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Conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)"S "cm^(-1). Calculate its molar conductivity and if wedge^(@) for acetic acid is 390.5 S cm^(2)mol^(-1), what is its dissociation constant? |
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Answer» SOLUTION :`wedge_(m)^(@)=(kappaxx1000)/("MOLARITY")=((7.896xx10^(-5)" S "cm^(-1))xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)` `alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(32.76)/(390.5)=8.4xx10^(-2),K_(a)=(CALPHA^(2))/(1-alpha)=(0.00241xx(8.4xx10^(-2))^(2))/(1-0.084)=1.86xx10^(-5)` |
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