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    (Original post by master y)
    ^^ Can i ask you to help me on my question please? Thank you!!
    4(e)
    Mark point 1: We chose Lactic acid as it has a pKa value closest to the pH.

    Mark point 2: To make the buffer we would need the acid: Lactic acid and a salt of the acid, Sodium Lactate. If you can't name them, just draw them.

    Mark point 3: PKa of lactic acid= 3.86
    Therefore the Ka = 10^-3.86 = 1.38 x 10^-4 mol dm-3

    Mark point 4: PH of sweet = 3.55
    Therefore the H+ = 10^-3.55 = 2.82 x 10^-4 mol dm-3

    Mark point 5: Ka = [H+] X ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    So 1.38 x 10^-4 = 2.82 x 10^-4 x ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    Rearrange
    (1.38 x 10^-4 / 2.82 x 10^-4) = ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    essentially (KA/H+) = (BUFFER SALT/BUFFER ACID)
    SO we have: 0.490/1 ratio to make the magic tang.

    Mark point 5: This is kind of a duff mark, my teacher thought this was a bit stupid. you can make a comment about how the magic tang may come from other chemicals in the sweet.
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    (Original post by H0ls)
    4(e)
    Mark point 1: We chose Lactic acid as it has a pKa value closest to the pH.

    Mark point 2: To make the buffer we would need the acid: Lactic acid and a salt of the acid, Sodium Lactate. If you can't name them, just draw them.

    Mark point 3: PKa of lactic acid= 3.86
    Therefore the Ka = 10^-3.86 = 1.38 x 10^-4 mol dm-3

    Mark point 4: PH of sweet = 3.55
    Therefore the H+ = 10^-3.55 = 2.82 x 10^-4 mol dm-3

    Mark point 5: Ka = [H+] X ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    So 1.38 x 10^-4 = 2.82 x 10^-4 x ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    Rearrange
    (1.38 x 10^-4 / 2.82 x 10^-4) = ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    essentially (KA/H+) = (BUFFER SALT/BUFFER ACID)
    SO we have: 0.490/1 ratio to make the magic tang.

    Mark point 5: This is kind of a duff mark, my teacher thought this was a bit stupid. you can make a comment about how the magic tang may come from other chemicals in the sweet.
    Your first point is wrong we cannot use the relationship that Pka=PH as they are not the same , you have to use buffer formula
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    (Original post by otrivine)
    Your first point is wrong we cannot use the relationship that Pka=PH as they are not the same , you have to use buffer formula
    I didn't use the relationship pKa=PH, but that's how you know what acid to use?
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    (Original post by H0ls)
    I didn't use the relationship pKa=PH, but that's how you know what acid to use?
    Not necessarily you have to use the H+=Ka x [HA]/[A-] or the other buffer formula and that would give you the rough indication of which acid to use
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    (Original post by AMIgirl)
    hi,can Anyone send me the jan 2013 paper?
    please and thank you
    Hi there, unfortunately I do not have the paper, but this link contains a unofficial paper.

    Check it out if you're interested

    http://www.thestudentroom.co.uk/show....php?t=2340955

    F322, F325 and F322 are also enclosed as well.

    Hope this helps!

    Another unofficial for f325

    https://docs.google.com/document/d/1...ilebasic?pli=1
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    (Original post by otrivine)
    Not necessarily you have to use the H+=Ka x [HA]/[A-] or the other buffer formula and that would give you the rough indication of which acid to use
    Are we looking at the same question? It doesn't say that anywhere on the mark scheme, as you're not calculating which one to use, you're working out the ratios in the buffer solution?
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    (Original post by reneetaylor)
    Hi there, unfortunately I do not have the paper, but this link contains a unofficial paper.

    Check it out if you're interested

    http://www.thestudentroom.co.uk/show....php?t=2340955

    F322, F325 and F322 are also enclosed as well.

    Hope this helps!
    Hello missy how are you


    Long time no revision
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    (Original post by otrivine)
    Hello missy how are you


    Long time no revision
    Haha, hello Otrivine.

    I've been revising hard so I can answer nearly any tricky questions you like to ask!

    Let me know when you ready!
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    (Original post by reneetaylor)
    Haha, hello Otrivine.

    I've been revising hard so I can answer nearly any tricky questions you like to ask!

    Let me know when you ready!
    I am ready any time

    give me ur tricky question please
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    (Original post by otrivine)
    I am ready any time

    give me ur tricky question please
    :laugh:

    Outline how a IR spectrum helps to identify a COOH bond in a molecule.
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    (Original post by H0ls)
    4d(ii) Well, we have two carboxylic acid groups on the molecule, in acid dissociation we have two possibility: Either one group dissociates, or both dissociates.

    Pka= 1.23 (this shows the strongest acid ph, therefore we can conclude that this is the molecule where we keep one carboxylic acid group.
    C2H2O4 ---> C2HO4- + H+

    Pka+ 4.19 (well this has a less acidic ph, so we know that is the molecule where we have lost both carboxylic acid groups)
    C2HO4- ----> C2O42- + H+

    (I was told to do the second equation as if the first carboxylic acid group had already been lost, i'm sure you can do it straight from the C2H2O4)



    (Original post by H0ls)
    4(e)
    Mark point 1: We chose Lactic acid as it has a pKa value closest to the pH.

    Mark point 2: To make the buffer we would need the acid: Lactic acid and a salt of the acid, Sodium Lactate. If you can't name them, just draw them.

    Mark point 3: PKa of lactic acid= 3.86
    Therefore the Ka = 10^-3.86 = 1.38 x 10^-4 mol dm-3

    Mark point 4: PH of sweet = 3.55
    Therefore the H+ = 10^-3.55 = 2.82 x 10^-4 mol dm-3

    Mark point 5: Ka = [H+] X ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    So 1.38 x 10^-4 = 2.82 x 10^-4 x ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    Rearrange
    (1.38 x 10^-4 / 2.82 x 10^-4) = ([CH3CHOHCOO-] / [CH3CHOHCOOH])
    essentially (KA/H+) = (BUFFER SALT/BUFFER ACID)
    SO we have: 0.490/1 ratio to make the magic tang.

    Mark point 5: This is kind of a duff mark, my teacher thought this was a bit stupid. you can make a comment about how the magic tang may come from other chemicals in the sweet.
    You are a godess! Thank you for the clear answers.... i STILL don;t know why and when you assume that pH = pKa though.... .
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    (Original post by master y)
    You are a godess! Thank you for the clear answers.... i STILL don;t know why and when you assume that pH = pKa though.... .
    I will ask my teacher and let you know by is evening
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    (Original post by otrivine)
    ...
    I know this isn't relevant to this thread, but I've got a question that I didn't get right in a mock I did for unit 1 and was hoping you could help me out with it?

    The question was "Predict the formula of the zinc salt that could be formed by adding excess of Zinc to phosphoric (V) acid, H3PO4.

    I put it as Zn3PO4 but its actually meant to be Zn3(PO4)2, why is this?


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    (Original post by otrivine)
    I will ask my teacher and let you know by is evening
    Thank you
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    (Original post by MathsNerd1)
    I know this isn't relevant to this thread, but I've got a question that I didn't get right in a mock I did for unit 1 and was hoping you could help me out with it?

    The question was "Predict the formula of the zinc salt that could be formed by adding excess of Zinc to phosphoric (V) acid, H3PO4.

    I put it as Zn3PO4 but its actually meant to be Zn3(PO4)2, why is this?


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    Because the ions are Zn^2+ and PO4^3- hence there has to be 3 of Zn and 2 of PO4
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    (Original post by master y)
    Because the ions are Zn^2+ and PO4^3- hence there has to be 3 of Zn and 2 of PO4
    Oh yeah, how stupid of me not to get that! Thanks!


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    Help with.
    f325 june 2011 paper.
    Q7 dii) about half equations "Predict the half equations at each electrode for the decomposition of the electrolyte (KOH) to form hydrogen adn oxygen.

    I know the answer, but HOW do you go about doing it?

    Any help much appreciated
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    (Original post by reneetaylor)
    :laugh:

    Outline how a IR spectrum helps to identify a COOH bond in a molecule.
    cOME ON man this is not hard


    In the IR spectrum if there is a peak between the wavelength number of 1640-1750cm-1 this indicates that a C=O functional group is present and a peak at 1000-1300cm-1 indicates that a C-O functional group is present and hence, from the function groups C=O and C-O bond we can see that a carboxylic acid (functional group) is present in the molecule
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    It sounds like a lot of people on here have done the Jan 13 paper as a Chemistry mock - how did everyone find it?
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    (Original post by zef1995)
    It sounds like a lot of people on here have done the Jan 13 paper as a Chemistry mock - how did everyone find it?
    :bricks: how about you, what did ya get?
 
 
 
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