A $0.050 M$ solution of a weak monobasic acid HA has molar conductivity $7.81 S c m^{2} mo l^{- 1}$ . The limiting molar conductivity of HA is $390.5 S c m^{2} mo l^{- 1}$ . Using $K_{a} = \frac{c α ^{2}}{1 - α}$ , the value of $K_{a} \times 1 0^{5}$ (two decimals) is

Numerical answer type, tolerance 0.03.

Answer: 2.04

α = Λ_{m} / Λ_{m}^{\circ} = 7.81/390.5 = 0.0200

K_{a} = \frac{c α ^{2}}{1 - α} = \frac{0.050 \times ( 0.0200 ) ^{2}}{1 - 0.0200} = \frac{2.00 \times 1 0 ^{- 5}}{0.980} = 2.04 \times 1 0^{- 5}

. So

K_{a} \times 1 0^{5} = 2.04