The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of `1 cm^(2)`. The conductance of this solution was found to be `5xx10^(-7) S`. The pH of the solution is 4. Calculate the value of limiting molar conductivity.

Explanation

Cell constant `((l)/(a)) = (120 cm)/(1 cm^(2)) = 120 cm^(-1)`
`kappa = G xx` Cell constant
`= (5 xx 10^(-7) S) xx (120 cm^(-1))`
`= 6 xx 10^(-5) S cm^(-1)`
`Lambda_(m) = (kappa xx 1000)/(M)`
`= (6 xx 10^(-5) xx 1000)/(0.0015)`
`=40 S cm^(2) mol^(-1)`
Now , pH = `- log [H^(+)] = 4`
`therefore [H^(+)] = 10^(-4)`
For dissociation of a weak monobasic acid , `[H^(+)] = c xx a `
where `alpha`= degree of dissociation
`10^(-4) - 0.0015 xx alpha`
or `alpha = (10^(-4))/(0.0015) = (1)/(15)`
Also `alpha = (Lambda_(m)^(c))/(Lambda_(m)^(@))`
`(1)/(15) = (40)/(Lambda_(m)^(@))`
`therefore Lambda_(m) ^(@) = 40 xx 15`
`= 6 xx 10^(2) S cm^(2) mol^(-1)`