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# Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

1. (Original post by Rosi M)
Hey i was wondering if you could explain something to me please?

If you've done the june 2010 paper, question 3d (the one on acid base pairs)

The equation is: HNO3 + CH3CH2COOH -----> ---- ------

I wrote: HNO3 + CH3CH2COOH ----> CH3CH2COO- + H2NO3

This was wrong and it was CH3CH2COOH2 + HNO3-

How do i know this? They both acids so how do i know which proton is going where?

Would you be able to explain how you would tackle question 7d in that same paper? I did it all the way till finding the concentration of undiluted hydrogen peroxide? I could do the oxygen part. Just that undiluted part please

Thank you, really appreciate this
at the start of the question it says that propanoic is a weak acid, that's all you really have to go on. you should know that nitric is a strong acid too though
so the stronger acid will act as the acid, making the other act as a base.

7b) so 0.02345dm3 of 0.02mol dm3 MnO4 2- was required to neutralise
so 0.02345x0.02 = moles of MnO4 2- =0.000469moles
ratio of MnO4 2- to H2O2 is 2:5, so you times the moles of MnO4 2- by 2.5 to get 0.0011725 - thats in 25cm3, so x10 to get the moles in 250cm3 which is 0.011725moles
n=c x v so c=n/v so 0.011725/0.025= concentration of H2O2 (because you want it undiluted and the question says there was 25cm3 of undiluted H2O2, so you divide by 0.025 not 0.25)= 0.469mol dm3
but you want grams per dm3
so m=n x M so 0.469 x 34 = 15.946 g dm-3

7d
2. (Original post by volvicstar)
also what do we need to know on neutralisation curves?
I think you need to know about the equivalence/end point (vertical part on graph) and how a suitable indicator would have a pH range that lies on the vertical part of the graph around the end point.

Also the shapes - strong acid/strong base goes from a very low pH to a very high pH;
weak acid/strong base goes from a pH of around 4-6 to a very high ph;
strong acid/weak base goes from a very low pH to a pH of around 8-10;
weak acid/weak base doesn't have an equivalence point due to there not being a vertical part of the curve on the graph so you don't use an indicator for it and goes from a pH of around 4-6 to a pH of around 8-10.

I don't think we need to remember names of indicators and their ranges but I'm not sure. We probably need to remember that the x axis is vol' of base added if the curve is going upwards not down and that the y axis is the pH.
3. (Original post by Rogercbinboy)
I think you need to know about the equivalence/end point (vertical part on graph) and how a suitable indicator would have a pH range that lies on the vertical part of the graph around the end point.

Also the shapes - strong acid/strong base goes from a very low pH to a very high pH;
weak acid/strong base goes from a pH of around 4-6 to a very high ph;
strong acid/weak base goes from a very low pH to a pH of around 8-10;
weak acid/weak base doesn't have an equivalence point due to there not being a vertical part of the curve on the graph so you don't use an indicator for it and goes from a pH of around 4-6 to a pH of around 8-10.

I don't think we need to remember names of indicators and their ranges but I'm not sure. We probably need to remember that the x axis is vol' of base added if the curve is going upwards not down and that the y axis is the pH.

Thanks
4. (Original post by Hydro.ZX)
Hi all, i'm finding the past exam questions too easy, does anyone here have a list of stretch and challenge questions that i'll struggle to do? thanks.
Small post so will probably get ignored, so i'm quoting it.
5. (Original post by Rosi M)
Hey i was wondering if you could explain something to me please?

If you've done the june 2010 paper, question 3d (the one on acid base pairs)

The equation is: HNO3 + CH3CH2COOH -----> ---- ------

I wrote: HNO3 + CH3CH2COOH ----> CH3CH2COO- + H2NO3

This was wrong and it was CH3CH2COOH2 + HNO3-

How do i know this? They both acids so how do i know which proton is going where?

Would you be able to explain how you would tackle question 7d in that same paper? I did it all the way till finding the concentration of undiluted hydrogen peroxide? I could do the oxygen part. Just that undiluted part please

Thank you, really appreciate this
Hey, no worries:
First question: You know the second one is right because if you look at the reactants, you know HNO3 is a strong acid, and propanoic is a weak acid, therefore in relative you know HNO3 is the acid that will donate H+, and CH3CH2COOH will accept H+

Does that make sense?

Now secondly, hold let me find the paper...

Okay so: moles of KMnO4 = 0.000469
because of the molar rations moles of H2O2 = 2.5 x moles of MnO4 = 0.0011725
now this is in 25cm3 so in order to get the 250cm3 sample, times the moles by ten...
but this is still diluted as the solution was made up to 250cm3 from an original 25, so you can times your moles again by 10 giving you: 0.11725 moles x this by 34 to give you g per 250cm3 (3.9865) x this by 4 to get into g per dm3 (15.9 g dm3)

Now for the O2: Volume = Moles x 24
because you know from the equation on the previous page you know that O2 is only produced in the titration, and therefore the moles to use in this equation is your diluted H202 = 0.0011725 x 24 = 0.028 dm3

does that make sense?
6. (Original post by Hydro.ZX)
Small post so will probably get ignored, so i'm quoting it.
Nice tactic, I found these redox titration practise questions quite difficult: http://www.docbrown.info/page07/SSqu...vol_calcs1.htm
7. CAN someone explain 6D part 3
i know what sterioisomers are as well as optical isomerism, but do you draw the optical isomer of Picolinate
and the answer could you just draw it like this

or do you have to include the ligands
8. (Original post by haydyb123)
Nice tactic, I found these redox titration practise questions quite difficult: http://www.docbrown.info/page07/SSqu...vol_calcs1.htm
Thank you
9. (Original post by voices1)
CAN someone explain 6D part 3
i know what sterioisomers are as well as optical isomerism, but do you draw the optical isomer of Picolinate
and the answer could you just draw it like this

or do you have to include the ligands
Mark scheme says you can draw it like that. So no, you do not have to include the ligands.
10. (Original post by haydyb123)
Haha!
Hmm, not to try and shun your questions, but I'm not sure if the spec contains; specifics of why you can't measure all of the enthalpies directly. Therefore I think it will be unlikely that it will be examined... Thank the real gods of Chemistry
1. Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar.
A sample of verdigris has the formula [(CH3COO)2Cu]2.Cu(OH)2.xH2O.?Analysis of the sample shows that it contains 16.3% water by mass.?Calculate the value of x in the formula.

Didnt know how to do it! Someone explain plss
Hope we get easier q than this!
11. (Original post by arvin_infinity)
1. Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar.
A sample of verdigris has the formula [(CH3COO)2Cu]2.Cu(OH)2.xH2O.?Analysis of the sample shows that it contains 16.3% water by mass.?Calculate the value of x in the formula.

Didnt know how to do it! Someone explain plss
Hope we get easier q than this!
so when you're doing empirical formulas and water of crystallization you can use % as g

so H20 16.3/18 = 0.91

[(CH3COO)2Cu]2.Cu(OH)2
mr = 460.5
83.7/460.4 = 0.18

0.91/0.18 = 5.05 so x = 5

make sense?
12. (Original post by arvin_infinity)
1. Artists between the 13th and the 19th Centuries used a green pigment called verdigris. The artists made the pigment by hanging copper foil over boiling vinegar.
A sample of verdigris has the formula [(CH3COO)2Cu]2.Cu(OH)2.xH2O.?Analysis of the sample shows that it contains 16.3% water by mass.?Calculate the value of x in the formula.

Didnt know how to do it! Someone explain plss
Hope we get easier q than this!
did this one the other day!
if its 16.3% water its 83.7% of of the rest of it
so 16.3/18 = 0.91
83.7/449.55 = 0.186

divide 0.91/0.186 = 4.8, so thats the ratio of of verdigris to water, 1:4.8, which can be rounded up to 5, so 5 H2O molecules
13. Can someone please explain this question to me of the january 2011 exam (2C).
6FeSO4 + 7H2SO4 + Na2Cr2O7 --> 3Fe2(SO4)3 + Cr2(SO4)3 + Na2SO4 + 7H2O and question is show that oxidation and reduction have both taken part in this redoz reaction? Now I understand you are targeting Cr and Fe as these are transition elements. But I am not getting the correct values. My oxidation states are coming in at like +10 and +15!!
14. (Original post by wilsea05)
did this one the other day!
if its 16.3% water its 83.7% of of the rest of it
so 16.3/18 = 0.91
83.7/449.55 = 0.186

divide 0.91/0.186 = 4.8, so thats the ratio of of verdigris to water, 1:4.8, which can be rounded up to 5, so 5 H2O molecules
Got the same offer for sheffield..maybe see u there if we didnt fail these exams lol
heydyb123
Thanks guys..got it
Here is another one

3. An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
(a) A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
Calculate the percentage yield, by mass, of sodium ferrate(VI).

got 46.xx %
15. (Original post by wilsea05)
at the start of the question it says that propanoic is a weak acid, that's all you really have to go on. you should know that nitric is a strong acid too though
so the stronger acid will act as the acid, making the other act as a base.

7b) so 0.02345dm3 of 0.02mol dm3 MnO4 2- was required to neutralise
so 0.02345x0.02 = moles of MnO4 2- =0.000469moles
ratio of MnO4 2- to H2O2 is 2:5, so you times the moles of MnO4 2- by 2.5 to get 0.0011725 - thats in 25cm3, so x10 to get the moles in 250cm3 which is 0.011725moles
n=c x v so c=n/v so 0.011725/0.025= concentration of H2O2 (because you want it undiluted and the question says there was 25cm3 of undiluted H2O2, so you divide by 0.025 not 0.25)= 0.469mol dm3
but you want grams per dm3
so m=n x M so 0.469 x 34 = 15.946 g dm-3

7d
Why do you multiply by ten again once you have got the solution up to 25cm3??
16. (Original post by arvin_infinity)
Got the same offer for sheffield..maybe see u there if we didnt fail these exams lol

Thanks guys..got it
Here is another one

3. An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
(a) A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
Calculate the percentage yield, by mass, of sodium ferrate(VI).

got 46.xx %
I'm getting 42.9%, what is the answer?
17. (Original post by voices1)
Why do you multiply by ten again once you have got the solution up to 25cm3??
25 titration sample x 10 to get 250cm3 sample, and x 10 again to get the moles of the original 25 cm3.
18. (Original post by haydyb123)
Hey, no worries:
First question: You know the second one is right because if you look at the reactants, you know HNO3 is a strong acid, and propanoic is a weak acid, therefore in relative you know HNO3 is the acid that will donate H+, and CH3CH2COOH will accept H+

Does that make sense?

Now secondly, hold let me find the paper...

Okay so: moles of KMnO4 = 0.000469
because of the molar rations moles of H2O2 = 2.5 x moles of MnO4 = 0.0011725
now this is in 25cm3 so in order to get the 250cm3 sample, times the moles by ten...
but this is still diluted as the solution was made up to 250cm3 from an original 25, so you can times your moles again by 10 giving you: 0.11725 moles x this by 34 to give you g per 250cm3 (3.9865) x this by 4 to get into g per dm3 (15.9 g dm3)

Now for the O2: Volume = Moles x 24
because you know from the equation on the previous page you know that O2 is only produced in the titration, and therefore the moles to use in this equation is your diluted H202 = 0.0011725 x 24 = 0.028 dm3

does that make sense?
Why do you multiply by ten again once you have got the solution up to 250cm3?
19. (Original post by voices1)
Why do you multiply by ten again once you have got the solution up to 250cm3?
20. (Original post by haydyb123)
Thanks.

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