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Equivalent conductivity of a weak acid HA at infinite dilution is390 S cm^(2) eq^(-1).Conductivity of 1xx10^(-3) N HA solution is 4.9xx10^(-5) S cm^(-1).Calculate the extent of dissociation and dissociation constant of the acid. |
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Answer» Solution :`Lambda_(c) = (4.9 xx 10^(-5) xx 1000)/(1xx10^(-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) = 1xx10^(-3) (0.126)^(2)=1.5 x 10^(-5) "mol" L^(-1)`. |
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