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

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Original post by medstudent1010
Can someone help with part d the answer is 0.13mol/dm^3 image.png
Attachment not found


need to rearrange the Ka expression to make A- the subject to find the conc of the salt
Original post by Dinasaurus
yh can someone explain, all I can see is E = E of reduced + E of oxidised.


The E of O2/OH is +0.40 and the overall is +2.71
We know the difference is 2.31

If i was positive it would be 2.31 - 0.40 which is 1.91 == not 2.71!!

therefore to get 2.71 = 0.40 - -2.31 which equals 2.71
Original post by lai812matthew
my last attempt on A* after i flopped f324 if this doesn't go well i'll have no uni to go to.......


I feel for you, I also made a huge number of silly mistakes in F324 (by the sound of it I've done worse than you!), and Chemistry was supposed to be my sure-fire A* for my uni offer. But luckily F325 counts for more UMS than F324, so if we smash it tomorrow it should make up for a bad F324.
Original post by Ezexamsalltheway
HI , guys how many of the transition metal aqueous ion colours do we actually need to know. all of them? thought i should better start my revision now :P rip


You need to know copper and cobalt with ammonia and chlorine. And then copper, cobalt, iron 2 and 3 with hydroxides. That's it as far as I'm aware. for copper and cobalt you need to know colour changes and formula
Original post by tcameron
it may be added NH3 like it was in last years paper but they still accepted the H+ + OH-
They'll accept it if you name the acid itself or just show the formation of water it's the same thing essentially

Ooh okay thanks :smile: so for last year's, would you talk in terms of the actual acid and base?
Original post by medstudent1010
Can someone help with part d the answer is 0.13mol/dm^3 image.png
Attachment not found

Phenol is a weak acid.

pH = - log the root of (Ka x HA).


Ka: calculate Ka using normal method

HA:
We're given an awkward concentration using grams rather than moles, so you need to convert it to moles first. Simple use m/Mr. m = 4.7, Mr = Mr phenol = 94. The answer to this is your conc, HA.

Rest is self explanatory
(edited 7 years ago)
Original post by MrZebraCookie
It doesn't say that the aluminium cell is positive or negative. However, since aluminium is being oxidised, you can assume that it is more negative than the oxygen cell


But if it's an addition positive or negative, 0.34 + 2.74 isn't = -2.31

-0.34 + 2.74 is 2.31 not -2.31

(I know the value wasn't 2.74 I am not in the computer now so can't see it but you get what I mean?)


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Original post by mechanism
Ooh okay thanks :smile: so for last year's, would you talk in terms of the actual acid and base?


well I put NH3 + H+ -> NH4+ which is correct but H+ + OH- -> H2O was also correct
the equlibrium still shifts the same way
Original post by tcameron
need to rearrange the Ka expression to make A- the subject to find the conc of the salt


That's what I did however got a completely different answer
Original post by medstudent1010
Can someone help with part d the answer is 0.13mol/dm^3 image.png
Attachment not found


I did that question and got 0.013 mol/dm^3. I'm pretty sure the answer is wrong by 1 d.p.
Original post by Jessica Mighall
The E of O2/OH is +0.40 and the overall is +2.71
We know the difference is 2.31

If i was positive it would be 2.31 - 0.40 which is 1.91 == not 2.71!!

therefore to get 2.71 = 0.40 - -2.31 which equals 2.71


Oh god I never thought that the 2.71 could've been the overall and not just another half cell.

Oh thanks


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Original post by medstudent1010
That's what I did however got a completely different answer


maybe you made a mistake somewhere putting the numbers in the wrong places because that's all you're meant to do
Original post by Dinasaurus
But if it's an addition positive or negative, 0.34 + 2.74 isn't = -2.31

-0.34 + 2.74 is 2.31 not -2.31

(I know the value wasn't 2.74 I am not in the computer now so can't see it but you get what I mean?)


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Ah I read the question wrong. The 2.7V given is the E of the entire cell. That means 2.7=0.4-x. So x has to be a negative
Original post by mamounaltayeb
ImageUploadedByStudent Room1466527008.813326.jpgCan someone explain this


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okay i'm not sure but i think it is because :
remember weak acids/buffers can only minimise changes in pH with SMALL additions of acid/alkali, so if you add enough alkali, to counteract this more H+ would be made (equilibrium shift to right) so all the H+ have to react.

we know it is definitely feasible because remember the titration curves: there is one that shows a weak acid can be titrated with a STRONG alkali ( a weak alkali would not work as no vertical line so indicator cannot show equivalence point clearly) so this is how i first figured out that the student was wrong and a titration was possible, then i just had to give a scientific reason why :smile:

hope this helps sorry i'm bad at explaining things :smile:
Original post by sunsri101
I did that question and got 0.013 mol/dm^3. I'm pretty sure the answer is wrong by 1 d.p.


Yes I got the same,however when looking at their concentration of H+ it's different to what 10^-8.71 would give,so the way they worked it out is different too
Original post by tcameron
well I put NH3 + H+ -> NH4+ which is correct but H+ + OH- -> H2O was also correct
the equlibrium still shifts the same way

Great thanks!
which transition metals will give hexaamine complex ions, which ones would give (NH3)4(H2O)2 complex ions?
What half equations do we need to know by heart for tommorow?
Manganate??
Chromium ??
Iodine ??
Sodium thiosulphate ??



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Original post by lai812matthew
which transition metals will give hexaamine complex ions, which ones would give (NH3)4(H2O)2 complex ions?


I thought any transition metal could give any complex?


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anyone have a proper definition of Kstab? The one in the book makes no sense to me

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