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15/01/2022 6:04 pm
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The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ°(H+) = 349.6 S cm2 mol-1 and λ°(HCOO-) = 54.6 S cm2 mol
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15/01/2022 6:16 pm
C = 0.025 mol L-1
Λm = 46.1 S cm2 mol-1
λ0(H+) = 349.6 S cm2 mol-1
λ0(HCOO-) = 54.6 S cm2 mol-1
Λ0m(HCOOH) = λ0(H+) + λ0(HCOO-)
= 349.6 + 54.6
= 404.2 S cm2 mol-1
Now, degree of dissociation
α = \(\frac{Λ_m(HCOOH}{Λ_m^0(HCOOH)}\)
= \(\frac{46.1}{404.2}\)
= 0.114 (approximately)
Thus, dissociation constant
K = \(\frac{c ∝^2}{(1 - ∝)}\)
= \(\frac{(0.025mol L^{-1})(0.114)^2}{(1-0.114)}\)
= 3.67 x 10-4 mol L-1