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

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Original post by BioStudentx
Haha, upon reflection I lost 4 marks to stupid mistakes (and 2 marks I didn't genuinely know).

Oh dear, this paper got a bit harder all of a sudden :laugh:

Concentration levels are bound to drop during a pressurised 2 hour exam tbf.


Yeah I for sure dropped 1 mark on that [OH-] question, and another for the equation with Cu+ and I-, I reckon I may or may not have dropped one for the question about the rate equations so am hoping (fingers crossed) for around 93/4 as I'm sure there was a few more marks of bs I lost in there. Sounds like you might be in the same boat, hopefully we still hit dat 1 fiddy
Original post by k.russell
Yeah I for sure dropped 1 mark on that [OH-] question, and another for the equation with Cu+ and I-, I reckon I may or may not have dropped one for the question about the rate equations so am hoping (fingers crossed) for around 93/4 as I'm sure there was a few more marks of bs I lost in there. Sounds like you might be in the same boat, hopefully we still hit dat 1 fiddy


Doubt we'll hit 150. I'll probably have dropped more marks (not sure about you). There's always something :frown:
there are 2 ways to prove that pH=7 is alkali in 60 degree celsius. is a bit easier to see the picture by neutral pH calculation imo. Kw=[H+(aq)][OH-(aq)], so square root of Kw equal to [H+(aq)], and find the pH equals 6.52. anything above 6.52 is alkaline, vice versa. 7>6.52, so it is alkali.
Do u remember if the Au equation was 1 or 2 marks?
Original post by BioStudentx
Doubt we'll hit 150. I'll probably have dropped more marks (not sure about you). There's always something :frown:


hmm my coursework is poor so need a 95% ums average (there abouts) from the exams to get an A*. It's gonna be really really close as to whether I've done enough, a 150 would help me out so much
deffo dropped 2 marks on both papers and I am sure there are more mistakes I can't remember. ****ing wish I had 40/40 coursework could've probs got that without too much sweat as well..
Original post by MilkFriend
Got 10.76 instead of 10.68 for one of the ph calculation... not sure where I went wrong tho...


Anyone got 10.76... no idea about what happened.
Original post by BioStudentx
Haha, upon reflection I lost 4 marks to stupid mistakes (and 2 marks I didn't genuinely know).

Oh dear, this paper got a bit harder all of a sudden :laugh:

Concentration levels are bound to drop during a pressurised 2 hour exam tbf.


dropped at least 7 marks, got cis and trans the wrong way round, charge of complex ion wrong way round (i written 2+), don't know that starch is used to give color change, don't know what is standard form, and the most idiotic mistake comes to......i written down that i am using redox system 5 and 6, but used redox system 6 and 7 and write equations and explain from it. !
Original post by Slenderman
In comparion to 2015. Harder or easier? Am I retarded to think the grade boundaries will be 75 for an A?


Not retarded but very optimistic my friend. I would say 80 = A
[H+]= square root of Kw at 60c
[H+]= square root 9.313*10^-14
[H+]= 3.05*10^-7
-log 3.05*10^-7= 6.5
Since 6.5<7 so pH 7 is alkaline
Original post by Beta14
I divided the conc by 5 then used the Kw of 60*C i think thats what you were supposed to do, a similar question came up last year 2015 paper as a 3 marker and thats what the mark scheme says to do


I tried to do that and failed so....
I ended up just putting that 11.73 or whatever. But I'm sure your method is right yeah.
what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
and it was only 3 marks so probably nothing tricky was needed?
Original post by lawlieto
what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
and it was only 3 marks so probably nothing tricky was needed?


Calculate moles of acid and conjugate base, divide by 1000 to get concentration, use Kaacidoversalt to get [H+], - log to get 3.43 IIRC


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Original post by GetOverHere
Calculate moles of acid and conjugate base, divide by 1000 to get concentration, use Kaacidoversalt to get [H+], - log to get 3.43 IIRC


Posted from TSR Mobile


that's exactly what i've done; as it was 1 dm^3 basically moles of acid and base was same as concentrations if i remember correctly
but people seem to have done 3 different things in this thread so i got unsure...
Original post by lawlieto
what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
and it was only 3 marks so probably nothing tricky was needed?


4 marks so it is defo not boring buffer question. if they give concentration in buffer there will only be 2 marks.
Original post by lai812matthew
4 marks so it is defo not boring buffer question. if they give concentration in buffer there will only be 2 marks.


you had to calculate concentrations of acid + base in the buffer solution (that could be 2 marks, each term worth 1 marks)

1 mark for for using the right formula

1 mark for answer?

:biggrin: what have you done in that question?
Would the marks be for calculating moles or the actuallt conc (I know they're the same but still)
Original post by lawlieto
what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
and it was only 3 marks so probably nothing tricky was needed?


Number of mole of HNO2 was 0.04
Number of mole of NaNO2 was 0.05
0.05-0.04=0.01 [NO2-]
[H+]= (Ka*0.04)/0.01
-log [H+] =pH
I don't remember the Ka
(edited 7 years ago)
Original post by pezhman
Number of mole of HNO2 was 0.04
Number of mole of NaNO2 was 0.05
0.05-0.04=0.01 [NO2-]
[H+]= (Ka*0.04)/0.01
-log [H+] =pH
I don't remember the Ka


Dont take off 0.04 Moles

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Has anyone done the paper? Unofficial mark scheme?


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Original post by k.russell
Isn't it to do with the black-grey precipitate (I2) and the yellow precipitate (AgI)?


idk, I reckon that was the hardest question on the paper though ngl

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