Question

Find the percent ionization of a 0.200 M HC2H3O2 solution at 25 C [Ka(HC2H3O2)=1.8x10^-5]

Answer 1

Lets write the dissociation equation of HC2H3O2

HC2H3O2 -----> H+ + C2H3O2-

0.2 0 0

0.2-x x x

Ka = [H+][C2H3O2-]/[HC2H3O2]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((1.8*10^-5)*0.2) = 1.897*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.897*10^-3 M

% dissociation = (x*100)/c

= 1.897*10^-3*100/0.200

= 0.9487 %

Answer: = 0.949 %