Home
Class 12
CHEMISTRY
The conductivity of 0.001028 M acetic ac...

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

Text Solution

Verified by Experts

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)`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    PRADEEP|Exercise Curiosity Question|5 Videos

  • ELECTROCHEMISTRY

    PRADEEP|Exercise Problem for Practice|80 Videos

  • ELECTROCHEMISTRY

    PRADEEP|Exercise IMPORTANT QUESTIONS FOR BOARD EXAMINATION|25 Videos

  • D- AND F-BLOCK ELEMENTS

    PRADEEP|Exercise IMPORTANT QUESTIONS|30 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    PRADEEP|Exercise Curiosity Questions|2 Videos

Similar Questions

Explore conceptually related problems

(a). The resistance of a decinormal solution of an electrolyte in a conductivity cell was found to be 245 ohms. Calculate the equilvalet conductivity of the solution if the electrodes in the cell were 2 cm apart and each has an area of 3.5 sq. cm. (b). The conductivity of 0.001028M acetic acid is 4.95xx10^(-5)Scm^(-1) . Calculate its dissociation constant if ^^_(m)^(@) for acetic acid is 390.5Scm^(2_mol^(-1)

The conductivity of 0.001028M acetic acid is 4.95xx10^(-5)Scm^(-1) . Calculate its dissociation constant if Lambda_(m)^(0) for acetic acid is 390.5Scm^(2)mol^(-1) .

Calculating the value of dissociation constant of weak electrolyte: The conductivity of 0.001028 mo L^(-1) acetic acid is 4.95 xx 10^(-5) S cm^(-1) . Calculate its dssociation constatnt if Lambda_(m)^(0) for acetic acid is 390.5 S cm^(2) mol^(-1) . Strategy: We can determine the value of the dissociation constant for week electrolytes once we know the Lambda_(m)^(0) and Lambda_(m) at any given concentration C .

The conductivity of 0.00241 M acetic acid is 7.896xx10^(-5)Scm^(-1) . Calculate its molar conductivity. If wedge_(m)^(@) for acetic acid is 390.5Scm^(2)mol^(-1) , what is its dissociation constant ?