Calculate the percent ionization of a 0.457 M solution of phenol (a weak acid),

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Calculate the percent ionization of a 0.457 M solution of phenol (a weak acid), C6H5OH


% Ionization = %
Jun 5th, 2015

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Jun 5th, 2015

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Jun 5th, 2015

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 %


Jun 5th, 2015

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Jun 5th, 2015

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