AQA CHEM5 A2 Chemistry - 19th June 2013

The issue with the calculations on Unit 5 is that there are different types of calculations that they may ask you. For example, they could ask you about redox titrations, or a back titration.

In this example, and I have it in front of me now, there are two main equations.

i) a periodicity equation. The reaction between P4O10 and 6H20 to form 4H3PO4

ii)acid-base neutralisation between NaOH and Phosphoric Acid.

You work out the moles of Sodium Hydroxide using the conc. and the volume given. You use the ratios in your Acid-base neutralisation equation between the sodium hydroxide and the phosphoric acid to work out the moles of phosphoric acid. From that you can work out the initial amount of P4O10 using the equation between P4O10 and 6H20.

There isn't really any universal step by step method on these kind of questions as they may be asking for different things. I just keep practising them.

I just did the June 2011 Chem 5 paper, and I was doing pretty well up until question 7. I still managed to get an A* though . I thought the grade boundaries were surprisingly low though as the majority of the paper wasn't that difficult.

Did you get the "water hydrolyses aren't reversible" one?

Thanks... btw do you know how to write ionic equations from a balanced equation?

I just did the June 2011 Chem 5 paper, and I was doing pretty well up until question 7. I still managed to get an A* though . I thought the grade boundaries were surprisingly low though as the majority of the paper wasn't that difficult.

In the examiners report to jun 12 paper for chem 5 (which is the hardest past paper so far on this spec according to my chemistry teacher) they refer to june 2011 stating something like the demanding nature of the June 2012 paper is not unique as June 2011 was equally as demanding. The grade boundaries are low for chem 5 in general, but were especially low for june 11/12 because a lot of people found it challenging.

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Hey, on the past paper's I've been doing there's a question to do with the colours of complexes which keeps coming up..

So basically its when [Co(NH3)6]2+ (which I think is straw coloured) is allowed to stand in air.

So I know that it forms [Co(NH3)6]3+ but I've been taught that this is yellow and the mark scheme says it is dark brown.

Does anyone know why or if I will still get the marks for putting yellow?

Thanks

Hey, on the past paper's I've been doing there's a question to do with the colours of complexes which keeps coming up..

So basically its when [Co(NH3)6]2+ (which I think is straw coloured) is allowed to stand in air.

So I know that it forms [Co(NH3)6]3+ but I've been taught that this is yellow and the mark scheme says it is dark brown.

Does anyone know why or if I will still get the marks for putting yellow?

Thanks

Hiya...

Sorry if someone else has already answered this question for you.

With 2+ hexa-aqua ions, basically (well my chemistry teacher told me because there is no distinction in the book(s))

Its 3en which forms [M(en)₃]²⁺ + 6H₂O

With 3+ hexa aqua ions it does something different.

en is H₂NCH₂CH₂NH₂

This is the following equation

2[M(H₂O)₆ ]³⁺ + 3H₂NCH₂CH₂NH₂ ----> 2[M(H₂O)₃(OH)₃] + 3H₃NCH₂CH₂NH₃

This is because 3+ ions are slightly more acidic and can donate H⁺ more easily. Think of the reactions between the hexa aqua ions and carbonate ions.

Hope this helps. I was completely muddled by it all initially. There is not actually any distinction in the book (I use Nelson Thornes) at all, but I think you are supposed to apply knowledge of the differences between the reactions betweeen carbonate with 2+ hexa aqua ions and 3+ hexa aqua ions. en is a bidentate ligand, as is the carbonate ion.

Almost all of the mark schemes I've come across have said yellow/straw oxidised to a much darker, rusty brown. I really do doubt they would accept yellow for [Co(H2O)6]3+ as it's never usually mentioned on the right hand side of the MS for BOTD answers, so suggest taking and using was the mark schemes say, if in doubt always copy the mark scheme

can someone explain 6) d) i)

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-MS-JUN12.PDF

dont understand why the ratio is 3:5 isnt it 2:5

thanks alot, the NT is really vague on this topic

You have to combine the bottom two half equations first

Fe²⁺ $\rightarrow$ Fe³⁺ + e^-
C₂O₄^2- $\rightarrow$ 2CO₂ + 2e^-

FeC₂O₄ $\rightarrow$ Fe³⁺ + 2CO₂ + 3e^-

Then combine this equation with the top equation

3MnO₄^- + 5FeC₂O₄ + 24H⁺ $\rightarrow$ 3Mn²⁺ + 5Fe³⁺ + 10CO₂ + 12H₂O

So for every 3 moles of MnO₄^- which react, 5 moles of FeC₂O₄ react i.e. a 3:5 ratio

I think NT are a bit rubbish in some places. I am eternally grateful to my lovely teacher.

Does the value of emf increase if more electrons are released?