Question

What is the % ionization of a 0.30 M solution of HOCl (K_a = 3.5 x 10^-8)?

( write answer to three significant figures)

What is the % ionization of a 0.30 M solution of HOCI (Kg = 3.5 x 10-8)? ( write answer to three significant figures)

Answer 1

HOCl dissociates as:

HOCl -----> H+ + OCl-

0.3 0 0

0.3-x x x

Ka = [H+][OCl-]/[HOCl]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3.5*10^-8)*0.3) = 1.025*10^-4

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

so, x = 1.025*10^-4 M

% dissociation = (x*100)/c

= 1.025*10^-4*100/0.3

= 3.416*10^-2 %

Answer: 3.42*10^-2 %