1.

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

Answer»

Solution :`wedge_(m)^(@)(HCOOH)=lamda^(@)(H^(+))+lamda^(@)(HCOO^(-))=349.6+54.6"S "cm^(2)MOL^(-1)=404.2"S "cm^(2)mol^(-1)`
`wedge_(m)^(c)=46.1" S "cm^(2)mol^(1)` (GIVEN)
`thereforealpha=(wedge_(m)^(c))/(wedge_(m)^(0))=(41.1)/(404.2)=0.114`
`{:(,HCOOH,hArr,HCOO^(-),+,H^(+)),("Initial conc.","c mol "L^(-1),,,,),("Conc. at eqm.",c(1-alpha),,calpha,,calpha):}`
`thereforeK_(a)=(calpha.calpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)=3.67xx10^(-4)`


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