Given that the specific conductance, k, is 5×10−5 S cm−1 and the concentration, C, is 0.0025 M, we can find the molar conductivity, λm, as follows:
λm=Ck×1000=0.00255×10−5×103=2.5×10−35×10−2=20 S cm2 mol−1
Next, we find the degree of dissociation, α, by dividing λm by the limiting molar conductivity, λmo:
α=40020=201
Finally, we use the formula for the dissociation constant of a weak acid, Ka:
Ka=1−αCα2=20190.0025×201×201=19×200.0025=6.6×10−6=66×10−7
So, the correct answer is 66.
