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    (Original post by Beta14)
    I divided the conc by 5 then used the Kw of 60*C i think thats what you were supposed to do, a similar question came up last year 2015 paper as a 3 marker and thats what the mark scheme says to do
    I tried to do that and failed so....
    I ended up just putting that 11.73 or whatever. But I'm sure your method is right yeah.
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    what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
    I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
    and it was only 3 marks so probably nothing tricky was needed?
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    (Original post by lawlieto)
    what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
    I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
    and it was only 3 marks so probably nothing tricky was needed?
    Calculate moles of acid and conjugate base, divide by 1000 to get concentration, use Kaacidoversalt to get [H+], - log to get 3.43 IIRC


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    (Original post by GetOverHere)
    Calculate moles of acid and conjugate base, divide by 1000 to get concentration, use Kaacidoversalt to get [H+], - log to get 3.43 IIRC


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    that's exactly what i've done; as it was 1 dm^3 basically moles of acid and base was same as concentrations if i remember correctly
    but people seem to have done 3 different things in this thread so i got unsure...
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    (Original post by lawlieto)
    what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
    I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
    and it was only 3 marks so probably nothing tricky was needed?
    4 marks so it is defo not boring buffer question. if they give concentration in buffer there will only be 2 marks.
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    (Original post by lai812matthew)
    4 marks so it is defo not boring buffer question. if they give concentration in buffer there will only be 2 marks.
    you had to calculate concentrations of acid + base in the buffer solution (that could be 2 marks, each term worth 1 marks)

    1 mark for for using the right formula

    1 mark for answer?

    what have you done in that question?
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    Would the marks be for calculating moles or the actuallt conc (I know they're the same but still)
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    (Original post by lawlieto)
    what did people do in that buffer solution pH question where HNO2 and NaNO2 were added?
    I thought you just use the buffer concentrations calculated from the given concentrations and volumes in the question? It's just simply mixing an acid and it's conjugate base? (Na+ ions are only spectator ions I think so nothing actually reacts)
    and it was only 3 marks so probably nothing tricky was needed?
    Number of mole of HNO2 was 0.04
    Number of mole of NaNO2 was 0.05
    0.05-0.04=0.01 [NO2-]
    [H+]= (Ka*0.04)/0.01
    -log [H+] =pH
    I don't remember the Ka
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    (Original post by pezhman)
    Number of mole of HNO2 was 0.04
    Number of mole of NaNO2 was 0.05
    0.05-0.04=0.01 [NO2-]
    [H+]= (Ka*0.04)/0.01
    -log [H+] =pH
    I don't remember the Ka
    Dont take off 0.04 Moles

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    Has anyone done the paper? Unofficial mark scheme?


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    (Original post by k.russell)
    Isn't it to do with the black-grey precipitate (I2) and the yellow precipitate (AgI)?
    idk, I reckon that was the hardest question on the paper though ngl
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    (Original post by tomlam)
    idk, I reckon that was the hardest question on the paper though ngl
    Nooo that wasn't too bad really.


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    Does anyone know how many marks the page on the gold complex was, I ran out of time and didn't answer any of it. Do you think it's still possible to get an A?
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    (Original post by tommason)
    Does anyone know how many marks the page on the gold complex was, I ran out of time and didn't answer any of it. Do you think it's still possible to get an A?
    think it was 6 or 8 marks that single page, im thinking 6
    and yeap it should be if everything else was okay
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    No 2015 was easier! I'm with you
    (Original post by Slenderman)
    In comparion to 2015. Harder or easier? Am I retarded to think the grade boundaries will be 75 for an A?
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    (Original post by MilkFriend)
    Anyone got 10.76... no idea about what happened.
    It was 20cm3 of 0.0270moldm^-3 NaOH (or whatever?) diluted up to 100cm3, and the Kw was 9.311*10^-14.

    New conc = 0.0270/5

    So -log(9.311*10^-14 / (0.0270/5)) = 10.76.

    I think what the other person may have done is taken it as them adding 100cm3 (I can't remember the question but I think it said diluted up to?) so divided 0.0270/6 and put it in to get 10.68.
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    (Original post by lawlieto)
    you had to calculate concentrations of acid + base in the buffer solution (that could be 2 marks, each term worth 1 marks)

    1 mark for for using the right formula

    1 mark for answer?

    what have you done in that question?
    same as what you do
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    (Original post by ThatPerson2)
    It was 20cm3 of 0.0270moldm^-3 NaOH (or whatever?) diluted up to 100cm3, and the Kw was 9.311*10^-14.

    New conc = 0.0270/5

    So -log(9.311*10^-14 / (0.0270/5)) = 10.76.

    I think what the other person may have done is taken it as them adding 100cm3 (I can't remember the question but I think it said diluted up to?) so divided 0.0270/6 and put it in to get 10.68.
    i did the same ha......
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    (Original post by lai812matthew)
    i did the same ha......
    I'm pretty sure it said diluted up to as opposed to added to. I would think that whichever it wasn't would still get most of the marks though.
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    (Original post by ThatPerson2)
    It was 20cm3 of 0.0270moldm^-3 NaOH (or whatever?) diluted up to 100cm3, and the Kw was 9.311*10^-14.

    New conc = 0.0270/5

    So -log(9.311*10^-14 / (0.0270/5)) = 10.76.

    I think what the other person may have done is taken it as them adding 100cm3 (I can't remember the question but I think it said diluted up to?) so divided 0.0270/6 and put it in to get 10.68.
    Ok, thanks, So does that means 10.76 is more likely to be the right answer?
 
 
 
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