The relation between percent ionization a , Ka and C is Ka =
Ca2
Then ,
  a = (Ka/C)1/2
Given that Ka of H3AsO3 = 5.1 x
10-10
a)Â Â Â C = 0.206 M
     a = (Ka/C)1/2
           Â
= (5.1 x 10-10/0.206)1/2
        = 0.000049
       = 0.0049 %
Therefore,
percent ionization = 0.0049 %
b)Â Â Â C = 0.556 M
     a = (Ka/C)1/2
           Â
= (5.1 x 10-10/0.556)1/2
        = 0.00003
       = 0.003 %
Therefore,
percent ionization = 0.003 %
c)Â Â Â C = 0.822 M
     a = (Ka/C)1/2
           Â
= (5.1 x 10-10/0.822)1/2
        = 0.000025
       = 0.0025 %
Therefore,
percent ionization = 0.0025 %
