OCR Chemistry A 2017 Exam Thread (New A Level)

Hiya folks, instead of arguing whether it was or wasn't 4.5 if someone can remember the details of the ph question i'll show my workings for 4.5 and you can point of where i've gone wrong/where you've gone wrong

to get 4.5 you just used the given concentration CH3COOH and worked out concentration of CH3COONa, then put them into the equation. To get 4.86 you subtracted the concentration of CH3COONa from the concentration of CH3COOH and put those values in the Ka equation. I've seen the latter method used in all the past paper questions I've done in buffers but I don't know whether it was the correct method in this question. I did both and plumped for 4.86 in the end.

Reply 781

euphorical

2

Original post by FusionNetworks

Think I got 6.02x10^20. Can't remember how

I ticked that aswell but it defo wasn't right, it's the only answer I managed to find! If you find the moles of Ne based on the g and MR & multiply by Avogadros constant (6.02x10^23) you get that answer (6.02x10^20). So that's how many particles of Ne there are. Nothing to do with electrons.

Reply 782

Jitesh

14

The pH q could be done in one line
pH=pKa+log(cb/a)

Reply 783

jayeshforce1

10

Original post by pkpiggy

I put potassium chlorate (v) because I remembered it being KClO3?

yh i got 5 too

Reply 784

Lucas.Kaiser

5

Original post by B1g Al

Hiya folks, instead of arguing whether it was or wasn't 4.5 if someone can remember the details of the ph question i'll show my workings for 4.5 and you can point of where i've gone wrong/where you've gone wrong

I think 250 cm3 of 0.8 molar CH3COOH and 9.086 grams of CH3COONa

Reply 785

Tuffyandtab

16

Grade boundaries predictions?

Reply 786

username1763965

16

Original post by Mufc2167

Anyone get [cucl4] 2- for anything
And ligand substitution?

Yep.

Reply 787

The A* Doctor

1

Who else got 4.85 for the ph of the buffer??

Reply 788

randyorton

21

Original post by euphorical

I ticked that aswell but it defo wasn't right, it's the only answer I managed to find! If you find the moles of Ne based on the g and MR & multiply by Avogadros constant (6.02x10^23) you get that answer (6.02x10^20). So that's how many particles of Ne there are. Nothing to do with electrons.

one electron from each atom.

Reply 789

EleanorAngel

4

Original post by euphorical

So that's how many particles of Ne there are. Nothing to do with electrons.

Surely each atom of Ne loses one electron, so the number of electrons lost is proportional to the number of Ne atoms in that many moles of Ne??

Reply 790

Jdc18

4

For the last question would I get credit for putting C2N2H8 (molecular formula) for the ligand instead of structural formula?

Reply 791

Flumpler

1

Original post by Nettled

So adding H20 has no effect on the moles?

Sorry for the late reply. I hope your exam went well! Moles are the number of molecules of the chemical (avogadro's constant). By adding water, no more of the chemical is being added so moles doesnt change. It is just more spread out through the solution so concentration decreases :smile:

Reply 792

B1g Al

6

Original post by EleanorAngel

to get 4.5 you just used the given concentration CH3COOH and worked out concentration of CH3COONa, then put them into the equation. To get 4.86 you subtracted the concentration of CH3COONa from the concentration of CH3COOH and put those values in the Ka equation. I've seen the latter method used in all the past paper questions I've done in buffers but I don't know whether it was the correct method in this question. I did both and plumped for 4.86 in the end.

Hiya, are you not thinking or buffer solution formation by partial neutralisation? Where you work out moles of acid and alkali, the reacting moles in the moles of salt formed, and the excess moles is used to work weak acid concentration?

In this question, the salt was already given so no need for subtraction?

Reply 793

jayeshforce1

10

Original post by Mufc2167

Anyone get [cucl4] 2- for anything
And ligand substitution?

yes [Cu(cl)2]power of 2-

Reply 794

Mufc2167

8

Original post by parrot16

Yep.

Thank god

Reply 795

si1239293

3

Original post by euphorical

I ticked that aswell but it defo wasn't right, it's the only answer I managed to find! If you find the moles of Ne based on the g and MR & multiply by Avogadros constant (6.02x10^23) you get that answer (6.02x10^20). So that's how many particles of Ne there are. Nothing to do with electrons.

Every partical of NE loses one electron

fIE - energy required to remove an electron from each atom in one mole of gaseous atoms

Reply 796

Student207

7

Original post by The A* Doctor

Who else got 4.85 for the ph of the buffer??

I did and rounded it to 4.9, but I'm not convinced I did it right

Reply 797

RainBear

9

Is there an unofficial mark scheme yet?
I really think I messed this up as I panicked

Reply 798

autumnblue

2

Original post by Tuffyandtab

And I got 4.85 too.

most people at my school and the teachers that have been told the question say that it is 4.5 so peak for u

Reply 799

Pinkycherry

6

Original post by Lucas.Kaiser

My school did this experiment with a gas syringe so I think either is exceptable

I said an inverted measuring cylinder cause the volume of gas was quite large for a gas syringe and it mentioned in the question something about the scale

OCR Chemistry A 2017 Exam Thread (New A Level)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you