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Conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)" S "cm^(-1). Calculate its molar conductivity and if Lamda_(m)^(@) for acetic acid is 390.5 S cm^(-2)mol^(-1), what is its dissociation constant ? |
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Answer» Solution :* The CALCULATION of molar conductivity `Lamda_(m)^(c)`: Where, `M=0.00241" "CH_(3)COOH`, `Lamda_(m)^(c)=(kxx1000)/("molarity")""k=7.896xx10^(-5)" S "CM^(-1)` `=(7.896xx10^(-5)(" S "cm^(-1))xx1000(cm^(3)L^(-1)))/(0.00241(mol" "L^(-1)))` * The calculation of dissociation degree `(alpha)` of `CH_(3)COOH`: `therefore alpha=(Lamda_(m)^(c))/(Lamda_(m)^(@)),""alpha=(32.763)/(390.5)=0.0839|" Where,"alpha="dissociation degree,"Lamda_(m)^(c)=32.763" S "cm^(2)mol^(-1),""Lamda_(m)^(@)=390.5" S "cm^(2)mol^(-1)` * The calculation of dissociation constant `(K_(a))`: The EQUILIBRIUM in acetic acid `(CH_(3)COOH)` solution is as follow: `CH_(3)COOH_((aq))HARR CH_(3)COO_((aq))^(-)+H_((aq))^(+)` `therefore K_(a)=(CalphaxxCalpha)/(C-Calpha)=(Calpha^(2))/((1-alpha))`,`=(0.00241xx(0.0839)^(2))/(1-0.0839),""=(0.00241xx0.0839xx0.0839)/(0.9161)|" Where, dissociation degree "alpha=0.0839" Molarity "C=0.00241" mol "L^(-1)` `=1.8518xx10^(-5)` |
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