OCR A2 Chemistry Unit 5, F325 15th June 2015

Our methods are pretty similar, and the answer checks out so hopefully we are right? I got the question from this video at 3:30

https://m.youtube.com/watch?list=PLbJbCQgCpBPumOYeiFH8IRpjqSAjX6dQn&v=VCMDm3ISkBY

And then I googled the question as I wasn't satisfied with that answer and found this (nice few questions)
http://www.stjoe.k12.in.us/ourpages/auto/2011/2/8/56759726/buffers_key.doc

I think you're right and perhaps they just forgot about ha being in excess
Number 7 is the question we're doing, and they give 0.2 as the answer!

Yeah it seems like we are in agreement.. Surely we can't both be wrong?!
I might ask my Chemistry teacher about it tomorrow though just to be certain as I am going to see her anyway :smile:

Reply 1781

Dr. Django

12

Original post by Octarine_Flame

Yeah it seems like we are in agreement.. Surely we can't both be wrong?!
I might ask my Chemistry teacher about it tomorrow though just to be certain as I am going to see her anyway :smile:

Yeah, thanks for the reassurance. It'd be nice if you could get your teacher to check it but no pressure :smile:

I'm pleased I'm finally getting the hang of these buffer calculations as I used to find them impossible!

IF ANYONE WANTS A GOOD REVISION RESOURCE:

Look up 'Machemguy' on YouTube, the buffers stuff is quality and he had loads of different content, all explained really well.

Reply 1782

the1akshay

8

Original post by Dr. Django

Yeah, thanks for the reassurance. It'd be nice if you could get your teacher to check it but no pressure :smile:

I'm pleased I'm finally getting the hang of these buffer calculations as I used to find them impossible!

IF ANYONE WANTS A GOOD REVISION RESOURCE:

Look up 'Machemguy' on YouTube, the buffers stuff is quality and he had loads of different content, all explained really well.

I did it again and I agree with you two. The crux is that there's an excess of acid in the buffer, which is something I always forget :-/

Reply 1783

Dr. Django

12

Original post by the1akshay

I did it again and I agree with you two. The crux is that there's an excess of acid in the buffer, which is something I always forget :-/

Thanks, that's really reassuring as I still wasn't 100% convinced!

Up until a few days ago the whole excess thing massively confused me but 'Machemguy' finally made me understand, on my 3rd year of a-level chemistry :tongue:

Reply 1784

CurtisDean

11

Original post by the1akshay

So the reaction they're talking about is the combination of the first and the last half equations from the data. The electrode potential as well as the question's wording (V2+ present) tells you that permanganate is reduced and vanadium oxidised. So combine the reactions, multiplying the top one by 5 to match electrons:

MnO4- + 8H+ + 5V2+ -> 5V3+ + Mn2+ + 4H2O.

Their specific questions: Vanadium has an oxidation state of +3, and the species is V3+. The half equation is the reverse of the equation at the top.

Is that correct?

Thank you so much for response! :smile:

// That's exactly what I thought, but if you look at the answer (which is the third image attachment), it is completely different. I just didn't understand how that could be the answer. I am thinking that it might be a mistake, but the resst of the answers follow suite, so I don't know. :s-smilie:

Reply 1785

AnshulK

1

Original post by Dr. Django

Yeah, thanks for the reassurance. It'd be nice if you could get your teacher to check it but no pressure :smile:

I'm pleased I'm finally getting the hang of these buffer calculations as I used to find them impossible!

IF ANYONE WANTS A GOOD REVISION RESOURCE:

Look up 'Machemguy' on YouTube, the buffers stuff is quality and he had loads of different content, all explained really well.

For buffer calculations;

[H+] = Ka x [acid] / [salt]

Just learn how to rearrange it

Reply 1786

AnshulK

1

Original post by kate8

How can you tell if a transition metal complex will be square planar shape or tetrahedral?Thanks! :smile:

Firstly most of your complexes will be octahedral [Cu(H2O)6] for example. In other words, a complex with a coordination number of 6 will be octahedral.

To figure out which are tetrahedral just look for ones which have a coordination number of 4; [CoCl4] etc.

However, there is an exception; cis-platin...this one is unique. You need to know that cis-platin has a shape of sqaure-planar. All the other with a coordination number of four are tetrahedral.

I hope this helped

Reply 1787

Dr. Django

12

Original post by AnshulK

For buffer calculations;

[H+] = Ka x [acid] / [salt]

Just learn how to rearrange it

Thanks, I know that, there's just often a whole lot more to it than that! Anyway I'm fine with buffers now :smile:

Reply 1788

AbiiZola

2

Guys in the textbook can anyone explain how to work out 2.2.9 question 2a? The answer in the textbook doesn't make sense to me because the number of hydrogen's aren't balanced or maybe it's me just going crazy 😓

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Reply 1789

username1155480

13

Original post by AbiiZola

Guys in the textbook can anyone explain how to work out 2.2.9 question 2a? The answer in the textbook doesn't make sense to me because the number of hydrogen's aren't balanced or maybe it's me just going crazy 😓

Posted from TSR Mobile