Chemistry question

[Awuah29]

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guys quick question
Application of Ka and Kb
if I have Ka= XX/2.0-X=1.8x10^(-5) how do I solve this. Quadatic equation thats what Kaplan says. Can I do here an aproximation. and how do I do it ? Am I expected to use quadratric equation without calculator? Please help

 

[Eddie830]

guys quick question
Application of Ka and Kb
if I have Ka= XX/2.0-X=1.8x10^(-5) how do I solve this. Quadatic equation thats what Kaplan says. Can I do here an aproximation. and how do I do it ? Am I expected to use quadratric equation without calculator? Please help

You can assume X is negligible, therefore your equation would look like: Ka= (X^2)/2.0=1.8x10^(-5) ...simplify and you get...
X^2=0.9x10^(-5) OR X^2=9x10^(-6).
so square root both sides and you have:
X=3x10^(-3) or .003

You'll notice that if you plug .003 back into the original equation, it will have pretty much no effect on the denomintor, that is why is was ok to make this assumption in the first place. For the sake of the DAT, you can always assume the problem will make X negligible.