Question

Molar conductivity of a weak acid HA at infinite dilution is 345.8 S cm2/mol. Calculate molar conductivity of 0.05 M HA solution. Given that alpha = 5.8 ?? 10-6.

Answer 1

Answer:

Molar Conductivity of 0.05 M HA solution is

$2.00564 \times {10 }^{ - 3} s \: {cm}^{2} \: {mol}^{ - 1}$

Answer 2

Answer:

Molar conductivity of 0.05M HA solution is 2×10⁻³ Scm²mol⁻¹.

Explanation:

Molar conductivity of a weak acid HA at infinite dilution $= 345.8 Scm^{2}mol^{-1}$

The degree of dissociation $\alpha =5.8*10^{-6}$

We know that   α = ∧/∧°

Molar Conductivity, ∧ = α×∧°

                                 ∧  =$5.8*10^{-6}*345.8Scm^{2}mol^{-1}$

                                ∧ = $2*10^{-3}Scm^{2}mol^{-1}$

Therefore for 0.05M HA solution the molar conductivity is 2×10⁻³ Scm²mol⁻¹.