Question

Determine the percent ionization of the following solutions of phenol at 25°C:   (a) 0.713 M   %   (b) 0.220 M   %   (c) 1.86 × 10−6 M   × 10% (Enter your answer in scientific notation.)

Answer 1

Ka of phenol = 1.6*10^-10

C6H5OH <-----> C6H5O- +   H+
C                 0         0
C-x               x         x

Ka = x*x/(C-x)
since x is small,
Ka = x*x/(C)
x = sqrt (Ka*C)

% ionisation = x*100/C
= sqrt (Ka*C) *100 / C

a)
% = sqrt (Ka*C) *100 / C
% = sqrt ((1.6*10^-10)*0.713)*100/0.713
=(1.068*10^-3)/0.713
= 1.5*10^-3 %

b)
% = sqrt (Ka*C) *100 / C
% = sqrt ((1.6*10^-10)*0.220)*100/0.220
= 2.7*10^-3 %

c)
% = sqrt (Ka*C) *100 / C
% = sqrt ((1.6*10^-10)*1.86*10^-6)*100/(1.86*10^-6)
= 0.93 %