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

I'm having a bit of trouble with the following question from Jan 15 IAL:

*(iii) An experiment was carried out to determine the value of Kc for this reaction.
0.120 mol of ethanoic acid was added to 0.220 mol of ethanol.
5.00 cm3 of 1.00 mol dm–3 hydrochloric acid was added as a catalyst.
Thiscontains 0.278 mol of water. The mixture was left to reach equilibrium.
The mixture was titrated with 1.00 mol dm–3 sodium hydroxide, whichreacted with both of the acids.
The titre was 45.0 cm3.
Use these data to determine the value for Kc.

Has anyone figured out how to do it yet? If please can someone explain?
Also sorry if this question has already been asked before.
MS is here: (it's 24 a(iii))

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Chemistry/2013/Exam%20materials/WCH04_01_msc_20150503.pdf

Thanks in advance!

I'm having a bit of trouble with the following question from Jan 15 IAL:

*(iii) An experiment was carried out to determine the value of Kc for this reaction.
0.120 mol of ethanoic acid was added to 0.220 mol of ethanol.
5.00 cm3 of 1.00 mol dm–3 hydrochloric acid was added as a catalyst.
Thiscontains 0.278 mol of water. The mixture was left to reach equilibrium.
The mixture was titrated with 1.00 mol dm–3 sodium hydroxide, whichreacted with both of the acids.
The titre was 45.0 cm3.
Use these data to determine the value for Kc.

Has anyone figured out how to do it yet? If please can someone explain?
Also sorry if this question has already been asked before.
MS is here: (it's 24 a(iii))

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Chemistry/2013/Exam%20materials/WCH04_01_msc_20150503.pdf

Thanks in advance!

Once you get around the wordy question, it's not so bad

The important piece of the information provided is that that both of the acids react with 45.0cm³ of 1.00 moldm^-3 NaOH.

From this, we want to find out the moles of just CH3COOH that react from our equilibrium mixture. So, we work out the moles of NaOH that are added, and subtract the number of moles that reacted from the 5.00cm³ of 1.00moldm^-3 HCl that was added (in bullet point 2). This gives us the moles of CH3COOH as:

45/1000 - 5/1000 = 0.04 mol CH3COOH at equilibrium.

^That's probably the hardest part of the question to get your head around. Once you've done that, you should be okay. They give you the initial moles of the reactants CH3COOH and CH3CH2OH, and also the initial moles of H2O, which is one of the products. Since you know that there are 0.04 mol CH3COOH left at equilibrium, and all the ratios are 1:1, you should be able to work out the number of moles of CH3CHOH left at equilibrium, as well as the mol of CH3COOCH2CH3 formed, and the combined (remember there were 0.278 mol H2O initially) number of moles of H2O as well. Hopefully you will be able to work the equilibrium moles out from this?

Then just plug in to your equation for Kc, happy days.

Hope you could follow this.

oops, seems like you answered the question whilst i was still typing mine haha my bad, but yeah your explanation is so much better.

in one of the pps, to go from CH₃CH(Br)COOH (2-bromopropanoic aicd) to CH₃CH(OH)COOH (2-hydroxypropanoic acid), we use 2 steps:

First steps reagent is NaOH or KOH
the second steps reagent is a strong acid. Why do we need that? would NaOH/KOH not take you straight to the desired product?

In a reversible reaction that has reached equilibrium, when we change the concentration of one of the reactants/products, this would not affect the value of equilibrium yield right (as the equilibrium will shift to the side to oppose the change) ?

Can anyone summarise what happens to the Fehlings solution itself when it reacts with an aldehyde, this was in a question but the mark scheme didn't make much sense to me . Thanks !

is anyone able to explain the whole pH of water thing?
I've been reading different stuff and it seems like they contradict i.e. water stays neutral, and the pH of water changes
I understand that water stays neutral because [H+]=[OH-], but I have also read in one source that its pH changes because as Kw changes with temperature, [H+] changes
I don't really get it

So basically, the dissociation of water into H+ ions and OH- ions is an equilibrium, and the forward reaction is endothermic (as bond breaking requires energy.) As with any equilibrium, increasing the temperature will favour the endothermic reaction, so increasing the temperature will result in more H2O molecules dissociating, increasing the concentration of both H+ ions and OH- ions. As pH = -log(H+), this would lower the pH, but the solution would still be neutral due to the fact that there is an equal concentration of H+ ions and OH- ions. So if [H+] is equal to [OH-], the solution is always neutral but doesn't necessarily have to be pH 7 since the temperature has to be 298K for that. Hope that helps

I read this and wasn't sure why will the phenolphthalein be exactly half the methyl orange? "Sodium carbonate solution and dilute hydrochloric acid titration If you use phenolphthalein or methyl orange, both will give a valid titration result - but the value with phenolphthalein will be exactly half the methyl orange one."

Isn't that from the question with the di-carboxylic acid?

It wasnt from a past paper but if it was a dicarboylic acid would that make a difference?

No problem - good for them to have two different explanations

Would someone be able to explain question 13 a (ii) from this paper - about the pH at the equivalence point being less than 7. I don't quite understand the mark scheme

http://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Chemistry/2013/Exam%20materials/WCH04_01_que_20150610.pdf

It wasnt from a past paper but if it was a dicarboylic acid would that make a difference?