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    How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
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    Anybody got a summary on ligands and complex ions?
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    june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.
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    (Original post by Dinasaurus)
    So Ka is actually on the fraction?
    So Ka/[HA] = [H][A]

    So Ka/HA x [A] = H?

    I am so lost
    It's just a rearrangement mate. You'll normally have 3 values and rearrange to find the unknown. Or 1 value and a ratio.
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    (Original post by Dinasaurus)
    So Ka is actually on the fraction?
    So Ka/[HA] = [H][A]

    So Ka/HA x [A] = H?

    I am so lost
    finding pH of a buffer: use 'kacidoversalt' which is [H+] = Ka x [HA]/[A-]
    then -log as normal
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    (Original post by HasanAlam)
    How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
    In theory if it's greater than 0 it's feasible. In practicality anything below 0.3 is unlikely to be feasible. You won't get docked marks for saying it's feasible if it's above 0.
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    (Original post by HasanAlam)
    How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
    it MAY not be feasible if it's below 0.4 V difference but doesn't mean it won't be
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    (Original post by HasanAlam)
    How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
    The 0.4V rule is just a general rule but it doesn't apply to everything so I'd just ignore it tbh.
    There are two main answers they look for when they say 'why may this not be feasible?':
    1) the conditions aren't standard
    2) The activation energy is high
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    (Original post by CalistaJupiter)
    Ka=([A][H])/[HA]
    Which rearranges to [H]= Ka*[HA]/[A]
    It's really confusing as my revision shows the fraction line underneath the part with Ka itself.

    Ok finally got 3.99 cheers a lot

    Also what is the formula for a pH of a strong acid if not the same?
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    >Feels confident with revision
    >Does reasonably well on lots of past papers
    >Decides to do January 2011 F325 as a last piece of revision
    >Dies inside
    >Sees the grade boundaries are A = 65
    >Reborn
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    (Original post by BioStudentx)
    It's just a rearrangement mate. You'll normally have 3 values and rearrange to find the unknown. Or 1 value and a ratio.
    My revision guide had Ka on the numerator not to the side of it, which made it only have 2 terms.
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    (Original post by lai812matthew)
    june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.
    It is diluted. They added water until volume was 100cm^3. The original volume was 25cm^3. Therefore, concentration is 4 x less (as c=n/v and v is 4x bigger). 0.480/4=0.120
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    (Original post by HasanAlam)
    How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?
    its feasible if E>0


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    anyone have the F325 June 2015 mark scheme? Or an unofficial one?
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    (Original post by Dinasaurus)
    My revision guide had Ka on the numerator not to the side of it, which made it only have 2 terms.
    Post a picture. You can rearrange the equation to have Ka on any side. It's never [H+]^2... You're just confusing weak acids and buffers.The only way you can have two terms is if you're given a ratio. Most Qs will work with ratios in big buffer qs. And ratios always involve the acid and conjugate base (HA and A-).
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    (Original post by BioStudentx)
    It is diluted. They added water until volume was 100cm^3. The original volume was 25cm^3. Therefore, concentration is 4 x less (as c=n/v and v is 4x bigger). 0.480/4=0.120
    haha sorry i misread q4eiii) answer to this question's answer thx
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    (Original post by phoebeg76)
    anyone have the F325 June 2015 mark scheme? Or an unofficial one?
    youtube f325 2015
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    (Original post by BioStudentx)
    youtube f325 2015
    thank you!
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    (Original post by BioStudentx)
    Post a picture. You can rearrange the equation to have Ka on any side. It's never [H+]^2... You're just confusing weak acids and buffers.The only way you can have two terms is if you're given a ratio. Most Qs will work with ratios in big buffer qs. And ratios always involve the acid and conjugate base (HA and A-).
    You are actually right I was on the weak acids page but what I meant was the Ka was literally on the fraction. So instead of say A=(B/C) it just had A=B/C
    With A and B being on the same line.
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    (Original post by Dinasaurus)
    You are actually right I was on the weak acids page but what I meant was the Ka was literally on the fraction. So instead of say A=(B/C) it just had A=B/C
    With A and B being on the same line.
    I'm struggling to understand! If you post a picture I'll understand better.

    A=(B/C) and A=B/C are the same thing? Haha.
 
 
 
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