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# Titration watch

1. A 25.0 cm3 sample of a solution containing both Fe2+ and Fe3+ ions was acidified and titrated with 0.022 mol dm-3 manganate (VII) solution (containing the MnO4- ion) requiring 15.0 cm3. A second sample of the same solution was treated with excess Zn to reduce all the Fe3+ to Fe2+ and then, after filtering, was titrated against the same permanganate solution requiring 19.0 cm3.

In the first titration only the Fe2+ reacts with acidified MnO4-. Use this information, and a balanced net overall redox reaction, to calculate the concentration of Fe2+ in the solution.

my answer is 0.066 mol dm-3. I got thi by using the equation
8H+ + Mno4- + 5fe2+ ----> Mn2+ + 5Fe3+ + 4h2o

are my correct?
2. (Original post by otrivine)
A 25.0 cm3 sample of a solution containing both Fe2+ and Fe3+ ions was acidified and titrated with 0.022 mol dm-3 manganate (VII) solution (containing the MnO4- ion) requiring 15.0 cm3. A second sample of the same solution was treated with excess Zn to reduce all the Fe3+ to Fe2+ and then, after filtering, was titrated against the same permanganate solution requiring 19.0 cm3.

In the first titration only the Fe2+ reacts with acidified MnO4-. Use this information, and a balanced net overall redox reaction, to calculate the concentration of Fe2+ in the solution.

my answer is 0.066 mol dm-3. I got thi by using the equation
8H+ + Mno4- + 5fe2+ ----> Mn2+ + 5Fe3+ + 4h2o

are my correct?
3. (Original post by charco)
thank you

d) After all the Fe3+ has been converted to Fe2+ it is then titrated again with acidified MnO4-. Calculate the concentration of Fe2+ at this point and use this value, and your answer to b), to calculate the concentration of Fe3+ present in the solution before the zinc was added.

this is the second part to the question
4. (Original post by otrivine)
thank you

d) After all the Fe3+ has been converted to Fe2+ it is then titrated again with acidified MnO4-. Calculate the concentration of Fe2+ at this point and use this value, and your answer to b), to calculate the concentration of Fe3+ present in the solution before the zinc was added.

this is the second part to the question
5. (Original post by charco)
the titre before you reduced the iron (III) was 15 cm3 of manganate and the titre after you reduced the iron (III) was 19 cm3. So the difference is the amount of manganate needed to oxidise the "new" iron (II) that came from the reduced iron (III)

difference in titre = 4 cm3

that's (4/1000) x 0.022 = 0.000 088 moles of manganate

times by 5 to get the iron (II) (which equals iron (III) )

so 0.000 44 moles of iron (III)

that was in 25cm3, so in a litre you have

0.000 44 x (1000/25) = 0.0176 mol dm-3 of iron (III)

is this correct for the above ?
6. (Original post by otrivine)
the titre before you reduced the iron (III) was 15 cm3 of manganate and the titre after you reduced the iron (III) was 19 cm3. So the difference is the amount of manganate needed to oxidise the "new" iron (II) that came from the reduced iron (III)

difference in titre = 4 cm3

that's (4/1000) x 0.022 = 0.000 088 moles of manganate

times by 5 to get the iron (II) (which equals iron (III) )

so 0.000 44 moles of iron (III)

that was in 25cm3, so in a litre you have

0.000 44 x (1000/25) = 0.0176 mol dm-3 of iron (III)

is this correct for the above ?
It most certainly is correct ...
7. (Original post by charco)
It most certainly is correct ...
should I also include the equation , its the same but you put the Fe2+ on LHS and Fe3+ on RHS

For each of the unbalanced redox equations below:

a) MnO4- + U3+ → Mn2+ + UO2+ in aqueous acid (H+ present) conditions
b) Cr2O72- + C2O42- → CO2 + Cr3+ in aqueous acid (H+ present) solutions

(i) Identify which species is the oxidizing agent and which is the reducing agent.

for this

a) oxidisng agnet = Mno4- and reducing agent = UO2+
b) oxidisng agent = Cr2o7 2- and reducing agent = C2O4 2-
8. (Original post by otrivine)
should I also include the equation , its the same but you put the Fe2+ on LHS and Fe3+ on RHS

For each of the unbalanced redox equations below:

a) MnO4- + U3+ → Mn2+ + UO2+ in aqueous acid (H+ present) conditions
b) Cr2O72- + C2O42- → CO2 + Cr3+ in aqueous acid (H+ present) solutions

(i) Identify which species is the oxidizing agent and which is the reducing agent.

for this

a) oxidisng agnet = Mno4- and reducing agent = UO2+
b) oxidisng agent = Cr2o7 2- and reducing agent = C2O4 2-
a) incorrect
b) correct
9. (Original post by charco)
a) incorrect
b) correct
oh no, why is part a wrong?
10. (Original post by charco)
a) incorrect
b) correct
IS IT U3+ ? FOR THE first one part a) for the reducing agent
11. (Original post by otrivine)
IS IT U3+ ? FOR THE first one part a) for the reducing agent
12. (Original post by charco)
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted.
i) Calculate the value of Kc at the lower temperature. Give your answer to 3 significant figures.
(6 marks)

this one?
13. Am stuck with question 6 b and d help
14. (Original post by bahonsi)
Am stuck with question 6 b and d help
Hi Bahonsi!

did you finish all the questions?
15. (Original post by otrivine)
Hi Bahonsi!

did you finish all the questions?
what question is it that you want help with?
16. (Original post by ZakRob)
what question is it that you want help with?
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted.
i) Calculate the value of Kc at the lower temperature. Give your answer to 3 significant figures.
(6 marks)

17. (Original post by otrivine)
Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) ---> CH3COOH(g)

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted.
i) Calculate the value of Kc at the lower temperature. Give your answer to 3 significant figures.
(6 marks)

firstly, you need a volume of the container which this is carried out in, to be able to calculate concentrations.

however if you do use moles for the Kc calculation (which is technically incorrect) the answer would be o.583
18. (Original post by ZakRob)
firstly, you need a volume of the container which this is carried out in, to be able to calculate concentrations.
is that the answer you got?
19. (Original post by ZakRob)
firstly, you need a volume of the container which this is carried out in, to be able to calculate concentrations.

however if you do use moles for the Kc calculation (which is technically incorrect) the answer would be o.583
So my answer is correct right
20. (Original post by otrivine)
is that the answer you got?
The answer if calculated with moles, is however though 0.583. so your method is of working is correct. You just need to technically calculate concentration, then use the Kc equation.

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Updated: April 11, 2013
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