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Question

The ionization constant of a weak acid is 1.6×105 and the molar conductivity at infinite dilution is 380×104Sm2mol1. If the cell constant is 0.01 m1, then conductance (G) of 0.01 M acid solution is:



A

1.52×105S

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B

1.52×104S

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C

1.52×103S

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D
1.52 S
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Solution

The correct option is D 1.52 S
By Ostwald dilution law,
Ka=Cα21α1.6×105
For weak acid,
α<<1
Ka=Cα2

α=KaC

α=1.6×1050.010.04

​​​​​​α=Molar conductivity at a given concentrationMolar conductivity at a infinite dilution

​​​​​​α=ΛmΛ0m

Λm=α×Λ0m

Λm=0.04×380×104 Sm2mol1

Λm=15.2×104Sm2mol1

1L=103 m3
1L1=103 m3

Molar concentration, C=0.01 mol L1
Molar concentration, C=0.01×103 mol m3

Specific conductance, κ=Λm×C
κ=15.2×104×103×102
κ=15.2×103 S m1

We know,
Specific conductance, κ=Conductance ×Cell constant

Conductance=Specific conductanceCell constant

Conductance=15.2×1030.01

Conductance, G=1.52 S


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