Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: mafagafo on June 12, 2014, 09:22:48 PM
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This is my attempt to solve a problem.
I would like to have someone to check it for me. Does it seem right to you?
Supposing that the dissociation constant for C6H5COOH is 6.5e-05, calculate the percentage of acid that does not ionize.
[tex]K_a=\frac{[A^-][H^+]}{[AH]}=\frac{x^2}{1-x}[/tex]
[tex]6.5\cdot10^{-5}\approx x^2[/tex]
[tex]x^2\approx 64\cdot10^{-6}[/tex]
[tex]x=8\cdot10^{-3}[/tex]
[tex]\frac{[AH]_{eq}}{[AH]_i}=\frac{1-8\cdot10^{-3}}{1}=0.992[/tex]
[tex]\Rightarrow 99.2\%[/tex]
Thanks in anticipate, mafagafo.
Note: I can NOT use a calculator during the exam. That is the reason why I made those approximations.
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Unless you misquoted the question it is rather ugly, as dissociation fraction is always a function of concentration, and concentration is not given, making the problem impossible to solve.
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"III) A constante de dissociação do ácido benzóico 6,5 x 10 -5.
Calcule a porcentagem desse ácido que permanece na forma não ionizada, no equilíbrio."
No, nothing about concentration.
This problem is from a chemistry olympiad.
This is a plot I did in Mathematica, probably useless to you, but that helped me see how the answer depends in the concentration.
Blue line is % ionized. Red line is % not ionized. X axis is mol/L concentration.
(http://s29.postimg.org/brxfh2vrr/acid_ionization.jpg)