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Aqa chem 4/ chem 5 june 2016 thread

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Original post by Superbubbles
URGENT help needed on chem 4 3bii june 2015 paper, if anyone can explain this I would be very grateful


Hi,

I had a look for you
When you look at curve H you can see the line starts from a PH above 12 and drops to 0-2 with its mid point being for the range of napthyl red indicator - now from the graph we can see that the curve is shown for adding acid to alkali as the ph is so high at the start - then if you look at the table in 3(b) it shows you that at high ph(10-14) ideally of a alkali to a low ph - with napthyl red indicator the colour change will be yellow (high ph) to red (low ph) thus that's the colour change - remember the colour change will still be in range of ph 3.7-5.0 as can be seen in table but instead of being red to yellow it'll be yellow to red

Hope I helped if you need anymore help am more than happy to help


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Original post by ahsan_ijaz
Hi,

I had a look for you
When you look at curve H you can see the line starts from a PH above 12 and drops to 0-2 with its mid point being for the range of napthyl red indicator - now from the graph we can see that the curve is shown for adding acid to alkali as the ph is so high at the start - then if you look at the table in 3(b) it shows you that at high ph(10-14) ideally of a alkali to a low ph - with napthyl red indicator the colour change will be yellow (high ph) to red (low ph) thus that's the colour change - remember the colour change will still be in range of ph 3.7-5.0 as can be seen in table but instead of being red to yellow it'll be yellow to red

Hope I helped if you need anymore help am more than happy to help


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thank you so much, please can you help me with 3biii as well?
Original post by man i
need some help on this, the june 2011 paper

question 2ci)

At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value 1.75 × 10–5 mol dm–3.2 (c) (i) Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassiumhydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C........................................................................................................................................

and 2cii)
(ii) Calculate the pH of the solution formed when 40.0 cm3 of 0.154 mol dm–3 potassiumhydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C. At 25 °C, Kw has the value 1.00 × 10–14 mol2 dm–6.

the first one i got right you just rearrange the formula etc. the second one says that you need to divide 0.00308 (XS) by 60x1000 now im not really sure why thats the case and the CGP book doesent explain it well either. any help will be appreciated.


basically in the first one, the CH3COOH was in XS obviously when you calculated the moles of both KOH and CH3COOH. However, in the second question, KOH is in XS because when you calculate the moles (note that now 40cm3 are being used) the moles of KOH are now 0.00616 whilst the moles of CH3COOH are only 0.00308 meaning that KOH is now in XS. So when you minus 0.00308 from 0.00616 you get 0.00308, this is the moles of OH-, then you must divide by the total volume of 60cm3 (40+20) and obviously divide by 1000 to make it dm3 rather than cm3. This gives you the [OH-] which you then put into the Kw equation and rearrange it to get [H+].

Hope this helps :smile:
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Original post by ssamarai
basically in the first one, the CH3COOH was in XS obviously when you calculated the moles of both KOH and CH3COOH. However, in the second question, KOH is in XS because when you calculate the moles (note that now 40cm3 are being used) the moles of KOH are now 0.00616 whilst the moles of CH3COOH are only 0.00308 meaning that KOH is now in XS. So when you minus 0.00308 from 0.00616 you get 0.00308, this is the moles of OH-, then you must divide by the total volume of 60cm3 (40+20) and obviously divide by 1000 to make it dm3 rather than cm3. This gives you the [OH-] which you then put into the Kw equation and rearrange it to get [H+].

Hope this helps :smile:


so you only work out the total volume when the alkali is in excess when added into acid. my main issue is WHEN you use the total volume and why that wasnt the case for the first is it because the acid was in excess.
Original post by man i
so you only work out the total volume when the alkali is in excess when added into acid. my main issue is WHEN you use the total volume and why that wasnt the case for the first is it because the acid was in excess.


Oh I see what you mean, you should always divide by the total volume, in the question provided (2ci) its not actually necessary to divide by the volume because it all cancels out. But i always use the total volume and get the same answer :smile: so you should just get into the habit of using it especially if you're not sure when to use it just do it, you get the same answer :smile:
please can anyone explain how this is a optical isomer? and which is the chiral carbon??
optical isomer .jpg4.1KB
Original post by Superbubbles
please can anyone explain how this is a optical isomer? and which is the chiral carbon??


I think there are two chiral carbons- the C with the CH3 attached the C with OH attached :smile:
Original post by PleaseHelpMe:(
Hi.guys I need a really really big favour from everyone here. I am resitting all 9 of AS exams this year along with 9 A2 exams for bio chem and maths. I am really struggling with chem 4 and I keep getting <30 in all my past papers :,( . No matter how much I revise, I can't get anything into my head. So can any of you bright clever people post ur revision notes on here please. I will be so grateful. I really want to pass chemistry Thank you


Okay these are my notes, I don't know if they'll make sense but they might help idk. It's everything EXCEPT chapter 11 (analytical techniques) and all the mechanisms
Attachment not found


ImageUploadedByStudent Room1465676160.741972.jpg

Also I really recommend printing off the spec and using it as a checklist since they can only ask you things that are on there, then do past papers to polish up your exam technique.

Good luck!



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http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JAN11.PDF
Hii, can anyobody explain to me how to work out 3a? I really don't get it. The answers are:
Methanol-0.07
Hydrogen-0.24
:smile:
Any predictions for unit 4? It's going to be hard


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Original post by Superbubbles
thank you so much, please can you help me with 3biii as well?


yeah can you please help us with 3bii. i'm having trouble with it as well.
Original post by jammypancake
When writing an equation for the reduction of nitrobenzene to phenylamine, how do you know how many [H]'s to use? I can never seem to balance these equations correctly?


for one nitro group it's 6H and 2H20 on the other side, if more than one nitro group simply multiply these values by the number of groups!
image.png196.6KB
Original post by JK11
You are wrong. you are thinking of Kc.


nope only temperature affects the value of the rate constant, K. Increase temp and K increases, decrease temp and K decreases. Nothing else affects.
Hi does anyone have all the organic synthesis we need to know? I've looked online and they're all varied and some I've seen the mark scheme does not accept
Original post by Signorina
Hi does anyone have all the organic synthesis we need to know? I've looked online and they're all varied and some I've seen the mark scheme does not accept

https://chemrevise.files.wordpress.com/2014/04/aqa-mechanisms-a21.pdf
hey, can someone explain when in a nucleophilic substitution reaction, what happens when you use excess NH3, and then when you use excess haloalkane
Can anyone explain how to do the ph questions when the acid is diluted, for example 'Calculate pH when 10cm3 of 0.154M HCL is added to 990cm3 of water'
Thanks in advance! 😀
Original post by bat_man
hey, can someone explain when in a nucleophilic substitution reaction, what happens when you use excess NH3, and then when you use excess haloalkane


excess ammonia---> primary amine
excess haloalkane---> quaternary ammonium salt
Original post by bat_man
hey, can someone explain when in a nucleophilic substitution reaction, what happens when you use excess NH3, and then when you use excess haloalkane

When excess NH3 is used:
Only one NH2 is added, so once one NH3 group has been added, it has a positive charge, one H donates its electron and leaves the molecule leaving -NH2
eg CH3CH2Br + 2NH3 --> CH3CH2NH2 + NH4+
When excess haloalkane is used:
The amine group on the molecule continues to act as a nucleophile and continues to replace the halogen on the molecule, goes to secondary, tertiary the quaternary ammonium salt
Original post by kattara6
http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JAN11.PDF
Hii, can anyobody explain to me how to work out 3a? I really don't get it. The answers are:
Methanol-0.07
Hydrogen-0.24
:smile:

These are really easy once you have a method to working it out:
1.) write out the equation
2.) Underneath this, write down the mol values you have for the CO and the H2 and AND the value you already have for the CO afterwards
3,) Work out the difference with the CO values (0.24-0.17=0.07)
4.) Now you know the difference for equilibrium CH3OH must be 0.07
5.) There are 2 mol of H2, multiple 0.07 by 2 (0.14), then minus the 0.014 from the original (0.38-0.14=0.24)
Sorry if this was confusing

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