Turn on thread page Beta

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

    Offline

    6
    ReputationRep:
    (Original post by dj3k)
    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/co...c_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.
    Offline

    2
    ReputationRep:
    (Original post by dj3k)
    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/co...c_20150503.pdf

    Thanks in advance!
    You have to work out total moles of acid that reacted NaOH = ( 45/1000 x1 ) = 0.045
    Then work out moles of HCL (which is 5x10^-3) and minus it from 0.045 as the NaOH reacts with hcl + ethanoic acid at equilibrium. This gives the moles of ethanoic acid at equilibrium (0.04)
    From here use the moles of ethanoic acid to work out ethanol, ethyl ethanoate and water at equilibrium ( but remember there was an inital moles of 0.278 of water). -ICE or which method you use to work out moles at equilibrium-
    So the equilibrium moles should be Ethanoic acid = 0.04, ethanol= 0.14, ethyl ethanoate= 0.08 and water= 0.358.
    plug the values in the kc expression for the reaction and the answer should be 5.114
    Hope this makes sense
    Offline

    2
    ReputationRep:
    (Original post by Funky_Giraffe)
    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.
    Offline

    6
    ReputationRep:
    (Original post by _H_V)
    oops, seems like you answered the question whilst i was still typing mine haha my bad, but yeah your explanation is so much better.
    No problem - good for them to have two different explanations
    Offline

    2
    ReputationRep:
    (Original post by imnoteinstein)
    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?
    I think the first step involves the removal of Br and addition of just O- from the OH, and the second step involves the H+ from acid protonating the O- to form the OH group.
    Offline

    16
    ReputationRep:
    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) ?
    Offline

    2
    ReputationRep:
    (Original post by PlayerBB)
    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) ?
    It would not affect Kp/Kc
    Offline

    3
    ReputationRep:
    (Original post by _H_V)
    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 !
    The main thing to remember is the Cu2+ ions in Fehling's
    The premise of the Fehling's/Benedict's test is that aldehydes can be oxidised easily but ketones can't
    When an aldehyde is added, it is oxidised. This means that the Cu2+ ions get reduced to Cu+, which is a different colour to Cu2+ (blue --> orange)
    From what I can recall anyway - I've only seen it on an international paper once!
    Offline

    3
    ReputationRep:
    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
    Offline

    3
    ReputationRep:
    (Original post by genevievelaw)
    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
    Offline

    3
    ReputationRep:
    (Original post by PythagorasRex)
    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
    Yeah I think I get it, thank you
    Offline

    20
    ReputationRep:
    Anyone know any unit 4 past papers with long titration questions?
    Offline

    2
    ReputationRep:
    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."
    Offline

    6
    ReputationRep:
    (Original post by ayvaak)
    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?
    Offline

    6
    ReputationRep:
    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/co...e_20150610.pdf
    Offline

    2
    ReputationRep:
    (Original post by Whizbox)
    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?
    Offline

    6
    ReputationRep:
    (Original post by ayvaak)
    It wasnt from a past paper but if it was a dicarboylic acid would that make a difference?
    Where's the question from, sounds a weird one to me
    Offline

    3
    ReputationRep:
    (Original post by Funky_Giraffe)
    No problem - good for them to have two different explanations
    Lol thanks so much! I fully understand it now.
    Offline

    6
    ReputationRep:
    (Original post by Funky_Giraffe)
    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/co...e_20150610.pdf
    Essentially HCl reacts with NH3 to form NH4+ ions, NH4+ can react with the water to donate a proton to turn back into NH3 and H3O+, hence there's oxonium ions in the solution so it will have a pH lower than 7 due to being slightly acidic
    Offline

    8
    ReputationRep:
    (Original post by ayvaak)
    It wasnt from a past paper but if it was a dicarboylic acid would that make a difference?
    Yes it does. Dicarboxylic acids can undergo deprotonation twice. After the first deprotonation, the Ka value of the acid changes which means the PH at which it will undergo its second deprotonation will change. Therefore, the titre values will be different depending on the indicator u use.
 
 
 
The home of Results and Clearing

1,102

people online now

1,567,000

students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Poll
Do you want your parents to be with you when you collect your A-level results?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.