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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Chemistry - Equilibria questions involving ICE tables

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