The relation between Molar conductance λm and specific conductance κ is
λm=Mk×1000
M = molarity
λm=(2.5×10−3)(5×10−5)×(103)=20Scm2mol−1
Now, the degree of dissociation, α=λ∞λm
α=40020=201
The acid dissociation constant, Ka=(1−α)Cα2=(2019)(2.5×10−3)(201×201)
=65.789×10–7
≈66×10–7