Calculating x at equilibrium

Acidity
Basicity
The Conjugate Seesaw

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Diana_Anum1G
Posts: 30
Joined: Wed Sep 21, 2016 2:59 pm

Calculating x at equilibrium

Postby Diana_Anum1G » Sat Dec 03, 2016 4:39 pm

So from the example that Dr. Lavelle did yesterday when we were calculating x the equation was: 5.55x10^-10 =x^2/0.0600- x then we made it approximately equal to x^2/0.0600 which assumes that x was zero right? Can we expect to do this on the final instead of using the quadratic equation? Or should I be on the safe side and use the quadratic equation to find x?

Emma_Green_2C
Posts: 21
Joined: Wed Sep 21, 2016 2:57 pm

Re: Calculating x at equilibrium

Postby Emma_Green_2C » Sat Dec 03, 2016 5:05 pm

I think my TA said that if the Ka value is 10^-5 or smaller then you can assume x is zero. So under that circumstance I'm pretty sure you would be safe making that assumption on the final.

giareyes_2B
Posts: 20
Joined: Wed Sep 21, 2016 2:58 pm

Re: Calculating x at equilibrium

Postby giareyes_2B » Sat Dec 03, 2016 7:42 pm

I believe that you can assume that x=0 is correct if the value you got for x is less than 5% of the value you were subtracting it from. For example, you would take whatever value you got for x, and then divide it by .0600. If the percentage you get is less than 5%, then you should be fine in assuming.


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