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Equivalent conductivity of a weak acid HA at infinite dilution is 390 S cm^(2) eq^(-1).Conductivity of 1 xx 10^(-3) N HA solution is 4.9 xx 10^(-5) S cm^(-1). Calculate the extent of dissociation and dissociation constant of the acid. |
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Answer» Solution :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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