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

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    I am looking forward to finishing it too, I've done all the past papers for unit 4 (except IAL jan 16 which will be tomorrow) and I know there is more to revise... You never know if you've done enough!
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    What does the Pkin value for an indicator mean and what does it tell you?
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    (Original post by Don Pedro K.)
    The rate equation involves only propanone and hydrogen ions. The concentration of iodine therefore has no effect on the rate. Hence, when the concentration of propanone is doubled, the rate will also double; not quadruple. The doubling of iodine concentration will have no effect on the rate since the order of reaction with respect to iodine is 0 (since it is not involved in the rate equation). Therefore, the answer is D

    Also, you wouldn't be able to tell whether C was correct or not due to the fact that we are not told whether the forward reaction is exothermic or endothermic, and thus cannot make inferences about what a change to temperature will do to the rate constant !
    Your last point isnt true, rate constant increases with temp
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    Just did the 2016 IAL, decent paper. Hopefully ours is as decent as that .
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    (Original post by _H_V)
    Help with 2016 IAL paper question 10, (answer is B) I cant understand why

    The pH of three solutions with concentration 1.0 mol dm–3 was measured.
    Solution 1 NH3
    Solution 2 CH3COONa
    Solution 3 NH4Cl

    Which of the following shows the three solutions in order of increasing pH?
    A 1, 2, 3
    B 3, 2, 1
    C 3, 1, 2
    D 2, 3, 1
    Can anyone explain this? Don't get this as well
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    Just did 2014 r and I got 55/90 I literally can't believe it, it wasn't even a hard paper


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    (Original post by RUNSran)
    Can anyone explain this? Don't get this as well
    I did it this way. I kind of considered them in the form of conjugate acid-bases. Example, NH4+ is the conjugate acid of NH3 meaning it will give out protons thereby lowering the pH. CH3COONa is a basic salt because if we consider the conjugate base, i.e ethanoate ions, it reacts with water to form OH ions thereby having a higher pH than NH4+ and NH3 is obviously higher than the rest so the answer is B.
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    (Original post by samb1234)
    Your last point isnt true, rate constant increases with temp
    Oh crap I was thinking about equilibrium constants xD Man hahaha sorry Neha121101!!!
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    (Original post by sabahshahed294)
    I did it this way. I kind of considered them in the form of conjugate acid-bases. Example, NH4+ is the conjugate acid of NH3 meaning it will give out protons thereby lowering the pH. CH3COONa is a basic salt because if we consider the conjugate base, i.e ethanoate ions, it reacts with water to form OH ions thereby having a higher pH than NH4+ and NH3 is obviously higher than the rest so the answer is B.
    Oh Nice! thank you very much
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    (Original post by dj3k)
    What does the Pkin value for an indicator mean and what does it tell you?
    Anyone???
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    (Original post by dj3k)
    Anyone???
    ok so basically an indicator has its own equilibrium during a titration and depending on whether the conditions are acidic or basic, its colour will change.Consider the equilibrium of an indicator: where 'in' is just an indicator

    Hin ⇌ in + H+ (Hin will be a different colour to in)

    the Kc value would be equal to [in][H+] / [Hin] but as it is an indicator, we call it 'Kin' which is the equilibrium constant for an indicator but it has the same meaning as Kc. Now Pkin = -log[Kin] and it gives an indication of the working range of an indicator. You don't really need to know abt this too much but the ph range of an indicator is usually + and - 1 unit from the Pkin value. So if an indicator had a Pkin value of 4, it's range would be about 3-5. Hope that makes sense, i'm not really that good at explaining things :\
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    (Original post by Don Pedro K.)
    Oh crap I was thinking about equilibrium constants xD Man hahaha sorry Neha121101!!!
    No worries, it makes sense now! thanks
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    does anyone know what the bonding in CH3COOH2+ ion would look like? there was a conjugate acid Question and I was curious to know i havent managed to find anything by googling it
    samb1234 Don Pedro K. TeachChemistry
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    (Original post by setarcos)
    ok so basically an indicator has its own equilibrium during a titration and depending on whether the conditions are acidic or basic, its colour will change.Consider the equilibrium of an indicator: where 'in' is just an indicator

    Hin ⇌ in + H+ (Hin will be a different colour to in)

    the Kc value would be equal to [in][H+] / [Hin] but as it is an indicator, we call it 'Kin' which is the equilibrium constant for an indicator but it has the same meaning as Kc. Now Pkin = -log[Kin] and it gives an indication of the working range of an indicator. You don't really need to know abt this too much but the ph range of an indicator is usually + and - 1 unit from the Pkin value. So if an indicator had a Pkin value of 4, it's range would be about 3-5. Hope that makes sense, i'm not really that good at explaining things :\
    Thanks for this. So does the Pkin value not have a real use then? It's always seemed a little useless considering we already know the exact pH ranges for the indicators. Surely the pKin values are inferior to them?
    Thanks so much again!
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    (Original post by ayvaak)
    does anyone know what the bonding in CH3COOH2+ ion would look like? there was a conjugate acid Question and I was curious to know i havent managed to find anything by googling it
    samb1234 Don Pedro K. TeachChemistry
    I reckon it might be a dative covalent bond, because that's what you get with H3O+ and NH3+ I think for example >.> Not entirely sure though!
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    (Original post by dj3k)
    Thanks for this. So does the Pkin value not have a real use then? It's always seemed a little useless considering we already know the exact pH ranges for the indicators. Surely the pKin values are inferior to them?
    Thanks so much again!
    The Pkin values are helpful to explain how the indicator actually works. Looking at it in terms of ticking the markscheme, might be worth mentioning it in the explanation questions but other than that, talking abt PH ranges will usually suffice.
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    (Original post by Don Pedro K.)
    I reckon it might be a dative covalent bond, because that's what you get with H3O+ and NH3+ I think for example >.> Not entirely sure though!
    (Original post by ayvaak)
    does anyone know what the bonding in CH3COOH2+ ion would look like? there was a conjugate acid Question and I was curious to know i havent managed to find anything by googling it
    samb1234 Don Pedro K. TeachChemistry
    Yeah, I think it is a dative covalent bond as well because the oxygen atom will donate its lone pair to the partially positive H atom in the acid and that's why it is CH3COOH2+ ( However, not entirely sure)
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    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!
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    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?
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    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 !
 
 
 

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