0.2 M solution of a weak acid HA is 1% ionized 250C.  Ka for the acid is equal to

HA  + H2O  ⇌  H3O+ + A-[HA] =C-Cα HA = 0.2 - 0.2 × 1100 = 0.198 MA-1 =H+=Cα=0.2×10-2= 0.002  MKa=[A-][H3O+][HA]=0.002×0.0020.198