The molar conductivity of 0.025 M methanoic acid (HCOOH) is `46.15" S "cm^(2)mol^(-1)`. Calculate its degree of dissociation and dissociation constant. Given `lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1)` and `lambda_((HCOO^(-)))^(@)=54.6" S " cm^(2)mol^(-1)`.

Explanation

Step I. Calculate of degree of dissociation (alpha) of methoic acid
`Lambda_(m)^(c )=46.15 " S "cm^(2)mol^(-1)`
`Lambda_(m(HCOOH))^(0)=lamda_(m(HCOO^(-)))^(@)+lambda_(m(H^(+)))^(@)`
`=54.6 " S "cm^(2)mol^(-1)+349.6" S "cm^(2)mol^(-1)=404.2 " S "cm^(2)mol^(-1)`
`alpha=(Lambda_(m)^(c ))/(Lambda_(m)^(0))=((46.15" S "cm^(2)mol^(-1)))/((404.2" S "cm^(2)mol^(-1)))=0.1140`
Step II. Calculate of dissociation constant `K_(a)`
`{:(,HCOOH(aq), harr ,HCOO^(-)(aq),+,H^(+)(aq)),("Initial conc."," "C,," "0,," "0),("Eqm.conc.",C(1-alpha),," "Calpha,," "Calpha):}`
Dissociation constant, `K_(a)=([HCOO^(-)][H^(+)])/([HCOOH])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
Substituting the values, `K_(alpha)=((0.0025" mol " L^(-1))xx(0.1140)^(2))/((1-0.1140))`
`=((3.249xx10^(-4)" mol "L^(-1)))/(0.886)=3.67xx10^(-4)" mol "L^(-1)`