Home
Class 12
CHEMISTRY
The conductivity of 0.001028 mol L ^(-1)...

The conductivity of `0.001028 mol L ^(-1)` acetic acid is `4.95 xx10^(-5)S cm ^(-1).` Calculate its dissociation constant if `^^(m)^(0)` for acetic acid id `390.5 S cm ^(2)mol ^(-1).`

Text Solution

Verified by Experts

`^^_(m)=k/c=(4.9510^(-5) S cm^(-1))/(0.001028 mol L ^(-1))xx(1000cm ^(3))/(L)`
`alpha =(^^_(m))/(^^_(m)^(0))=(48.15 S cm ^(2) mol ^(-1))/(390.5 S cm ^(2) mol ^(-1))=0.1233`
`k=(c alpha^(2))/( (1-alpha))=(0.001020 mol L ^(-1)xx(0.1233)^(2))/(1-0.1233)=1.78xx10^(-5) mol L ^(-1)`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY & CHEMICAL KINETICS

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise PROBLEMS|55 Videos

  • ELECTROCHEMISTRY & CHEMICAL KINETICS

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise VERY SHORT ANSWER QUESTIONS|63 Videos

  • D AND F- BLOCK ELEMENTS AND CO-ORDINATION COMPOUNDS

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise INTEXT QUESTIONS|33 Videos

  • GENERAL PRINCIPLES OF MMETALLURGY

    VIKRAM PUBLICATION ( ANDHRA PUBLICATION)|Exercise DAM SURE SAQ|1 Videos

Similar Questions

Explore conceptually related problems

The conductivity of 0.001028 mol L^(-1) acetic acid is 4.95 xx 10^(-5) S cm^(-1) . Calculate its dissociation constant if Lambda_(m)^(0) for acetic is 390.5 S cm^(2) mol^(-1) ?

Conductivity of 0.00241 M acetic acid is 7.896 xx 10^(-4) S cm^(-1) . Calculate its moar conductivity and if ^^_(m)^(@) for acetic is 390.5 S cm^(2) mo^(-1) , what is its dissociation constant ?

The molar conductivity of 0.025mol L^(-1) methanoic acid is 46.1 S cm^(2)mol ^(-1). Calculate its degree of dissociation and dissociation constant. Given, lamda^(0)(H^(+))=349.6S cm^(2) mol ^(-1) and lamda^(0)(HCOO^(-))=54.6S cm^(2)mol^(-1)

The molar conductivity of 0.024 "mol" L^(-1) methanoic acid is 46.1 S cm^(-2)"mol"^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(0)(H^(+) = 349.6 S cm^(2) mol^(-1) and lambda^(0)(HCOO^(-)) = 54.6 S cm^(2) mol^(-1) .

The conductivity of 0.001 M acetic acid is 5 xx 10^(-5) S cm^(-1) and ^^""^(0) is 390.5 S cm^(2) "mol" ^(-1) then the calculated value of dissociation constant of acetic acid would be

The conductivity of 0.20M solution of KCl at 298 K is 0.0248S cm^(-1). Calculate molar conductance.

The pH of 0.1 M solution of an organic acid is 4.0. Calculate the dissociation constant of the acid.

The pH of 0.05 M acetic acid is (K_(a) = 2 xx 10^(-5))