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    (Original post by AtomicMan)
    Sorry I dont really understand your question I am just saying what Edexcel define as the assumption for Ka calculations. In reality things are probably different, but I imagine for A-Level they dont want to overcomplicate things for us.
    Haha I was just teasing

    The way it's done is that you set a mass balance, CHA=[HA]+[A-] (i.e. both forms of the acid have to add up to a total concentration, which is the concentration of acid you originally put in), and a charge balance (charges have to be equal in the solution so [H+]=[A-]+[OH-]). So now the condition for writing [H+]=[A-] should be obvious: [OH-]=0.
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    (Original post by Big-Daddy)
    Haha I was just teasing

    The way it's done is that you set a mass balance, CHA=[HA]+[A-] (i.e. both forms of the acid have to add up to a total concentration, which is the concentration of acid you originally put in), and a charge balance (charges have to be equal in the solution so [H+]=[A-]+[OH-]). So now the condition for writing [H+]=[A-] should be obvious: [OH-]=0.
    Ah, that makes sense, thanks
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    (Original post by AtomicMan)
    Ah, that makes sense, thanks
    was ur assumption wrong
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    (Original post by otrivine)
    was ur assumption wrong
    No ... I don't know if you know this but Kw=[OH-]*[H+] and Kw being fixed (it's an equilibrium constant like any other) means that [OH-]=Constant/[H+] so it's obvious that, in a very acidic solution (to be fair, even a decently acidic one), [OH-] will be small so you can make the approximation [H+]=[A-]. This is what is meant when people say "neglecting the auto-dissociation of water" (because you come to the same result by setting up mass balance, charge balance and equilibrium expressions without Kw as you do by eliminating the [OH-] term at the end setting it to 0 - if you don't understand this, read over my last post again or just forget about it as you won't need it at this stage). I only intervened to explain the wrong logic behind using the approximation [H+]=[A-].
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    (Original post by AtomicMan)
    For B I get a pH of 2.3, how did you work it out?

    The rest are correct
    Hi, could you please explain to me how you got the value for C I keep on getting 1.22 :s
 
 
 
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