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

Chemistry
Tutor: None Selected Time limit: 1 Day

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


% Ionization = %
Jun 5th, 2015

Dear your question is under process and shall be available to you in next few minutes.


Jun 5th, 2015

OKay

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

it was wrong

Jun 5th, 2015

Are you studying on the go? Check out our FREE app and post questions on the fly!
Download on the
App Store
...
Jun 5th, 2015
...
Jun 5th, 2015
Dec 4th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer