1.

The conductivity of 0.001028 M acetic acid is 4.95xx10^(-5)" S "cm^(-1). Calculate its dissociation constant if wedge^(@) for acetic acid is 390.5 S cm^(2)mol^(-1).

Answer»

Solution :For the given concentration of acetic acid solution,
`wedge_(m)=(kappaxx1000)/(c)=(4.95xx10^(-5)S" "CM^(-1)xx1000cm^(3)L^(-1))/(0.001028" mol "L^(-1))=48.15" S "cm^(2)mol^(-1)`
`alpha=(wedge_(m))/(wedge_(m)^(@))=(48.15" S "cm^(2)mol^(-1))/(390.5" S "cm^(2)mol^(-1))=0.1233`
`{:(,CH_(3)COOH,HARR,CH_(3)COO^(-),+,H^(+)),("INITIAL conc.",c,,0,,0),("Equilibrium conc.",c-calpha=c(1-alpha),,calpha,,calpha):}`
`K=(calpha.calpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)=((0.001028" "molL^(-1))(0.1233)^(2))/(1-0.1233)=1.78xx10^(-5)mol" "L^(-1)`


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