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)`.
Text Solution
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We know, `Lambda_(m)=kxx(1000)/(M)`.....(i) Given : `"k=4.95 x 10"^(-5)"S cm"^(-1),M=0.001028` `therefore"From"(i)Lambda_(m)=4.95xx10^(-5)xx(1000)/(0.001028)` `="48.15 ohm"^(-1)cm^(2)mol^(-1)` Degree of dissociation,
(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 0.001028 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) .
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 ?
