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# Aqa chem 4/ chem 5 june 2016 thread

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1. Advantage - can melt/burn the waste, produce heat energy to generate electricity.
The other two main ones would be recycling and landfill.
2. GOOD LUCK guys!!!
I AM GOING TO FAIL!!
Basically I am good with everything apart from when they ask you to deduce a structure of an isomer with the molecular formula "C6H12O2" i take too long on the questions, and I always end up getting them wrong! Does anyone have a quick method of working these out?

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Exactly the same with me
4. Why does water only very slightly dissociate?
5. can someone help me with )#

2cii_ jan 2010

Calculate the pH of the buffer solution formed when 10.00 cm3 of0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3ethanoic acid

i worked out the moles and and the numbers of mols after the XS etc divided each by the total volume and that came out as being wrong. Why is it that you dont divide by the total volume?
6. can someone please explain june 11 2ci and 2cii i think they were the two buffer questions dont make sense to me thanks
7. Guys how do you draw R2CH2 and R3CH? I've never understood how to draw them.
8. (Original post by man i)
can someone help me with )#

2cii_ jan 2010

Calculate the pH of the buffer solution formed when 10.00 cm3 of0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3ethanoic acid

i worked out the moles and and the numbers of mols after the XS etc divided each by the total volume and that came out as being wrong. Why is it that you dont divide by the total volume?
even though its a buffer you look at the reactants its a weak acid with a strong base, determine which is in excess and decide what to do; if excess acid its a slightly longer method than if the base was in excess
9. (Original post by Sniperdon227)
even though its a buffer you look at the reactants its a weak acid with a strong base, determine which is in excess and decide what to do; if excess acid its a slightly longer method than if the base was in excess
i just looked at what i done and i misread what my numbers,due to my writing not being neat so placed the wrong numbers in so yeah i get what i done wrong thanks by the way.
10. (Original post by heyitskim)
Guys how do you draw R2CH2 and R3CH? I've never understood how to draw them.
So for R2CH2 - the R means any other group however the most important thing is to look at the splitting pattern - if had has a triplet - it must be attacthed so a CH2 so it will become RCH2CH2
11. Let's see how this exam goes with a sleep deprivation headache from 4 hrs sleep (exam stress stops me sleeping)

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12. When the concentrations of A and B were both doubled, the initial rate increased by afactor of 4. Deduce the overall order of the reaction.

13. (Original post by Lilly1234567890)
When the concentrations of A and B were both doubled, the initial rate increased by afactor of 4. Deduce the overall order of the reaction.

is it the overall order2?
14. in amino acids like H2N-c-cooh and the other groups does it have ionic bonding or is it only when a zwitterion it has ionic???
15. (Original post by man i)
can someone help me with )#

2cii_ jan 2010

Calculate the pH of the buffer solution formed when 10.00 cm3 of0.100 mol dm–3 potassium hydroxide are added to 25.00 cm3 of 0.410 mol dm–3ethanoic acid

i worked out the moles and and the numbers of mols after the XS etc divided each by the total volume and that came out as being wrong. Why is it that you dont divide by the total volume?
this isnt too hard, so a buffer is a acid and its salt .
potassium hydroxide makes the salt from the acid, its not the salt of ethanoic acid.
so we 0.100 Mol DM-3 x ( 10/1000 ) ( to save time you can do 10x10-3) = 1x10-3 Moles = 0.001 molw
0.410 Mol DM-3 x ( 25/1000 ) = 0.01025 molw

The ethanoic acid is in excess since its 0.01025 compared to the 0.001 moles of KOH, so all thw KOH reacts to form the salt.

so we get 0.001 of salt, which in this case will be CH3COO- ,
we also used 0.001 of the acid to form the salt.
so we have
0.001 salt from the KOH reacting with the CH3COOH
And we lose 0.001 moles of acid as its reacted above, so we have 0.01025-0.001=0.00925

so now we have the moles of both salt and acid. This is the first step really.

now we rearange the Equation to get . [H]+= Ka x Acid / salt

so in this case
Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3 from the Q.

so 1.74x10-5 x0.00925 / 0.001 =1.6095

-log ( 1.6095) = 3.79
16. (Original post by Lilly1234567890)
When the concentrations of A and B were both doubled, the initial rate increased by afactor of 4. Deduce the overall order of the reaction.

2, since Both are first order,
17. (Original post by rimstone)
this isnt too hard, so a buffer is a acid and its salt .
potassium hydroxide makes the salt from the acid, its not the salt of ethanoic acid.
so we 0.100 Mol DM-3 x ( 10/1000 ) ( to save time you can do 10x10-3) = 1x10-3 Moles = 0.001 molw
0.410 Mol DM-3 x ( 25/1000 ) = 0.01025 molw

The ethanoic acid is in excess since its 0.01025 compared to the 0.001 moles of KOH, so all thw KOH reacts to form the salt.

so we get 0.001 of salt, which in this case will be CH3COO- ,
we also used 0.001 of the acid to form the salt.
so we have
0.001 salt from the KOH reacting with the CH3COOH
And we lose 0.001 moles of acid as its reacted above, so we have 0.01025-0.001=0.00925

so now we have the moles of both salt and acid. This is the first step really.

now we rearange the Equation to get . [H]+= Ka x Acid / salt

so in this case
Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3 from the Q.

so 1.74x10-5 x0.00925 / 0.001 =1.6095

-log ( 1.6095) = 3.79

In what circumstances do you divide them both by the total volume, or the excess by total volume?
18. (Original post by Sniperdon227)
in amino acids like H2N-c-cooh and the other groups does it have ionic bonding or is it only when a zwitterion it has ionic???
isnt it an amide bond /Peptide bond.... or something like that. like how condensation occures in polymers ?
amino acids = non mentals =no ionic bond between two non-mental.
19. (Original post by rimstone)
this isnt too hard, so a buffer is a acid and its salt .
potassium hydroxide makes the salt from the acid, its not the salt of ethanoic acid.
so we 0.100 Mol DM-3 x ( 10/1000 ) ( to save time you can do 10x10-3) = 1x10-4 Moles = 0.001 molw
0.410 Mol DM-3 x ( 25/1000 ) = 0.01025 molw

The ethanoic acid is in excess since its 0.01025 compared to the 0.001 moles of KOH, so all thw KOH reacts to form the salt.

so we get 0.001 of salt, which in this case will be CH3COO- ,
we also used 0.001 of the acid to form the salt.
so we have
0.001 salt from the KOH reacting with the CH3COOH
And we lose 0.0001 moles of acid as its reacted above, so we have 0.01025-0.001=0.00925

so now we have the moles of both salt and acid. This is the first step really.

now we rearange the Equation to get . [H]+= Ka x Acid / salt

so in this case
Ka, for ethanoic acid in aqueous solution is 1.74 × 10–5 mol dm–3 from the Q.

so 1.74x10-5 x0.00925 / 0.001 =1.6095

-log ( 1.6095) = 3.79
yeah i sort of messed up the equation because of my bad set up, i got the salt etc and done molex1000/v to get conc for both the left over solution and salt and then messed up my number by using 9.25x10-3 rather than my worked out concentration. so i shouldve got it right either way.

just another question i wanted to ask for KC if the volume of the container increases say 1.5dm3 to 3.0dm3 it does not effect the kc does it or the conc right or does it effect the conc only. im really confused on these for some reason
20. (Original post by Charco63)
In what circumstances do you divide them both by the total volume, or the excess by total volume?
When the ACID is in excess, it can also be base, but they have never asked for a base, since its much harder to do, and if it comes up, well then **** me. But anyways when you have an acid and a OH- ( not its salt ), and the acid is in excess, which it almost always will be, you take away from the acid and add to the salt, salt is 0, since it has none of its own salt.

And you can use moles or mol DM-3 for the equation, but both acid and salt have to be either Mol, or Mol DM-3, they cant be both at once. Use MOL, when they give stuff at different volumes, in this case the acid was in 25 cm3, and the KOH in 10cm3, so We should convert it to MOL, to make them both have no volume at all. To convert to MOL, we do CONCXDM3 ( CM3/1000 = DM3 ) , on other Q, they might give the same volume or it might be easier to use MOL DM-3. This is really a skill you need to pick up, if im being honest, i did 3 Buffer Q and figured it out, i find buffers extremely easy, since its all just basic maths. Anyways my explainations are crap, since i just somehow know how to do it... i would recommend learning it and doing a Few Q, once you do a Few Q, its easy as hell.

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