Given:
[H2AsO3]=0.2 M
Percent dissociation is 0.0051 %
Solution
Reaction:
H2AsO3 (aq) + HÂ2O (l) --- >
HAsOÂÂ3- (aq) + H3O+
(aq)
Acid dissociation constant expression:
ka = [HAsO3-] [
H3ÂO+] / [H2AsO3]
In order to get the ka value we must know the equilibrium
concentrations of these all the species involved.
ICE
H2AsO3 (aq) + HÂ2O (l) --- >
HAsOÂÂ3- (aq) + H3O+
(aq)
IÂ Â Â Â Â Â Â Â Â Â
0.200Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â
-x                                           Â
+x                              Â
+x
EÂ Â Â Â Â Â Â Â Â
(0.200-x)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
x                                Â
x
Ka = x2/ (0.200-x)
Percent dissociation
0.00051 = (x/0.200) * 100
x = 1.02 E-5 lets plug this value of x in ka expression
ka = ( 1.02 E-5)2/ (0.200-1.02 E-5)
ka = 5.20 x 10-10
