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?

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Chemistry - Equilibria questions involving ICE tables

Loading...

Similar Threads - Chemistry Equilibria questions | Date |
---|---|

Is there a reaction for which the product is the catalyst? | Tuesday at 11:58 AM |

How can I stop condensation (not fog) build-up on my goggles? | Feb 9, 2018 |

Effect of impurities on the boiling point of ethyl ethanoate | Feb 7, 2018 |

Glycolic acid functional group(s) | Jan 21, 2018 |

Equilibrium Constants | Oct 19, 2014 |

**Physics Forums - The Fusion of Science and Community**