Titration

OP

Original post by ZakRob

Yes, since it's the same question and they havent said anything otherwise.

Does that clear everything up now? :smile:

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?

Reply 41

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

Reply 42

OP

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

Reply 43

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

Reply 44

OP

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!

Reply 45

bahonsi

16

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. ;'(

Reply 46

bahonsi

16

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.

Reply 47

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?

(edited 11 years ago)

Reply 48

OP

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 ?

Reply 49

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+

Reply 50

OP

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?

Reply 51

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.

Reply 52

OP

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

Reply 53

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 :smile:

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

Reply 54

OP

Original post by ZakRob

Seems all correct, I'm going to do some physics revision now :smile:

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

thank you so much +reps :biggrin:

I will , post some more , is that ok :colondollar:

Reply 55

OP

Original post by ZakRob

Seems all correct, I'm going to do some physics revision now :smile:

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

This A2 work right?

OP

Original post by yodawg321

This A2 work right?

yes

Reply 58

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

Reply 59

OP