Calculate the percent ionization of hypochlorous acid (HClO) in
solutions of each of the following concentrations (Ka =
3.0e-08.)
(a) 0.144 M
%
(b) 0.568 M
%
(c) 0.896 M
%
a)
HClO dissociates as:
HClO
-----> H+ + ClO-
0.144
0 0
0.144-x
x x
Ka = [H+][ClO-]/[HClO]
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*10^-8)*0.144) = 6.573*10^-5
since c is much greater than x, our assumption is correct
so, x = 6.573*10^-5 M
% dissociation = (x*100)/c
= 6.573*10^-5*100/0.144
= 4.564*10^-2 %
Answer: 4.6*10^-2 %
b)
HClO dissociates as:
HClO
-----> H+ + ClO-
0.568
0 0
0.568-x
x x
Ka = [H+][ClO-]/[HClO]
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*10^-8)*0.568) = 1.305*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.305*10^-4 M
% dissociation = (x*100)/c
= 1.305*10^-4*100/0.568
= 2.298*10^-2 %
Answer: 2.3*10^-2 %
c)
HClO dissociates as:
HClO
-----> H+ + ClO-
0.896
0 0
0.896-x
x x
Ka = [H+][ClO-]/[HClO]
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*10^-8)*0.896) = 1.64*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.64*10^-4 M
% dissociation = (x*100)/c
= 1.64*10^-4*100/0.896
= 1.83*10^-2 %
Answer: 1.8*10^-2 %
Calculate the percent ionization of hypochlorous acid (HClO) in solutions of each of the following concentrations...
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