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# Titration Watch

1. (Original post by otrivine)
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

ai) 5e- + 8H+ + MnO4- -> + Mn2+ 4H2O
ii) Zn -> Zn2+ + 2e-
iii) e- + Fe3+ -> Fe2+
iv) Fe2+ -> Fe3+ + e-

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

e) the zinc causes Fe3+ to be reduced, the titration will oxidise Fe2+ into Fe3+ if this is then reduced there is no end point for the titration
2. (Original post by ZakRob)

ai) 5e- + 8H+ + MnO4- -> + Mn2+ 4H2O
ii) Zn -> Zn2+ + 2e-
iii) e- + Fe3+ -> Fe2+
iv) Fe2+ -> Fe3+ + e-

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

e) the zinc causes Fe3+ to be reduced, the titration will oxidise Fe2+ into Fe3+ if this is then reduced there is no end point for the titration
Yup I got those too! Actually you know the Kc calculation we did last night! I am still stuck on it, the 77.6%

3. (Original post by otrivine)
Yup I got those too! Actually you know the Kc calculation we did last night! I am still stuck on it, the 77.6%

Yeah

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4. Are you on OCR?Are you doing F324 in summer then?
5. the first calculation was 1.53 and second was 2.91, why would then the reaction be exothermic? cause if forward reaction is exothermic ,Kc decreases?
6. (Original post by otrivine)
Are you on OCR?Are you doing F324 in summer then?
Yh im doing f324 in the summer

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7. (Original post by ZakRob)
Yh im doing f324 in the summer

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Me too, what to revise

Define:zwitterion
8. (Original post by otrivine)
the first calculation was 1.53 and second was 2.91, why would then the reaction be exothermic? cause if forward reaction is exothermic ,Kc decreases?
For simplicity you can't look so deep into it. Yes I see where your coming from, the system is trying to move equilibrium to try to oppose the change, so any temp increase maybe very small

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9. (Original post by otrivine)
Me too, what to revise

Define:zwitterion
Im still doing my physics, but I'll try answering. It's an ion which can have a positive and negative charge at the same time

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10. (Original post by ZakRob)
Im still doing my physics, but I'll try answering. It's an ion which can have a positive and negative charge at the same time

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yes but also state that it is a dipolar ionic form of an amino acid by the donation of hydrogen ion from carboxyl group to amino group.

My turn, wow Physics is so hard, I just did GCSE physics and got B. No more
11. (Original post by otrivine)
the first calculation was 1.53 and second was 2.91, why would then the reaction be exothermic? cause if forward reaction is exothermic ,Kc decreases?
HI sorry am new on here...pls how did you get 2.91? thanks
12. Can you help me question 3 and show me step by step please, I am struggling. thank you.

3.Ethanoic acid can be manufactured by the followingreaction, which is carried out between150 °C and 200 °C.
CH3OH(g)+ CO(g) CH3COOH(g)

a) Amixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed toreach equilibrium at 175°C in a container with volume 5 dm3. It wasfound that 12.2 moles of ethanoic acid had been formed.
Write down anexpression for Kc for this reaction

i) Calculate the concentrations of methanol andcarbon monoxide present at equilibrium. Show your working.
(4 marks)
ii) Hence calculate Kc (to 3significant figures) including its units
(2 marks)

i) The value of the equilibrium constant forformation of ethanoic acid
(2 marks)
ii) The equilibrium yield of ethanoic acid.

c) Anothersample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide wasallowed to reach equilibrium, but at a lower temperature. This time it wasfound that 77.6% of methanol had reacted.
(i)Calculate the valueof Kc at the lower temperature. Give your answer to 3 significantfigures

13. (Original post by kayd)
HI sorry am new on here...pls how did you get 2.91? thanks
why? what course are you doing
14. (Original post by otrivine)
why? what course are you doing
Am doing a foundation degree and i happen to have the same questions...
15. (Original post by kayd)
Am doin a foundation degree ad i happe to have the same questions...
what is ur name?
16. (Original post by otrivine)
what is ur name?
Kate
17. (Original post by kayd)
Kate
my name is otrivine

Literally, if you see an example above , it explains how to do the question
18. (Original post by otrivine)
my name is otrivine

Literally, if you see an example above , it explains how to do the question
Nice name
I worked my way to 1.53 but i got stuck on my way abouts the 77.6%.
19. (Original post by kayd)
Nice name
I worked my way to 1.53 but i got stuck on my way abouts the 77.6%.
what you do is 77.6/100 x 17.6=13.something , this is the equilibrium amount/mole of the acid.

then you use this value for substarting the initial amount of CH3OH and CO. then get concentration and sub the values into the Kc expression.
20. (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-
Otrivine, what is U?

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Updated: April 11, 2013
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