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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 - Chemistry

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प्रश्न

The molar conductivity of 0.025 mol L−1 methanoic acid is 46.1 S cm2 mol1. Calculate its degree of dissociation and dissociation constant. Given \[\ce{λ^0_{(H^+)}}\] = 349.6 S cm2 mol1 and \[\ce{λ^0_{(HCOO^-)}}\] = 54.6 S cm2 mol1.

संख्यात्मक

उत्तर

\[\ce{∧^0_{{m}(HCOOH)} = λ^0_{(H^+)} + λ^0_{(HCOO^-)}}\]

= 349.6 + 54.6

= 404.2 S cm2 mol−1

Given: \[\ce{∧_{{m}(HCOOH)}}\] = 46.1 S cm2 mol−1

Degree of dissociation, α = `∧_"m"/(∧_"m"^0)`

= `(46.1  "S cm"^2  "mol"^-1)/(404.2  "S cm"^2  "mol"^-1)`

= 0.114

\[\ce{HCOOH ⇌ HCOO^- + H^+}\]

Initial concentration c mol L−1 0 0
Concentration at equilibrium c(1 − α)

∴ Kα = `("c"α . "c"α)/("c"(1 - α))`

= `("c"α^2)/(1 - α)`

= `(0.025 xx (0.114)^2)/(1 - 0.114)`

= 3.67 × 10−4

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अध्याय 3: Electrochemistry - Intext Questions [पृष्ठ ८३]

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एनसीईआरटी Chemistry [English] Class 12
अध्याय 3 Electrochemistry
Intext Questions | Q 9 | पृष्ठ ८३

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