Calculate the percent ionization of a 0.457 M solution of phenol (a weak acid), C6H5OH.
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Phenol (C6H5OH, also called carbolic acid) has a pKa of 9.89.
pKa of 9.89 means pK = 10⁻⁹ ⁸⁹ = 1.29E-010 Ka = [C6H5-O-] [H3O+]/[C6H5OH] (equilibrium concentrations) [C6H5-O-] = [H3O+] and [C6H5OH] = 0.457 M - [C6H5-O-] [C6H5-O-]²/(0.457 - [C6H5-O-]) = 1.29E-010 x²/(0.457-x) = 1.29E-010 x~~3.55912x10^-6 M = [C6H5-O-] percent ionized: 3.55912x10^-6*100/0.0982 = 0,0036 %
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