percentage ionization

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Leslie Romo 1C
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

percentage ionization

Postby Leslie Romo 1C » Thu Nov 26, 2015 5:09 pm

In the problem "What is the PH and percentage ionization of acetic acid, in .10M CH3COOH(aq)? (Ka=1.8x10^-5 for acetic acid)"
How do you find the percentage ionization of the acetic acid??

I found the pH to be -log[1.3x10^-3]= 2.9

Kelly Sun 3E
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

Re: percentage ionization

Postby Kelly Sun 3E » Thu Nov 26, 2015 8:37 pm

To find percent ionization, you divide (the concentration of the conjugate base)/(the original concentration). So using the concentration you found-->1.3x10^-3 M

[CH3COO-] / [CH3COOH]
(1.3x10^-3) / (.10) x 100%= 1.3%


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