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The molar conductivity of acetic acid solution at infinite dilution is 390.7 Omega^(-1)cm^(2)mol^(-1). Calculate the moalr conductivity of 0.01M acetic acid solution, given that the dissociation of acetic acid is 1.8xx10^(-5) |
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Answer» Solution :`UNDERSET(C(1-alpha))(CH_(3)COOH)hArrunderset(calpha)(CH_(3)COO^(-))+underset(calpha)(H^(+))` `K_(eq)=calpha^(2)"or"alpha=sqrt((K_(eq))/(c))=(wedge_(m)^(c))/(wedge_(m)^(OO))` `THEREFORE(wedge_(m)^(c))/(390.7)=sqrt((1.8xx10^(-5))/(0.01))=sqrt(18xx10^(-4))=4.243xx10^(-2)` or `wedge_(m)^(c)=16.57Omega^(-1)cm^(2)`. |
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