This problem is providing us with the pKa of phenol and its initial concentration. It is asking for the percent ionization it exhibits in water. At the end, the answer turns out to be 0.00405 %.
In chemistry, when acids and bases dissolve in water, they undergo ionization, a process whereby they split into a cation and an anion. In this case, since phenol has a pKa of 9.89, one can write the following ionization reaction:
[tex]C_6H_5OH(aq)+H_2O\rightleftharpoons C_6H_5O^-(aq)+H_3O^+(aq)[/tex]
And the equilibrium constant can be written as:
[tex]K=\frac{[C_6H_5O^-][H_3O^+]}{[C_6H_5OH]} =\frac{x^2}{0.0786} =10^{-9.89}\\\\1.29x10^{-10}=\frac{x^2}{0.0786}[/tex]
Which can be solved for x with:
[tex]x=\sqrt{1.29x10^{-10}*0.0786} \\\\x=3.19x10^{-6}M[/tex]
Finally, the percent ionization:
[tex]\%ioniz=\frac{3.18x10^{-6}M}{0.0786M}*100\%\\ \\\%ioniz=0.00405\%[/tex]
Learn more about percent ionization: https://brainly.com/question/9173942
