OCR Chemistry A F325 - Equilibria, Energetics and Elements - 12th June 2013

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Amz1217
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Oxford, Cambridge and RSA Examinations (OCR)

~Equilibria, Energetics and Elements F325~
Date of exam: 12th June F215 (50% of A2)
Specification Statement: http://www.ocr.org.uk/images/81089-specification.pdf
Past Papers: http://www.ocr.org.uk/i-want-to/prep...papers-finder/
Duration: 2 Hours
Maximum Mark: 100

~Modules~
* Rates, Equilibrium and pH
* Energy
* Transition Elements

A2 Endorsed Textbook

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Namod
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who is gonnaa start now?
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Amz1217
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(Original post by Namod)
who is gonnaa start now?
Obviously no-one is gonna start straight away but I thought I would just make it so it'll be around for then
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Gkt1378312870
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Doing well in today's F324 has actually motivated me for this exam! And we've hardly started learning about it yet! haha
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Shika222
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Hello again everyone


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Cyclohexane
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Ah, this unit's gonna be HELLLLLLLLLL.

Hello everybody!
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bashedup
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F325 my nemesis, I hope next week is the last time I ever see you!

If you guys see me in here again it means I didn't claw together the 10 measly UMS I need from my re-sits....
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Gold Phoenix
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Chemistry just gets better doesn't it.

Lol
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niki135
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Is anyone also doing F324 in June?
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chiny94
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(Original post by niki135)
Is anyone also doing F324 in June?
I think i might have to :L

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Thatasianguy
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Hey guys, can someone post a pH for buffer solutions question pack, or link me.
Any other question packs are much appreciated thanks!
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D4rth
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F325 LEGACY PAPERS
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ItsDatGuy
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What are these?
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D4rth
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(Original post by ItsDatGuy)
What are these?
The old chemistry papers before they changed the specification.

Not all of the questions will be useful for F325 so pick the relevant ones.
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master y
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Does anyone have the mark scheme for january 2013 PLease, this would help a lot!
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Treacle2
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Anyone got the Jan 2013 F325 paper and mark scheme???
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TomTennant11
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I feel like I am in a hole for revising for this exam and can't see a way out.. Anyone got any revision tips? Any good revision guides I could buy??
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Could somebody please help me with the following question from the June 2012 paper. I know how to find the pH however am unable to calculate the right concentrations and can't find it in the textbook, the mark scheme is very vague. I would really appreciate it. Thanks in advance
I keep on getting 4.12 instead of the correct 4.22
7
(c) The student adds 50.0cm3 of 0.250moldm–3 butanoic acid to 50.0cm3 of 0.0500moldm–3
sodium hydroxide. A buffer solution forms. (i) Explain why a buffer solution forms.
................................ ................................ ................................ ................................ ........... ................................ ................................ ................................ ................................ ........... ................................ ................................ ................................ ................................ ........... ................................ ................................ ................................ ................................ ...... [2]
(ii) Calculate the pH of the buffer solution.
The Ka of butanoic acid is 1.51 × 10–5 mol dm–3. Give your answer to two decimal places.
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teester
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I really struggled with this question too, but after looking it up in my revision guide I've cracked it!
I think it forms a buffer solution because it's a weak acid mixed with a strong base of a lowish concentration. The acid is in excess because of it's concentration- meaning some of the acid is neutralised to make its salt but some is left un-neutralised. Therefore the reacting mixture would have a weak acid and it's salt- a buffer solution.
For calculating the pH:
CH3CH2CH2COOH+NaOH(in eqm)-->CH3CH2CH2COONa+H2O
Number of moles of CH3CH2CH2COOH= c x v/1000= 0.25 x (50/1000)= 0.0125 mol
Number of moles of NaOH= 0.05 x (50/1000)= 0.0025 mol
As there is so much more acid than alkali, the acid is in excess- meaning at equilibrium, all of the alkali will have reacted, so your molar quantities at equilibrium are:























CH3CH2CH2COOH
NaOH
CH3CH2CH2COONa
H2O
Initial amount
0.0125
0.0025
0
0
Equilibrium amount
0.01
0
0.0025
0.0025


The total volume of your solution is 50cm^3+50cm^3= 100cm^3
Concentration of CH3CH2CH2COOH at eqm= (1000xn)/v=(1000x0.1)/100=0.1 mol dm^-3
Concentration of CH3CH2CH2COONa at eqm= (1000x0.0025)/100= 0.025

Ka=[H+][CH3CH2CH2COO-]
[CH3CH2CH2COOH]

So [H+]= Ka x [CH3CH2CH2COOH]
[CH3CH2CH2COO-]
=(1.51 x 10^-5)
0.025
=0.0000604, and pH= -log[H+] so pH= 4.218963061= 4.22

Hope this helps!
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mabz123
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can somebody tell me if the following questions come from the ocr endorsed book, posted above:

33. Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3–. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid.
(a) 25.0 cm3 of 0.100 mol dm–3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm–3 MnO4–(aq) and 30.0 cm3 is required to oxidise the V2+(aq) back to VO3–(aq).
The half equation for acidified MnO4– acting as an oxidising agent is shown below.
MnO4– + 8H+ + 5e– ----> Mn2+ + 4H2O
Show that the vanadium has changed oxidation state from +2 to +5 in this titration.






[4]

(b) Suggest an equation for the oxidation of V2+(aq) to VO3–(aq) by MnO4–(aq) under acid conditions.





................................ ................................ ................................ ........................
[2]
[Total 6 marks]

33. (a) Moles V2+ = 25.0 × 0.100 / 1000 = 0.0025 mols 1
Moles MnO4– = 30.0 × 0.0500 / 1000 = 0.00150 mols 1
1 mole of MnO4- changes its Oxidation State by 5 to change
the Oxidation State of 1.67 moles of V2+ 1
Oxidation State of V2+ changes by 5 / 1.67 = 3 1


(b) 3MnO4– + 5V2+ + 3H2O → 3Mn2+ + 5VO3– + 6H+
(1 mark for correct species, 1 mark for balanced) 2
[6]




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