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    Can anyone help me with this?

    A 0.150 moldm^-3 solution of a weak acid, HX, has a pH of 2.34. Calculate the Ka for this acid.

    So obviously, Ka=[H][X]/[HA]

    I thought it should be Ka=([10^-2.34]^2)/(0.15-10^-2.34)

    but in the mark scheme, it doesnt take away [H] from [HA]. Why not?

    Help please!
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    Because [HA] is a weak acid it is assumed that it dissociates very very little and so its equilibrium concentration is the same as its original concentration - so you dont take away [H+] from [HA]

    So Ka=([10^-2.34]^2)/(0.150)
    = 1.39 x 10^-4 moldm^-3
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    (Original post by xarcul)
    Can anyone help me with this?

    A 0.150 moldm^-3 solution of a weak acid, HX, has a pH of 2.34. Calculate the Ka for this acid.

    So obviously, Ka=[H][X]/[HA]

    I thought it should be Ka=([10^-2.34]^2)/(0.15-10^-2.34)

    but in the mark scheme, it doesnt take away [H] from [HA]. Why not?

    Help please!
    The assumption is made that the concentration of the weak acid is hardly affected at all by the dissociation. This allows us to do the calculation without resorting to quadratic equations.

    EDIT: Beaten to the draw by Bhups ...
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    (Original post by charco)
    The assumption is made that the concentration of the weak acid is hardly affected at all by the dissociation. This allows us to do the calculation without resorting to quadratic equations.
    Well thats what I thought if you had Ka and wanted to work out pH for example. But if you already have [H+] then surely you just take it away? I'm sure that in other questions I've done I was required to do so.
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    For the Edexcel exam board we always make the assumptions that [H+]=[A-] and that [HA] initial = [HA] equilibrium
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    Incidentally, chemists are cr*p at maths.......
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    (Original post by xarcul)
    Incidentally, chemists are cr*p at maths.......
    Bit of a Generalisation there... What evidence do you have?
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    (Original post by xarcul)
    Incidentally, chemists are cr*p at maths.......
    physical chemists get very "physical" with math, computer modelling, and a lot of calculus, surely you are generalising there.
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    I'm not generalising! I'm making a bloody joke...
    I was referring to the fact they unnecessarily assume something just to avoid a quadratic equation...
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    (Original post by xarcul)
    I'm not generalising! I'm making a bloody joke...
    I was referring to the fact they unnecessarily assume something just to avoid a quadratic equation...
    A lot of problems in math comes down to simple assumptions too, and the fact that chemists use these assumptions(approximations) is because chemists must be capable in math too!

    what a joker!
 
 
 
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