(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)`

Explanation

Given `R=245Omega, N=(1)/(10),l=2cm, A =3.5cm^(2)` `lamda_(eq)^(infty)=390.5Scm^(2)mol^(-1)`
Asked `lamda_(eq)and k_(a)=?`
formula used `k=Gxx(l)/(a)=(1)/(R)xx(l)/(a).^^_(eq)=Kxx(1000)/("normality"),alpha=(^^_(m))/(^^_(m)^(@))`
Explanation `k=` conductivity, `N=` normality `lamda_(eq)=` equivalence conductance `lamda_(eq)^(infty)=` equivalence conductance at infinite dilution.
substitution ad calculation:
(a). `k=Gxx(1)/(a)xx(1)/(R)xx(l)/(a)=(1)/(245Omega)xx(2cm)/(3.5cm^(2))`
`^^_(eq)=Kxx(1000)/("normality")=((1)/(245)xx(2)/(3.5)Omega^(-1)cm^(-1))xx(1000cm^(3)L^(-1))/(0.1molL^(-1))=23.32Omega^(-1)cm^(2)eq^(-1)`
(b). for the given concentration of acetic acid solution:
`^^_(m)=(kxx1000)/(c)=(4.95xx10^(-5)Scm^(-1)xx1000cm^(3)L^(-1))/(0.001028molL^(-1))=48.15Scm^(2)mol^(1)`
`alpha=(^^_(m))/(^^_(m)^(@))=(48.15Scm^(2)mol^(-1))/(390.5Scm^(2)mol^(-1))=0.1233`
`CH_(3)COOHhArrunderset(0)(CH_(3))COO^(-)underset(0)(H^(+))`
initial conc. Equilibrium conc.
`{:("equilibrium conc",c-calpha=c(1-alpha),calpha,calpha):}`
`K=(calpha.calpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)=((0.001028 mol l^(-1))(0.1233)^(2))/(1-0.1233)=178xx10^(-5)molL^(-1)`