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Question
Conductivity of 0.00241M acetic is 7.896×10−5S cm−1
Calculate its molar conductivity. If λ∘m for acetic acid is 390.5 S cm2mol−1, what is its dissociation constant ?
Conductivity of 0.00241M acetic is 7.896×10−5S cm−1
Calculate its molar conductivity. If λ∘m for acetic acid is 390.5 S cm2mol−1, what is its dissociation constant ?
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Solution
Step I : Calculation of molar conductivity (λcm)
k=7.896×10−5 S cm−1
C=0.00241mol L−1=0.00241103mol cm−3
=241×10−8mol cm−3
λcm=kC=(7.896×10−5Scm−1)(241×10−8mol cm−3)=32.76 S cm2 mol−1
Step II : Calculatio of dissociation constant (Ka)
(a) Calculation of degree of dissociation of CH3COOH(α)
α=λcmλ∘m=(32.76 S cm2mol−1)(390.5 S cm2 mol−1)=0.084=8.4×10−2
(b) Ka=[CH3COO−][H+][CH3COOH]=Cα×CαC(1−α)=Cα2(1−α)
=(0.00241 mol L−1)×(0.084)2(1−0.084)=0.0000185 mol L−1=1.85×10−5mol L−1
Molar conductivity (λcm)=32.76 S cm2mol−1
Dissociation constant (Ka)=1.85×10−5mol−1
Step I : Calculation of molar conductivity (λcm)
k=7.896×10−5 S cm−1
C=0.00241mol L−1=0.00241103mol cm−3
=241×10−8mol cm−3
λcm=kC=(7.896×10−5Scm−1)(241×10−8mol cm−3)=32.76 S cm2 mol−1
Step II : Calculatio of dissociation constant (Ka)
(a) Calculation of degree of dissociation of CH3COOH(α)
α=λcmλ∘m=(32.76 S cm2mol−1)(390.5 S cm2 mol−1)=0.084=8.4×10−2
(b) Ka=[CH3COO−][H+][CH3COOH]=Cα×CαC(1−α)=Cα2(1−α)
=(0.00241 mol L−1)×(0.084)2(1−0.084)=0.0000185 mol L−1=1.85×10−5mol L−1
Molar conductivity (λcm)=32.76 S cm2mol−1
Dissociation constant (Ka)=1.85×10−5mol−1
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