What is the % ionization of a 0.30 M solution of HOCl (Ka = 3.5 x 10-8)?
( write answer to three significant figures)
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 %
What is the % ionization of a 0.30 M solution of HOCl (Ka = 3.5 x...
What is the % ionization of a 0.30 M solution of HOCl (Ka = 3.5 x 10^-8)? ( write answer to three significant figures) please help , show work will be great appreciated.
What is the % ionization of a 0.30 M solution of HOCl (Ka = 3.5 x 10-8)?
What is the % ionization of a 0.30 M solution of HOCl (Ka = 3.5 x 10^-8)? PLEASE HELP , MAY YOU PLEASE SHOW ANSWER WITH STEPS THANK YOU IN ADVANCE.
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HOCL Ka value in textbook 4.0x10^-8
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