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# Titration Watch

1. (Original post by ZakRob)
Yes, since it's the same question and they havent said anything otherwise.

Does that clear everything up now?

Posted from TSR Mobile
Yes sorry , ! but that is quite tricky from the examiner cause people may assume its 1dm3!

e) [PF4]+[PF6]-

by the way the one which are underline what is the oxidation number? do you get +5?
2. (Original post by otrivine)
Yes sorry , ! but that is quite tricky from the examiner cause people may assume its 1dm3!

e)[PF4]+[PF6]-

by the way the one which are underline what is the oxidation number? do you get +5?
Yes, are you doing f325 in June?

Posted from TSR Mobile
3. (Original post by ZakRob)
Yes, are you doing f325 in June?

Posted from TSR Mobile
yes I am

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.

b) 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.

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.

what do u get
4. (Original post by otrivine)
yes I am

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.

b)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.

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.

what do u get
I'l do it tommorow:P are you retaking or first time round?

Posted from TSR Mobile
5. (Original post by ZakRob)
I'l do it tommorow:P are you retaking or first time round?

Posted from TSR Mobile
Retaking , our paper in jan was hard!
6. (Original post by otrivine)
Hi Bahonsi!

did you finish all the questions?
Am stuck with this Assignment i get the conversion but every other question is just a blur to me. So worried am not gonna do well. ;'(
7. (Original post by ZakRob)
what question is it that you want help with?
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.

b) 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.

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.
8. (Original post by bahonsi)
A 25.0 cm3 sample was added.

(Original post by otrivine)
yes I am

what do u get
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.

b) 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.

0.066 moldm-3

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.

0.0176moldm-3

Do you want an explanation?
9. (Original post by ZakRob)
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.

b) 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.

0.066 moldm-3

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.

0.0176moldm-3

Do you want an explanation?
Yep got the same as you for the first one, the 0.0176 mol dm-3 is that the concentration of Fe3+ you got, it also mentioned to calculate the concentration of Fe2+ which I got 0.0836 mol dm-3 ?
10. (Original post by otrivine)
Yep got the same as you for the first one, the 0.0176 mol dm-3 is that the concentration of Fe3+ you got, it also mentioned to calculate the concentration of Fe2+ which I got 0.0836 mol dm-3 ?
Yh, 0.0176 moldm-3 is for the Fe3+ concentration and I also got 0.0836 for the concentraion of Fe2+
11. (Original post by ZakRob)
Yh, 0.0176 moldm-3 is for the Fe3+ concentration and I also got 0.0836 for the concentraion of Fe2+
so we got the same answers, you sure yes, i thought I made a mistake somewhere?

by the way ,the Kc question we did last night, is the reaction exothermic because first the Kc=1.53 at higher temp and lower temp value was 2.83 something, so as temperature lowers Kc value increases which means exothermic?
12. (Original post by otrivine)
so we got the same answers, you sure yes, i thought I made a mistake somewhere?

by the way ,the Kc question we did last night, is the reaction exothermic because first the Kc=1.53 at higher temp and lower temp value was 2.83 something, so as temperature lowers Kc value increases which means exothermic?
I'm pretty certain I got the correct answers :P

For the Kc thing your correct, but if you wanted a full explanation it would be: when the reaction takes place at a lower temperature then according to La Chatliers principle the system oppose any changes. Therefore the equilibrium would move to the exothermic side to oppose the decrease in temp. the Kc is a higher value compared to the first reaction at a higher temp, so this means the equilibrium has moved more to the right. So with both pieces of information i.e. move to exothermic side and move to the right we can deduce it is a exothermic reaction.
13. (Original post by ZakRob)
I'm pretty certain I got the correct answers :P

For the Kc thing your correct, but if you wanted a full explanation it would be: when the reaction takes place at a lower temperature then according to La Chatliers principle the system oppose any changes. Therefore the equilibrium would move to the exothermic side to oppose the decrease in temp. the Kc is a higher value compared to the first reaction at a higher temp, so this means the equilibrium has moved more to the right. So with both pieces of information i.e. move to exothermic side and move to the right we can deduce it is a exothermic reaction.
oh I GOT it!! thanks

For each of the following species calculate the oxidation state of the underlined atom
I got

a) UO3- = +5
b) HCHO =0
c) P4O10 = +5
d) AgNO3 = +5
e) [PF4]+[PF6]- = +5
14. (Original post by otrivine)
oh I GOT it!! thanks

For each of the following species calculate the oxidation state of the underlined atom
I got

a) UO3- = +5
b) HCHO =0
c) P4O10 = +5
d) AgNO3 = +5
e) [PF4]+[PF6]- = +5
Seems all correct, I'm going to do some physics revision now

If you've got any more Q's post them, and i'l do them later.
15. (Original post by ZakRob)
Seems all correct, I'm going to do some physics revision now

If you've got any more Q's post them, and i'l do them later.
thank you so much +reps

I will , post some more , is that ok
16. (Original post by ZakRob)
Seems all correct, I'm going to do some physics revision now

If you've got any more Q's post them, and i'l do them later.
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.

(ii) Write the balanced half-equations for the oxidation half reaction and the reduction half reaction

(iii) Balance the redox equation
17. This A2 work right?
18. (Original post by yodawg321)
This A2 work right?
yes
19. (Original post by otrivine)
yes
For each of the unbalanced redox equations below:

a) MnO4- + U3+ → Mn2+ + UO2+ in aqueous acid (H+ present) conditions

(i) MnO4- is the oxidizing agent U3+ is the reducing agent

(iii) MnO4- + 2U3+ → Mn2+ + 2UO2+

b) Cr2O72- + C2O42- → CO2 + Cr3+ in aqueous acid (H+ present) solutions

(i) Cr2O72- is the oxidizing agent C2O42- is the reducing agent
20. (Original post by ZakRob)
For each of the unbalanced redox equations below:

a) MnO4- + U3+ → Mn2+ + UO2+ in aqueous acid (H+ present) conditions

(i) MnO4- is the oxidizing agent U3+ is the reducing agent

(iii) MnO4- + 2U3+ → Mn2+ + 2UO2+

b) Cr2O72- + C2O42- → CO2 + Cr3+ in aqueous acid (H+ present) solutions

(i) Cr2O72- is the oxidizing agent C2O42- is the reducing agent
yep that is what I got

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.

a) Write down balanced half equations for the following

i) MnO4- is reduced to Mn2+ in acid conditions
ii) Zn is oxidized to Zn2+
iii) Fe3+ is reduced to Fe2+
iv) Fe2+ is oxidized to Fe3+ (4 marks)

b) 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. (4 marks)

c) With the second sample the zinc reacts with Fe3+ to Fe2+. Write a balanced equation for this reaction. (1 mark)

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. (5 marks)
e) Why is it necessary to filter off the zinc before the second titration takes place? (1 mark)

we did the calculation part , can you do a)c)e)

to see if my answer is same as urs

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