm having a lot of trouble with part d), because I don't understand firstly how the Fe3+ ions come about and I also don't really know what to do with the information I've been given. Thank you!

(1) Fe³⁺  + 1e     Fe²⁺
     0.0019             0.0019 
(2) Zn                                Zn²⁺  + 2e
(3) MnO₄^- + 8H⁺ + 5Fe²⁺  Mn²⁺ + 4H₂) + 5Fe³⁺
     0.00038            0.0019     

These above equation shows you the order of all reactions happen to the secon 25 cm3 solution.
From these equations, Fe²⁺ is being oxidized by potassium manganate(VII). Therefore, the oxidant solution mentioned in the question is solution of KMnO₄.

V_KMnO4 = 19 cm³ = 0.019 dm³
Concentration: 0.02 M
n_KMnO4 = 0.019*0.02 = 0.00038 moles

Plug it in the above equations, have
Number of moles of Fe³⁺ = 0.0019 moles

The aliquot has 2 part, the second part contains half of the amount of Fe³⁺ in the whole solution. However, the concentration of the aliquot is the same as in any part of it

The concentration of Fe³⁺
C=n/V = 0.0019/0.025 = 0.076 M

[doctorwho23]

Thank you!!