Question

Which of the following aqueous solutions will have the largest percent ionization if the pka of phenol is 10.00 and the pka of propylammonium ion is 10.71? A 0.300 M phenol phenols O B 0.300 M propylammonium ion C. This cannot be predicted from the given information OD 0.500 M phenol E. 0.500 M propylammonium ion Reset Selection

Answer 1

Ka for phenol = 10^(-pKa) = 10^-10

Ka for propylammonium = 10^-(10.71) = 1.95*10^-11

A) 0.300 M phenol

[H+]^2 = 0.300 * 10^-10  

[H+] = 5.48*10^-6 M

% ionization= 5.48*10^-6 * 100/0.300 = 0.00183%

B) 0.300 M propyl ammonium, it will have less % ionization as Ka of it is less than phenol

D) [H+]^2 = 0.500 * 10^-10

[H+] = 7.07*10^-6 M

% ionization = 7.07*10^-6 * 100/0.500 = 0.00141 %

E) it will have lesser ionization than 0.500 M phenol.

Thus, option (A) i.e. 0.300 M phenol is the answer.