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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

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Original post by k.russell
Could someone drop me a pdf of the 2015 paper xxx
Did it as a mock in class ages ago - it has very high ums right?


https://www.dropbox.com/sh/kwj0elokp2hco0c/AADgT1LXHwXXFXE6S-r2mI-Pa?dl=0
Original post by itsConnor_


<3
Original post by RetroSpectro
Can I double check my answer to this question with anyone, I think the mark scheme is wrong:

Calculate the pH of a buffer solution that is 0.100 mol dm–3 with respect to sodium methanoate and 0.200 mol dm–3 with respect to methanoic acid. The Ka of methanoicacid is 1.8 × 10–4 mol dm–3.


Could you explain this please?
Guys, can someone please explain part (ii)

I understand that in solution, the ions will react with the water. So this will decrease their concentration, which will shift the equilibrium to the left, to make more Cu2+.

My initial thought was: the electrons combine with the cu2+ which will increase the emf, but that's wrong :frown:
Screen Shot 2016-06-19 at 23.48.38.png70.9KB
And I don't know if i'm confusing myself because i'm stressed, but just to double check

when calculating the Mr of say 7Fe(So4)2 it's (Mr of Fe) + (2xMr of S) + (8xMr of O) in other words, the 7 doesn't get factored in?
Can someone please explain the mark scheme to this answer, have no idea how to get to it X=Mg
Picture11.png162.4KB
Original post by RayMasterio
Guys, can someone please explain part (ii)

I understand that in solution, the ions will react with the water. So this will decrease their concentration, which will shift the equilibrium to the left, to make more Cu2+.

My initial thought was: the electrons combine with the cu2+ which will increase the emf, but that's wrong :frown:


Water added to the cu half cell
Concentration of cu2+ decreases
Equilibrium moves to left
More oxidation taking place
E value of half cell decreases
Leading to a greater difference in E values between two half cells
Thus electrode potential increase
Hope this helps

Posted from TSR Mobile
Original post by ImNervous
Water added to the cu half cell
Concentration of cu2+ decreases
Equilibrium moves to left
More oxidation taking place
E value of half cell decreases
Leading to a greater difference in E values between two half cells
Thus electrode potential increase
Hope this helps

Posted from TSR Mobile


Cell potential increases*

Posted from TSR Mobile
Original post by RayMasterio
And I don't know if i'm confusing myself because i'm stressed, but just to double check

when calculating the Mr of say 7Fe(So4)2 it's (Mr of Fe) + (2xMr of S) + (8xMr of O) in other words, the 7 doesn't get factored in?


You are correct

Posted from TSR Mobile
Original post by tcameron
Can someone please explain the mark scheme to this answer, have no idea how to get to it X=Mg


First right an equation for what happens between two half cells
Then find moles of Cr formed
Using molar ratios from the equation you found find moles of the metal that is lost. Mass/moles gives you Mr to find the metal. Hope this helps.

Posted from TSR Mobile
Original post by ImNervous
First right an equation for what happens between two half cells
Then find moles of Cr formed
Using molar ratios from the equation you found find moles of the metal that is lost. Mass/moles gives you Mr to find the metal. Hope this helps.

Posted from TSR Mobile


why just the moles of chromium metal? I found the moles of Cr2(SO4)3
Original post by tcameron
Can someone please explain the mark scheme to this answer, have no idea how to get to it X=Mg


X----->X2+ +2e-
Cr3+ +3e- ----->Cr
Add the two half equations, work out moles of Cr gained, Mole ratio is 2:3 then mr=mass/moles and you should get 24.3?
Original post by tcameron
why just the moles of chromium metal? I found the moles of Cr2(SO4)3


How do you do that?
Original post by Kamara7
I literally just did this paper and finished marking it XD I also got the exact same answer as you :lol: Did you get to the 4th step in @itsConnor_'s working out above? If so, then you don't need to divide by two afterwards because you're finding the % mass of chromium and not the chromium-containing compound (Fe(CrO2)2) and as you can see in the equation, the moles of Cr are the same on both sides of the equation (4 moles on each side), so you don't need to multiply/divide by anything as you already got the moles of Cr (0.034)! I don't know if you did the same mistake as me though, so it could be something else.

Ah thankyou! Yeah I had divided by two because the ratio of fe(cro)2 to the sodium compound was 1:2, but now I can see that the moles of Cr are actually the same
Original post by AqsaMx
Could you explain this please?


Its fairly straightforward,

First write out the Ka equation for the dissociation of methanoic acid

Ka = [H+][HCOO-]/[HCOOH]

and from the question you have been given the concentrations of the acid, the salt and the Ka value

0.100 mol dm–3 with respect to sodium methanoate
0.200 mol dm–3 with respect to methanoic
The Ka of methanoic acid is 1.8 × 10–4 mol dm–3

Rearrange the equation to get [H+] and plug in the given values, then use -log([H+])
(edited 7 years ago)
Original post by RetroSpectro
Its fairly straightforward,

First write out the Ka equation for the dissociation of methanoic acid

Ka = [H+][HCOO-]/[HCOOH]

and from the question you have been given the concentrations of the acid, the salt and the Ka value

0.100 mol dm–3 with respect to sodium methanoate
0.200 mol dm–3 with respect to methanoic
The Ka of methanoic acid is 1.8 × 10–4 mol dm–3

Rearrange the equation to get [H+] and plug in the given values, then use -log([H+])


I think I misred the question, I thought It said making a buffer solution so I started doing the excess-limiting thing oops, thanks for clearing that up :smile:

How do you know when you're supposed to do the limiting calculation, will the question be phrased differently?
Original post by AqsaMx
I think I misred the question, I thought It said making a buffer solution so I started doing the excess-limiting thing oops, thanks for clearing that up :smile:

How do you know when you're supposed to do the limiting calculation, will the question be phrased differently?


In those questions you are usually given the volume and concentration of both an acid and the base. This allows you to calculate the moles, to find that the acid moles will be in excess.

In this question you had no information about the volume of the solution or any other reagent.

I think the best way to go about it is to see if it is possible to calculate the moles of each reagent
image.jpeg

Could someone explain the actual answer of this question please?
Original post by RetroSpectro
In those questions you are usually given the volume and concentration of both an acid and the base. This allows you to calculate the moles, to find that the acid moles will be in excess.

In this question you had no information about the volume of the solution or any other reagent.

I think the best way to go about it is to see if it is possible to calculate the moles of each reagent


Oh okay thank you v much :smile:
Original post by AqsaMx
image.jpeg

Could someone explain the actual answer of this question please?


You are unable to determine the enthalpy changes that occur in the CO3 2- ion as it consists of more than one element

I think thats one way you can interpret it but im sure theres more answers

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