Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread

Which one's the answer?

Yes but that also gives OH^- ions in solution. NH3 is not an acid.

B (3,2,1)

In this question volumes are different but I still used moles to get the correct answer :s, I've been trying using moles for a lot of questions now and they give the same answer even when volume us different.

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Am I being stupid
Why is NH3 stronger base than ethanoate ion
I mean my gut feeling does tell me that it has a higher pH but I thought ammonia was a weak base but ethanoate ions were strong base (conjugate base of a weak acid is a strong base?)

Yeah I know it isn't but since I thought it's in a solution then it would dissociate and and CH3COONa would dissociate as well to give OH- ions so that's why I am confused

can someone explain why C is the answer?

multiply?

Yeah same here!!! I am feeling really stupid like my gut feeling tells me that NH3 is definitely a stronger base but then I thought ethanoate ions are the conjucate base and that they;re stronger and also if they dissociate they would give CH3COOH while ammonia will give NH4+ (NH3 + H+) so I can't figure it out!

The salt of a weak acid and strong base is slightly basic. This is because ethanoate ions react with water CH3COO- + H2O --> CH3COOH + OH-.The salt of a strong acid and weak base is slightly acidic. Again because the salt is hydrolysed NH4+ + H2O --> NH3 + H3O+.We know ammonia is a weak base so it has to have the highest pH out of the 3 as it would dissociate the most. Hence the answer is B (Copied this from my earlier post)

General form for straight line is y=mx+c
y=lnrate
m=-Ea/R
x=1/T
c=constant

So you've got gradient=-Ea/R and rearranging this gives Ea=-( gradient) R

thanks!

I do not understand part b at all, would be really helpful if anyone could explain it if they have time. I know its super late!

I bet you we're going to get a question like this tomorrow:

https://www.youtube.com/watch?v=mb9v7NIaSoo

You heard it here first peeps!!!1onetwo!!

Okay I'm kinda of going to answer this all together instead of in the separate parts because I think it'll make more sense that way.

Basically, the job of sodium thiosulfate is to reduce any iodine formed back into iodide ions:

2S₂O₃^2- + I₂ --> S₄O₆^2- + 2I^-

So, the solution initially will be colourless, since the persulfate ions oxidise iodide ions to iodine, but the sodium thiosulfate reduces the iodine immediately back to iodide ions, so a black colour doesn't form straight away (which is what would happen if sodium thiosulfate wasn't used --> answer to part ii). The final colour change occurs when all of the sodium thiosulfate has reacted, so any iodine formed will no longer be reduced back down to iodide ions. Therefore, the final colour change is colourless to black (answer to part i). In relation to part iii), the iodide concentration initially remains constant due to persulfate ions oxidising iodide ions to iodine but then sodium thiosulfate reducing iodine immediately back down to iodide ions again.

Hope that makes da sense