[mafagafo]

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.

[Borek]

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.

[mafagafo]

"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.