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Molar conductivity of a weak acid HA at ...

Molar conductivity of a weak acid HA at infinite dilution is 345.8 `cm^(2) mol^(-)` calculate molar conductivity of 0.05 M HA solution `(alpha=5.8` x `10^(-6))`

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Equivalent conductivity of `1 xx 10^(-3) N HA` solution `(Lambda_m) = k//"normality"`
`Lambda_c = (4.9 xx 10^(-5) xx 1000)/(1 xx 10^(-3)) = 49 S cm^(2) eq^(-1)`
Equivalent conductance at infinite dilution `(Lambda_0) = 390 S cm^(2) eq^(-1)`
Extent of dissociation = `alpha = (Lambda_c)/(Lambda_0) = 49/390 = 0.126`
Dissociation constant of acid = `C alpha^(2) = 1 xx 10^(-3) (0.126)^(2) = 1.5 xx 10^(-5) mol L^(-1)`.
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