Question

Calculate the percent ionization of phenol (HC6H5O) in solutions of each of the following concentrations (Ka = 1.3e-10.)

(a) 0.271 M

(b) 0.510 M

(c) 0.727 M

Answer 1

phenol dissociate as

C₆H₅OH C₆H₅O^-  + H⁺

thus, Ka = [ C₆H₅O^- ] [H⁺] / [C₆H₅OH]

C₆H₅O^- and H⁺ have same concetration

then consider [ C₆H₅O^- ] = [H⁺] = X

above equation become

Ka = [ X ] [X] / [C₆H₅OH]

Ka = [ X ]² / [C₆H₅OH] substitute value in this equation

(a) 0.271 M

1.3 10^-10 =  [ X ]² / (0.271)

[ X ]² = 1.3 10^-10 0.271 = 3.523 10^-11

[ X ] = 5.93 10^-6 M =  [ C₆H₅O^- ] = [H⁺]

% dissociation = [H⁺] 100/ [C₆H₅OH] =  5.93 10^-6 100 / (0.271 ) = 0.002188%

% dissociation of  0.271 M C₆H₅OH = 0.002188%

(b)

0.510 M

1.3 10^-10 =  [ X ]² / (0.510)

[ X ]² = 1.3 10^-10 0.510 = 6.63 10^-11

[ X ] = 8.14 10^-6 M =  [ C₆H₅O^- ] = [H⁺]

% dissociation = [H⁺] 100/ [C₆H₅OH] = 8.14 10^-6 100 / (0.510 ) = 0.001596%

% dissociation of 0.510M C₆H₅OH = 0.001596%

(c)

0.727 M

1.3 10^-10 =  [ X ]² / (0.727)

[ X ]² = 1.3 10^-10 0.727 = 9.451 10^-11

[ X ] = 9.72 10^-6 M =  [ C₆H₅O^- ] = [H⁺]

% dissociation = [H⁺] 100/ [C₆H₅OH] = 9.72 10^-6 100 / (0.727 ) = 0.001337%

% dissociation of 0.727M C₆H₅OH = 0.001337%