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    (Original post by ahsan_ijaz)
    Hi,

    Well since your adding extra oh- ions they react with the H+ when acid dissociate
    So HX= H+ + X-

    When you add extra OH- ions the OH- will react with the H+ to form water thus the H+ conc is reduced at the time. However, We know the H+ conc has been reduced so the ph should drop but since we know a buffer solution is one that resists changes in Ph when small ammount of alkali/acid is added. The equilibrium will shift to the right to replace the lost H+ which will bring the pH back close to its original value

    hope it helps
    ahh thanks very much!!
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    (Original post by jammypancake)
    ahh thanks very much!!
    I believe its best to have the understanding rather than to try memorise where it shifts etc as then you wont get caught out in the exam - and as far as i know im anticipating an hard paper this year
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    can anyone help on Q5 eii - i dont know where it got that ppm value from i got 5-40 ppm
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    http://filestore.aqa.org.uk/subjects...4-QP-JUN15.PDF

    Can someone help me with question 3biii

    I literally have no idea how to do it
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    when showing a dipeptide should you show linking bonds at the end or put NH2 and COOH at the end? confused
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    (Original post by Lilly1234567890)
    http://filestore.aqa.org.uk/subjects...4-QP-JUN15.PDF

    Can someone help me with question 3biii

    I literally have no idea how to do it
    Only one of the indicators has pH 6 within its indicating range and that's 4-nitrophenol. At lower pH it is colourless and as pH increases it becomes yellow. The result is a mix between this and hence light yellow.
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    (Original post by jammypancake)
    when showing a dipeptide should you show linking bonds at the end or put NH2 and COOH at the end? confused
    NH2 and COOH at the end. Trailing bonds only come into play when asked for repeating unit!


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    (Original post by ahsan_ijaz)
    I believe its best to have the understanding rather than to try memorise where it shifts etc as then you wont get caught out in the exam - and as far as i know im anticipating an hard paper this year
    Just curious, why do you predict a hard exam this year?
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    (Original post by ssamarai)
    Just curious, why do you predict a hard exam this year?
    Well as they know that their AS papers were relatively easier - thus i have made an assumption they wouldnt want even the B grade students to get an A overall as people would have took a lot of ums from as exams so even if they underperform in the A2 exams they can get A overall - thus they will try to make it hard to distinguish between the actual A grade students rather than those who have overachieved in As

    this may all be wrong and we may get the easiest paper possible lol
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    (Original post by ahsan_ijaz)
    Well as they know that their AS papers were relatively easier - thus i have made an assumption they wouldnt want even the B grade students to get an A overall as people would have took a lot of ums from as exams so even if they underperform in the A2 exams they can get A overall - thus they will try to make it hard to distinguish between the actual A grade students rather than those who have overachieved in As

    this may all be wrong and we may get the easiest paper possible lol
    Ah fair enough but from what i've seen online i don't think many people found the resists that good, at least especially from what i've seen. However i do agree
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    (Original post by SubwayLover1)
    There are two hydrogen's attached to each one aswell
    yes but they are not two different groups?
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    (Original post by Superbubbles)
    yes but they are not two different groups?
    I mean normally..... The Carbon would have two hydrogen's attached. But one of the H's has been swapped for a different group e.g CH3 or OH
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    (Original post by SubwayLover1)
    I mean normally..... The Carbon would have two hydrogen's attached. But one of the H's has been swapped for a different group e.g CH3 or OH
    i see what you mean now, is the benzene ring part counted as 1 group or 2?
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    At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value 1.75 × 10–5 mol dm–3.
    (c) (i) Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassiumhydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.

    Help please
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    (Original post by Medibananas)
    NH2 and COOH at the end. Trailing bonds only come into play when asked for repeating unit!


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    ahh so always do that unless they state and ask for a repeating unit?
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    (Original post by string210)
    At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value 1.75 × 10–5 mol dm–3.
    (c) (i) Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassiumhydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.

    Help please
    1. Find the moles of the KOH and CH3COOH
    2. Put into equation [H+]= Ka x [CH3COOH] / [KOH]
    3. Find pH
    I think that's right
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    Anyone predict what I would need to get this year to get an A overall?
    Unit 1: B
    Unit 2: A
    Unit 3X: A
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    (Original post by jammypancake)
    Anyone predict what I would need to get this year to get an A overall?
    Unit 1: B
    Unit 2: A
    Unit 3X: A

    depends on your UMS
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    (Original post by jammypancake)
    1. Find the moles of the KOH and CH3COOH
    2. Put into equation [H+]= Ka x [CH3COOH] / [KOH]
    3. Find pH
    I think that's right
    thats what i did, checked the mark scheme, it wasnt right!
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    (Original post by string210)
    thats what i did, checked the mark scheme, it wasnt right!
    hmm that's strange how many marks was it?
 
 
 
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