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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
    Nice explanation!

    I'm always tempted to write that the oh- reacts with HX to form X-, but because the [HX] and [X-] are so high relative to the [oh-], that the ratio of the [HX]/[X-] effectively remains constant, so the solution resists a change in pH despite a small amount of alkali being added... do you think this would be accepted too?
    It's rarely on, not many mark schemes to see
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    For acid base questions if the acid is in XS do you divide the XS by the total volume or do you only do that if the OH is in XS?
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    (Original post by DanMargetts)
    Nice explanation!

    I'm always tempted to write that the oh- reacts with HX to form X-, but because the [HX] and [X-] are so high relative to the [oh-], that the ratio of the [HX]/[X-] effectively remains constant, so the solution resists a change in pH despite a small amount of alkali being added... do you think this would be accepted too?
    It's rarely on, not many mark schemes to see
    Hi,

    M1 extra/added OH–removed by reaction with H+or the acid

    M2 correct discussion of equm shift i.e. HXH++ X– movesto right
    ORratio[X ][HX]-remains almost constant

    This is legit the marks scheme i found regarding this Q so its what the examiner expects from us - However , I think what you have said is on this mark scheme and their is no reason to believe that it wont be on it in the future so it should be accepted as it makes sense in terms of chemistry- Rememebr the examiners most of them are qualified teachers etc so if they think that your talking sense you shall be definitely awarded for it as i dont see anything in particular underlined on this ms answer.
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    Can someone explain 3e and 3eii to me.

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    Also 3a, Why cant you have c3h7?
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    (Original post by ahsan_ijaz)
    Hi,

    M1 extra/added OH–removed by reaction with H+or the acid

    M2 correct discussion of equm shift i.e. HXH++ X– movesto right
    ORratio[X ][HX]-remains almost constant

    This is legit the marks scheme i found regarding this Q so its what the examiner expects from us - However , I think what you have said is on this mark scheme and their is no reason to believe that it wont be on it in the future so it should be accepted as it makes sense in terms of chemistry- Rememebr the examiners most of them are qualified teachers etc so if they think that your talking sense you shall be definitely awarded for it as i dont see anything in particular underlined on this ms answer.

    Thank you for the advice, I appreciate the hopeful words! All the best with the exams
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    (Original post by string210)
    thats what i did, checked the mark scheme, it wasnt right!
    If you work out the moles of both the acid and the base you'll find that the moles of OH- = moles of H+. This is the half equivalence point.

    Here's my working if you're interested. It's a nice property to remember which will save you some time when working but the procedure is exactly the same for any neutralization reaction.

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    (Original post by DanMargetts)
    Nice explanation!

    I'm always tempted to write that the oh- reacts with HX to form X-, but because the [HX] and [X-] are so high relative to the [oh-], that the ratio of the [HX]/[X-] effectively remains constant, so the solution resists a change in pH despite a small amount of alkali being added... do you think this would be accepted too?
    It's rarely on, not many mark schemes to see
    hey wat do you mean when you say that the concentration of HX and X- are so high?
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    (Original post by Lilly1234567890)
    hey wat do you mean when you say that the concentration of HX and X- are so high?
    Relative to the amount of H+ added

    Or am I wrong? It could be that their moles are a lot higher, in buffer questions, the mol of H+ added is always really small, but often the concentration is too, so I guess there the mol and concentration of HX and X- are higher than the mol and conc of H+, if that makes sense..
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    (Original post by Super199)
    Can someone explain 3e and 3eii to me.

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    Also 3a, Why cant you have c3h7?
    3e) a primary amine and the carbon skeleton has 3 different environments for carbon so

    H2N - CH2 - C (CH3)3

    This has three different carbon environments and is a primary amine.

    3ei) this has 3 hydrogen environments (the two methyl groups are in the same environment) and the H on the carbon causes doublet splitting on the adjacent methyl groups

    (CH3)2-C-H
    I
    CH3 - N - CH3
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    (Original post by emma_1111)
    For acid base questions if the acid is in XS do you divide the XS by the total volume or do you only do that if the OH is in XS?
    always divide by volume
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    (Original post by Parallex)
    If you work out the moles of both the acid and the base you'll find that the moles of OH- = moles of H+. This is the half equivalence point.

    Here's my working if you're interested. It's a nice property to remember which will save you some time when working but the procedure is exactly the same for any neutralization reaction.

    Thank you so much for this! Need to look out for this on Tuesday
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    (Original post by lahigueraxxx)
    Thank you so much for this! Need to look out for this on Tuesday
    There's no need to remember specifics like the half equivalence point, that's just a little niche thing that can save you a bit of time. Just know that a neutralisation reaction will occur if OH- is added so you need to work out the new moles.
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    Could somebody explain to me why the answer is P on June 2015 Q10c)? At first I thought that it could be S or T - don't S and T also have 5 peaks? The number of peaks for S and T must be greater than 5 but I don't understand why.

    http://filestore.aqa.org.uk/subjects...4-QP-JUN15.PDF
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    4b

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    Can someone explain how to work out the number of molecular ion peaks?
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    When do you minus and find the XS in ph calculations? Plus how do you qualitatively explain how buffers work? I know it has something to do with eqbm and water and concentrations and the mark schemes arent much help in understanding it
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    Hi any tricks or tips as to how to easily work out structures? definitely my weakest point, takes me forever and most of the time can't even work them out anyway!! any help anyone!!!
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    (Original post by Aerosmith)
    Could somebody explain to me why the answer is P on June 2015 Q10c)? At first I thought that it could be S or T - don't S and T also have 5 peaks? The number of peaks for S and T must be greater than 5 but I don't understand why.

    http://filestore.aqa.org.uk/subjects...4-QP-JUN15.PDF
    S and T both have 6 peaks. For S, there are 5 non-equivalent H peaks in the ring structure and 1 on the OH. For T, the same applies for the ring and the methyl group.
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    (Original post by Super199)
    4b

    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF

    Can someone explain how to work out the number of molecular ion peaks?
    There are 4 chlorine atoms in the compound which exist as Cl-35 and Cl-37. There's 5 different combinations of the isotopes:

    1) Cl-35 - Cl-35 - Cl-35 - Cl-35
    2) Cl-37 - Cl-37 - Cl-37 - Cl-37
    3) Cl-35 - Cl-35 - Cl-35 - Cl-37
    4) Cl-35 - Cl-35 - Cl-37 - Cl-37
    5) Cl-35 - Cl-37 - Cl-37 - Cl-37
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    (Original post by Parallex)
    There are 4 chlorine atoms in the compound which exist as Cl-35 and Cl-37. There's 5 different combinations of the isotopes:

    1) Cl-35 - Cl-35 - Cl-35 - Cl-35
    2) Cl-37 - Cl-37 - Cl-37 - Cl-37
    3) Cl-35 - Cl-35 - Cl-35 - Cl-37
    4) Cl-35 - Cl-35 - Cl-37 - Cl-37
    5) Cl-35 - Cl-37 - Cl-37 - Cl-37
    I see, how would you work out the number of carbon peak 4dii
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    (Original post by Super199)
    I see, how would you work out the number of carbon peak 4dii


    Equivalent carbon groups are the same colour. The compound is symmetrical about the O bonds so you only need to consider one half of the structure. Splitting the compound horizontally there is also a symmetry, so the groups below are equivalent to the groups above. There are 3 non-equivalent carbon groups in the molecule.
 
 
 
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