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Chemistry - Equilibria questions involving ICE tables

  1. May 5, 2010 #1
    In my book for all these equilibria questions involving ice tables when I end up with [tex]Ka = \frac{(x)(x)}{initial amount - x}[/tex] they always say that if the Ka is tiny compared to the initial amounts then we can assume x will be too and we can omit as many x's as we like from this equation. This obviously makes solving for x a whole lot easier but I don't get how this works.

    First off a weak acid/base example
    [tex]Ka = \frac{(x)(x)}{initial\ amount - x}[/tex]
    if I omit 1 x I get [tex]x = \sqrt{Ka(initial\ amount)}[/tex]
    if I omitted 2 x's I'd get [tex]x = Ka(initial\ amount)[/tex] a completely different answer. Does this only apply if I have a [tex](constant \pm x)[/tex] then I just omit the plus/minus x?
     
  2. jcsd
  3. May 5, 2010 #2

    Borek

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    Staff: Mentor

    x can be omitted only in sums.

    There is a rule of thumb (so called 5% rule) which says that a+x can be approximated by a if x is less than 5% of a. See here.

    --
    methods
     
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